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+\section{T3 1/2 Space is Uniformizable}
+Tags: Uniformities, T3 1: 2 Spaces
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Then $T$ is [[Definition:Uniformizable Space|uniformizable]] {{iff}} $T$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+\end{theorem}<|endoftext|>
+\section{Pseudometric Space generates Uniformity}
+Tags: Pseudometric Spaces, Uniformities
+
+\begin{theorem}
+Let $P = \struct {A, d}$ be a [[Definition:Pseudometric Space|pseudometric space]].
+Let $\UU$ be the [[Definition:Set of Sets|set of sets]] defined as:
+:$\UU := \set {u \in A \times A: \exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u}$
+where:
+:$\R_{>0}$ is the set of [[Definition:Strictly Positive Real Number|strictly positive real numbers]]
+:$u_\epsilon$ is defined as:
+::$u_\epsilon := \set {\tuple {x, y}: \map d {x, y} < \epsilon}$
+Then $\UU$ is a [[Definition:Uniformity|uniformity]] on $A$ which generates a [[Definition:Uniform Space|uniform space]] with the same [[Definition:Topology|topology]] as the [[Pseudometric induces Topology|topology induced by $d$]].
+\end{theorem}
+
+\begin{proof}
+We check whether the [[Definition:Uniformity Axioms|Uniformity Axioms]] are satisfied.
+=== $\text U 1$ ===
+Let $x \in A$.
+Then $\tuple {x, x} \in \Delta_A$.
+Let $u \in \UU$.
+From definition:
+:$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$
+By [[Definition:Metric Space Axioms|Metric Space Axiom M1]]:
+:$\map d {x, x} = 0 < \epsilon$
+Hence we have $\tuple {x, x} \in u_\epsilon$.
+By definition of [[Definition:Subset|subset]], $\Delta_A \subseteq u_\epsilon \subseteq u$.
+{{qed|lemma}}
+=== $\text U 2$ ===
+Let $u, v \in \UU$.
+From definition:
+:$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$
+:$\exists \delta \in \R_{>0}: u_\delta \subseteq v$
+{{WLOG}} assume that $\delta \ge \epsilon$.
+Let $\tuple {x, y} \in u_\epsilon$.
+Then $\map d {x, y} < \epsilon \le \delta$.
+Thus $\tuple {x, y} \in u_\delta$.
+By definition of [[Definition:Subset|subset]], $u_\epsilon \subseteq u_\delta$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = u_\epsilon
+      | r = u_\epsilon \cap u_\delta
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{eqn | o = \subseteq
+      | r = u \cap v
+      | c = [[Set Intersection Preserves Subsets]]
+}}
+{{eqn | ll = \leadsto
+      | l = u \cap v
+      | o = \in
+      | r = \UU
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== $\text U 3$ ===
+Let $v \in A \times A$.
+Suppose $u \in \UU$ and $u \subseteq v$.
+From definition:
+:$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$
+Thus $u_\epsilon \subseteq v$.
+Therefore $v \in \UU$ as well.
+{{qed|lemma}}
+=== $\text U 4$ ===
+Let $u \in \UU$.
+From definition:
+:$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$
+Let $v = u_{\epsilon / 2}$.
+Since $u_{\epsilon / 2} \subseteq v$ and $\dfrac \epsilon 2 > 0$, $v \in \UU$.
+From definition:
+{{begin-eqn}}
+{{eqn | l = v \circ v
+      | r = \set {\tuple {x, z}: \exists y \in A: \tuple {x, y} \in v, \tuple {y, z} \in v}
+}}
+{{eqn | r = \set {\tuple {x, z}: \exists y \in A: \map d {x, y} < \frac \epsilon 2, \map d {y, z} < \frac \epsilon 2}
+}}
+{{end-eqn}}
+Let $\tuple {x, z} \in v \circ v$.
+Then:
+:$\exists y \in A: \map d {x, y} < \frac \epsilon 2, \map d {y, z} < \frac \epsilon 2$
+By [[Definition:Metric Space Axioms|Metric Space Axiom M2]]:
+{{begin-eqn}}
+{{eqn | l = \map d {x, z}
+      | o = \le
+      | r = \set {\tuple {x, z}: \exists y \in A: \tuple {x, y} \in v, \tuple {y, z} \in v}
+}}
+{{eqn | o = <
+      | r = \frac \epsilon 2 + \frac \epsilon 2
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+Hence $\tuple {x, z} \in u_\epsilon \subseteq u$.
+By definition of [[Definition:Subset|subset]], $v \circ v \subseteq u$.
+{{qed|lemma}}
+=== $\text U 5$ ===
+Let $u \in \UU$.
+From definition:
+:$\exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u$
+Suppose $\tuple {b, a} \in u_\epsilon$.
+Then $\map d {b, a} < \epsilon$.
+By [[Definition:Metric Space Axioms|Metric Space Axiom M3]]:
+:$\map d {a, b} = \map d {b, a} < \epsilon$
+Hence $\tuple {a, b} \in u_\epsilon \subseteq u$.
+Consider the set $u^{-1} = \set {\tuple {y, x}: \tuple {x, y} \in u}$.
+From the above, $\tuple {b, a} \in u^{-1}$.
+By definition of [[Definition:Subset|subset]], $u_\epsilon \subseteq u^{-1}$.
+Hence $u^{-1} \in \UU$.
+{{qed|lemma}}
+{{finish|It remains to show that the topology induced by the uniformity $\UU$ is the same as the one induced by the pseudometric $d$}}
+\end{proof}<|endoftext|>
+\section{Definition:Pseudometrizable Uniformity}
+Tags: Definitions: Metric Spaces, Definitions: Uniformities
+
+\begin{theorem}
+Let $P = \struct {A, d}$ be a [[Definition:Pseudometric Space|pseudometric space]].
+Let $\UU$ be the [[Definition:Uniformity|uniformity]] on $A$ defined as:
+:$\UU := \set {u_\epsilon: \epsilon \in \R_{>0} }$
+where:
+:$\R_{>0}$ is the set of [[Definition:Strictly Positive Real Number|strictly positive real numbers]]
+:$u_\epsilon$ is defined as:
+::$u_\epsilon := \set {\tuple {x, y}: \map d {x, y} < \epsilon}$
+Then $\UU$ is defined as '''pseudometrizable'''.
+\end{theorem}<|endoftext|>
+\section{Metric Space generates Uniformity}
+Tags: Metric Spaces, Uniformities
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $\UU$ be the [[Definition:Set of Sets|set of sets]] defined as:
+:$\UU := \set {u \in A \times A: \exists \epsilon \in \R_{>0}: u_\epsilon \subseteq u}$
+where:
+:$\R_{>0}$ is the set of [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real numbers]]
+:$u_\epsilon$ is defined as:
+::$u_\epsilon := \set {\tuple {x, y}: \map d {x, y} < \epsilon}$
+Then $\UU$ is a [[Definition:Uniformity|uniformity]] on $A$ which generates a [[Definition:Uniform Space|uniform space]] with the same [[Definition:Topology|topology]] as the [[Metric Induces Topology|topology induced by $d$]].
+\end{theorem}
+
+\begin{proof}
+From definition it is clear that a [[Definition:Metric Space|metric space]] is a [[Definition:Pseudometric Space|pseudometric space]].
+The result then follows from [[Pseudometric Space generates Uniformity]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multiplication is Superfunction}
+Tags: Superfunctions
+
+\begin{theorem}
+The [[Definition:Function|function]] $f: \C \to \C$, defined as:
+:$\map f z = z \times c$
+is a [[Definition:Superfunction|superfunction]] for any [[Definition:Complex Number|complex number]] $c$.
+\end{theorem}
+
+\begin{proof}
+Define $h: \C \to \C$ by $\map h z = z + c$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map h {\map f z}
+      | r = \map h {z \times c}
+      | c = 
+}}
+{{eqn | r = z \times c + c
+      | c = 
+}}
+{{eqn | r = \paren {z + 1} \times c
+      | c = 
+}}
+{{eqn | r = \map f {z + 1}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\map f z = z \times c$ is a [[Definition:Superfunction|superfunction]] and $\map h z = z + c$ is the corresponding [[Definition:Transfer Function (Number Theory)|transfer function]].
+{{qed}}
+[[Category:Superfunctions]]
+cab3ee5qz40q2bqeshg3oe1373qvkwn
+\end{proof}<|endoftext|>
+\section{Discrete Topology is Topology}
+Tags: Discrete Topology
+
+\begin{theorem}
+:$\tau$ is a [[Definition:Topology|topology]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be the [[Definition:Discrete Space|discrete space]] on $S$.
+Then by definition $\tau = \powerset S$, that is, is the [[Definition:Power Set|power set]] of $S$.
+We confirm the criteria for $T$ to be a [[Definition:Topology|topology]]:
+:$(1): \quad$ By definition of [[Definition:Power Set|power set]], $\O \in \powerset S$ and $S \in \powerset S$.
+:$(2): \quad$ From [[Power Set with Union is Monoid]], $\powerset S$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Set Union|set union]].
+:$(3): \quad$ From [[Power Set with Intersection is Monoid]], $\powerset S$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Set Intersection|set intersection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Topology is Finest Topology}
+Tags: Discrete Topology
+
+\begin{theorem}
+:$\tau$ is the [[Definition:Finer Topology|finest topology]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Let $\phi$ be any [[Definition:Topology|topology]] on $S$.
+Let $U \in \phi$.
+Then, by the definition of [[Definition:Topology|topology]], $U \subseteq S$.
+Then, by the definition of [[Definition:Discrete Topology|discrete topological space]], $U \in \tau$.
+Hence by definition of [[Definition:Subset|subset]], $\phi \subseteq \tau$.
+Hence by definition of [[Definition:Finer Topology|finer topology]], $\tau$ is [[Definition:Finer Topology|finer]] than $\phi$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set in Discrete Topology is Clopen}
+Tags: Discrete Topology, Clopen Sets
+
+\begin{theorem}
+:$\forall U \subseteq S: U$ is both [[Definition:Closed Set (Topology)|closed]] and [[Definition:Open Set (Topology)|open]] in $\struct {S, \tau}$.
+\end{theorem}
+
+\begin{proof}
+Let $U \subseteq S$.
+By definition of [[Definition:Discrete Topology|discrete topological space]], $U \in \tau$.
+By definition of [[Definition:Closed Set (Topology)|closed set]], $\relcomp S U$ is [[Definition:Closed Set (Topology)|closed]] in $T$, where $\relcomp S U$ is the [[Definition:Relative Complement|relative complement]] of $U$ in $S$. 
+But from [[Set Difference is Subset]]:
+:$\relcomp S U = S \setminus U \subseteq S$
+and so:
+:$\relcomp S U \in \tau$
+That is, $\relcomp S U$ is both [[Definition:Closed Set (Topology)|closed]] and [[Definition:Open Set (Topology)|open]] in $T$. 
+Then by [[Relative Complement of Relative Complement]]:
+:$\relcomp S {\relcomp S U} = U$
+which is seen to be both [[Definition:Closed Set (Topology)|closed]] and [[Definition:Open Set (Topology)|open]] in $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Space is Discrete iff All Points are Isolated}
+Tags: Isolated Points, Discrete Topology
+
+\begin{theorem}
+:$\tau$ is the [[Definition:Discrete Topology|discrete topology]] on $S$ {{iff}} all points in $S$ are [[Definition:Isolated Point (Topology)|isolated points of $T$]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $T = \struct {S, \tau}$ be the [[Definition:Discrete Space|discrete space]] on $S$.
+Then by definition $\tau = \powerset S$, that is, $\tau$ is the [[Definition:Power Set|power set]] of $S$.
+Let $x \in S$.
+Then from [[Set in Discrete Topology is Clopen]] it follows that $\set x$ is [[Definition:Open Set (Topology)|open]] in $T$.
+Thus by definition $x \in S$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $T$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $T$ be such that $x \in S$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $T$ for all $x \in T$.
+{{AimForCont}} $T = \struct {S, \tau}$ is not the [[Definition:Discrete Space|discrete space]] on $S$.
+Then from [[Basis for Discrete Topology]] it follows that:
+:$\BB := \set {\set x: x \in S}$
+is not a [[Definition:Basis (Topology)|basis]] for $T$.
+Thus:
+:$\exists x \in S: \set x \notin \tau$
+Let $U \in \tau$ such that $x \in U$.
+Then as $U \ne \set x$ it follows that $\exists y \in U: y \ne x$.
+That is, there are no [[Definition:Open Set (Topology)|open sets]] of $T$ which contain only $x$.
+So $x$ is not an [[Definition:Isolated Point (Topology)|isolated point]].
+It follows by [[Proof by Contradiction]] that $T$ is the [[Definition:Discrete Space|discrete space]] on $S$.
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Point in Discrete Space is Adherent Point}
+Tags: Discrete Topology, Adherent Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
+Let $U \subseteq S$.
+Then $x$ is an [[Definition:Adherent Point|adherent point]] of $U$ {{iff}} $x \in U$.
+\end{theorem}
+
+\begin{proof}
+Let $U \subseteq S$.
+From [[Set in Discrete Topology is Clopen]] it follows that $U$ is [[Definition:Open Set (Topology)|open]] in $T$.
+Let $x \in U$.
+Then $U$ is an [[Definition:Open Neighborhood|open neighborhood]] of $x$ containing (trivially) an element of $U$, that is, $x$.
+So, by definition, $x$ is an [[Definition:Adherent Point|adherent point]] of $U$.
+Now suppose $x \notin U$.
+Then $\set x$ is an [[Definition:Open Neighborhood|open neighborhood]] of $x$ such that $\set x \cap U = \O$.
+So $x$ can not be an [[Definition:Adherent Point|adherent point]] of $U$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Interior Equals Closure of Subset of Discrete Space}
+Tags: Discrete Topology
+
+\begin{theorem}
+Let $A \subseteq S$.
+Then:
+:$A = A^\circ = A^-$
+where:
+:$A^\circ$ is the [[Definition:Interior (Topology)|interior]] of $A$
+:$A^-$ is the [[Definition:Closure (Topology)|closure]] of $A$.
+\end{theorem}
+
+\begin{proof}
+Let $A \subseteq S$.
+Then from [[Set in Discrete Topology is Clopen]] it follows that $A$ is both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] in $T$.
+From [[Closed Set Equals its Closure]] we have that $A = A^-$.
+From [[Set Interior is Largest Open Set]], we have that $A^\circ = A$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boundary of Subset of Discrete Space is Null}
+Tags: Discrete Topology, Boundaries
+
+\begin{theorem}
+Let $A \subseteq S$.
+Then:
+:$\partial A = \O$
+where:
+:$\partial A$ is the [[Definition:Boundary (Topology)|boundary]] of $A$ in $T$.
+\end{theorem}
+
+\begin{proof}
+Let $A \subseteq S$.
+Then from [[Set in Discrete Topology is Clopen]] it follows that $A$ is both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] in $T$.
+The result follows from [[Set Clopen iff Boundary is Empty]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping from Discrete Space is Continuous}
+Tags: Discrete Topology
+
+\begin{theorem}
+Let $T_1 = \struct {S_1, \tau_1}$ be the [[Definition:Discrete Space|discrete topological space]] on $S_1$.
+Let $T_2 = \struct {S_2, \tau_2}$ be any other [[Definition:Topological Space|topological space]].
+Let $\phi: S_1 \to S_2$ be a [[Definition:Mapping|mapping]].
+Then $\phi$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Everywhere Continuous Mapping (Topology)|continuous]]:
+:$U \in \tau_2 \implies \phi^{-1} \sqbrk U \in \tau_1$
+But as $\phi^{-1} \sqbrk U \subseteq S_1$ it follows from the definition of [[Definition:Discrete Space|discrete space]] that $\phi^{-1} \sqbrk U \in \tau_1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponential Function is Superfunction}
+Tags: Superfunctions
+
+\begin{theorem}
+The [[Definition:Function|function]] $f : \C \to \C$, defined as
+:$f \left({ z }\right) = c^z$
+is a [[Definition:Superfunction|superfunction]] for any [[Definition:Complex Number|complex number]] $c$.
+\end{theorem}
+
+\begin{proof}
+Define $h : \C \to \C$ by $h \left({ z }\right) = z \times c$.  Then
+{{begin-eqn}}
+{{eqn | l=h \left({ f \left({ z }\right) }\right)
+      | r=h \left({ c^z }\right)
+      | c=
+}}
+{{eqn | r=c^z \times c
+      | c=
+}}
+{{eqn | r=c^{z + 1}
+      | c=
+}}
+{{eqn | r=f \left({ z + 1 }\right)
+      | c=
+}}
+{{end-eqn}}
+Thus  $f \left({z}\right) = c^z$ is a superfunction and $h \left({z}\right) = z \times c$ is the corresponding [[Definition:Transfer Function (Number Theory)|transfer function]].
+{{qed}}
+[[Category:Superfunctions]]
+2z342fbx3ysenws7x3br3acdq21fska
+\end{proof}<|endoftext|>
+\section{Standard Discrete Metric induces Discrete Topology}
+Tags: Discrete Metrics, Discrete Topology, Metrizable Topologies
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be the [[Definition:Standard Discrete Metric|(standard) discrete metric space]] on $A$.
+Then [[Metric Induces Topology|$d$ induces]] the [[Definition:Discrete Topology|discrete topology]] on $A$.
+Thus the [[Definition:Discrete Topology|discrete topology]] is [[Definition:Metrizable Topology|metrizable]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in A$.
+From [[Subset of Standard Discrete Metric Space is Open]], a set $U \subseteq A$ is [[Definition:Open Set (Metric Space)|open]] in $M$.
+So, in particular, $\set a$ is [[Definition:Open Set (Metric Space)|open]] in $\struct {A, d}$.
+This holds for all $a \in A$.
+From [[Metric Induces Topology]] it follows that $\set a$ is an [[Definition:Open Set (Topology)|open set]] in $\struct {A, \tau_{A, d} }$.
+The result follows from [[Basis for Discrete Topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space satisfies all Separation Properties}
+Tags: Discrete Topology, Separation Axioms
+
+\begin{theorem}
+Let $T = \struct {S, \powerset S}$ be the [[Definition:Discrete Space|discrete topological space]] on $S$.
+Then $T$ fulfils all [[Definition:Separation Axioms|separation axioms]]:
+:$T$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]]
+:$T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]]
+:$T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]]
+:$T$ is a [[Definition:Semiregular Space|semiregular space]]
+:$T$ is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]] 
+:$T$ is a [[Definition:T3 Space|$T_3$ space]] 
+:$T$ is a [[Definition:Regular Space|regular space]]
+:$T$ is an [[Definition:Urysohn Space|Urysohn space]]
+:$T$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]]
+:$T$ is a [[Definition:Tychonoff Space|Tychonoff (completely regular) space]]
+:$T$ is a [[Definition:T4 Space|$T_4$ space]]
+:$T$ is a [[Definition:Normal Space|normal space]]
+:$T$ is a [[Definition:T5 Space|$T_5$ space]] 
+:$T$ is a [[Definition:Completely Normal Space|completely normal space]]
+:$T$ is a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]]
+:$T$ is a [[Definition:Perfectly Normal Space|perfectly normal space]]
+\end{theorem}
+
+\begin{proof}
+We have that a [[Standard Discrete Metric induces Discrete Topology]].
+Then we use [[Metric Space fulfils all Separation Axioms]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Point in Discrete Space is Neighborhood}
+Tags: Discrete Topology, Neighborhoods
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
+Let $x \in S$.
+Then $\set x$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$ in $T$.
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Neighborhood of Point|neighborhood]] $N_x$ of $x$ is any [[Definition:Subset|subset]] of $S$ containing an [[Definition:Open Set (Topology)|open set]] which itself contains $x$.
+That is:
+:$\exists U \in \tau: x \in U \subseteq N_x \subseteq S$
+From [[Set in Discrete Topology is Clopen]] we have that $\set x$ is [[Definition:Open Set (Topology)|open set]] in $S$.
+So by [[Set is Subset of Itself]], $\set x$ is a [[Definition:Subset|subset]] of $S$ containing an [[Definition:Open Set (Topology)|open set]] $\set x$ which itself contains $x$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space is Strongly Locally Compact}
+Tags: Strongly Locally Compact Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
+Then $T$ is [[Definition:Strongly Locally Compact Space|strongly locally compact]].
+\end{theorem}
+
+\begin{proof}
+From [[Point in Discrete Space is Neighborhood]], every point $x \in S$ is contained in an [[Definition:Open Set (Topology)|open set]] $\set x$ of $T$.
+Then from [[Interior Equals Closure of Subset of Discrete Space]] we have that $\set x$ equals its [[Definition:Closure (Topology)|closure]] in $T$.
+From [[Singleton Set in Discrete Space is Compact]], we have that $\set x$ is [[Definition:Compact Topological Subspace|compact]] in $T$.
+Hence the result by definition of [[Definition:Strongly Locally Compact Space|strongly locally compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space is First-Countable}
+Tags: First-Countable Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
+Then $T$ is [[Definition:First-Countable Space|first-countable]].
+\end{theorem}
+
+\begin{proof}
+From [[Point in Discrete Space is Neighborhood]], every point $x \in S$ is contained in an [[Definition:Open Set (Topology)|open set]] $\set x$.
+From the definition of [[Definition:Local Basis|local basis]], it is clear that $\set {\set x}$ is (trivially) a [[Definition:Local Basis|local basis]] at $x$.
+That is, that every [[Definition:Open Set (Topology)|open set]] of $S$ containing $x$ also contains at least one of the [[Definition:Set|sets]] of $\set {\set x}$.
+Equally trivially, we have that $\set {\set x}$ is [[Definition:Countable Set|countable]].
+Hence the result by definition of [[Definition:First-Countable Space|first-countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space has Open Locally Finite Cover}
+Tags: Covers, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
+Consider the [[Definition:Set|set]] $\CC$ of all [[Definition:Singleton|singleton]] [[Definition:Subset|subsets]] of $S$:
+:$\CC := \set {\set x: x \in S}$
+Then $\CC$ is an [[Definition:Open Cover|open cover]] of $T$ which is [[Definition:Locally Finite Cover|locally finite]].
+This [[Definition:Cover of Set|cover]] is the [[Definition:Finer Cover|finest cover]] on $S$.
+That is, if $\VV$ is a [[Definition:Cover of Set|cover]] of $T$, then $\CC$ is a [[Definition:Refinement of Cover|refinement]] of $\VV$.
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$\forall x \in S: \exists \set x \in \CC: x \in \set x$
+and so $\CC$ is a [[Definition:Cover of Set|cover]] for $S$.
+Then from [[Set in Discrete Topology is Clopen]], it follows that $\CC$ is an [[Definition:Open Cover|open cover]] of $T$.
+From [[Point in Discrete Space is Neighborhood]], every point $x \in S$ has a [[Definition:Neighborhood of Point|neighborhood]] $\set x$.
+This [[Definition:Neighborhood of Point|neighborhood]] $\set x$ [[Definition:Set Intersection|intersects]] exactly one [[Definition:Element|element]] of $\CC$, that is: $\set x$ itself.
+As $1$ is a [[Definition:Finite Set|finite number]], the result follows from definition of [[Definition:Locally Finite Cover|locally finite]].
+Now consider any [[Definition:Cover of Set|cover]] $\VV$ of $S$.
+By definition:
+:$\forall x \in S: \exists V \in \VV: x \in V$
+That is:
+:$\forall x \in S: \exists V \in \VV: \set x \subseteq V$
+That is, every [[Definition:Element|element]] of $\CC$ is contained in some [[Definition:Element|element]] of $\VV$.
+Thus by definition, $\CC$ is a [[Definition:Refinement of Cover|refinement]] of $\VV$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space is Paracompact}
+Tags: Paracompact Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
+Then $T$ is [[Definition:Paracompact Space|paracompact]].
+\end{theorem}
+
+\begin{proof}
+Let $\VV$ be any [[Definition:Open Cover|open cover]] of $S$.
+Consider the [[Definition:Set|set]] $\CC$ of all [[Definition:Singleton|singleton]] [[Definition:Subset|subsets]] of $S$:
+:$\CC := \set {\set x: x \in S}$
+From [[Discrete Space has Open Locally Finite Cover]], $\CC$ is an [[Definition:Open Cover|open cover]] which is [[Definition:Locally Finite Cover|locally finite]].
+This result also shows that $\CC$ is the [[Definition:Finer Cover|finest cover]] on $T$.
+So $\CC$ is an [[Definition:Open Refinement|open refinement]] of $\VV$ which is [[Definition:Locally Finite Cover|locally finite]].
+So $T$ is [[Definition:Paracompact Space|paracompact]], by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Singleton Set in Discrete Space is Compact}
+Tags: Compact Spaces, Discrete Topology, Singleton Set in Discrete Space is Compact
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] where $\tau$ is the [[Definition:Discrete Topology|discrete topology]] on $S$.
+Let $x \in S$.
+Then $\set x$ is [[Definition:Compact Topological Subspace|compact]].
+\end{theorem}
+
+\begin{proof}
+From [[Point in Discrete Space is Neighborhood]], every point $x \in S$ is contained in an [[Definition:Open Set (Topology)|open set]] $\set x$.
+Then from [[Interior Equals Closure of Subset of Discrete Space]] we have that $\set x$ equals its [[Definition:Closure (Topology)|closure]].
+As $\set {\set x}$ is (trivially) an [[Definition:Open Cover|open cover]] of $\set x$, it follows by definition that $\set x$ is [[Definition:Compact Topological Space|compact]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Follows directly from [[Finite Topological Space is Compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Discrete Space is Sigma-Compact}
+Tags: Sigma-Compact Spaces, Discrete Topology, Countable Discrete Space is Sigma-Compact
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Discrete Topology|countable discrete topological space]].
+Then $T$ is [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+\end{theorem}
+
+\begin{proof}
+We have that [[Singleton Set in Discrete Space is Compact]].
+We also have that $S$ is the [[Definition:Set Union|union]] of all its [[Definition:Singleton|singleton sets]]:
+:$\displaystyle S = \bigcup_{x \mathop \in S} \set x$
+As $S$ is [[Definition:Countable Set|countable]], it is the [[Definition:Set Union|union]] of [[Definition:Countable Set|countably]] many [[Definition:Compact Topological Subspace|compact]] sets.
+Hence the result, by definition of [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+A direct application of [[Countable Space is Sigma-Compact]].
+\end{proof}<|endoftext|>
+\section{Countable Discrete Space is Lindelöf}
+Tags: Lindelöf Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Discrete Topology|countable discrete topological space]].
+Then $T$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Countable Discrete Space is Sigma-Compact|Countable Discrete Space is $\sigma$-Compact]]
+:[[Sigma-Compact Space is Lindelöf|$\sigma$-Compact Space is Lindelöf]]
+So if $S$ is [[Definition:Countable Set|countable]], $T$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis for Discrete Topology}
+Tags: Discrete Topology, Topological Bases
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\tau$ be the [[Definition:Discrete Topology|discrete topology]] on $S$.
+Let $\BB$ be the set of all [[Definition:Singleton|singleton]] subsets of $S$:
+:$\BB := \set {\set x: x \in S}$.
+Then $\BB$ is a [[Definition:Basis (Topology)|basis]] for $T$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be the [[Definition:Discrete Space|discrete space]] on $S$. 
+Let $U \in \tau$.
+Then:
+:$\displaystyle U = \bigcup_{x \mathop \in U} \set x$
+Hence:
+:$\forall x \in U: \exists \set x \in \BB: \set x \subseteq U$
+Thus $U$ is the [[Definition:Set Union|union]] of [[Definition:Element|elements]] of $\BB$.
+Hence by definition $\BB$ is a [[Definition:Basis (Topology)|basis]] for $T$.
+{{qed}}
+[[Category:Discrete Topology]]
+[[Category:Topological Bases]]
+q66i6e4itpusdv0jhwo5feqfpte6yfm
+\end{proof}<|endoftext|>
+\section{Countable Discrete Space is Second-Countable}
+Tags: Second-Countable Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Discrete Topology|countable discrete topological space]].
+Then $T$ is [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+From [[Basis for Discrete Topology]], the set:
+:$\BB := \set {\set x: x \in S}$
+is a [[Definition:Basis (Topology)|basis]] for $T$.
+There is a trivial [[Definition:Bijection|one-to-one correspondence]] $\phi: S \leftrightarrow \BB$ between $S$ and $\BB$:
+:$\forall x \in S: \map \phi x = \set x$
+Let $S$ be [[Definition:Countable Set|countable]].
+Then $\BB$ is also [[Definition:Countable Set|countable]] by definition of [[Definition:Countable Set|countability]].
+So we have that $T$ has a [[Definition:Countable Basis|countable basis]], and so is [[Definition:Second-Countable Space|second-countable]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Discrete Space is Separable}
+Tags: Separable Spaces, Discrete Topology, Countable Discrete Space is Separable
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Discrete Topology|countable discrete topological space]].
+Then $T$ is [[Definition:Separable Space|separable]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Countable Discrete Topology|countable discrete topological space]].
+From [[Countable Discrete Space is Second-Countable]]:
+:$T$ is [[Definition:Second-Countable Space|second-countable]].
+From [[Second-Countable Space is Separable]]:
+:$T$ is [[Definition:Separable Space|separable]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Follows immediately from [[Countable Space is Separable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Discrete Space is not Separable}
+Tags: Separable Spaces, Discrete Topology, Uncountable Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be an [[Definition:Uncountable Discrete Topology|uncountable discrete topological space]].
+Then $T$ is not [[Definition:Separable Space|separable]].
+\end{theorem}
+
+\begin{proof}
+By definition, $T$ is [[Definition:Separable Space|separable]] {{iff}} there exists a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$ which is [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+Let $H \subseteq S$ be [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+Then by definition of [[Definition:Everywhere Dense|everywhere dense]], $H^- = S$ where $H^-$ denotes the [[Definition:Closure (Topology)|closure]] of $H$.
+However, as $T$ is a [[Definition:Discrete Space|discrete space]], $H^- = H$ from [[Interior Equals Closure of Subset of Discrete Space]].
+So $H^- = S \implies H = S$.
+But $S$ is [[Definition:Uncountable Set|uncountable]].
+So there exists no $H \subseteq S$ such that $H$ is both [[Definition:Countable Set|countable]] and [[Definition:Everywhere Dense|everywhere dense]].
+Hence by definition of [[Definition:Separable Space|separable space]], if $T$ is an [[Definition:Uncountable Discrete Topology|uncountable discrete space]] it can not be [[Definition:Separable Space|separable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Discrete Space is not Second-Countable}
+Tags: Second-Countable Spaces, Discrete Topology, Uncountable Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be an [[Definition:Uncountable Discrete Topology|uncountable discrete topological space]].
+Then $T$ is not [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+We have that an [[Uncountable Discrete Space is not Separable]].
+From [[Second-Countable Space is Separable]], it follows that $T$ can not be [[Definition:Second-Countable Space|second-countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Discrete Space is not Lindelöf}
+Tags: Lindelöf Spaces, Discrete Topology, Uncountable Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be an [[Definition:Uncountable Discrete Topology|uncountable discrete topological space]].
+Then $T$ is not a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Set|set]] $\CC$ of all [[Definition:Singleton|singleton]] [[Definition:Subset|subsets]] of $S$:
+:$\CC := \set {\set x: x \in S}$
+From [[Discrete Space has Open Locally Finite Cover]], $\CC$ is an [[Definition:Open Cover|open cover]] of $S$ which is [[Definition:Finer Cover|finer]] than any other [[Definition:Open Cover|open cover]] of $S$.
+That is, $\CC$ is an [[Definition:Open Cover|open cover]] of $S$ which is [[Definition:Uncountable Set|uncountable]] and has no [[Definition:Countable Subcover|countable subcover]].
+(Note that a [[Subcover is Refinement of Cover|subcover is a refinement of a cover]].)
+So by definition $T$ can not be a [[Definition:Lindelöf Space|Lindelöf space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Discrete Space is not Sigma-Compact}
+Tags: Sigma-Compact Spaces, Discrete Topology, Uncountable Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be an [[Definition:Uncountable Discrete Topology|uncountable discrete topological space]].
+Then $T$ is not [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+\end{theorem}
+
+\begin{proof}
+We have that an [[Uncountable Discrete Space is not Lindelöf]].
+But a [[Sigma-Compact Space is Lindelöf|$\sigma$-compact space is Lindelöf]].
+So an [[Definition:Uncountable Discrete Topology|uncountable discrete space]] can not be [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space is Compact iff Finite}
+Tags: Finite Topological Spaces, Compact Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]] where $\tau$ is the [[Definition:Discrete Topology|discrete topology]] on $S$.
+Then $T$ is [[Definition:Compact Topological Space|compact]] {{iff}} $S$ is a [[Definition:Finite Set|finite set]], thereby making $\tau$ the [[Definition:Finite Discrete Topology|finite discrete topology]] on $S$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $T$ be a [[Definition:Compact Topological Space|compact]] [[Definition:Discrete Topology|discrete space]].
+{{AimForCont}} $T$ is [[Definition:Infinite Discrete Topology|infinite]].
+Let $\mathcal C$ be the [[Definition:Cover of Set|cover]] for $S$ defined as:
+:$\mathcal C = \left\{{\left\{{x}\right\}: x \in S}\right\}$
+As $S$ is an [[Definition:Infinite Set|infinite set]] then so is $\mathcal C$.
+Let $\mathcal C'$ be a [[Definition:Proper Subset|proper subset]] of $\mathcal C$.
+Then:
+: $\exists y \in S: \left\{{y}\right\} \notin \mathcal C'$
+and so $\mathcal C'$ is not a [[Definition:Cover of Set|cover]] for $S$.
+So by definition $\mathcal C'$ is not a [[Definition:Subcover|subcover]] of $\mathcal C$.
+So $\mathcal C$ can have no [[Definition:Finite Subcover|finite subcover]].
+Hence by definition $T$ can not be [[Definition:Compact Topological Space|compact]].
+By [[Proof by Contradiction]] it follows that of $T$ is [[Definition:Compact Space|compact]] then $S$ must be [[Definition:Finite Set|finite]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $T$ be a [[Definition:Finite Discrete Topology|finite discrete space]].
+Then from [[Finite Topological Space is Compact]] it follows that $T$ is a [[Definition:Compact Topological Space|compact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Space is Sequentially Compact}
+Tags: Sequentially Compact Spaces, Finite Topological Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] where $S$ is a [[Definition:Finite Set|finite set]].
+Then $T$ is [[Definition:Sequentially Compact Space|sequentially compact]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Finite Topological Space is Compact]]
+:[[Compact Space is Countably Compact]]
+:[[Finite Space is Second-Countable]]
+:[[Second-Countable Space is First-Countable]]
+:[[First-Countable Space is Sequentially Compact iff Countably Compact]].
+Hence the result.
+{{qed}}
+[[Category:Sequentially Compact Spaces]]
+[[Category:Finite Topological Spaces]]
+ax16keb5z9otsvz85ygtjga65ksbolz
+\end{proof}<|endoftext|>
+\section{Discrete Space is Fully Normal}
+Tags: Fully Normal Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
+Then $T$ is [[Definition:Fully Normal Space|fully normal]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Discrete Space is Fully T4|Discrete Space is fully $T_4$]].
+Then we note that from [[Discrete Space satisfies all Separation Properties]], a [[Definition:Discrete Space|discrete space]] is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+Therefore, by definition, $T$ is [[Definition:Fully Normal Space|fully normal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space is Fully T4}
+Tags: Fully T4 Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
+Then $T$ is [[Definition:Fully T4 Space|fully $T_4$]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Set|set]] $\CC$ of all [[Definition:Singleton|singleton]] [[Definition:Subset|subsets]] of $S$:
+:$\CC := \set {\set x: x \in S}$
+From [[Discrete Space has Open Locally Finite Cover]]:
+:$\CC$ is an [[Definition:Open Cover|open cover]] of $T$ which is [[Definition:Locally Finite Cover|locally finite]].
+Let $x \in S$.
+The [[Definition:Star (Topology)|star of $x$ with respect to $\CC$]] is defined as:
+:$\displaystyle x^* := \bigcup \set {U \in \CC: x \in U}$
+That is, the [[Definition:Set Union|union]] of all sets in $\CC$ which contain $x$.
+But here the only set of $\CC$ containing $x$ is $\set x$.
+Now let $\VV$ be an [[Definition:Open Cover|open cover]] of $T$.
+From the above, it follows that $\CC$ is a [[Definition:Cover of Set|cover]] for $S$ such that:
+:$\forall x \in S: \exists U \in \VV: x^* \subseteq U$
+where $x^* = \set x$ is the [[Definition:Star (Topology)|star of $x$ with respect to $\CC$]].
+That is, $\CC$ is a [[Definition:Star Refinement|star refinement]] of $\VV$.
+Therefore, by definition, $T$ is [[Definition:Fully T4 Space|fully $T_4$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Topological Space is Compact}
+Tags: Finite Topological Spaces, Compact Spaces, Finite Topological Space is Compact
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] where $S$ is a [[Definition:Finite Set|finite set]].
+Then $T$ is [[Definition:Compact Topological Space|compact]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathcal V$ be an [[Definition:Open Cover|open cover]] of $T$.
+For each $x \in S$, define $\mathcal V_x$ to be $\left\{{V \in \mathcal V : x \in V}\right\}$
+Since $S$ is [[Definition:Finite Set|finite]], and since by definition of a cover, each $x\in S$ is contained in at least one $V$ in $\mathcal V$, we have that $\left\{{\mathcal V_x : x \in S}\right\}$ is a [[Definition:Finite Set|finite collection]] of nonempty sets.
+From [[Principle of Finite Choice]], there is a choice function which selects one $V_x$ from $\mathcal V_x$ for each $x \in S$.  By definition of $\mathcal V_x$, such a $V_x$ contains $x$.
+Since there were only [[Definition:Finite Set|finitely many]] $\mathcal V_x$, this provides [[Definition:Finite Set|finitely many]] open sets $V_x \in \mathcal V$ such that $\displaystyle S \subseteq \bigcup_{x \in S} V_x$.
+Thus $\left\{{\mathcal V_x : x \in S}\right\}$ is a [[Definition:Finite Subcover|finite subcover]] of $\mathcal V$.
+The result follows by definition of [[Definition:Compact Topological Space|compact]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Power Set of Finite Set is Finite]], the [[Definition:Power Set|power set]] of $S$ is [[Definition:Finite Set|finite]].
+Since the [[Definition:Topology|topology]] $\tau$ is by definition a [[Definition:Set of Sets|set]] of [[Definition:Subset|subsets]] of $S$, it follows that $\tau$ is [[Definition:Finite Set|finite]] as well.
+Let $\VV$ be an [[Definition:Open Cover|open cover]] of $S$.
+By definition $\VV \subseteq \tau$ and so is also a [[Definition:Finite Set|finite set]].
+From [[Set is Subset of Itself]], $\VV \subseteq \VV$.
+Thus by definition $\VV$ is a [[Definition:Finite Subcover|finite subcover]] of $\VV$.
+The result follows by definition of [[Definition:Compact Topological Space|compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Space is Second-Countable}
+Tags: Second-Countable Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] where $S$ is a [[Definition:Finite Set|finite set]].
+Then $T$ is a [[Definition:Second-Countable Space|second-countable space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] where $S$ is [[Definition:Finite Set|finite]].
+As $S$ is [[Definition:Finite Set|finite]], then so is its [[Definition:Power Set|power set]] $\powerset S$.
+So is $\tau \subseteq \powerset S$.
+Now $\tau$ is a [[Definition:Basis (Topology)|basis]] for itself, as every set in $\tau$ is a [[Definition:Set Union|union]] of sets from $\tau$ (just one).
+But $\tau$ is [[Definition:Finite Set|finite]], and so [[Definition:Countable Set|countable]].
+The result follows from definition of a [[Definition:Second-Countable Space|second-countable space]].
+{{qed}}
+[[Category:Second-Countable Spaces]]
+2qgy61t10hcpmo8ryzugg9d0grhnhzb
+\end{proof}<|endoftext|>
+\section{Finite Discrete Space satisfies all Compactness Properties}
+Tags: Finite Topological Spaces, Compact Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Discrete Topology|finite discrete topological space]].
+Then $T$ satisfies the following [[Definition:Compact Topological Space|compactness properties]]:
+:$T$ is [[Definition:Compact Topological Space|compact]].
+:$T$ is [[Definition:Sequentially Compact Space|Sequentially Compact]].
+:$T$ is [[Definition:Countably Compact Space|Countably Compact]].
+:$T$ is [[Definition:Weakly Countably Compact Space|Weakly Countably Compact]].
+:$T$ is a [[Definition:Lindelöf Space|Lindelöf Space]]
+:$T$ is [[Definition:Pseudocompact Space|Pseudocompact]].
+:$T$ is [[Definition:Sigma-Compact Space|$\sigma$-Compact]].
+:$T$ is [[Definition:Locally Compact Space|Locally Compact]].
+:$T$ is [[Definition:Weakly Locally Compact Space|Weakly Locally Compact]].
+:$T$ is [[Definition:Strongly Locally Compact Space|Strongly Locally Compact]].
+:$T$ is [[Definition:Sigma-Locally Compact Space|$\sigma$-Locally Compact]].
+:$T$ is [[Definition:Weakly Sigma-Locally Compact Space|Weakly $\sigma$-Locally Compact]].
+:$T$ is [[Definition:Fully Normal Space|Fully Normal]].
+:$T$ is [[Definition:Fully T4 Space|Fully $T_4$]].
+:$T$ is [[Definition:Paracompact Space|Paracompact]].
+:$T$ is [[Definition:Countably Paracompact Space|Countably Paracompact]].
+:$T$ is [[Definition:Metacompact Space|Metacompact]].
+:$T$ is [[Definition:Countably Metacompact Space|Countably Metacompact]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Finite Discrete Topology|finite discrete space]] is by definition a [[Definition:Topology|topology]] on a [[Definition:Finite Set|finite set]].
+A [[Discrete Space is Fully Normal]].
+A [[Definition:Fully Normal Space|fully normal space]] is [[Definition:Fully T4 Space|fully $T_4$]] by definition.
+The rest of the results follow directly from [[Finite Space Satisfies All Compactness Properties]].
+\end{proof}<|endoftext|>
+\section{Finite Space Satisfies All Compactness Properties}
+Tags: Finite Topological Spaces, Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] where $S$ is a [[Definition:Finite Set|finite set]].
+Then $T$ satisfies the following [[Definition:Compact Topological Space|compactness properties]]:
+: $T$ is [[Definition:Compact Topological Space|compact]].
+: $T$ is [[Definition:Sequentially Compact Space|sequentially compact]].
+: $T$ is [[Definition:Countably Compact Space|countably compact]].
+: $T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+: $T$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+: $T$ is [[Definition:Pseudocompact Space|pseudocompact]].
+: $T$ is [[Definition:Sigma-Compact Space|$\sigma$-compact]].
+: $T$ is [[Definition:Strongly Locally Compact Space|strongly locally compact]].
+: $T$ is [[Definition:Sigma-Locally Compact Space|$\sigma$-locally compact]].
+: $T$ is [[Definition:Weakly Sigma-Locally Compact Space|weakly $\sigma$-locally compact]].
+: $T$ is [[Definition:Locally Compact Space|locally compact]].
+: $T$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+: $T$ is [[Definition:Paracompact Space|paracompact]].
+: $T$ is [[Definition:Countably Paracompact Space|countably paracompact]].
+: $T$ is [[Definition:Metacompact Space|metacompact]].
+: $T$ is [[Definition:Countably Metacompact Space|countably metacompact]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+: A [[Finite Topological Space is Compact]].
+: A [[Finite Space is Sequentially Compact]].
+The remaining properties are demonstrated in:
+: [[Sequence of Implications of Global Compactness Properties]]
+: [[Sequence of Implications of Local Compactness Properties]]
+: [[Sequence of Implications of Paracompactness Properties]]
+{{qed}}
+[[Category:Finite Topological Spaces]]
+[[Category:Compact Spaces]]
+41h380j70f2hy79vyo0sjcjwk1rof12
+\end{proof}<|endoftext|>
+\section{Discrete Space is Non-Meager}
+Tags: Non-Meager Spaces, Discrete Topology, Discrete Space is Non-Meager
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Discrete Topology|discrete topological space]].
+Then $T$ is [[Definition:Non-Meager Space|non-meager]].
+\end{theorem}
+
+\begin{proof}
+Let $U \subseteq S$ such that $U \ne \O$.
+From [[Interior Equals Closure of Subset of Discrete Space]], we have:
+:$U^\circ = U = U^-$
+where $U^\circ$ is the [[Definition:Interior (Topology)|interior]] and $U^-$ the [[Definition:Closure (Topology)|closure]] of $U$.
+So:
+:$\paren {U^-}^\circ = U \ne \O$
+Thus, by definition, no [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$ is [[Definition:Nowhere Dense|nowhere dense]].
+So $S$ can not be the [[Definition:Set Union|union]] ([[Definition:Countable Set|countable]] or otherwise) of [[Definition:Nowhere Dense|nowhere dense]] [[Definition:Subset|subsets]].
+So by definition $S$ can not be [[Definition:Meager Space|meager]].
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $x \in S$ be an arbitrary point of $T$.
+From [[Set in Discrete Topology is Clopen]] it follows that $\left\{{x}\right\}$ is [[Definition:Open Set (Topology)|open]] in $T$.
+The result follows from [[Space with Open Point is Non-Meager]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space is Complete Metric Space}
+Tags: Discrete Metrics, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
+Then $T$ is a [[Definition:Complete Metric Space|complete metric space]].
+\end{theorem}
+
+\begin{proof}
+Let $d: S \times S \to \R$ be the [[Definition:Standard Discrete Metric|standard discrete metric]].
+From [[Standard Discrete Metric induces Discrete Topology]] we have that $\struct {S, d}$ is a [[Definition:Metric Space|metric space]] whose [[Definition:Topology Induced by Metric|induced topology]] is $\tau$.
+Consider now a [[Definition:Cauchy Sequence|Cauchy sequence]] $\sequence {x_n}_{n \mathop \in \N}$ in $S$.
+By the definition of [[Definition:Cauchy Sequence|Cauchy sequence]]:
+:$\forall \epsilon > 0: \exists N \in \N$ such that $\forall n, m > N: \map d {x_n, x_m} < \epsilon$
+Take $\epsilon = \dfrac 1 2$.
+Then:
+:$\exists N \in \N: \forall n, m > N: \map d {x_n, x_m} < \epsilon = \dfrac 1 2$
+From the definition of [[Definition:Standard Discrete Metric|standard discrete metric]]:
+:$\exists N \in \N: \forall n, m > N: \map d {x_n, x_m} = 0 \implies x_n = x_m$
+Thus the [[Definition:Sequence|sequence]] $\sequence {x_n}_{n \mathop \in \N}$ is [[Definition:Eventually Constant Sequence|constant]] from some $N \in \N$.
+From [[Eventually Constant Sequence is Convergent]]:
+:$x_n \to x \in S$
+Thus every [[Definition:Cauchy Sequence|Cauchy sequence]] in $S$ [[Definition:Convergent Sequence (Metric Space)|converges]].
+The result follows by definition of [[Definition:Complete Metric Space|complete metric space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space is not Dense-In-Itself}
+Tags: Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
+Then $T$ is not [[Definition:Dense-in-itself|dense-in-itself]].
+\end{theorem}
+
+\begin{proof}
+By definition, $T$ is [[Definition:Dense-in-itself|dense-in-itself]] {{iff}} it contains no [[Definition:Isolated Point (Topology)|isolated points]].
+The result follows from [[Topological Space is Discrete iff All Points are Isolated]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Trivial Discrete Space is not Connected}
+Tags: Discrete Topology, Connected Spaces, Non-Trivial Discrete Space is not Connected
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a non-[[Definition:Trivial Topological Space|trivial]] [[Definition:Discrete Topology|discrete topological space]].
+Then $T$ is not [[Definition:Connected Topological Space|connected]].
+Thus also:
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a non-[[Definition:Trivial Topological Space|trivial]] [[Definition:Discrete Space|discrete space]].
+Let $a \in S$.
+Let $A = \set a$ and $B = \relcomp S {\set a}$, where $\relcomp S {\set a}$ is the [[Definition:Relative Complement|complement]] of $A$ in $S$.
+As $T$ is not [[Definition:Trivial Topological Space|trivial]], $B \ne \O$.
+Then from [[Set in Discrete Topology is Clopen]], $A$ and $B$ are both [[Definition:Open Set (Topology)|open]].
+So $A \mid B$ is a [[Definition:Separation (Topology)|separation]] of $S$.
+It follows, by definition, that $T$ is not [[Definition:Connected Topological Space|connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space is Locally Path-Connected}
+Tags: Locally Path-Connected Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Topology|discrete topological space]].
+Then $T$ is [[Definition:Locally Path-Connected Space|locally path-connected]].
+\end{theorem}
+
+\begin{proof}
+From [[Set in Discrete Topology is Clopen]], $\set a$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Basis for Discrete Topology]], the set:
+:$\BB := \set {\set x: x \in S}$
+is a [[Definition:Basis (Topology)|basis]] for $T$.
+Let $\set x \in \BB$.
+From [[Point is Path-Connected to Itself]], it follows that $\set x$ is [[Definition:Path-Connected Set (Topology)|path-connected]].
+Hence $T$ has a [[Definition:Basis (Topology)|basis]] consisting entirely of [[Definition:Path-Connected Set (Topology)|path-connected sets]].
+So by definition $T$ is [[Definition:Locally Path-Connected Space|locally path-connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Point is Path-Connected to Itself}
+Tags: Path-Connected Spaces
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $a \in S$.
+Then $a$ is [[Definition:Path-Connected Points|path-connected]] to itself.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Constant Mapping|constant mapping]] on the [[Definition:Closed Unit Interval|closed unit interval]] $\mathbb I = \left[{0 \,.\,.\, 1}\right]$:
+:$\forall x \in \mathbb I: f_a \left({x}\right) = a$
+Thus, in particular:
+:$f_a \left({0}\right) = a$
+:$f_a \left({1}\right) = a$
+As a [[Constant Mapping is Continuous]], it follows that $f_a$ is a [[Definition:Path (Topology)|path]] in $X$.
+Thus $a$ is [[Definition:Path-Connected Points|path-connected]] to itself.
+{{qed}}
+[[Category:Path-Connected Spaces]]
+9qyrlejgzgz62ovjgvu2scn2ca7ud1j
+\end{proof}<|endoftext|>
+\section{Discrete Uniformity is Uniformity}
+Tags: Discrete Uniformities
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\UU$ be the [[Definition:Discrete Uniformity|discrete uniformity]] on $S$
+Then $\UU$ is indeed a [[Definition:Uniformity|uniformity]].
+\end{theorem}
+
+\begin{proof}
+We examine the [[Definition:Uniformity Axioms|uniformity axioms]] in turn:
+;$\text U 1$: $\forall u \in \UU: \Delta_S \subseteq u$
+This follows by definition of the [[Definition:Discrete Uniformity|discrete uniformity]]:
+:$\UU := \set {u \subseteq S \times S: \Delta_S \subseteq u}$
+;$\text U 2$: $\forall u, v \in \UU: u \cap v \in \UU$
+We have that $\forall u, v \subseteq S \times S: \Delta_S \subseteq u, \Delta_S \subseteq v$
+So $\Delta_S \subseteq u \cap v$ from [[Intersection is Largest Subset]].
+So $u \cap v \in \UU$, and $\text U 2$ holds.
+;$\text U 3$: $u \in \UU, u \subseteq v \subseteq S \times S \implies v \in \UU$
+Let $u \subseteq v \subseteq S \times S$.
+We have that $\Delta_S \subseteq u$ and so $\Delta_S \subseteq v$ from [[Subset Relation is Transitive]].
+So $v \in \UU$ by definition of $\UU$, and $\text U 3$ holds.
+;$\text U 4$: $\forall u \in \UU: \exists v \in \UU: v \circ v \subseteq u$ where $\circ$ is defined as:
+::$u \circ v := \set {\tuple {x, z}: \exists y \in S: \tuple {x, y} \in v, \tuple {y, z} \in u}$
+We have by definition of $\UU$ that $\Delta_S \in \UU$.
+We note from [[Diagonal Relation is Equivalence]] that $\Delta_S$ is an [[Definition:Equivalence Relation|equivalence relation]] and so by definition [[Definition:Transitive Relation|transitive]].
+Therefore from [[Relation contains Composite with Self iff Transitive]] we have that $\Delta_S \circ \Delta_S \subseteq \Delta_S$.
+So:
+:$\forall u \in \UU: \exists \Delta_S \in \UU: \Delta_S \circ \Delta_S \subseteq \Delta_S \subseteq u$
+and so $\text U 4$ holds.
+;$\text U 5$: $\forall u \in \UU: \exists u^{-1} \in \UU$ where $u^{-1}$ is defined as:
+::$u^{-1} := \set {\tuple {y, x}: \tuple {x, y} \in u}$
+Let $u \in \UU$, so $\Delta_S \subseteq u \subseteq S \times S$.
+Thus:
+:$\forall x \in u: \tuple {x, x} \in u \implies \tuple {x, x} \in u^{-1}$
+and so $\Delta_S \subseteq u^{-1}$
+Thus $u^{-1} \in \UU$.
+{{qed|lemma}}
+All of the [[Definition:Uniformity Axioms|uniformity axioms]] are satisfied, and $\UU$ is demonstrated to be a [[Definition:Uniformity|uniformity]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Uniformity generates Discrete Topology}
+Tags: Discrete Uniformities, Discrete Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\UU$ be the [[Definition:Discrete Uniformity|discrete uniformity]] on $S$.
+Then [[Quasiuniformity Induces Topology|the topology generated by $\UU$]] is the [[Definition:Discrete Topology|discrete topology]] on $S$.
+The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$ generates the [[Definition:Basis (Topology)|basis]] for this [[Definition:Discrete Topology|discrete topology]].
+\end{theorem}
+
+\begin{proof}
+From [[Quasiuniformity Induces Topology|the construction]], let $\tau \subseteq \powerset S$ be a [[Definition:Subset|subset]] of the [[Definition:Power Set|power set]] of $S$, created from $\UU$ by:
+:$\tau := \set {\map u x: u \in \UU, x \in X}$
+where:
+:$\forall x \in X: \map u x = \set {y: \tuple {x, y} \in u}$
+We need to show that $\tau$ is the [[Definition:Discrete Topology|discrete topology]].
+Consider $\Delta_S \in \UU$.
+Let $x \in S$.
+Then the set:
+:$U_x := \set {y: \tuple {x, y} \in \Delta_S}$
+We have that:
+$\forall x \in S: U_x = \set x \in \UU$
+Thus we have:
+:$\BB := \set {U_x: x \in S} = \set {\set x: x \in S}$
+From [[Basis for Discrete Topology]], we have that $\BB$ is a [[Definition:Basis (Topology)|basis]] for the [[Definition:Discrete Topology|discrete topology]] on $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Topology is Topology}
+Tags: Indiscrete Topology
+
+\begin{theorem}
+:$\tau$ is a [[Definition:Topology|topology]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \set {\O, S} }$ be the [[Definition:Indiscrete Space|indiscrete space]] on $S$.
+Confirming the criteria for $T$ to be a [[Definition:Topology|topology]]:
+:$(1): \quad$ Trivially, by definition, $\O \in \tau$ and $S \in \tau$.
+:$(2): \quad \O \cup \O = \O \in \tau$, $\O \cup S = S \in \tau$ and $S \cup S = S \in \tau$ from [[Union with Empty Set]] and [[Union is Idempotent]].
+:$(3): \quad \O \cap \O = \O \in \tau$, $\O \cap S = \O \in \tau$ and $S \cap S = S \in \tau$ from [[Intersection with Empty Set]] and [[Intersection is Idempotent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Topology is Coarsest Topology}
+Tags: Indiscrete Topology
+
+\begin{theorem}
+:$\tau$ is the [[Definition:Coarser Topology|coarsest topology]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Let $\phi$ be any [[Definition:Topology|topology]] on $S$.
+Then by definition of [[Definition:Topology|topology]], $\O \in \phi$ and $S \in \phi$
+Hence by definition of [[Definition:Subset|subset]], $\tau \subseteq \phi$. 
+Hence by definition of [[Definition:Coarser Topology|coarser topology]], $\tau$ is [[Definition:Coarser Topology|coarser]] than $\phi$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open and Closed Sets in Indiscrete Topology}
+Tags: Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Let $H \subseteq S$.
+\end{theorem}<|endoftext|>
+\section{Subset of Indiscrete Space is Compact and Sequentially Compact}
+Tags: Indiscrete Topology, Compact Spaces, Sequentially Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Let $H \subseteq S$.
+\end{theorem}<|endoftext|>
+\section{Limit Points of Indiscrete Space}
+Tags: Limit Points, Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]] consisting of at least two points.
+Let $H$ be a [[Definition:Subset|subset]] of $T$ such that $H \ne \O$.
+Then every point of $T$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+\end{theorem}
+
+\begin{proof}
+By definition, $x \in \struct {S, \tau}$ is a [[Definition:Limit Point of Set|limit point of $H$]] if every [[Definition:Open Set (Topology)|open set]] $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$.
+Here, of course, there is only one [[Definition:Open Set (Topology)|open set]] that contains any points at all, and that is $S$.
+So as $S$ contains more than one point, it follows that every point of $T$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence in Indiscrete Space converges to Every Point}
+Tags: Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Let $\sequence {s_n}$ be a [[Definition:Sequence|sequence]] in $T$.
+Then $\sequence {s_n}$ [[Definition:Convergent Sequence (Topology)|converges]] to every point of $S$.
+\end{theorem}
+
+\begin{proof}
+Let $\alpha \in S$.
+By definition, $\sequence {s_n}$ [[Definition:Convergent Sequence (Topology)|converges]] to $\alpha$ if every [[Definition:Open Set (Topology)|open set]] in $T$ containing $\alpha$ contains all but a [[Definition:Finite Set|finite number]] of terms of $\sequence {s_n}$.
+But as $T$ has only one [[Definition:Open Set (Topology)|open set]] containing any points at all, '''every''' point of $\sequence {s_n}$ is contained in every [[Definition:Open Set (Topology)|open set]] in $T$ containing $\alpha$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Indiscrete Space is Dense-in-itself}
+Tags: Indiscrete Topology, Denseness
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Let $H \subseteq S$ be a [[Definition:Subset|subset]] of $S$ containing more than one point.
+Then $H$ is [[Definition:Dense-in-itself|dense-in-itself]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in H$.
+Then as $H$ is not [[Definition:Singleton|singleton]], $\exists y \in H: y \ne x$.
+Then every [[Definition:Neighborhood (Topology)|neighborhood]] of $x$ contains $y$, as the only [[Definition:Open Set (Topology)|open set]] of $T$ is $S$, which also contains both $x$ and $y$.
+Hence $x$ is not [[Definition:Isolated Point (Topology)|isolated]] by definition.
+$x$ is general, so all points in $H$ are similarly not [[Definition:Isolated Point (Topology)|isolated]].
+Hence the subset $H$ is [[Definition:Dense-in-itself|dense-in-itself]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Space is Non-Meager}
+Tags: Indiscrete Topology, Non-Meager Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Then $T$ is [[Definition:Non-Meager Space|non-meager]].
+\end{theorem}
+
+\begin{proof}
+Let $H \subseteq S$ such that $H \ne \O$.
+From [[Limit Points of Indiscrete Space]], every point of $S$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+So, by definition:
+:$H^- = S$
+where $H^-$ is the [[Definition:Closure (Topology)|closure]] of $H$ in $T$.
+Now the [[Definition:Interior (Topology)|interior]] of $S$ is $S$ itself (trivially, by definition).
+So:
+:$\paren {H^-}^\circ = S \ne \O$
+where $\paren {H^-}^\circ$ is the [[Definition:Interior (Topology)|interior]] of the [[Definition:Closure (Topology)|closure]] of $H$.
+So, by definition, $H$ is not [[Definition:Nowhere Dense|nowhere dense]].
+(Note that if $H = \O$, then [[Empty Set is Nowhere Dense]] applies, which is why the stipulation above that $H \ne \O$.)
+It follows that $H$ can not be the [[Definition:Countable Union|countable union]] of [[Definition:Nowhere Dense|nowhere dense]] [[Definition:Subset|subsets]] of $S$.
+So $H$ is, by definition, not [[Definition:Meager Space|meager]], so must be [[Definition:Non-Meager Space|non-meager]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Empty Set is Nowhere Dense}
+Tags: Empty Set, Denseness
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then the [[Definition:Empty Set|empty set]] $\O$ is [[Definition:Nowhere Dense|nowhere dense]] in $T$.
+\end{theorem}
+
+\begin{proof}
+From [[Empty Set is Closed in Topological Space]], $\O$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+From [[Closed Set Equals its Closure]]:
+:$\O^- = \O$
+where $\O^-$ is the the [[Definition:Closure (Topology)|closure]] of $\O$.
+From [[Empty Set is Element of Topology]], $\O$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From the definition (trivially) we also have that:
+:$\O^\circ = \O$
+where $\O^\circ$ is the [[Definition:Interior (Topology)|interior]] of $\O$.
+So:
+:$\struct {\O^-}^\circ = \O$
+and so by definition $\O$ is [[Definition:Nowhere Dense|nowhere dense]] in $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Interior of Open Set}
+Tags: Set Interiors, Open Sets
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $U \subseteq T$ be [[Definition:Open Set (Topology)|open]] in $T$.
+Then:
+: $U^\circ = U \iff U \in \tau$
+where $U^\circ$ is the [[Definition:Interior (Topology)|interior]] of $U$.
+That is, a [[Definition:Subset|subset]] of $S$ is [[Definition:Open Set (Topology)|open]] in $T$ {{iff}} it equals its [[Definition:Interior (Topology)|interior]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $U \subseteq T$ be [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Set Interior is Largest Open Set]]:
+: $U \subseteq U^\circ$
+But by definition of [[Definition:Interior (Topology)|interior]]:
+: $U^\circ \subseteq U$
+Hence $U^\circ = U$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose $U^\circ = U$.
+By definition $U^\circ$ is the [[Definition:Set Union|union]] of all [[Definition:Subset|subsets]] of $U$ which are [[Definition:Open Set (Topology)|open]] in $U$.
+By the definition of a [[Definition:Topology|topology]], that means $U$ itself must also be [[Definition:Open Set (Topology)|open]].
+{{qed}}
+[[Category:Set Interiors]]
+[[Category:Open Sets]]
+f6uidcbbematexftq7jgr2afde1q33s
+\end{proof}<|endoftext|>
+\section{Interior of Subset of Indiscrete Space}
+Tags: Indiscrete Topology, Set Interiors
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Let $H \subset S$ (that is, let $H$ be a [[Definition:Proper Subset|proper subset]] of $T$).
+Then:
+:$H^\circ = H^{\circ -} = H^{\circ - \circ} = \O$
+where:
+:$H^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of $H$
+:$H^-$ denotes the [[Definition:Closure (Topology)|closure]] of $H$.
+\end{theorem}
+
+\begin{proof}
+As $H \subset S$, it follows that $H \ne T$.
+So the only [[Definition:Open Set (Topology)|open subset]] of $H$ is $\O$.
+So by definition:
+:$H^\circ = \O$
+From [[Empty Set is Closed in Topological Space]], $\O$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+From [[Closed Set Equals its Closure]]:
+:$\O^- = \O$
+From [[Empty Set is Element of Topology]], $\O$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Interior of Open Set]]:
+:$\O^\circ = \O$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Subset of Indiscrete Space}
+Tags: Set Closures, Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Let $\O \subsetneq H \subseteq S$ (that is, let $H$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $T$).
+Then:
+:$H^- = H^{- \circ} = H^{- \circ -} = S$
+where:
+:$H^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of $H$
+:$H^-$ denotes the [[Definition:Closure (Topology)|closure]] of $H$.
+\end{theorem}
+
+\begin{proof}
+From [[Limit Points of Indiscrete Space]], every point in $S$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+So from the definition of [[Definition:Closure (Topology)|closure]], $H \ne \O \implies H^-= S$.
+From the [[Definition:Open Set Axioms|open set axioms]], $S$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Interior of Open Set]], $S^\circ = S$.
+From [[Underlying Set of Topological Space is Clopen]], $S$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+From [[Closed Set Equals its Closure]], $S^- = S$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boundary of Subset of Indiscrete Space}
+Tags: Indiscrete Topology, Boundaries
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Let $\O \subsetneq H \subsetneq S$ (that is, let $H$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Proper Subset|proper subset]] of $T$).
+Then:
+:$\partial H = S$
+where $\partial H$ denotes the [[Definition:Boundary (Topology)|boundary]] of $H$.
+If $H = \O$ or $H = S$ then $\partial H = \O$.
+\end{theorem}
+
+\begin{proof}
+From [[Closure of Subset of Indiscrete Space]], $H^- = S$, where $H^-$ denotes [[Definition:Closure (Topology)|set closure]].
+From [[Interior of Subset of Indiscrete Space]], $H^\circ = \O$, where $H^\circ$ denotes [[Definition:Interior (Topology)|set interior]].
+By definition:
+:$\partial H = H^- \setminus H^\circ = S \setminus \O = S$
+From [[Open and Closed Sets in Topological Space]], $\O$ and $S$ are both [[Definition:Closed Set (Topology)|closed]] and [[Definition:Open Set (Topology)|open]] in $T$.
+So from [[Set Clopen iff Boundary is Empty]]:
+:$\partial H = \O$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boundary of Boundary of Subset of Indiscrete Space}
+Tags: Indiscrete Topology, Boundaries
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Let $H \subseteq S$.
+Then:
+:$\map \partial {\partial H} = \O$
+where $\partial H$ denotes the [[Definition:Boundary (Topology)|boundary]] of $H$.
+\end{theorem}
+
+\begin{proof}
+From [[Boundary of Subset of Indiscrete Space]], either $\partial H = S$ or $\partial H = \O$, depending on whether $H = \O$ or $H = S$ or not.
+From [[Open and Closed Sets in Topological Space]], $\O$ and $S$ are both [[Definition:Closed Set (Topology)|closed]] and [[Definition:Open Set (Topology)|open]] in $T$.
+So from [[Set Clopen iff Boundary is Empty]]:
+:$\map \partial {\partial H} = \O$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Indiscrete Space is Everywhere Dense}
+Tags: Indiscrete Topology, Denseness
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Let $H \subseteq S$ such that $H \ne \O$.
+Then $H$ is [[Definition:Everywhere Dense|everywhere dense]].
+\end{theorem}
+
+\begin{proof}
+From [[Limit Points of Indiscrete Space]], every point of $T$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+Hence $H$ is [[Definition:Everywhere Dense|everywhere dense]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Space is Separable}
+Tags: Indiscrete Topology, Separable Spaces, Indiscrete Space is Separable
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]] such that $S$ has more than one element.
+Then $T$ is [[Definition:Separable Space|separable]].
+\end{theorem}<|endoftext|>
+\section{Indiscrete Space is Second-Countable}
+Tags: Second-Countable Spaces, Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Then  $T$ is a [[Definition:Second-Countable Space|second-countable space]].
+\end{theorem}
+
+\begin{proof}
+The only [[Definition:Basis (Topology)|basis]] for $T$ is $\set S$ which is trivially [[Definition:Countable Set|countable]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping to Indiscrete Space is Continuous}
+Tags: Discrete Topology
+
+\begin{theorem}
+Let $T_1 = \struct {S_1, \tau_1}$ be any [[Definition:Topological Space|topological space]].
+Let $T_2 = \struct {S_2, \tau_2}$ be the [[Definition:Indiscrete Space|indiscrete topological space]] on $S_2$.
+Let $\phi: S_1 \to S_2$ be a [[Definition:Mapping|mapping]].
+Then $\phi$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Everywhere Continuous Mapping (Topology)|continuous]]:
+:$U \in \tau_2 \implies \phi^{-1} \sqbrk U \in \tau_1$
+The only [[Definition:Element|elements]] of $\tau_2$ are $S_2$ and $\O$, from which:
+:$\phi^{-1} \sqbrk {S_2} = S_1 \in \tau_1$
+:$\phi^{-1} \sqbrk \O = \O \in \tau_1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Space is Path-Connected}
+Tags: Path-Connected Spaces, Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Then $T$ is [[Definition:Path-Connected Space|path-connected]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in S$.
+Consider any [[Definition:Mapping|mapping]] $f: \closedint 0 1 \to S$ such that $\map f 0 = a$ and $\map f 1 = b$.
+From [[Mapping to Indiscrete Space is Continuous]], we have that $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+The result follows by definition of [[Definition:Path-Connected Space|path-connectedness]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Space is Connected}
+Tags: Connected Spaces, Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Then $T$ is [[Definition:Connected Topological Space|connected]].
+\end{theorem}
+
+\begin{proof}
+We have that an [[Indiscrete Space is Path-Connected]].
+Then we have that a [[Path-Connected Space is Connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Space is Arc-Connected iff Uncountable}
+Tags: Arc-Connected Spaces, Indiscrete Topology, Uncountable Sets
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Then $T$ is [[Definition:Arc-Connected Space|arc-connected]] {{iff}} $S$ is an [[Definition:Uncountable Set|uncountable set]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]] such that $S$ is [[Definition:Uncountable Set|uncountable]].
+Let $a, b \in S$.
+Consider an [[Definition:Injection|injection]] $f: \closedint 0 1 \to S$ such that $\map f 0 = a$ and $\map f 1 = b$.
+This can always be found because $S$ is itself [[Definition:Uncountable Set|uncountable]].
+From [[Mapping to Indiscrete Space is Continuous]], we have that $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Thus $T$ is [[Definition:Arc-Connected Space|arc-connected]].
+Now suppose $S$ is an [[Definition:Indiscrete Space|indiscrete topological space]] which is [[Definition:Arc-Connected Space|arc-connected]].
+Then there exists an [[Definition:Injection|injection]] $f: \closedint 0 1 \to S$ such that $\map f 0 = a$ and $\map f 1 = b$.
+This can only exist if $S$ is [[Definition:Uncountable Set|uncountable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Space is Irreducible}
+Tags: Irreducible Spaces, Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Then $T$ is [[Definition:Irreducible Space|irreducible]].
+\end{theorem}
+
+\begin{proof}
+There is only one [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open set]] in $T$.
+So there can be no two [[Definition:Open Set (Topology)|open sets]] in $T$ which are [[Definition:Disjoint Sets|disjoint]].
+Hence (trivially) $T$ is [[Definition:Irreducible Space|irreducible]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Space is Ultraconnected}
+Tags: Ultraconnected Spaces, Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Then $T$ is [[Definition:Ultraconnected Space|ultraconnected]].
+\end{theorem}
+
+\begin{proof}
+There is only one [[Definition:Non-Empty Set|non-empty]] [[Definition:Closed Set (Topology)|closed set]] in $T$.
+So there can be no two [[Definition:Closed Set (Topology)|closed sets]] in $T$ which are [[Definition:Disjoint Sets|disjoint]].
+Hence (trivially) $T$ is [[Definition:Ultraconnected Space|ultraconnected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Non-Singleton Space is not T0}
+Tags: T0 Spaces, Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]] which has more than one [[Definition:Element|element]].
+Then $T$ is not a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in S$.
+By definition of [[Definition:Indiscrete Space|indiscrete space]], $S$ is the only [[Definition:Open Set (Topology)|open set]] in $T$ which is [[Definition:Non-Empty Set|non-empty]].
+So (trivially) there is no [[Definition:Open Set (Topology)|open set]] in $T$ containing $a$ and not $b$, or $b$ and not $a$.
+Hence the result, by definition of [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Space is T5}
+Tags: T5 Spaces, Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Then $T$ is a [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+By definition, the only two [[Definition:Separated Sets|separated sets]] in $T$ are $S$ and $\O$.
+Then there exist two [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $S$ and $\O$ containing $S$ and $\O$ respectively.
+Hence (trivially) $T$ is a [[Definition:T5 Space|$T_5$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Space is T4}
+Tags: T4 Spaces, Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Then $T$ is a [[Definition:T4 Space|$T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+We have that an [[Indiscrete Space is T5|indiscrete space is a $T_5$ space]].
+Then we have that a [[T5 Space is T4 Space|$T_5$ Space is a $T_4$ Space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Space is T3}
+Tags: T3 Spaces, Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]] .
+Then $T$ is a [[Definition:T3 Space|$T_3$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $F \subseteq S$ be a [[Definition:Closed Set (Topology)|closed set]] in $S$.
+Let $y \in S$ such that $y \notin F$.
+The only way this can happen is if $F = \O$.
+So there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.
+That is, $U = \O$ and $V = S$.
+Hence (trivially) the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Indiscrete Space is Pseudometrizable}
+Tags: Pseudometric Spaces, Indiscrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Then $T$ is [[Definition:Pseudometrizable Topology|pseudometrizable]].
+\end{theorem}
+
+\begin{proof}
+Let $d: S \times S \to \R$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x, y \in S: \map d {x, y} = 0$
+Then clearly $d$ is a [[Definition:Pseudometric|pseudometric]].
+Let $\struct {S, \tau_{\struct {S, d} } }$ be the [[Pseudometric induces Topology|topological space induced]] by $d$.
+Since $\struct {S, \tau_{\struct {S, d} } }$ is a [[Definition:Topological Space|topological space]], by [[Empty Set is Element of Topology]] we have that $\O$ is [[Definition:Open Set (Topology)|open]].
+For any $\epsilon \in \R_{>0}$ we have:
+:$\map {B_\epsilon} a := \set {x \in S: \map d {x, a} < \epsilon} = S$
+where $\map {B_\epsilon} a$ is the [[Definition:Open Ball of Pseudometric Space|open $\epsilon$-ball]] of $a$.
+Let $U\subset S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open set]].
+By [[Open Sets in Pseudometric Space]] we see that for every $x \in U$, there must exist an $\epsilon > 0$ such that $\map {B_\epsilon} x \subset U$.
+However, $S \subset \map {B_\epsilon} x$.
+Thus $U = S$.
+Hence it is seen that $\O$ and $S$ are the only [[Definition:Open Set (Topology)|open sets]] of this resulting [[Definition:Pseudometric Space|pseudometric space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partition Topology is Topology}
+Tags: Partition Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\PP$ be a [[Definition:Partition (Set Theory)|partition]] of $S$.
+Let $\tau$ be the [[Definition:Set of Sets|set of subsets]] of $S$ defined as:
+:$a \in \tau \iff a$ is the [[Definition:Set Union|union]] of sets of $\PP$
+Then $\tau$ is a [[Definition:Topology|topology]] on $S$.
+\end{theorem}
+
+\begin{proof}
+From [[Basis for Partition Topology]], we have that $\PP$ is a [[Definition:Basis (Topology)|basis]] for the [[Definition:Partition Topology|partition topology]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis for Partition Topology}
+Tags: Partition Topology, Topological Bases
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\PP$ be a [[Definition:Partition (Set Theory)|partition]] of $S$.
+Let $\tau$ be the [[Definition:Partition Topology|partition topology]] on $S$ defined as:
+:$a \in \tau \iff a$ is the [[Definition:Set Union|union]] of [[Definition:Set|sets]] of $\mathcal P$
+Then $\PP$ forms a [[Definition:Basis (Topology)|basis]] of $\tau$.
+\end{theorem}
+
+\begin{proof}
+Checking the criteria for $\PP$ to be a [[Definition:Synthetic Basis|synthetic basis]] for $\tau$:
+We have that $\displaystyle S = \bigcup \PP$ from the definition of a [[Definition:Partition (Set Theory)|partition]].
+Therefore, $\displaystyle S \subseteq \bigcup \PP$ and $\PP$ is a [[Definition:Cover of Set|cover]] for $S$.
+Next, let $B_1, B_2 \in \PP$.
+Then as $\PP$ is a [[Definition:Partition (Set Theory)|partition]] of $S$, we have that $B_1 \cap B_2 = \O$.
+But from [[Union of Empty Set]] we have that $\O$ is the (vacuous) [[Definition:Set Union|union]] of [[Definition:Set|sets]] of $\PP$.
+Hence $\PP$ is a [[Definition:Synthetic Basis|synthetic basis]] for $\tau$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Set in Partition Topology is also Closed}
+Tags: Partition Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then $T$ is a [[Definition:Partition Space|partition space]] {{iff}}:
+:$\forall U \subseteq S: U \in \tau \iff \relcomp S U \in \tau$
+That is, a [[Definition:Topological Space|topological space]] is a [[Definition:Partition Space|partition space]] {{iff}} all [[Definition:Open Set (Topology)|open sets]] are [[Definition:Closed Set (Topology)|closed]], and all [[Definition:Closed Set (Topology)|closed sets]] are also [[Definition:Open Set (Topology)|open]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+=== Necessary Condition ===
+Let $T$ be a [[Definition:Partition Space|partition space]].
+Then there exists a [[Definition:Partition (Set Theory)|partition]] $\PP$ which forms the [[Basis for Partition Topology|basis for $T$]].
+Let $U \in \tau$.
+Then $U$ is the [[Definition:Set Union|union]] of elements of $\PP$.
+Then by definition $\relcomp S U$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+Then $\relcomp S U$ is the [[Definition:Set Union|union]] of all the other elements of $\PP$.
+That is, $\relcomp S U$ is also [[Definition:Open Set (Topology)|open]] in $T$.
+By definition, then, $\relcomp S {\relcomp S U} = U$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Now suppose that every [[Definition:Open Set (Topology)|open set]] in $T$ is [[Definition:Closed Set (Topology)|closed]], and every [[Definition:Closed Set (Topology)|closed set]] in $T$ is [[Definition:Open Set (Topology)|open]].
+{{AimForCont}} $T$ is not a [[Definition:Partition Space|partition space]].
+Thus, by definition, there exists an [[Definition:Open Set (Topology)|open set]] of $T$ which is not the [[Definition:Set Union|union]] of [[Definition:Disjoint Sets|disjoint sets]].
+{{finish|Anyone?}}
+\end{proof}<|endoftext|>
+\section{Open Set in Partition Topology is Component}
+Tags: Partition Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Partition Space|partition topological space]].
+Then each of its [[Definition:Open Set (Topology)|open sets]] are [[Definition:Component (Topology)|components]] of $T$.
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Partition (Topology)|partition]] $\PP$ be a [[Definition:Basis (Topology)|basis]] of $T$.
+From [[Open Set in Partition Topology is also Closed]], open sets are in fact [[Definition:Clopen Set|clopen]].
+So the elements of $\PP$ are [[Definition:Clopen Set|clopen]].
+The result follows from the definition of [[Definition:Component (Topology)|components]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Topology of Partition Topology is Discrete Space}
+Tags: Discrete Topology, Partition Topology
+
+\begin{theorem}
+Let $\PP$ be a [[Definition:Partition (Set Theory)|partition]] of a [[Definition:Set|set]] $S$.
+Let $T = \struct {S, \tau}$ be the [[Definition:Partition Space|partition space]] formed from $\PP$.
+Let $S / \PP$ be the [[Definition:Quotient Set|quotient set]] of $S$ by $\PP$.
+Then the [[Definition:Quotient Topology|quotient topology]] $\tau_{S / \PP}$ is a [[Definition:Discrete Topology|discrete topology]].
+\end{theorem}
+
+\begin{proof}
+Let $\BB$ be the [[Definition:Set|set]] defined as:
+:$\BB = \set {\set A: A \in S / \PP}$
+From [[Basis for Partition Topology]], $\BB$ forms a [[Definition:Basis (Topology)|basis]] for a [[Definition:Partition Space|partition space]] on $S$.
+From [[Basis for Discrete Topology]], $\BB$ forms a [[Definition:Basis (Topology)|basis]] for the [[Definition:Discrete Topology|discrete topology]] on $S / \PP$.
+Hence the result, by definition of [[Definition:Quotient Topology|quotient topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Singleton Partition yields Indiscrete Topology}
+Tags: Indiscrete Topology, Partition Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] which is [[Definition:Non-Empty Set|not empty]].
+Let $\PP$ be the [[Definition:Singleton Partition|(trivial) singleton partition]] $\set S$ on $S$.
+Then the [[Definition:Partition Topology|partition topology]] on $\PP$ is the [[Definition:Indiscrete Topology|indiscrete topology]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Partition Topology|partition topology]] on $\PP$ is the set of all [[Definition:Set Union|unions]] from $\PP$.
+This is (trivially, and from [[Union of Empty Set]]) $\set {\O, S}$ which is the [[Definition:Indiscrete Topology|indiscrete topology]] on $S$ by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partition of Singletons yields Discrete Topology}
+Tags: Discrete Topology, Partition Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] which is [[Definition:Non-Empty Set|non-empty]].
+Let $\PP$ be the [[Definition:Partition of Singletons|(trivial) partition of singletons]] on $S$:
+:$\PP = \set {\set x: x \in S}$
+Then the [[Definition:Partition Topology|partition topology]] on $\PP$ is the [[Definition:Discrete Topology|discrete topology]].
+\end{theorem}
+
+\begin{proof}
+From [[Basis for Discrete Topology]] it is shown that $\PP$ as defined here forms the [[Definition:Basis (Topology)|basis]] of the [[Definition:Discrete Topology|discrete topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partition Topology is not T0}
+Tags: T0 Spaces, Partition Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] and let $\PP$ be a [[Definition:Partition (Set Theory)|partition]] on $S$ which is not the [[Definition:Partition of Singletons|(trivial) partition of singletons]].
+Let $T = \struct {S, \tau}$ be the [[Definition:Partition Space|partition space]] whose [[Basis for Partition Topology|basis]] is $\PP$.
+Then $T$ is not a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+\end{theorem}
+
+\begin{proof}
+As $\PP$ is not the [[Definition:Partition of Singletons|partition of singletons]], there exists some $H \in \PP$ such that $a, b \in H: a \ne b$.
+Any [[Definition:Set Union|union]] of sets from $\PP$ which includes $H$ will therefore contain both $a$ and $b$.
+Therefore, any element of $\tau$ containing $a$ will also contain $b$, and similarly, any element of $\tau$ containing $b$ will also contain $a$.
+So there is no [[Definition:Open Set (Topology)|open set]] in $T$ containing $a$ and not $b$, or $b$ and not $a$.
+Hence the result, by definition of [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+{{qed}}
+Note the reason for the exception in the statement as given.
+If $\PP$ is the [[Definition:Partition of Singletons|partition of singletons]]:
+:$\PP = \set {\set x: x \in S}$
+then $T$ as constructed is the [[Partition of Singletons yields Discrete Topology|discrete topology]] on $S$.
+From [[Discrete Space satisfies all Separation Properties]], in that case $T$ ''does'' satisfy the [[Definition:Kolmogorov Space|$T_0$ property]].
+\end{proof}<|endoftext|>
+\section{Partition Topology is T5}
+Tags: T5 Spaces, Partition Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] and let $\PP$ be a [[Definition:Partition (Set Theory)|partition]] on $S$ which is not the [[Definition:Partition of Singletons|(trivial) partition of singletons]].
+Let $T = \struct {S, \tau}$ be the [[Definition:Partition Space|partition space]] whose [[Basis for Partition Topology|basis]] is $\PP$.
+Then $T$ is a [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $A$ and $B$ be subsets of $S$ such that $A^- \cap B = A \cap B^- = \O$.
+From [[Open Set in Partition Topology is also Closed]], we get that $A^-$ and $B^-$ are both [[Definition:Clopen Set|clopen]]. 
+From [[Set is Subset of its Topological Closure]]:
+:$A \subseteq A^-$
+From $A^- \cap B = \O$ it follows from [[Intersection with Complement is Empty iff Subset]] that $B \subseteq S \setminus \paren {A^-}$.
+Since $A^-$ is [[Definition:Clopen Set|clopen]], it follows that $A^-$ and $S \setminus \paren {A^-}$ are both [[Definition:Open Set (Topology)|open]].
+Also, they are [[Definition:Disjoint Sets|disjoint]] by definition.
+Thus we have $A^-$, an [[Definition:Open Set (Topology)|open set]] containing $A$, and $S \setminus \paren {A^-}$, an [[Definition:Open Set (Topology)|open set]] containing $B$.
+Thus, by definition, $T$ is a [[Definition:T5 Space|$T_5$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partition Topology is T3 1/2}
+Tags: T3 1: 2 Spaces, Partition Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] and let $\PP$ be a [[Definition:Partition (Set Theory)|partition]] on $S$.
+Let $T = \struct {S, \tau}$ be the [[Definition:Partition Space|partition space]] whose [[Basis for Partition Topology|basis]] is $\PP$.
+Then $T$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $F \subseteq S$ be [[Definition:Closed Set (Topology)|closed]], and let $x \in S$, $x \notin F$. 
+Denote by $S \setminus F$ the [[Definition:Relative Complement|relative complement]] of $F$ in $S$.
+Define a [[Definition:Mapping|mapping]] $f: S \to \closedint 0 1$ as:
+:$\map f s := \begin{cases}
+ 1 & : \text { if $s \in F$} \\
+ 0 & : \text { if $s \in S \setminus F$}
+\end{cases}$
+Then $f$ is identically $1$ on $F$, and identically $0$ on $\set x$.
+Now if $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]], it will be a [[Definition:Urysohn Function|Urysohn function]] for $F$ and $\set y$, and $T$ will be a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+Now for any $V \subseteq \closedint 0 1$, we have:
+:$f^{-1} \sqbrk V = \begin{cases}
+ \O            & : \text{ if $0,1 \notin V$} \\
+ F             & : \text{ if $0 \notin V$ and $1 \in V$} \\
+ S \setminus F & : \text{ if $0 \in V$ and $1 \notin V$} \\
+ S             & : \text{ if $0,1 \in V$}
+\end{cases}$
+By definition of $\tau$, $F$ is [[Definition:Open Set (Topology)|open]] in $T$.
+By [[Open Set in Partition Topology is also Closed]], $F$ is also [[Definition:Closed Set (Topology)|closed]], and so $S \setminus F$ is [[Definition:Open Set (Topology)|open]] in $T$.
+Thus, the [[Definition:Preimage of Subset under Mapping|preimage]] of any [[Definition:Subset|subset]] $V$ of $\closedint 0 1$ is [[Definition:Open Set (Topology)|open]] in $T$.
+In particular, this holds for the [[Definition:Open Set (Topology)|open sets]] of $\closedint 0 1$.
+It follows that $f$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]], and so a [[Definition:Urysohn Function|Urysohn function]].
+Hence $T$ is [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Odd-Even Topology is Second-Countable}
+Tags: Second-Countable Spaces, Odd-Even Topology
+
+\begin{theorem}
+Let $T = \struct {\Z_{>0}, \tau}$ be a [[Definition:Topological Space|topological space]] where $\tau$ is the [[Definition:Odd-Even Topology|odd-even topology]] on the [[Definition:Strictly Positive Integer|strictly positive integers]] $\Z_{>0}$.
+Then $T$ is [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+From [[Basis for Partition Topology]], the set:
+:$\BB := \set {\set {2 k - 1, 2 k}: k \in \Z, k > 0}$
+is a [[Definition:Basis (Topology)|basis]] for $T$.
+There is an obvious [[Definition:Bijection|one-to-one correspondence]] $\phi: \Z_{>0} \leftrightarrow \BB$ between $\Z_{>0}$ and $\BB$:
+:$\forall x \in \Z_{>0}: \map \phi x = \set {2 x - 1, 2 x}$
+But $\Z_{>0} \subseteq \Z$, and [[Integers are Countably Infinite]].
+So from [[Subset of Countably Infinite Set is Countable]], $\Z_{>0}$ is [[Definition:Countable Set|countable]].
+Thus $\BB$ is also [[Definition:Countable Set|countable]] by definition of [[Definition:Countable Set|countability]].
+So we have that $T$ has a [[Definition:Countable Basis|countable basis]], and so is [[Definition:Second-Countable Space|second-countable]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Odd-Even Topology is Weakly Countably Compact}
+Tags: Weakly Countably Compact Spaces, Odd-Even Topology
+
+\begin{theorem}
+Let $T = \struct {\Z_{>0}, \tau}$ be a [[Definition:Topological Space|topological space]] where $\tau$ is the [[Definition:Odd-Even Topology|odd-even topology]] on the [[Definition:Strictly Positive Integer|strictly positive integers]] $\Z_{>0}$.
+Then $T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+\end{theorem}
+
+\begin{proof}
+Let $H \subseteq \Z_{>0}$ such that $H$ is [[Definition:Infinite Set|infinite]].
+Let $x \in H$.
+By definition, the [[Definition:Odd-Even Topology|odd-even topology]] is a [[Definition:Partition Topology|partition topology]].
+So $U$ is a [[Definition:Set Union|union]] of sets of the form $\set {2 k - 1, 2 k}$.
+Now if $x \in U$, it will be of the form $2 k - 1$ or $2 k$.
+So there will exist $y \in U$ of the form $2 k$ or $2 k - 1$.
+So, by definition, $x$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+So, by definition, $T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Odd-Even Topology is not Countably Compact}
+Tags: Countably Compact Spaces, Odd-Even Topology
+
+\begin{theorem}
+Let $T = \struct {\Z_{>0}, \tau}$ be a [[Definition:Topological Space|topological space]] where $\tau$ is the [[Definition:Odd-Even Topology|odd-even topology]] on the [[Definition:Positive Integer|strictly positive integers]] $\Z_{>0}$.
+Then $T$ is not [[Definition:Countably Compact Space|countably compact]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Odd-Even Topology|odd-even topology]] is a [[Definition:Partition Topology|partition topology]].
+Let $\PP$ be the [[Definition:Partition (Set Theory)|partition]] which is the [[Basis for Partition Topology|basis for $T$]]:
+:$\PP = \set {\set {2 k - 1, 2 k}: k \in \Z_{>0} }$
+Then $\PP$ is a [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $S$ which has no [[Definition:Finite Subcover|finite subcover]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Discrete Space is not Weakly Countably Compact}
+Tags: Discrete Topology, Weakly Countably Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Discrete Topology|countable discrete space]].
+Then $T$ is not [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+\end{theorem}
+
+\begin{proof}
+Let $A \subseteq S$ be an [[Definition:Infinite Set|infinite]] subset of $S$.
+Then as [[Set in Discrete Topology is Clopen]], $A$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+From [[Closed Set Equals its Closure]], $A = A^-$ where $A^-$ is the [[Definition:Closure (Topology)|closure]] of $A$.
+From the definition, $x$ is a [[Definition:Limit Point of Set|limit point of $A$]] if it belongs to the [[Definition:Closure (Topology)|closure]] of $A$ but is not an [[Definition:Isolated Point (Topology)|isolated point]] of $A$.
+But then we have [[Topological Space is Discrete iff All Points are Isolated]].
+So $A$ has no [[Definition:Limit Point of Set|limit points]] in $T$.
+So $T$ is not [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Weak Countable Compactness is not Preserved under Continuous Maps}
+Tags: Weakly Countably Compact Spaces
+
+\begin{theorem}
+Let $T_A = \struct {S_A, \tau_A}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+Let $T_B = \struct {S_B, \tau_B}$ be another [[Definition:Topological Space|topological space]].
+Let $\phi: T_A \to T_B$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+Then $T_B$ is not necessarily [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+\end{theorem}
+
+\begin{proof}
+Let $\Z_{>0}$ be the [[Definition:Strictly Positive Integer|strictly positive integers]]:
+:$\Z_{>0} = \set {1, 2, 3, \ldots}$
+Let $T_A = \struct {\Z_{>0}, \tau_A}$ be the [[Definition:Odd-Even Topology|odd-even topology]].
+Let $T_B = \struct {\Z_{>0}, \tau_B}$ be the [[Definition:Discrete Topology|discrete topology]] on $\Z_{>0}$.
+Let $\phi: T_A \to T_B$ be the [[Definition:Mapping|mapping]]:
+:$\map \phi {2 k} = k, \map \phi {2 k - 1} = k$
+Then:
+:$\map {\phi^{-1} } k = \set {2 k, 2 k - 1} \in \tau_A$
+demonstrating that $\phi$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Now we have that the [[Odd-Even Topology is Weakly Countably Compact]].
+But we also have that a [[Countable Discrete Space is not Weakly Countably Compact]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Deleted Integer Topology is not Countably Compact}
+Tags: Countably Compact Spaces, Deleted Integer Topology
+
+\begin{theorem}
+Let $S = \R_{\ge 0} \setminus \Z$, and let $\tau$ be the [[Definition:Deleted Integer Topology|deleted integer topology]] on $S$.
+Then the [[Definition:Topological Space|topological space]] $T = \struct {S, \tau}$ is not [[Definition:Countably Compact Space|countably compact]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Deleted Integer Topology|deleted integer topology]] is a [[Definition:Partition Topology|partition topology]].
+Let $\PP$ be the [[Definition:Partition (Set Theory)|partition]] which is the [[Basis for Partition Topology|basis for $T$]], that is:
+:$\PP = \set {\openint {n - 1} n: n \in \Z_{> 0} }$
+Then $\PP$ is a [[Definition:Countable Set|countable]] [[Definition:Open Cover|open cover]] of $S$ which has no [[Definition:Finite Subcover|finite subcover]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Deleted Integer Topology is Second-Countable}
+Tags: Second-Countable Spaces, Deleted Integer Topology
+
+\begin{theorem}
+Let $S = \R_{\ge 0} \setminus \Z$.
+Let $\tau$ be the [[Definition:Deleted Integer Topology|deleted integer topology]] on $S$.
+Then the [[Definition:Topological Space|topological space]] $T = \struct {S, \tau}$ is [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+Let $\Z_{>0}$ be understood as the set of [[Definition:Positive Integer|strictly positive integers]]:
+:$\Z_{>0} = \set {x \in \Z: x > 0} = \set {1, 2, 3, \ldots}$
+From [[Basis for Partition Topology]], the set:
+:$\BB = \set {\openint {n - 1} n: n \in Z_{>0} }$
+is a [[Definition:Basis (Topology)|basis]] for $T$.
+There is an obvious [[Definition:Bijection|one-to-one correspondence]] $\phi: \Z_{> 0} \leftrightarrow \BB$ between $\Z_{> 0}$ and $\BB$:
+:$\forall x \in \Z_{>0}: \map \phi x = \openint {x - 1} x$
+But $\Z_{>0} \subseteq \Z$ and [[Integers are Countably Infinite]].
+So from [[Subset of Countably Infinite Set is Countable]], $\Z_{>0}$ is [[Definition:Countable Set|countable]].
+Thus $\BB$ is also countable by definition of [[Definition:Countable Set|countability]].
+So we have that $T$ has a [[Definition:Countable Basis|countable basis]], and so is [[Definition:Second-Countable Space|second-countable]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Deleted Integer Topology is Weakly Countably Compact}
+Tags: Weakly Countably Compact Spaces, Deleted Integer Topology
+
+\begin{theorem}
+Let $S = \R_{\ge 0} \setminus \Z$.
+Let $\tau$ be the [[Definition:Deleted Integer Topology|deleted integer topology]] on $S$.
+Then the [[Definition:Topological Space|topological space]] $T = \struct {S, \tau}$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+\end{theorem}
+
+\begin{proof}
+Let $A \subseteq S$ such that $A$ is [[Definition:Infinite Set|infinite]].
+Let $x \in A$.
+Let $n \in \Z$ such that $n < x < n + 1$.
+By the definition of the [[Definition:Deleted Integer Topology|deleted integer topology]], $\openint n {n + 1}$ is [[Definition:Open Set (Topology)|open]] in $T$.
+We have that $\openint n {n + 1}$ is [[Definition:Infinite Set|infinite]].
+Take some $y \in \openint n {n + 1}$ such that $x \ne y$.
+Now we claim that $y$ is a [[Definition:Limit Point of Set|limit point]] of $A$.
+Let $N_y$ be an [[Definition:Open Neighborhood|open neighbourhood]] of $y$.
+To be [[Definition:Open Set (Topology)|open]] in $T$, $N_y$ needs to be the [[Definition:Set Union|union]] of [[Definition:Open Real Interval|open intervals]] of the form $\openint k {k + 1}$ for some $k \in \Z_{\ge 0}$.
+Therefore to contain $y$ and be [[Definition:Open Set (Topology)|open]] at the same time, $\openint n {n + 1} \subseteq N_y$.
+Thus $x \in N_y$ and $y \ne x \in A \cap N_y \ne \O$.
+This is the definition of a [[Definition:Limit Point of Set|limit point]].
+So, by definition, $T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pseudometric induces Topology}
+Tags: Topology, Pseudometric Spaces
+
+\begin{theorem}
+Let $S \ne \O$ be a [[Definition:Non-Empty Set|non-empty set]].
+Consider a [[Definition:Pseudometric Space|pseudometric space]] $\struct {S, d}$ where $d: S \times S \to \R_{\ge 0}$ is a [[Definition:Pseudometric|pseudometric]].
+Then $\struct {S, d}$ gives rise to a [[Definition:Topological Space|topological space]] $\struct {S, \tau_d}$ whose [[Definition:Topology|topology]] $\tau_d$ is '''defined''' (or '''induced''') by $d$.
+\end{theorem}
+
+\begin{proof}
+Let $\tau_d$ be the set of all $X \subseteq S$ which are [[Definition:Open Set (Pseudometric Space)|open]] in the sense that:
+:$\forall y \in X: \exists \epsilon > 0: \map {B_\epsilon} y \subseteq X$
+where $\map {B_\epsilon} y$ is the [[Definition:Open Ball of Pseudometric Space|open $\epsilon$-ball]] of $y$.
+Equivalently:
+:$\forall x \in X: \exists \epsilon \in \R_{>0}: \forall y \in S: \map d {x, y} < \epsilon \implies y \in X$
+We need to show that $\tau_d$ forms a [[Definition:Topology|topology]] on $S$.
+We examine each of the [[Definition:Open Set Axioms|open set axioms]] in turn.
+=== $(\text O 1)$: Union of Open Sets ===
+Let $\left \langle{U_i}\right \rangle_{i \in I}$ be an [[Definition:Indexed Family of Sets|indexed family]] of [[Definition:Open Set (Pseudometric Space)|open sets]] of $S$.
+Let $\displaystyle V = \bigcup_{i \mathop \in I} U_i$ be the [[Definition:Union of Family|union]] of $\family {U_i}_{i \mathop \in I}$.
+Then by the definition of [[Definition:Union of Family|union]]:
+:$\forall x \in V: \exists i \in I: x \in U_i$
+and so by the definition of [[Definition:Open Set (Pseudometric Space)|open set]]:
+:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq U_i \subseteq V$
+Hence $V$ is [[Definition:Open Set (Pseudometric Space)|open]] by definition.
+{{qed|lemma}}
+=== $(\text O 2)$: Intersection of Open Sets ===
+Let $U$ and $V$ be [[Definition:Open Set (Pseudometric Space)|open sets]] of $S$.
+Let $x \in U \cap V$.
+Then:
+:$\exists \epsilon_U \in \R_{>0}: \map {B_{\epsilon_U} } x \subseteq U$
+:$\exists \epsilon_V \in \R_{>0}: \map {B_{\epsilon_V} } x \subseteq V$
+Let $\epsilon := \min \set {\epsilon_U, \epsilon_V}$.
+Then:
+:$\map {B_\epsilon} x \subseteq U \cap V$
+Hence $U \cap V$ is [[Definition:Open Set (Pseudometric Space)|open]] by definition.
+{{qed|lemma}}
+=== $(\text O 3)$: Empty Set and Set Itself ===
+From [[Open Sets in Pseudometric Space]]:
+:$\O \in \tau_d$
+and:
+:$S \in \tau_d$
+{{qed|lemma}}
+All the [[Definition:Open Set Axioms|open set axioms]] are fulfilled, and the result follows.
+{{qed}}
+{{proofread}}
+\end{proof}<|endoftext|>
+\section{Open Sets in Pseudometric Space}
+Tags: Pseudometric Spaces
+
+\begin{theorem}
+Let $P = \left({A, d}\right)$ be a [[Definition:Pseudometric Space|pseudometric space]].
+Then $\varnothing$ and $A$ are both [[Definition:Open Set (Pseudometric Space)|open]] in $P$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Open Set (Pseudometric Space)|open set]]:
+: $\forall y \in U: \exists \epsilon \in \R_{>0}: B_\epsilon \left({y}\right) \subseteq U$
+where $U$ is an [[Definition:Open Set (Pseudometric Space)|open set]] of $P$.
+That is:
+:''One cannot get out of $U$ by moving an arbitrarily small distance from any point in $U$.''
+Take the case where $U = \varnothing$.
+The [[Definition:Empty Set|empty set]] $\varnothing$ is [[Definition:Open Set (Pseudometric Space)|open]] by dint of the fact that there are no points $y$ in $\varnothing$ for which the condition is false.
+Thus for $\varnothing$:
+: $\forall y \in U: \exists \epsilon \in \R_{>0}: B_\epsilon \left({y}\right) \subseteq U$ is [[Definition:Vacuous Truth|vacuously true]].
+When $U = A$, there are no points in $A$ to which one could get to by leaving $U$, an arbitrarily short distance or no, because there are no points in $A$ that ''are'' outside of $U$.
+{{qed}}
+[[Category:Pseudometric Spaces]]
+hpnso1ai2e9gfv4bwkqj43qqfxx0g9r
+\end{proof}<|endoftext|>
+\section{Partition Space is Pseudometrizable}
+Tags: Pseudometric Spaces, Partition Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Partition Space|partition space]].
+Then $T$ is [[Definition:Pseudometrizable Topology|pseudometrizable]].
+\end{theorem}
+
+\begin{proof}
+Let $\PP$ be the [[Definition:Partition (Set Theory)|partition]] which is the [[Basis for Partition Topology|basis for $T$]].
+Let us define $d: S^2 \to \R$ by:
+:$\forall x, y \in S: \map d {x, y} = \begin{cases}
+0 & : \exists U \in \PP: x, y \in U \\
+1 & : \text{otherwise} \end{cases}$
+That is, $\map d {x, y} = 0$ when $x$ and $y$ are in the same set in the [[Definition:Partition (Set Theory)|partition]] $\PP$, and $1$ otherwise.
+This is easily established as being a [[Definition:Pseudometric|pseudometric]] on $S$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Double Pointed Topology is not T0}
+Tags: Double Pointed Topologies, T0 Spaces, Double Pointed Topology is not T0
+
+\begin{theorem}
+Let $T_1 = \struct {S, \tau_S}$ be a [[Definition:Topological Space|topological space]].
+Let $D = \struct {A, \set {\O, A} }$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on an arbitrary [[Definition:Doubleton|doubleton]] $A = \set {a, b}$.
+Let $T = \struct {T_1 \times D, \tau}$ be the [[Definition:Double Pointed Topology|double pointed topological space]] on $T_1$.
+Then $T$ is not a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Double Pointed Topology|double pointed topology]] $\tau$ on $T_1$ is the [[Definition:Product Topology|product topology]] on $T_1 \times D$.
+Let $x \in S$, and consider the point $\tuple {x, a} \in S \times A$.
+Then:
+:$\forall U \in \tau: \tuple {x, a} \in U \implies \tuple {x, b} \in U$
+:$\forall U \in \tau: \tuple {x, b} \in U \implies \tuple {x, a} \in U$
+as $D$ is an [[Definition:Indiscrete Topology|indiscrete space]].
+Hence the result, by definition of [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition, the [[Definition:Double Pointed Topology|double pointed topology]] $\tau$ on $T_1$ is the [[Definition:Product Topology|product topology]] on $T_1 \times D$.
+By definition, $D$ is the [[Definition:Indiscrete Topology|indiscrete space]] on a [[Definition:Doubleton|doubleton]].
+{{AimForCont}} $T$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+Then from [[Product Space is T0 iff Factor Spaces are T0|Product Space is $T_0$ iff Factor Spaces are $T_0$]] it follows that $D$ is also a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+But from [[Indiscrete Non-Singleton Space is not T0|Indiscrete Non-Singleton Space is not $T_0$]], $D$ is not a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+The result follows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Double Pointed Discrete Real Number Space is Weakly Countably Compact}
+Tags: Weakly Countably Compact Spaces, Partition Topology, Double Pointed Topologies, Discrete Topology
+
+\begin{theorem}
+Let $T_\R = \struct {\R, \tau_\R}$ be the [[Definition:Uncountable Discrete Topology|(uncountable) discrete space]] on the [[Definition:Real Numbers|set of real numbers]].
+Let $T_D = \struct {D, \tau_D}$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on the [[Definition:Doubleton|doubleton]] $D = \set {a, b}$.
+Let $T = T_\R \times T_D$ be the [[Definition:Double Pointed Topology|double pointed]] [[Definition:Uncountable Discrete Topology|(uncountable) discrete space]] which is the [[Definition:Product Space (Topology)|product space]] of $T_\R$ and $T_D$.
+Then $T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+\end{theorem}
+
+\begin{proof}
+We have that $T$ is a [[Definition:Partition Topology|partition topology]], whose [[Basis for Partition Topology|basis]] $\PP$ is defined as:
+:$\PP = \set {\set {\tuple {s, a}, \tuple {s, b} }: s \in \R}$
+Let $A \subseteq \R \times D$ such that $A$ is [[Definition:Infinite Set|infinite]].
+Let $x \in A$.
+Let $U$ be the [[Definition:Set Union|union]] of [[Definition:Set|sets]] of the form $\set {\tuple {s, a}, \tuple {s, b} }$, and hence [[Definition:Open Set (Topology)|open in $T$]].
+Now if $x \in U$, it will be an [[Definition:Element|element]] in some $\set {\tuple {s, a}, \tuple {s, b} }$.
+So there will exist $y \in U$ which will also be an [[Definition:Element|element]] in that $\set {\tuple {s, a}, \tuple {s, b} }$.
+So, by definition, $x$ is a [[Definition:Limit Point of Set|limit point]] of $A$.
+So, by definition, $T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Double Pointed Discrete Real Number Space is not Lindelöf}
+Tags: Lindelöf Spaces, Partition Topology, Double Pointed Topologies, Discrete Topology
+
+\begin{theorem}
+Let $T_\R = \struct {\R, \tau_\R}$ be the [[Definition:Uncountable Discrete Topology|(uncountable) discrete space]] on the [[Definition:Real Numbers|set of real numbers]].
+Let $T_D = \struct {D, \tau_D}$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on the [[Definition:Doubleton|doubleton]] $D = \set {a, b}$.
+Let $T = T_\R \times T_D$ be the [[Definition:Double Pointed Topology|double pointed]] [[Definition:Uncountable Discrete Topology|(uncountable) discrete space]] which is the [[Definition:Product Space (Topology)|product space]] of $T_\R$ and $T_D$.
+Then $T$ is '''not''' a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+We have that $T$ is a [[Definition:Partition Topology|partition topology]], whose [[Basis for Partition Topology|basis]] $\PP$ is defined as:
+:$\PP = \set {\set {\paren {s, a}, \paren {s, b} }: s \in \R}$
+We have that $\PP$ is an [[Definition:Open Cover|open cover]] of $T$.
+But $\PP$ has no [[Definition:Countable Subcover|countable subcover]].
+Hence the result, by definition of [[Definition:Lindelöf Space|Lindelöf space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Point Topology is Topology}
+Tags: Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Space|particular point space]].
+Then $\tau_p$ is a [[Definition:Topology|topology]] on $S$, and $T$ is a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+We have [[Definition:Particular Point Topology|by definition]] that $\O \in \tau_p$, and as $p \in S$ we have that $S \in \tau_p$.
+Now let $U_1, U_2 \in \tau_p$.
+By definition $p \in U_1$ and $p \in U_2$, and so $p \in U_1 \cap U_2$ by definition of [[Definition:Set Intersection|set intersection]].
+So $U_1 \cap U_2 \in \tau_p$.
+Now let $\UU \subseteq \tau_p$.
+We have that $\forall U \in \UU: p \in U$.
+Hence from [[Subset of Union]] $p \in \bigcup \UU$.
+So all the properties are fulfilled for $\tau_p$ to be a [[Definition:Topology|topology]] on $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Accumulation Points of Sequence of Distinct Terms in Infinite Particular Point Space}
+Tags: Limit Points, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be an [[Definition:Infinite Particular Point Topology|infinite particular point space]].
+Let $U \in \tau_p$ be a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Open Set (Topology)|open set]] of $T$.
+Let the [[Definition:Element|elements]] of $U$ be arranged into a [[Definition:Sequence of Distinct Terms|sequence $\sequence {a_i}$ of distinct terms]] in $T$.
+Then while every [[Definition:Element|element]] $x$ of $U$ such that $x \ne p$ is a [[Definition:Limit Point of Set|limit point]] of $U$, there exists no $x \in U$ such that $x$ is an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {a_i}$.
+\end{theorem}
+
+\begin{proof}
+We note from [[Limit Points in Particular Point Space]] that every $x \in U$ such that $x \ne p$ is a [[Definition:Limit Point of Set|limit point]] of $U$.
+Let $\sequence {a_i}$ be the [[Definition:Infinite Sequence|(infinite) sequence]] in $T$ whose [[Definition:Term of Sequence|terms]] are the [[Definition:Element|elements]] of $U$.
+{{AimForCont}} $\beta \in S$ is an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {a_i}$.
+Then from [[Accumulation Points for Sequence in Particular Point Space]], $\beta$ is such that an [[Definition:Infinite Set|infinite number]] of [[Definition:Term of Sequence|terms]] of $\sequence {a_i}$ are equal either to $\beta$ or to $p$.
+But this [[Definition:Contradiction|contradicts]] the assertion that $\sequence {a_i}$ is a [[Definition:Sequence of Distinct Terms|sequence of distinct terms]].
+Hence no [[Definition:Element|element]] of $U$ is an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {a_i}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Points in Particular Point Space}
+Tags: Limit Points, Particular Point Topology, Limit Points in Particular Point Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Let $x \in S$ such that $x \ne p$.
+Then $x$ is a [[Definition:Limit Point of Point|limit point]] of $p$.
+\end{theorem}<|endoftext|>
+\section{Closure of Open Set of Particular Point Space}
+Tags: Open Sets, Set Closures, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Let $U \in \tau_p$ be [[Definition:Open Set (Topology)|open]] in $T$ such that $U \ne \O$.
+Then:
+:$U^- = S$
+where $U^-$ denotes the [[Definition:Closure (Topology)|closure]] of $U$.
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+:[[Particular Point Topology is Closed Extension Topology of Discrete Topology]]
+:[[Closure of Open Set of Closed Extension Space]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Interior of Closed Set of Particular Point Space}
+Tags: Closed Sets, Set Interiors, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Let $V \subseteq S$ be [[Definition:Closed Set (Topology)|closed]] in $T$ such that $V \ne S$.
+Then:
+:$V^\circ = \O$
+where $V^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of $V$.
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\forall U \in \tau_p, U \ne \O: p \in U$
+Thus if $V$ is [[Definition:Closed Set (Topology)|closed]] in $T$:
+:$\exists U \subseteq T: V = \relcomp S U$
+So $p \notin V$.
+Hence no [[Definition:Open Set (Topology)|open set]] of $T$ can be a [[Definition:Subset|subset]] of $V$ unless it is $\O$.
+Hence the result, by definition of [[Definition:Interior (Topology)|interior]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Point in Particular Point Space is not Omega-Accumulation Point}
+Tags: Omega-Accumulation Points, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Space|particular point space]].
+Let $x \in S$ such that $x \ne p$.
+Let $H \subseteq S$ such that $p \in H$.
+Then $x$ is not an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] of $H$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in S, x \ne p$.
+By [[Limit Points in Particular Point Space]], $x$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+Consider the set $U = \set {x, p} \subseteq S$.
+By definition of the [[Definition:Particular Point Topology|particular point topology]], $U$ is [[Definition:Open Set (Topology)|open]] in $T$.
+But as $U$ contains only $x$ and $p$, it is clear that $U$ does not contain an [[Definition:Infinite|infinite]] number of points of $H$.
+Hence, by definition, $x$ can not be an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] of $H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Point Space is T0}
+Tags: T0 Spaces, Particular Point Topology, Particular Point Space is T0
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Space|particular point space]].
+Then $T$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T$ be a [[Definition:Trivial Topological Space|trivial space]], so that $S = \set p$.
+Then the result holds [[Definition:Vacuous Truth|vacuously]] as there are no two [[Definition:Distinct Elements|distinct points]] in $T$.
+Now suppose $T$ is not [[Definition:Trivial Topological Space|trivial]].
+Then $\exists x \in S: x \ne p$.
+Now we have that $\set p \subseteq T$ is [[Definition:Open Set (Topology)|open]] in $T$ such that $p \in \set p$ but $x \notin \set p$.
+Finally, suppose that $x, y \in S: x \ne y, x \ne p, y \ne p$.
+Then we have that $\set {p, x} \subseteq T$ is [[Definition:Open Set (Topology)|open]] in $T$ such that $x \in \set {p, x}$ but $y \notin \set {p, x}$.
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have:
+:[[Particular Point Topology is Closed Extension Topology of Discrete Topology]]
+:[[Discrete Space satisfies all Separation Properties]] (including being a [[Definition:Kolmogorov Space|$T_0$ space]])
+Then by [[Condition for Closed Extension Space to be T0 Space|Condition for Closed Extension Space to be $T_0$ Space]], as a [[Definition:Discrete Space|discrete space]] is $T_0$ then so is its [[Definition:Closed Extension Topology|closed extension]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Trivial Particular Point Topology is not T1}
+Tags: T1 Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Space|particular point space]] such that $S$ is not a [[Definition:Singleton|singleton]].
+Then $T$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+:[[Particular Point Topology is Closed Extension Topology of Discrete Topology]]
+:[[Closed Extension Topology is not T1|Closed Extension Topology is not $T_1$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Point Topology with Three Points is not T4}
+Tags: T4 Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]] such that $S$ is not a [[Definition:Singleton|singleton]] or a [[Definition:Doubleton|doubleton]].
+That is, such that $S$ has more than two [[Definition:Distinct Elements|distinct elements]].
+Then $T$ is not a [[Definition:T4 Space|$T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+We have that there are at least three [[Definition:Element|elements]] of $S$.
+So, consider $x, y, p \in S: x \ne y, x \ne p, y \ne p$.
+Then $X = \set x, Y = \set y$ are [[Definition:Closed Set (Topology)|closed]] in $T$ and $X \cap Y = \O$.
+Suppose $U, V \in \tau_p$ are [[Definition:Open Set (Topology)|open sets]] in $T$ such that $X \subseteq U, Y \subseteq V$.
+But as $p \in U, p \in V$ we have that $U \cap V \ne \O$.
+So $T$ is not a [[Definition:T4 Space|$T_4$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Trivial Particular Point Topology is not T3}
+Tags: T3 Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]] such that $S$ is not a [[Definition:Singleton|singleton]].
+Then $T$ is not a [[Definition:T3 Space|$T_3$ space]].
+\end{theorem}
+
+\begin{proof}
+We have that there are at least two [[Definition:Distinct Elements|distinct elements]] of $S$.
+So, consider $x, p \in S: x \ne p$.
+Then $X = \set x$ is [[Definition:Closed Set (Topology)|closed]] in $T$ and $p \notin X$.
+Suppose $U \in \tau_p$ is an [[Definition:Open Set (Topology)|open set]] in $T$ such that $X \subseteq U$.
+We have that $\set p \in \tau_p$ such that $p \in \set p$.
+But as $p \in U, p \in \set p$ we have that $U \cap \set p \ne \O$.
+So $T$ is not a [[Definition:T3 Space|$T_3$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Space in Particular Point Space}
+Tags: Compact Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Then $\set p$ is [[Definition:Compact Topological Subspace|compact]] in $T$.
+\end{theorem}
+
+\begin{proof}
+Any [[Definition:Open Cover|open cover]] of $\set p$ has a [[Definition:Finite Subcover|finite subcover]]: any single set that contains $p$ is a [[Definition:Cover of Set|cover]] for $\set p$.
+So $\set p$ is [[Definition:Compact Topological Subspace|compact]] in $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Multiplicative Subgroup of Field is Cyclic}
+Tags: Field Theory, Cyclic Groups
+
+\begin{theorem}
+Let $\left({F, +, \times}\right)$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\left({F^*, \times}\right)$ denote the [[Definition:Multiplicative Group|multiplicative group]] of $F$.
+Let $C$ be a [[Definition:Finite Group|finite]] [[Definition:Subgroup|subgroup]] of $\left({F^*, \times}\right)$.
+Then $C$ is [[Definition:Cyclic Group|cyclic]].
+\end{theorem}
+
+\begin{proof}
+As C is a finite [[Definition:Abelian Group|abelian group]], due to the [[Fundamental Theorem of Finite Abelian Groups]], $C$ is the [[Definition:Internal Group Direct Product|Internal Group Direct Product]] of [[Definition:Cyclic Group|Cyclic Groups]] $H_1, \ldots, H_r$ with orders $p_1^{e_1}, \ldots, p_r^{e_r}$ respectively.
+As [[Internal and External Group Direct Products are Isomorphic]],
+:$C \cong H_1 \times \ldots \times H_r$
+where $\cong$ denotes [[Definition:Group Isomorphism|(group) isomorphism]].
+As [[Powers of Coprime Numbers are Coprime]], the orders of $H_1, \ldots, H_r$ are coprime.
+By [[Group Direct Product of Cyclic Groups]], $H_1 \times \ldots \times H_r$ is cyclic, so $C$ is cyclic.
+{{qed}}
+[[Category:Field Theory]]
+[[Category:Cyclic Groups]]
+88txuy93qmfwfwqcmp2083jevg4kga2
+\end{proof}<|endoftext|>
+\section{Induced Homomorphism of Polynomial Forms}
+Tags: Polynomial Theory
+
+\begin{theorem}
+Let $R$ and $S$ be [[Definition:Commutative and Unitary Ring|commutative rings with unity]].
+Let $\phi : R \to S$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
+Let $R \left[{X}\right]$ and $S \left[{X}\right]$ be the [[Definition:Ring of Polynomial Forms|rings of polynomial forms]] over $R$ and $S$ respectively in the indeterminate $X$.
+Then the map $\overline{\phi}: R \left[{X}\right] \to S \left[{X}\right]$ given by:
+:$\overline{\phi} \left({a_0 + a_1 X + \cdots + a_n X^n}\right) = \phi \left({a_0}\right) + \phi \left({a_1}\right) X + \cdots + \phi \left({a_n}\right) X^n$
+is a [[Definition:Ring Homomorphism|ring homomorphism]].
+\end{theorem}
+
+\begin{proof}
+Let $f = a_0 + \cdots + a_n X^n$, $g = b_0 + \cdots + b_m X^m \in R \left[{X}\right]$.
+We have:
+{{begin-eqn}}
+{{eqn|l=\overline{\phi} \left({f}\right) \overline{\phi} \left({g}\right)
+     |r=\left[{\phi \left({a_0}\right) + \cdots + \phi \left({a_n}\right) X^n}\right] \left[{\phi \left({b_0}\right) + \cdots + \phi \left({b_m}\right) X^m}\right]
+     |c=
+}}
+{{eqn|r=\sum_{r \mathop = 0}^{m+n} \sum_{i \mathop + j \mathop = r} \phi \left({a_i}\right) \phi \left({b_j}\right) X^r
+     |c=by the definition of [[Definition:Multiplication of Polynomial Forms|multiplication of polynomial forms]]
+}}
+{{eqn|r=\sum_{r \mathop = 0}^{m+n} \phi \left({\sum_{i \mathop + j \mathop = r} a_i b_j}\right) X^r
+     |c=because $\phi$ is a [[Definition:Ring Homomorphism|homomorphism]]
+}}
+{{eqn|r=\overline{\phi} \left({f g}\right)
+     |c=
+}}
+{{end-eqn}}
+and
+{{begin-eqn}}
+{{eqn|l=\overline{\phi} \left({f}\right) + \overline{\phi} \left({g}\right)
+     |r=\left[{\phi \left({a_0}\right) + \cdots + \phi \left({a_n}\right) X^n}\right] + \left[{\phi \left({b_0}\right) + \cdots + \phi \left({b_m}\right) X^m}\right]
+     |c=
+}}
+{{eqn|r=\sum_{i \mathop = 0}^{\max \left\{ {m, n}\right\} } \left[ \phi \left({a_i}\right) + \phi \left({b_i}\right) \right] X^i
+     |c=by the definition of [[Definition:Addition of Polynomial Forms|addition of polynomial forms]]
+}}
+{{eqn|r=\sum_{i \mathop = 0}^{\max \left\{ {m, n}\right\} } \phi \left({a_i + b_i}\right) X^i
+     |c=because $\phi$ is a [[Definition:Ring Homomorphism|homomorphism]]
+}}
+{{eqn|r=\overline{\phi} \left({f + g}\right)
+     |c=
+}}
+{{end-eqn}}
+Thus $\overline{\phi}$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Structure of Simple Algebraic Field Extension}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $F / K$ be a [[Definition:Field Extension|field extension]].
+Let $\alpha \in F$ be [[Definition:Algebraic Field Extension|algebraic]] over $K$.
+Let $\mu_\alpha$ be the [[Definition:Minimal Polynomial|minimal polynomial]] of $\alpha$ over $K$.
+Let $K \left[{\alpha}\right]$ (resp. $K \left({\alpha}\right)$) be the [[Definition:Subring|subring]] (resp. [[Definition:Subfield|subfield]]) of $F$ [[Definition:Generator of Subfield|generated]] by $K \cup \left\{ {\alpha}\right\}$.
+Then:
+:$K \left[{\alpha}\right] = K \left({\alpha}\right) \simeq K \left[{X}\right] / \left\langle{\mu_\alpha}\right\rangle$
+where $\left\langle{\mu_\alpha}\right\rangle$ is the [[Definition:Ideal of Ring|ideal]] of the [[Definition:Ring of Polynomial Functions|ring of polynomial functions]] [[Definition:Generator|generated]] by $\mu_\alpha$.
+{{explain|Link to Generator needs refining}}
+Moreover:
+:$n := \left[{K \left({\alpha}\right) : K}\right] = \deg \mu_\alpha$
+and:
+:$1, \alpha, \dotsc, \alpha^{n - 1}$ is a [[Definition:Basis (Linear Algebra)|basis]] of $K \left({\alpha}\right)$ over $K$.
+\end{theorem}
+
+\begin{proof}
+Define $\phi: K \left[{X}\right] \to K \left[{\alpha}\right]$ by:
+:$\phi \left({f}\right) = f \left({\alpha}\right)$.
+We have
+:$\phi \left({f}\right) = 0 \iff f \left({\alpha}\right) = 0 \iff \mu_\alpha | f$
+where the last equivalence is proved in [[Minimal Polynomial]].
+{{explain|Presumably $\mu_\alpha \vert f$ means divisibility -- clarification needed.}}
+{{refactor|The above needs to be extracted into a different page, as [[Minimal Polynomial]] is to be refactored: merged, probably, into somewhere else.}}
+Thus:
+:$\ker \phi = \left\{ {f \in K \left[{X}\right]: \mu_\alpha | f}\right\} =: \left\langle{\mu_\alpha}\right\rangle$
+By the corollary to [[Field Adjoined Set]] $\phi$ is [[Definition:Surjective|surjective]].
+So by the [[First Isomorphism Theorem for Rings|First Isomorphism Theorem]],
+:$K \left[{X}\right] / \left\langle{\mu_\alpha}\right\rangle \simeq K \left[{\alpha}\right]$
+By [[Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal]]:
+:$\left\langle{\mu_\alpha}\right\rangle$ is [[Definition:Maximal Ideal of Ring|maximal]].
+So $K \left[{\alpha}\right]$ is a field by [[Maximal Ideal iff Quotient Ring is Field]].
+Also $K \left[{\alpha}\right]$ is the smallest ring containing $K \cup \left\{ {\alpha}\right\}$.
+So because a [[Field is Integral Domain|field is a ring]] it is also the smallest field containing $K \cup \left\{ {\alpha}\right\}$.
+This shows that $K \left[{\alpha}\right] = K \left({\alpha}\right)$.
+By [[Field Adjoined Set]]:
+: $K \left[{\alpha}\right] = \left\{ {f \left({\alpha}\right) : f \in K \left[{X}\right] }\right\}$
+where $K \left[{X}\right]$ is the [[Definition:Ring of Polynomial Functions|ring of polynomial functions]] over $K$.
+From [[Division Theorem for Polynomial Forms over Field]], for each $f \in K \left[{X}\right]$ there are $q, r \in K \left[{X}\right]$ such that:
+:$f = q \mu_\alpha + r$
+and:
+:$\deg r < d =: \deg \mu_\alpha$
+Therefore, since $\mu_\alpha \left({\alpha}\right) = 0$, we have 
+{{begin-eqn}}
+{{eqn | l = K \left[{\alpha}\right]
+      | r = \left\{ {r \left({\alpha}\right): r \in K \left[{X}\right], \ \deg r < \deg \mu_\alpha}\right\}
+}}
+{{eqn | r = \left\{ {a_0 + a_1 \alpha + \dotsb + a_{d - 1} \alpha^{d - 1}: \alpha_0, \dotsc, \alpha_d \in K}\right\}
+}}
+{{end-eqn}}
+Therefore $1, \dotsc, \alpha^{d - 1}$ [[Definition:Spanning Set|span]] $K \left[{\alpha}\right] / K$ as a [[Definition:Vector Space|vector space]].
+Moreover if $s \in K \left[{X}\right]$, then $s \left({\alpha}\right) = 0$ and $\deg s < \deg \mu_\alpha$ implies that $s = 0$.
+Thus $1, \dotsc, \alpha^{d - 1}$ are linearly independent.
+Therefore no non-zero [[Definition:Linear Combination|$K$-combination]] of $1, \dotsc, \alpha^{d-1}$ is zero.
+Therefore $\left\{ {1, \dotsc, \alpha^{d-1} }\right\}$ is a [[Definition:Basis (Linear Algebra)|basis]] of $K \left[{\alpha}\right] / K$.
+The [[Definition:Degree of Field Extension|degree]] of $K \left[{\alpha}\right] / K$ is by definition the number of elements of a basis for $K \left[{\alpha}\right] / K$.
+Therefore:
+:$d = \deg \mu_\alpha = \left[{K \left[{\alpha}\right] : K}\right]$
+{{qed}}
+[[Category:Field Extensions]]
+im9j193ofrijralo77nx3lswqxz3h8p
+\end{proof}<|endoftext|>
+\section{Subset of Particular Point Space is either Open or Closed}
+Tags: Open Sets, Closed Sets, Clopen Sets, Particular Point Topology
+
+\begin{theorem}
+Let $T = \left({S, \tau_p}\right)$ be a [[Definition:Particular Point Topology|particular point space]].
+Let $H \subseteq S$ be any [[Definition:Subset|subset]] of $T$.
+Then $H$ is either [[Definition:Open Set (Topology)|open]] or [[Definition:Closed Set (Topology)|closed]] in $T$.
+The only sets which are [[Definition:Clopen Set|both closed and open]] in $T$ are $S$ and $\varnothing$.
+\end{theorem}
+
+\begin{proof}
+Let $H \subseteq S$.
+There are two cases to consider:
+:$p \in H$
+:$p \notin H$
+If $p \in H$ then by definition of a [[Definition:Particular Point Topology|particular point topology]], $H$ is [[Definition:Open Set (Topology)|open]].
+If $p \notin H$, then $p \in \complement_S \left({H}\right)$, where $\complement_S \left({H}\right)$ is the [[Definition:Relative Complement|relative complement of $H$ in $S$]].
+So $\complement_S \left({H}\right)$ is [[Definition:Open Set (Topology)|open]] by definition of a [[Definition:Particular Point Topology|particular point topology]].
+From the definition of a [[Definition:Closed Set (Topology)|closed set]], $\complement_S \left({\complement_S \left({H}\right)}\right)$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+From [[Relative Complement of Relative Complement]] we have that $\complement_S \left({\complement_S \left({H}\right)}\right) = H$ and so $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+Now suppose $H \subseteq T$ is [[Definition:Clopen Set|both closed and open]] in $T$.
+From [[Open and Closed Sets in Topological Space]], if $H = \varnothing$ or $H = S$ then H is [[Definition:Clopen Set|both closed and open]] in $T$.
+So, suppose $H \ne \varnothing$ or $H \ne S$.
+As $H \subseteq T$ is [[Definition:Clopen Set|both closed and open]] in $T$, $\complement_S \left({H}\right)$ is also [[Definition:Clopen Set|both closed and open]] in $T$.
+From [[Boundary is Intersection of Closure with Closure of Complement]] we have:
+:$\partial H = H^- \cap \left({\complement_S \left({H}\right)}\right)^-$
+where $H^-$ is the [[Definition:Closure (Topology)|closure]] of $H$.
+By [[Closure of Open Set of Particular Point Space]] we have that $H^- = S$, and of course $\left({\complement_S \left({H}\right)}\right)^- = S$.
+So:
+:$\partial H = H^- \cap \left({\complement_S \left({H}\right)}\right)^- = S$
+However, from [[Set Clopen iff Boundary is Empty]] we have that $\partial H = \varnothing$.
+Hence if $H$ is [[Definition:Clopen Set|both closed and open]] in $T$, then $H = \varnothing$ or $H = S$.
+{{qed}}
+[[Category:Open Sets]]
+[[Category:Closed Sets]]
+[[Category:Clopen Sets]]
+[[Category:Particular Point Topology]]
+braurrhwyos5294fdt69nsx7if5n7fv
+\end{proof}<|endoftext|>
+\section{Particular Point Space is Weakly Locally Compact}
+Tags: Weakly Locally Compact Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Then $T$ is [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+\end{theorem}
+
+\begin{proof}
+By [[Compact Space in Particular Point Space]], $\set p$ is compact in $T$.
+But by definition $\set p$ is [[Definition:Open Set (Topology)|open]] in $T$.
+So $p$ is contained in a [[Definition:Compact Topological Subspace|compact]] [[Definition:Neighborhood of Point|neighborhood]], that is, $\set p$.
+Now let $x \in S: x \ne p$.
+Then $\set {x, p}$ is [[Definition:Open Set (Topology)|open]] in $T$.
+We have that $\set {x, p}$ trivially has an [[Definition:Open Cover|open cover]], that is, $\set {\set {x, p} }$ itself.
+Hence any open cover of $\set {x, p}$ has a [[Definition:Finite Subcover|finite subcover]]: any one [[Definition:Set|set]] that contains $x$ and $p$ is a cover for $\set {x, p}$.
+So $\set {x, p}$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$ which is [[Definition:Compact Topological Subspace|compact]] in $T$.
+So $x$ is contained in a [[Definition:Compact Topological Subspace|compact]] [[Definition:Neighborhood of Point|neighborhood]].
+Hence the result, by definition of [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Particular Point Space is not Strongly Locally Compact}
+Tags: Strongly Locally Compact Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be an [[Definition:Infinite Particular Point Topology|infinite particular point space]].
+Then $T$ is not [[Definition:Strongly Locally Compact Space|strongly locally compact]].
+\end{theorem}
+
+\begin{proof}
+By definition, $T$ is [[Definition:Strongly Locally Compact Space|strongly locally compact]] {{iff}} every point of $S$ is contained in an [[Definition:Open Set (Topology)|open set]] whose [[Definition:Closure (Topology)|closure]] is [[Definition:Compact Topological Subspace|compact]].
+Let $x \in S: x \ne p$.
+Let $x \in U$ where $U$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Closure of Open Set of Particular Point Space]] we have:
+:$U^- = S$
+where $U^-$ is the [[Definition:Closure (Topology)|closure]] of $U$.
+But from [[Infinite Particular Point Space is not Compact]] we have that $S$ is not [[Definition:Compact Topological Space|compact]].
+Hence the result, by definition of [[Definition:Strongly Locally Compact Space|strongly locally compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Particular Point Space is not Compact}
+Tags: Compact Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be an [[Definition:Infinite Particular Point Topology|infinite particular point space]].
+Then $T$ is not [[Definition:Compact Topological Space|compact]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Open Cover|open cover]] of $T$:
+:$\CC = \set {\set {x, p}: x \in S, x \ne p}$
+As $S$ is [[Definition:Infinite|infinite]], then so is $\CC$, as we can set up a [[Definition:Bijection|bijection]] from $\phi: S \setminus \set p \leftrightarrow \CC$:
+:$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$
+Hence $\CC$ has no [[Definition:Finite Subcover|finite subcover]].
+The result follows by definition of [[Definition:Compact Topological Space|compactness]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Particular Point Space is not Lindelöf}
+Tags: Lindelöf Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be an [[Definition:Uncountable Particular Point Topology|uncountable particular point space]].
+Then $T$ is not a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Open Cover|open cover]] of $T$:
+:$\CC = \set {\set {x, p}: x \in S, x \ne p}$
+As $S$ is [[Definition:Uncountable Set|uncountable]], then so is $\CC$, as we can set up a [[Definition:Bijection|bijection]] $\phi: S \setminus \set p \leftrightarrow \CC$:
+:$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$
+Hence $\CC$ has no [[Definition:Countable Subcover|countable subcover]].
+The result follows by definition of [[Definition:Lindelöf Space|Lindelöf space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Particular Point Space is Lindelöf}
+Tags: Lindelöf Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Countable Particular Point Topology|countable particular point space]].
+Then $T$ is a [[Definition:Lindelöf Space|Lindelöf apace]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Open Cover|open cover]] of $T$:
+:$\CC = \set {\set {x, p}: x \in S, x \ne p}$
+As $S$ is [[Definition:Countable|countable]], then so is $\CC$, as we can set up a [[Definition:Bijection|bijection]] from $\phi: S \setminus \set p \leftrightarrow \CC$:
+:$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$
+Hence $\CC$ is its own [[Definition:Countable Subcover|countable subcover]].
+The result follows by definition of [[Definition:Lindelöf Space|Lindelöf apace]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Point Space is Separable}
+Tags: Separable Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Then $T$ is [[Definition:Separable Space|separable]].
+\end{theorem}
+
+\begin{proof}
+By definition, $T$ is [[Definition:Separable Space|separable]] {{iff}} there exists a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$ which is [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+Consider $U := \set p \subseteq S$.
+By definition, $U$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Closure of Open Set of Particular Point Space]] we have that $U^- = S$, where $U^-$ is the [[Definition:Closure (Topology)|closure]] of $U$.
+By definition, $U$ is [[Definition:Everywhere Dense|everywhere dense in $T$]].
+$U$ is (trivially) [[Definition:Countable Set|countable]].
+Hence the result, by definition of a [[Definition:Separable Space|separable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field Adjoined Set}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $S \subseteq F$ be a [[Definition:Subset|subset]] of $F$.
+Let $K \le F$ be a [[Definition:Subfield|subfield]] of $F$.
+The [[Definition:Subring|subring]] $K \sqbrk S$ of $F$ [[Definition:Generator of Subfield|generated]] by $K \cup S$ is the [[Definition:Set|set]] of all [[Definition:Finite Set|finite]] [[Definition:Linear Combination|linear combinations]] of powers of elements of $S$ with coefficients in $K$.
+The [[Definition:Subfield|subfield]] $\map K S$ of $F$ [[Definition:Generator of Subfield|generated]] by $K \cup S$ is the [[Definition:Set|set]] of all $x y^{-1} \in F$ with $a, b \in K \sqbrk S$, $b \ne 0$.
+$\map K S$ is [[Definition:Field Isomorphism|isomorphic]] to the [[Definition:Field of Quotients|field of quotients]] $Q$ of $K \sqbrk S$.
+\end{theorem}
+
+\begin{proof}
+{{explain|Most of this needs to be explained in more detail. Suffers from coherence issues.}}
+Let $\set {X_s: s \in S}$ be a family of indeterminates [[Definition:Indexing Set|indexed]] by $S$.
+Let $\phi$ be the [[Evaluation Homomorphism]] such that $\phi \sqbrk {X_s} = s$.
+{{explain|Can the above [[Evaluation Homomorphism]] link be replaced by [[Definition:Polynomial Evaluation Homomorphism]]?}}
+By [[Ring Homomorphism Preserves Subrings]], $\Img \phi$ is a subring of $F$
+This subring contains $\phi \sqbrk {X_s} = s$ for each $s \in S$ and $\map \phi k = k$ for all $k \in K$.
+Moreover since it is obtained by evaluating [[Definition:Polynomial|polynomials]] on $S$ it is the set of all [[Definition:Finite|finite]] [[Definition:Linear Combination|linear combinations]] of powers elements of $S$ with coefficients in $K$ as claimed.
+All these linear combinations must belong to any subring of $F$ that contains $K$ and $S$ (otherwise it is not [[Definition:Closed Operation|closed]]), so $\Img \phi$ is the smallest such subring.
+By [[Universal Property for Field of Quotients]], the inclusion $K \sqbrk S \to F$ extends uniquely to a [[Definition:Field Homomorphism|homomorphism]] $\psi : Q \to F$, given by $\map \psi {a / b} = a b^{-1}$.
+We have:
+:$\Img \psi = \set {a b^{-1} \in F: a, b \in K \sqbrk S, \ b \ne 0} \simeq Q$
+The isomorphism comes from the fact that a field homomorphism is [[Definition:Injective|injective]] and the [[First Isomorphism Theorem for Rings|First Isomorphism Theorem]].
+Thus $\Img \phi$ is a subfield of $F$ containing $K \sqbrk S$ and $K \cup S$.
+Any subfield of $F$ containing $K$ and $S$ must contain $K \cup S$, $K \sqbrk S$ and all $a b^{-1}$ with $a, b \in K \sqbrk S$ (otherwise it would not be [[Definition:Closed Operation|closed]]).
+Therefore $Q \simeq \Img \phi$ is the smallest such subfield.
+{{qed}}
+[[Category:Field Extensions]]
+6huc5ywotbfw61onr0njwdp9aclt5dp
+\end{proof}<|endoftext|>
+\section{Separability in Uncountable Particular Point Space}
+Tags: Separable Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be an [[Definition:Uncountable Particular Point Topology|uncountable particular point space]].
+Let $H = S \setminus \set p$ where $\setminus$ denotes [[Definition:Set Difference|set difference]].
+Then $H$ is not [[Definition:Separable Space|separable]].
+\end{theorem}
+
+\begin{proof}
+By definition, $H$ is [[Definition:Separable Space|separable]] {{iff}} there exists a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$ which is [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+Let $V \subseteq H$ where $V$ is [[Definition:Countable Set|countable]].
+$V$ is not [[Definition:Open Set (Topology)|open]] in $T$ as it does not contain $p$.
+From [[Subset of Particular Point Space is either Open or Closed]] it follows that $V$ is [[Definition:Closed Set (Topology)|closed]].
+From [[Closed Set Equals its Closure]], $V^- = V$.
+But $V^- \ne H$ as $V$ is [[Definition:Countable Set|countable]] and $H$ is [[Definition:Uncountable Set|uncountable]].
+So whatever $V$ is, if it is [[Definition:Countable Set|countable]] it is not [[Definition:Everywhere Dense|everywhere dense]].
+The result follows from definition of [[Definition:Separable Space|separable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Particular Point Space is not Second-Countable}
+Tags: Second-Countable Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be an [[Definition:Uncountable Particular Point Topology|uncountable particular point space]].
+Then $T$ is not [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+Let $H = S \setminus \set p$ where $\setminus$ denotes [[Definition:Set Difference|set difference]].
+Every subset $V \subseteq H$ is a [[Definition:Closed Set (Topology)|closed set]] from [[Subset of Particular Point Space is either Open or Closed]].
+Thus we can consider $H$ as an [[Definition:Uncountable Discrete Topology|uncountable discrete space]].
+The result follows from [[Uncountable Discrete Space is not Second-Countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Point Space is First-Countable}
+Tags: First-Countable Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Then $T$ is [[Definition:First-Countable Space|first-countable]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in S: x \ne p$.
+Consider the set $U_x = \set {x, p} \subseteq S$.
+Now let $V \in \tau_p$ be an [[Definition:Open Set (Topology)|open set]] in $S$ such that $x \in V$.
+So $x \in V$, by definition of $V$, and $p \in V$ as $V$ is [[Definition:Open Set (Topology)|open]].
+It follows directly that $U_x \subseteq V$.
+So $\set {U_x}$ is a [[Definition:Local Basis|local basis]] at $x$ which is (trivially) [[Definition:Countable Set|countable]].
+If $x = p$ then $U_x = \set p$ and the same argument still applies.
+So every element of $T$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]].
+Hence $T$ is [[Definition:First-Countable Space|first-countable]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kronecker’s Theorem}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $f$ be a [[Definition:Polynomial (Abstract Algebra)|polynomial]] over $K$ of [[Definition:Degree (Polynomial)|degree]] $n \ge 1$.
+Then there exists a [[Definition:Finite Field Extension|finite extension]] $L / K$ of $K$ such that $f$ has at least one [[Definition:Root of Polynomial|root]] in $L$.
+Moreover, we can choose $L$ such that the [[Definition:Degree of Field Extension|degree]] $\index L K$ of $L / K$ satisfies $\index L K \le n$.
+{{explain|Work out exactly which definitions of Polynomial and Degree are appropriate here.}}
+\end{theorem}
+
+\begin{proof}
+Let $K \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over $K$.
+By [[Polynomial Forms over Field form Unique Factorization Domain]], $K \sqbrk X$ is a [[Definition:Unique Factorization Domain|unique factorisation domain]].
+Therefore, we can write $f = u g_1 \cdots g_r$, where $u$ a [[Definition:Unit of Ring|unit]] of $K \sqbrk X$ and $g_i$ is [[Definition:Irreducible Polynomial|irreducible]] for $i = 1, \ldots, r$.
+Clearly it is sufficient to find an extension of $K$ in which the irreducible factor $g_1$ of $f$ has a root.
+Let $L = K \sqbrk X / \gen {g_1}$, where $\gen {g_1}$ is the [[Definition:Generated Ideal of Ring|ideal generated]] by $g_1$.
+By [[Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal]], $\gen {g_1}$ is [[Definition:Maximal Ideal of Ring|maximal]].
+Therefore by [[Maximal Ideal iff Quotient Ring is Field]], $L$ is a field.
+Moreover, writing $\overline {\map p x} = \map p X + \gen {g_1}$ for $\map p x \in K \sqbrk X$:
+{{begin-eqn}}
+{{eqn | l = \map {g_1} {\overline X}
+      | r = \overline {\map {g_1} X}
+      | c = 
+}}
+{{eqn | r = \overline {0_{K \sqbrk X} }
+      | c = 
+}}
+{{eqn | r = 0_L
+      | c = 
+}}
+{{end-eqn}}
+So $\overline X$ is a root of $g_1$ in $L$.
+It remains to show that $\index L K \le n$.
+By the [[Euclidean Algorithm]] every polynomial $\map p X \in K \sqbrk X$ can be written as:
+:$\map p X = \map q X \map {g_1} X + \map r X$
+with $\map \deg {\map r X} < \map \deg {\map {g_1} X} \le \map \deg {\map f X} = n$.
+Now we have by the definition of the quotient ring:
+{{begin-eqn}}
+{{eqn | l = \overline {\map p X}
+      | r = \overline {\map q X \map {g_1} X + \map r X}
+      | c = 
+}}
+{{eqn | r = \overline {\map r X}
+      | c = 
+}}
+{{end-eqn}}
+So if $\map r X = r_0 + r_1 X + \cdots + r_{n - 1} X^{n - 1}$, we have:
+:$\overline {\map p X} = \overline {r_0 + r_1 X + \cdots + r_{n - 1} X^{n - 1} }$
+Our choice of $p$ was arbitrary, so every polynomial can be written in this form.
+In particular the [[Definition:Set|set]] $\set {\overline 1, \overline X, \ldots, \overline {X^{n - 1} } }$ [[Definition:Spanning Set|spans]] $L$ over $K$.
+Thus by [[Spanning Set Contains Basis]], a [[Definition:Basis (Linear Algebra)|basis]] of $L$ has at most $n$ [[Definition:Element|elements]].
+That is, $\index L K \le n$.
+{{qed}}
+{{Namedfor|Leopold Kronecker|cat = Kronecker}}
+[[Category:Field Extensions]]
+p81n7mn6mchiyahli6djva9jcnq4ctj
+\end{proof}<|endoftext|>
+\section{Homeomorphic Non-Comparable Particular Point Topologies}
+Tags: Homeomorphisms, Particular Point Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] with at least two [[Definition:Element|elements]].
+Let $p, q \in S: p \ne q$.
+Let $\tau_p$ and $\tau_q$ be the [[Definition:Particular Point Topology|particular point topologies]] on $S$ by $p$ and $q$ respectively.
+Then the [[Definition:Topological Space|topological spaces]] $T_p = \struct {S, \tau_p}$ and $T_q = \struct {S, \tau_q}$ are [[Definition:Homeomorphism|homeomorphic]].
+However, $\tau_p$ and $\tau_q$ are not [[Definition:Comparable Topologies|comparable]].
+\end{theorem}
+
+\begin{proof}
+We can set up the [[Definition:Mapping|mapping]] $\phi: S \to S$:
+:$\forall x \in S: \map \phi x = \begin{cases}
+q & : x = p \\
+p & : x = q \\
+x & : \text {otherwise} \end{cases}$
+It is straightforward to show that $\phi$ is a [[Definition:Homeomorphism|homeomorphism]].
+However, we have, for example, that $\set q \notin \tau_p$ and $\set p \notin \tau_q$.
+So neither $\tau_p$ nor $\tau_q$ are a [[Definition:Subset|subset]] of the other.
+Hence by definition $\tau_p$ and $\tau_q$ are not [[Definition:Comparable Topologies|comparable]]. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Universal Property for Simple Field Extensions}
+Tags: Field Extensions, Universal Properties
+
+\begin{theorem}
+Let $F / K$ be a [[Definition:Field Extension|field extension]].
+Let $\alpha \in F$ be an [[Definition:Algebraic Field Extension|algebraic]] over $K$.
+Let $\mu_\alpha$ be the [[Definition:Minimal Polynomial|minimal polynomial]] over $\alpha$ over $K$.
+Let $\psi : K \left({\alpha}\right) \to L$ be a [[Definition:Field Homomorphism|homomorphism]].
+Let $\phi = \psi \restriction_K$.
+Let $\overline \phi: K \left[{X}\right] \to L \left[{X}\right]$ be the [[Induced Homomorphism of Polynomial Forms]].
+Then $\psi\left({\alpha}\right)$ is a [[Definition:Root of Polynomial|root]] of $\overline \phi \left({\mu_\alpha}\right)$ in $L$.
+Conversely let $L$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\phi: K \to L$ be a [[Definition:Field Homomorphism|homomorphism]].
+Let $\beta$ be a [[Definition:Root of Polynomial|root]] of $\overline \phi  \left({\mu_\alpha}\right)$ in $L$.
+Then there exists a unique field homomorphism $\psi : K \left({\alpha}\right) \to L$ [[Definition:Extension of Mapping|extending]] $\phi$ that sends $\alpha$ to $\beta$.
+\end{theorem}
+
+\begin{proof}
+Let $\psi: K \left({\alpha}\right) \to L$ be a [[Definition:Field Homomorphism|homomorphism]].
+Let $\phi = \psi \big|_K$.
+Let $\overline \phi: K \left[{X}\right] \to L \left[{X}\right]$ be the [[Induced Homomorphism of Polynomial Forms]].
+For any $f = a_0 + \dotsb + a_n X^n \in K \left[{X}\right]$ we have:
+{{begin-eqn}}
+{{eqn | l = \psi \left({f \left({\alpha}\right)}\right)
+      | r = \psi \left({a_0 + \cdots + a_n \alpha^n}\right)
+      | c = 
+}}
+{{eqn | r = \phi \left({a_0}\right) + \cdots + \phi \left({a_n}\right) \psi \left({\alpha}\right)^n
+      | c = by the homomorphism property, and the fact that $\phi = \psi \big\vert_K$
+}}
+{{eqn | r = \overline \phi \left({f}\right) (\psi\left({\alpha}\right))
+      | c = 
+}}
+{{end-eqn}}
+Since $\mu_\alpha \left({\alpha}\right) = 0$ and a [[Ring Homomorphism Preserves Zero]], the above yields:
+:$\overline \phi \left({f}\right) \left({\psi \left({\alpha}\right)}\right) = 0$
+as required.
+{{qed|lemma}}
+Conversely let $L$ be a field.
+Let $\phi: K \to L$ be a homomorphism.
+Let $\beta$ be a root of $\overline \phi \left({\mu_\alpha}\right)$ in $L$.
+By [[Structure of Simple Algebraic Field Extension]], there exists a unique [[Definition:Field Isomorphism|isomorphism]]:
+:$\Delta: K \left[{\alpha}\right] \to K \left[{X}\right] / \langle \mu_\alpha \rangle$
+such that:
+:$\Delta \left({\alpha}\right) = X + \left\langle{\mu_\alpha}\right\rangle$
+and:
+:$\Delta \big|_K$ is the [[Definition:Identity Mapping|identity mapping]].
+By [[Universal Property of Polynomial Ring]] there exists a unique homomorphism:
+:$\chi: K \left[{X}\right] \to L$
+such that:
+:$\chi \left({X}\right) = \beta$
+and:
+:$\chi \big|_K = \phi$
+By [[Universal Property of Quotient Ring]] there exists a unique homomorphism:
+:$\psi: K \left[{X}\right] / \left\langle{\mu_\alpha}\right\rangle \to L$
+such that:
+:$\chi = \psi \circ \pi$
+So we have the following diagram:
+::$\begin{xy}\xymatrix@L+2mu@+1em {
+ K \left[{X}\right] \ar[r]^*{\pi}
+                    \ar@{-->}[rd]_*{\exists ! \chi}
+&
+ \dfrac {K \left[{X}\right]} {\left\langle{\mu_\alpha}\right\rangle} \ar@{-->}[d]^*{\exists ! \psi}
+                                                                     \ar[r]^*{\Delta}
+&
+ K \left[{\alpha}\right]
+\\ 
+&
+ L
+&
+}\end{xy}$
+Now we have:
+:$\chi \left({X}\right) = \beta$
+and:
+:$\pi \left({X}\right) = X + \left\langle{\mu_\alpha}\right\rangle$
+Therefore, because:
+:$\chi = \psi \circ \pi$
+we have:
+:$\psi \left({X + \left\langle{\mu_\alpha}\right\rangle}\right) = \beta$
+Also:
+:$\Delta^{-1} \left({\alpha}\right) = X + \left\langle{\mu_\alpha}\right\rangle$
+so by the above:
+: $\psi \circ \Delta^{-1}\left({ \alpha }\right) = \beta$
+: $\psi \circ \Delta^{-1} \big|_K =  \psi\big|_K = \phi$
+: the choice of $\Delta, \psi$ is unique.
+{{qed}}
+[[Category:Field Extensions]]
+[[Category:Universal Properties]]
+drpnzgmm7nkwt879f4v4gzg4wolq1pw
+\end{proof}<|endoftext|>
+\section{Finite Field Extension is Algebraic}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $L / K$ be a [[Definition:Finite Field Extension|finite field extension]].
+Then $L / K$ is [[Definition:Algebraic Field Extension|algebraic]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in L$ be arbitrary.
+Let $n = \left[{L : K}\right]$ be the [[Definition:Degree of Field Extension|degree]] of $L$ over $K$.
+Since [[any set of $n+1$ vectors in $L$ is linearly dependent]], there is a $K$-[[Definition:Linear Combination|linear combination]] of $\left\{{1, \ldots, x^n}\right\}$ equal to $0$.
+Say $a_n x^n + \cdots + a_1 x + a_0 = 0$, $a_i \in K$, $i = 0, \ldots, n$.
+Therefore $x$ satisfies a [[Definition:Polynomial Function (Abstract Algebra)|polynomial]] with [[Definition:Coefficient of Polynomial|coefficients]] in $K$.
+{{explain|Clarify whether the above link is what it is supposed to be}}
+That is, $x$ is [[Definition:Algebraic over Field|algebraic]].
+Since $x \in L$ was chosen arbitrarily, $L / K$ is [[Definition:Algebraic Field Extension|algebraic]].
+{{qed}}
+{{MissingLinks|I believe that "any set of $n+1$ vectors in $L$ is linearly dependent" may already exist. If not, this link needs to be renamed as there's $\LaTeX$ in the title, which renders it invalid.}}
+[[Category:Field Extensions]]
+sm8umfdxlne2wj3wixstqy0vuq2zwsz
+\end{proof}<|endoftext|>
+\section{Finitely Generated Algebraic Extension is Finite}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $L/K$ be a [[Definition:Field Extension|field extension]] and $\alpha_1,\ldots,\alpha_n \in L$ [[Definition:Algebraic Element of Field Extension|algebraic]] over $K$.
+Then $K \left({\alpha_1, \ldots, \alpha_n}\right) / K$ is a [[Definition:Finite Field Extension|finite field extension]].
+\end{theorem}
+
+\begin{proof}
+Let $S = \left\{{\alpha_1, \ldots, \alpha_n}\right\}$.
+We show by [[Principle of Mathematical Induction|induction]] on $n$ that $K \left({S}\right) / K$ is [[Definition:Finite Field Extension|finite]].
+Clearly $K$ is finite over itself, so the result holds when $n = 0$.
+Now suppose that for all sets $T \subseteq L$ with $\left|{T}\right| \leq n - 1$ and each element of $T$ [[Definition:Algebraic Element of Field Extension|algebraic]] over $K$, $K \left({T}\right) / K$ is finite. We have:
+:$K \left({S}\right) = K \left({\alpha_1, \ldots, \alpha_{n-1}}\right) \left({\alpha_n}\right)$
+By the induction hypothesis $K \left({\alpha_1, \ldots, \alpha_{n-1}}\right) / K$ is finite.
+By the [[Structure of Simple Algebraic Field Extension]] $K \left({S}\right) / K \left({\alpha_1, \ldots, \alpha_{n-1}}\right)$ is finite.
+Therefore, by the [[Tower Law]], $K \left({S}\right) / K$ is finite.
+{{qed}}
+[[Category:Field Extensions]]
+dmf7bqm08ocbjnyhee81auwcc05dz5u
+\end{proof}<|endoftext|>
+\section{Transitivity of Algebraic Extensions}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $E / F / K$ be a [[Definition:Tower of Fields|tower]] of [[Definition:Field Extension|field extensions]].
+Let $E$ be [[Definition:Algebraic Field Extension|algebraic]] over $F$.
+Let $F$ be [[Definition:Algebraic Field Extension|algebraic]] over $K$.
+Then $E$ is [[Definition:Algebraic Field Extension|algebraic]] over $K$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in E$.
+There are $a_0, \ldots, a_n \in F$ such that $a_0 + \cdots + a_n x^n = 0$.
+Let $L = K \left({a_0, \ldots, a_n}\right)$.
+We have that $L / K$ is [[Definition:Finitely Generated Field Extension|finitely generated]] and [[Definition:Algebraic Field Extension|algebraic]].
+Therefore by [[Finitely Generated Algebraic Extension is Finite]] this [[Definition:Field Extension|extension]] is [[Definition:Finite Field Extension|finite]].
+We have that $L \left({x}\right) / L$ is [[Definition:Simple Field Extension|simple]] and [[Definition:Algebraic Field Extension|algebraic]].
+So by [[Structure of Simple Algebraic Field Extension]], this extension is also [[Definition:Finite Field Extension|finite]].
+Therefore, by the [[Tower Law]]:
+:$L \left({x}\right) / K$ is [[Definition:Finite Field Extension|finite]].
+That is, $x$ is contained in a [[Definition:Finite Field Extension|finite extension]] of $K$.
+Therefore because a [[Finite Field Extension is Algebraic]], it follows that $x$ is [[Definition:Algebraic Element of Field Extension|algebraic]] over $K$, as was to be proved.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Maximal Algebraic Extension is Subfield}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $L/K$ be a [[Definition:Field Extension|extension of fields]].
+Let $K^a$ be the [[Definition:Maximal Algebraic Extension|maximal algebraic extension]] of $K$ contained in $L$.
+Then $K^a$ is a [[Definition:Subfield|subfield]] of $L$.
+\end{theorem}
+
+\begin{proof}
+Let $\alpha,\beta \in K^a$.
+By [[Field Adjoined Algebraic Elements is Algebraic]], $K(\alpha,\beta)/K$ is [[Definition:Algebraic Field Extension|algebraic]].
+By definition $K(\alpha,\beta)$ is a [[Definition:Field (Abstract Algebra)|field]].
+Therefore $\alpha\beta$, $\alpha^{-1}$ and $\alpha - \beta$ all lie in $K(\alpha,\beta)$, hence are [[Definition:Algebraic Element of Field Extension|algebraic]] over $K$.
+Also $K \subseteq K^a$ so $K^a \neq \emptyset$, and by the [[Subfield Test]], $K^a \leq L$ is a [[Definition:Subfield|subfield]].
+{{qed}}
+[[Category:Field Extensions]]
+lavzxgixxbb8jiuzry8vjewgbuieyv9
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Algebraically Closed Field}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $K$ be a [[Definition:Field (Abstract Algebra)|field]]. 
+{{TFAE|def = Algebraically Closed Field}}
+\end{theorem}
+
+\begin{proof}
+=== Definition $(1)$ implies Definition $(2)$ ===
+Let $K$ be [[Definition:Algebraically Closed Field/Definition 1|algebraically closed by definition 1]].
+Let $f$ be an [[Definition:Irreducible Polynomial|irreducible polynomial]] over $K$.
+By [[Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal]], the [[Definition:Ideal of Ring|ideal]] $\left\langle{f}\right\rangle$ [[Definition:Generator of Ideal|generated]] by $f$ is [[Definition:Maximal Ideal of Ring|maximal]].
+So by [[Maximal Ideal iff Quotient Ring is Field]]:
+: $L = K \left[{X}\right] / \left\langle{f}\right\rangle$ is a [[Definition:Field (Abstract Algebra)|field]]
+where $L$ is a [[Definition:Field Extension|field extension]] over $K \left[{X}\right]$.
+Now:
+:$L = \left\{{g + \left\langle{f}\right\rangle: g \in K \left[{X}\right]}\right\}$
+From [[Division Theorem for Polynomial Forms over Field]]:
+:$\forall g \in K \left[{X}\right]: \exists q, r \in K \left[{X}\right]: g = q f + r, \operatorname{deg} r < \operatorname{deg} f =: n$
+Therefore:
+:$L = \left\{{r + \left\langle{f}\right\rangle: r \in K \left[{X}\right],\ \operatorname{deg} r < n}\right\}$
+By [[Basis for Quotient of Polynomial Ring]], this has [[Definition:Basis for Field Extension|basis]]:
+:$1 + \left\langle{f}\right\rangle, \ldots, X^{n-1} + \left\langle{f}\right\rangle$ [[Definition:Span|span]] $L$.
+Thus $L$ is [[Definition:Finite Field Extension|finite]].
+By [[Finite Field Extension is Algebraic]], $L$ is [[Definition:Algebraic Field Extension|algebraic]].
+Also $K \subseteq L$.
+So by hypothesis $K = L$.
+This implies:
+: $\left[{L : K}\right] = 1$
+where $\left[{L : K}\right]$ is the [[Definition:Degree of Field Extension|degree of $L$ over $K$]].
+Hence $n = \operatorname{deg}f = 1$.
+Thus $K$ is [[Definition:Algebraically Closed Field/Definition 2|algebraically closed by definition 2]].
+{{qed|lemma}}
+=== Definition $(2)$ implies Definition $(3)$ ===
+Let $K$ be [[Definition:Algebraically Closed Field/Definition 2|algebraically closed by definition 2]].
+Let $f$ be a [[Definition:Polynomial|polynomial]] in $K \left[{X}\right]$ of [[Definition:Strictly Positive Integer|strictly positive]] [[Definition:Degree of Polynomial|degree]].
+From [[Polynomial Forms over Field form Principal Ideal Domain]], $K \left[{X}\right]$ is a [[Definition:Principal Ideal Domain|principal ideal domain]].
+From [[Principal Ideal Domain is Unique Factorization Domain]], $K \left[{X}\right]$ is a [[Definition:Unique Factorization Domain|unique factorization domain]].
+So $f$ can be [[Definition:Factorization|factorized]] $f = u g_1 \cdots g_r$ such that:
+: $u$ is a [[Definition:Unit of Ring|unit]]
+and:
+: $g_i$ are [[Definition:Irreducible Polynomial|irreducible]] for $i = 1, \ldots, r$.
+By hypothesis, $g_1$ has [[Definition:Degree (Polynomial)|degree]] $1$.
+Therefore by the [[Polynomial Factor Theorem]] $g_1$, and hence $f$, has a [[Definition:Root of Polynomial|root]] in $K$.
+Thus $K$ is [[Definition:Algebraically Closed Field/Definition 3|algebraically closed by definition 3]].
+{{qed|lemma}}
+=== Definition $(3)$ implies Definition $(1)$ ===
+Let $K$ be [[Definition:Algebraically Closed Field/Definition 3|algebraically closed by definition 3]].
+Let $L / K$ be an [[Definition:Algebraic Field Extension|algebraic field extension]] of $K$.
+Let $\alpha \in L$.
+By hypothesis, the [[Definition:Minimal Polynomial|minimal polynomial]] $\mu_\alpha$ of $\alpha$ over $K$ has a [[Definition:Root of Polynomial|root]] $\beta$ in $K$.
+Therefore by the [[Polynomial Factor Theorem]]:
+: $\mu_\alpha = \left({X - \beta}\right) g$ for some $g \in K \left[{X}\right]$.
+Since $\mu_\alpha$ is [[Definition:Irreducible Polynomial|irreducible]] and [[Definition:Monic Polynomial|monic]] (see [[Minimal Polynomial]]) it follows that:
+:$\mu_\alpha = X - \beta$
+{{explain|Review the link to [[Minimal Polynomial]] as the material inside may be moving elsewhere.}}
+Also:
+: $\mu_\alpha \left({\alpha}\right) = \alpha - \beta = 0$
+so $\alpha = \beta$.
+Therefore $\alpha \in K$.
+Therefore $L = K$ as required.
+Thus $K$ is [[Definition:Algebraically Closed Field/Definition 1|algebraically closed by definition 1]].
+{{qed}}
+[[Category:Field Extensions]]
+ndwyuryxd6gwp2e6nmhtoqqe6ue081p
+\end{proof}<|endoftext|>
+\section{Closed Set in Particular Point Space has no Limit Points}
+Tags: Closed Sets, Limit Points, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Let $H \subsetneq S$ be [[Definition:Closed Set (Topology)|closed]] in $T$.
+Then $H$ has no [[Definition:Limit Point of Set|limit points]].
+\end{theorem}
+
+\begin{proof}
+Let $H$ be [[Definition:Closed Set (Topology)|closed]] in $T$.
+Then by definition $p \notin H$.
+Let $x \in H$.
+By definition, $x$ is a [[Definition:Limit Point of Set|limit point of $H$]] if every [[Definition:Open Set (Topology)|open set]] $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$.
+Consider the set $U_x := \set {x, p} \subseteq S$.
+As $p \in U_x$ we have that $U_x$ is [[Definition:Open Set (Topology)|open]] in $T$.
+But there is no $y \in U_x: y \in H, y \ne x$ and so $x$ is not a [[Definition:Limit Point of Set|limit point]] of $H$.
+So $H$ has no [[Definition:Limit Point of Set|limit points]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Point Space is Scattered}
+Tags: Scattered Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Then $T$ is [[Definition:Scattered Space|scattered]].
+\end{theorem}
+
+\begin{proof}
+Let $H \subseteq S$ such that $p \notin H$.
+From [[Subset of Particular Point Space is either Open or Closed]], $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+We have that [[Closed Set in Particular Point Space has no Limit Points]].
+So if $p \notin H$ then $H$ has no [[Definition:Limit Point of Set|limit points]].
+So by definition, every [[Definition:Element|element]] of such a [[Definition:Set|set]] $H$ is an [[Definition:Isolated Point of Subset|isolated point]].
+So if $p \notin H$ then $H$ is by definition not [[Definition:Dense-in-itself|dense-in-itself]].
+Now let $H \subseteq S$ such that $p \in H$.
+We have that $\set p \in \tau_p$.
+That is, $p$ is an [[Definition:Open Point|open point]] of $T$.
+From [[Point in Topological Space is Open iff Isolated]], $p$ is an [[Definition:Isolated Point (Topology)|isolated point]].
+So if $p \in H$ then $H$ is by definition not [[Definition:Dense-in-itself|dense-in-itself]].
+Hence the result, by definition of a [[Definition:Scattered Space|scattered space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cook-Levin Theorem}
+Tags: Computer Science, Complexity Theory
+
+\begin{theorem}
+The [[Definition:Boolean Satisfiability Problem|Boolean Satisfiability Problem]] is [[Definition:NP-Complete|NP-Complete]].
+\end{theorem}
+
+\begin{proof}
+=== The Boolean Satisfiability Problem is NP ===
+Given a [[Definition:Boolean Satisfiability Problem|Boolean Satisfiability Problem]] with a set of variables $X$ and clauses $L$ and a possible solution to the problem, it is a trivial matter to evaluate all the clauses in $L$ to verify the solution in polynomial time.
+{{Handwaving|The above needs to be justified with a page demonstrating this.}}
+[[NP Problem iff Solution Verifiable in Polynomial Time|Because a potential solution can be verified or rejected in polynomial time]] it is [[Definition:NP Complexity Class|NP]].
+=== All NP-complete problems are polynomially reducible to the Boolean Satisfiability problem ===
+The objective is, given a non-deterministic Turing Machine $M$ with $k$ internal states and an input $x$, to construct a set of variables and  clauses that has a solution iff $M$ accepts the input $x$.
+From  the definition of [[Definition:NP Complexity Class|NP]] we know that $M$ either accepts the input $x$ within $p \left({\left\vert{x}\right\vert}\right)$ steps, or it does not accept the input at all, where $p$ is some polynomial.
+From [[Turing Machine cannot use More Squares of Memory than the Number of Steps that it Runs]], we only need to concern  ourselves with the first $p \left({\left\vert{x}\right\vert}\right)$ squares of memory.
+Let $\Sigma$ denote the finite alphabet that the machine recognizes.
+Let the variables for the Boolean Satisfiability Problem be given by:
+{| border="1"
+|-
+! Variable
+! Interpretation
+! Number
+|-
+| $q_{j n}$
+| $M$ is in state $j$ on step $n$
+| $k * p \left({\left\vert{x}\right\vert}\right)$
+|-
+| $T_{m \ \alpha \ n}$
+| The $m$th square contains the symbol $\alpha \in \Sigma$ on step $n$
+| $|\Sigma| * p \left({x}\right)^2$
+|-
+| $H_{m \ n}$
+| The $m$th square is being looked at during step $n$
+| $p \left({\left\vert{x}\right\vert}\right)^2$
+|}
+{{explain|The symbol $*$ in the above -- while it is frequently used in computer science to denote multiplication, this is a page on a mathematics website, and so a standard mathematics symbol is required.}}
+Let the clauses for the Boolean Satisfiability Problem be given by:
+{| border="1"
+|-
+! Clause
+! Interpretation
+! Number
+! Length
+|-
+| $q_{0 0}$
+| The Machine is in the initial state $\left({q_0}\right)$ on step 0.
+| 1
+| Constant
+|-
+| $if \ m < \left\vert{x}\right\vert \  T_{m x_m 0}$
+| Let $x_m$ denote the $m$'th symbol of the input.  This represents the initial state of the work tape.
+| $\left\vert{x}\right\vert$
+| Constant
+|- 
+| $H_{00}$
+| The Tape is in the initial position on step 0.
+| 1
+| Constant
+|-
+| $if \ a \ne b \ q_{a n} \implies \neg q_{b n}$
+| Only one internal state at a time
+| $\left({k^2 - k}\right) * n$
+| Constant
+|-
+| $if \ a \ne b \ T_{m \ a \ n} \implies \neg T_{m \ b \ n}$
+| Only one symbol per square at a time
+| $\left({\left\vert{\Sigma}\right\vert^2 - \left\vert{\Sigma}\right\vert}\right) * p \left({\left\vert{x}\right\vert}\right)^2$
+| Constant
+|-
+| $if \ a \ne b \ H_{a n} \implies \neg H_{b n}$
+| Only one square is being looked at during any cycle.
+| $p \left({\left\vert{x}\right\vert}\right)^2 - p \left({\left\vert{x}\right\vert}\right)$
+| Constant
+|-
+| $\neg H_{m \ n} \implies \left({T_{m \ \alpha \ n} = T_{m \ \alpha \ n+1} }\right)$
+| A square that was not looked at cannot change.
+| $\left\vert{\Sigma}\right\vert *  p \left({\left\vert{x}\right\vert}\right)^2$
+| Constant
+|-
+| $q_{accept \  p \left({\left\vert{x}\right\vert}\right)}$
+| $M$ is in the accepting state at the end of the computation.
+| 1
+| Constant
+|-
+| $\left({q_{j \ n} \land T_{m \ \alpha \ n} \land H_{m \ n} }\right) \implies \bigvee \left({q_{l \ n + 1} \land T_{m \ \beta \ n + 1} \land H_{m \pm 1 \ n + 1} }\right)$
+| These clauses encode the rules of $M$ into logical expressions.
+See the explanation below.
+| $k * \left\vert{\Sigma}\right\vert *  p \left({\left\vert{x}\right\vert}\right)^2$
+| Varies (see below)
+|}
+A production rule in a non-deterministic Turing machine can be written in the form:
+: $\left({q_a, \alpha}\right) \to \left({q_b, \beta, D_1}\right) \lor \left({q_c, \gamma, D_2}\right)$
+meaning:
+:if the machine is in state $q_a$ and reading $\alpha$ on the tape,
+::either:
+:::replace $\alpha$ with $\beta$
+:::move one square in direction $D_1$ (either left or right)
+:::change the internal state to $q_b$
+::or:
+:::replace $\alpha$ with $\gamma$
+:::move one square in the $D_2$ direction
+:::go to internal state $q_c$.
+If $D_1$ is left and $D_2$ is right then this rule would translate to:
+: $\left({q_a \land T_{m \ \alpha \ n} \land H_{m \ n} }\right) \implies \left({\left({q_b \land T_{m \ \beta \ n+1} \land H_{m-1 \ n+1} }\right) \lor \left({q_c \land T_{m \ \gamma \ n+1} \land H_{m+1 \ n+1}}\right)}\right)$
+If the rule in the machine allows for more then two choices then this rule can be modified by adding more triplets to the right hand side of the implication rule.
+The length of this clause is determined by the number of choices that $M$ gives for a given internal state and a given input.
+Because this number of choices is bounded for any given machine, the total space used for this group of clauses is $O \left({p \left({\left\vert{x}\right\vert}\right)^2}\right)$.
+In total, the size of the the Boolean Satisfiability problem is $O \left({p \left({ \left\vert{x}\right\vert }\right)^2 }\right)$, with the constant depending on $M$.
+The conversion from a description of $M$ to the Boolean Satisfiability problem is straightforward and can be done in polynomial time.
+The problem described has a solution {{iff}} $M$ accepts $x$ as its input.
+All NP problems are polynomially reducible to the Boolean Satisfiability problem.
+Therefore the Boolean Satisfiability is [[Definition:NP-hard|NP-hard]].
+The Boolean Satisfiability Problem is NP-complete.
+{{qed}}
+{{Namedfor|Stephen Arthur Cook|name2 = Leonid Anatolievich Levin|cat = Cook|cat2 = Levin}}
+\end{proof}<|endoftext|>
+\section{Particular Point Space is Irreducible}
+Tags: Irreducible Spaces, Particular Point Topology, Particular Point Space is Irreducible
+
+\begin{theorem}
+Let $T = \left({S, \tau_p}\right)$ be a [[Definition:Particular Point Topology|particular point space]].
+Then $T$ is [[Definition:Irreducible Space|irreducible]].
+\end{theorem}
+
+\begin{proof}
+By definition, $T = \struct {S, \tau_p}$ is [[Definition:Irreducible Space|irreducible]] {{iff}} every two [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open sets]] of $T$ have [[Definition:Non-Empty Set|non-empty]] [[Definition:Set Intersection|intersection]].
+Let $U_1$ and $U_2$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open sets]] of $T$.
+By definition of [[Definition:Particular Point Topology|particular point space]], $p \in U_1$ and $p \in U_2$.
+Thus:
+:$p \in U_1 \cap U_2$
+and so:
+:$U_1 \cap U_2 \ne \O$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Point Space is not Ultraconnected}
+Tags: Ultraconnected Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]] with at least three points.
+Then $T$ is not [[Definition:Ultraconnected Space|ultraconnected]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S: x \ne p, y \ne p, x \ne y$.
+Consider $\set x$ and $\set y$.
+Neither are [[Definition:Open Set (Topology)|open]] as neither contain $p$.
+So from [[Subset of Particular Point Space is either Open or Closed]] they are both [[Definition:Closed Set (Topology)|closed]].
+We have that $\set x \cap \set y = \O$.
+The result follows by definition of [[Definition:Ultraconnected Space|ultraconnected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dispersion Point in Particular Point Space}
+Tags: Dispersion Points, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Then $p$ is [[Definition:Dispersion Point|dispersion point]] of $T$.
+\end{theorem}
+
+\begin{proof}
+Let $H = S \setminus \set p$.
+Let $T_H = \struct {H, \tau_H}$ be the [[Definition:Topological Subspace|topological subspace induced on $H$ by $\tau_p$]].
+From [[Particular Point Space less Particular Point is Discrete]], the space $T_H$ is [[Definition:Discrete Space|discrete]].
+We have [[Discrete Space is Locally Connected]].
+Thus from [[Totally Disconnected and Locally Connected Space is Discrete]] we have that $S \setminus \set p$ is [[Definition:Totally Disconnected Space|totally disconnected]].
+Hence the result, from definition of [[Definition:Dispersion Point|dispersion point]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Particular Point Space is not Weakly Countably Compact}
+Tags: Weakly Countably Compact Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be an [[Definition:Infinite Particular Point Topology|infinite particular point space]].
+Then $T$ is not [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+\end{theorem}
+
+\begin{proof}
+By definition, $T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]] {{iff}} every [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$ has a [[Definition:Limit Point of Set|limit point]] in $S$.
+Let $H \subseteq S$ be an [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$ where $p \notin H$.
+$H$ is not [[Definition:Open Set (Topology)|open]] in $T$ by definition.
+So from [[Subset of Particular Point Space is either Open or Closed]], $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+Then we have that a [[Closed Set in Particular Point Space has no Limit Points]].
+The result follows from definition of [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+{{qed}}
+If $T = \struct {S, \tau_p}$ is a [[Definition:Finite Particular Point Topology|finite particular point space]], then [[Finite Space Satisfies All Compactness Properties]] applies.
+\end{proof}<|endoftext|>
+\section{Fundamental Theorem of Symmetric Polynomials}
+Tags: Symmetric Polynomials, Fundamental Theorems
+
+\begin{theorem}
+Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $f$ a [[Definition:Symmetric Polynomial|symmetric polynomial]] over $K$.
+Then $f$ can be written uniquely as a [[Definition:Polynomial|polynomial]] in the [[Definition:Elementary Symmetric Polynomial|elementary symmetric polynomials]].
+\end{theorem}
+
+\begin{proof}
+Let $p \in R := K \sqbrk {x_1, \dotsc, x_n}$ be a [[Definition:Symmetric Polynomial|symmetric polynomial]].
+Let $\sigma_1, \dotsc, \sigma_n$ denote the $n$ [[Definition:Elementary Symmetric Polynomial|elementary symmetric polynomials]].
+Let $<$ be the [[Definition:Degree Reverse Lexicographic Monomial Order|degree reverse lexicographic monomial order]].
+Let us denote by $\map {LM} f$ the [[Definition:Leading Monomial|leading monomial]] of any $f \in R$, that is to say the unique $<$-largest [[Definition:Monomial|monomial]] appearing in $f$.
+Then it is straightforward to verify that $\map {LM} {\sigma_k} = x_1 \cdots x_k$ for $1 \le k \le n$:
+For $k < n$, any other [[Definition:Monomial|monomial]] appearing in $\sigma_k$ must contain a factor $x_j^{\alpha_1}$ with $j > k$ and $\alpha_1 > 0$.
+This makes it smaller than $x_1 \cdots x_k$ with respect to $<$.
+The case $k = n$ is trivial because $\sigma_n$ consists of only one [[Definition:Monomial|monomial]].
+It follows the proof that $p$ may be written as a [[Definition:Polynomial|polynomial]] in $\sigma_1, \dotsc, \sigma_k$.
+Firstly, since $p$ is [[Definition:Symmetric Polynomial|symmetric]], it must have the property:
+:$\map {LM} p = x_1^{a_1} \dotsm x_n^{a_n}$
+with $a_1 \ge a_2 \ge \ldots \ge a_n$.
+To see this, assume that $a_i > a_j$ for some $1 \le i < j \le n$.
+The [[Definition:Monomial|monomial]]:
+:$x_1^{a_1} \dotsm x_{i - 1}^{a_{i - 1} } \cdot x_i^{a_j} x_j^{a_i} \cdot x_{j + 1}^{a_{j + 1} } \cdots x_n^{a_n}$
+must also appear in $p$, and if $a_i > a_j$ then it would be larger than $\map {LM} p$, a [[Definition:Contradiction|contradiction]].
+Now consider:
+:$\sigma :=  c \sigma_1^{a_1 - a_2} \cdot \sigma_2^{a_2 - a_3} \cdot \dotsm \cdot \sigma_{n - 1}^{a_{n - 1} - a_n} \cdot \sigma_n^{a_n}$
+where $c$ is the coefficient of $\map {LM} p$ in $p$.
+Define $\bar p := p - \sigma$.
+Because:
+{{begin-eqn}}
+{{eqn | l = \map {LM} \sigma
+      | r = \prod_{i \mathop = 1}^n \map {LM} {\sigma_i^{a_i - a_{i + 1} } }
+      | c = 
+}}
+{{eqn | r = x_1^{a_1 - a_2} \cdot \paren {x_1 x_2}^{a_2 - a_3} \cdot \dotsm \cdot \paren {x_1 \ldots x_{n - 1} }^{a_{n - 1} - a_n} \cdot \paren {x_1 \ldots x_n}^{a_n}
+      | c = 
+}}
+{{eqn | r = x_1^{a_1} \dotsm x_n^{a_n}
+      | c = 
+}}
+{{eqn | r = \map {LM} p
+      | c = 
+}}
+{{end-eqn}}
+the [[Definition:Leading Monomial|leading monomials]] in the difference $\bar p$ cancel.
+Therefore:
+:$\map {LM} {\bar p} < \map {LM} p$
+Iterate this method, and after [[Definition:Finite Set|finitely many]] steps one has decomposed $p$ into a [[Definition:Polynomial|polynomial]] $F \in K \sqbrk {y_1, \dotsc, y_n}$ with $\map F {\sigma_1, \dotsc, \sigma_n} = p$.
+The termination of this algorithm follows immediately from [[Dickson's Lemma]].
+{{qed}}
+[[Category:Symmetric Polynomials]]
+[[Category:Fundamental Theorems]]
+dbvjw2s6ppn77j0avsq5cmja8ca59hz
+\end{proof}<|endoftext|>
+\section{Particular Point Space is Pseudocompact}
+Tags: Pseudocompact Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Then $T$ is [[Definition:Pseudocompact Space|pseudocompact]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:a [[Particular Point Space is Irreducible]]
+:an [[Irreducible Space is Pseudocompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Point Space is Path-Connected}
+Tags: Path-Connected Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Then $T$ is [[Definition:Path-Connected Space|path-connected]].
+\end{theorem}
+
+\begin{proof}
+Let $q \in S$.
+Let $\mathbb I$ be the [[Definition:Closed Unit Interval|closed unit interval]] in $\R$.
+Let $f: \mathbb I \to S$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in \mathbb I: \map f x = \begin{cases}
+p & : x \in \hointr 0 1 \\
+q & : x = 1
+\end{cases}$
+Let $U \in \tau_p$.
+Then by definition of [[Definition:Particular Point Topology|particular point space]], $p \in U$
+Either $q \in U$ or $q \notin U$.
+If $q \in U$ then $f^{-1} \sqbrk U = \closedint 0 1$ which is [[Definition:Open Set (Topology)|open]] in $\mathbb I$ because $\closedint 0 1 = \mathbb I$.
+If $q \notin U$ then $f^{-1} \sqbrk U = \hointr 0 1$ which is [[Definition:Half-Open Real Interval|half open]] in $\R$ but [[Definition:Open Set (Topology)|open]] in $\mathbb I$.
+So $f: \mathbb I \to S$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] and so a [[Definition:Path (Topology)|path]] from $p$ to $q$.
+As $q$ is any point in $S$, it follows from [[Path-Connected iff Path-Connected to Point]] that $T$ is [[Definition:Path-Connected Space|path-connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Point Space is not Arc-Connected}
+Tags: Connected Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Then $T$ is not [[Definition:Arc-Connected Space|arc-connected]].
+\end{theorem}
+
+\begin{proof}
+Let $q \in S$ be such that $q \ne p$.
+Let $f: \closedint 0 1 \to T$ be an [[Definition:Injection|injection]] such that:
+:$\map f 0 = q$
+:$\map f 1 = p$
+Because $f$ is an [[Definition:Injection|injection]], it must be that:
+:$\map {f^{-1} } {\set p} = \set 1$
+where $f^{-1}$ denotes the [[Definition:Preimage of Subset under Mapping|preimage]] under $f$.
+Now $\set p$ is [[Definition:Open Set (Topology)|open]] in $\tau_p$ by definition of [[Definition:Particular Point Topology|particular point topology]].
+By [[Closed Real Interval is not Open Set]], $\set 1 = \closedint 1 1$ is not [[Definition:Open Set (Topology)|open]].
+Thus we have exhibited an [[Definition:Open Set (Topology)|open set]] whose [[Definition:Preimage of Subset under Mapping|preimage]] under $f$ is not [[Definition:Open Set (Topology)|open]].
+Hence $f$ is not [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Since $f$ was an arbitrary [[Definition:Injection|injection]], no [[Definition:Arc (Topology)|arc]] between $p$ and $q$ can exist.
+Hence $T$ is not [[Definition:Arc-Connected Space|arc-connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis for Particular Point Space}
+Tags: Particular Point Topology, Topological Bases
+
+\begin{theorem}
+Let $T = \left({S, \tau_p}\right)$ be a [[Definition:Particular Point Topology|particular point space]].
+Consider the set $\mathcal B$ defined as:
+:$\mathcal B = \left\{{\left\{{x, p}\right\}: x \in S}\right\} \cup \left\{{p}\right\}$
+Then $B$ is a [[Definition:Basis (Topology)|basis]] for $S$.
+\end{theorem}
+
+\begin{proof}
+Let $H \in \tau_p$ be [[Definition:Open Set (Topology)|open]] in $T$.
+Then:
+: $\forall y \in H: \exists \left\{{y, p}\right\} \in \mathcal B$
+which also holds when $y = p$ as $\left\{{y, p}\right\} = \left\{{p}\right\} \in \mathcal B$.
+Thus:
+:$\displaystyle H = \bigcup_{y \mathop \in H} \left\{{y, p}\right\}$
+So $\mathcal B$ is an [[Definition:Analytic Basis|analytic basis]] for $T$.
+{{qed}}
+It could equally well be shown that $\mathcal B$ is also a [[Definition:Synthetic Basis|synthetic basis]] for $T$.
+[[Category:Particular Point Topology]]
+[[Category:Topological Bases]]
+5l71n059ph20ei9lqy9eqgkkcte2dxf
+\end{proof}<|endoftext|>
+\section{Particular Point Space is Locally Path-Connected}
+Tags: Locally Path-Connected Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Then $T$ is [[Definition:Locally Path-Connected Space|locally path-connected]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Set|set]] $\BB$ defined as:
+:$\BB = \set {\set {x, p}: x \in S}$
+Then $\BB$ is a [[Basis for Particular Point Space|basis for $T$]].
+Now consider the [[Definition:Open Set (Topology)|open set]] $\set {p, q} \in \BB$.
+Let $\mathbb I$ be the [[Definition:Closed Unit Interval|closed unit interval]] in $\R$.
+Let $f: \mathbb I \to S$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in \mathbb I: \map f x = \begin{cases}
+p & : x \in \hointr 0 1 \\
+q & : x = 1
+\end{cases}$
+Let $U \in \tau_p$.
+Then either $q \in U$ or $q \notin U$.
+If $q \in U$ then $f^{-1} \sqbrk U = \closedint 0 1$ which is [[Definition:Open Set (Topology)|open]] in $\mathbb I$ because $\closedint 0 1 = \mathbb I$.
+If $q \notin U$ then $f^{-1} \sqbrk U = \hointr 0 1$ which is [[Definition:Half-Open Real Interval|half open]] in $\R$ but [[Definition:Open Set (Topology)|open]] in $\mathbb I$.
+So $f: \mathbb I \to S$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] and so a [[Definition:Path (Topology)|path]] from $p$ to $q$.
+As $\set {p, q}$ is any element of $\BB$ it follows that $T$ is [[Definition:Locally Path-Connected Space|locally path-connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Point Space is Non-Meager}
+Tags: Non-Meager Spaces, Particular Point Topology, Particular Point Space is Non-Meager
+
+\begin{theorem}
+Let $T = \left({S, \tau_p}\right)$ be a [[Definition:Particular Point Topology|particular point space]].
+Then $T$ is [[Definition:Non-Meager Space|non-meager]].
+\end{theorem}
+
+\begin{proof}
+Suppose $T$ were [[Definition:Meager Space|meager]].
+Then it would be a [[Definition:Countable Union|countable union]] of [[Definition:Subset|subsets]] which are [[Definition:Nowhere Dense|nowhere dense]] in $T$.
+Let $H \subseteq S$.
+From [[Closure of Open Set of Particular Point Space]], the [[Definition:Closure (Topology)|closure]] of $H$ is $S$.
+From the definition of [[Definition:Interior (Topology)|interior]], the [[Definition:Interior (Topology)|interior]] of $S$ is $S$.
+So the [[Definition:Interior (Topology)|interior]] of the [[Definition:Closure (Topology)|closure]] of $H$ is not [[Definition:Empty Set|empty]].
+So $T$ can not be the [[Definition:Set Union|union]] of a [[Definition:Countable Set|countable set]] of [[Definition:Subset|subsets]] which are [[Definition:Nowhere Dense|nowhere dense]] in $T$.
+Hence $T$ is not [[Definition:Meager Space|meager]] and so by definition must be [[Definition:Non-Meager Space|non-meager]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of [[Definition:Particular Point Topology|particular point space]], any [[Definition:Subset|subset]] of $S$ which contains $p$ is [[Definition:Open Set (Topology)|open]] in $T$.
+So $\left\{{p}\right\}$ itself is [[Definition:Open Set (Topology)|open]] in $T$.
+That is, $p$ is an [[Definition:Open Point|open point]].
+The result follows from [[Space with Open Point is Non-Meager]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of [[Definition:Particular Point Topology|particular point space]], any [[Definition:Subset|subset]] of $S$ which contains $p$ is [[Definition:Open Set (Topology)|open]] in $T$.
+{{AimForCont}} $T$ is [[Definition:Meager Space|meager]].
+By definition, $T$ is [[Definition:Meager Space|meager]] {{iff}} it is a [[Definition:Countable Union|countable union]] of [[Definition:Subset|subsets]] of $S$ which are [[Definition:Nowhere Dense|nowhere dense]] in $T$.
+At least one such [[Definition:Nowhere Dense|nowhere dense]] [[Definition:Subset|subset]] $U$ of $S$ must contain $p$.
+By definition, $U$ is [[Definition:Nowhere Dense|nowhere dense]] in $T$ {{iff}}:
+:$U^-$ contains no [[Definition:Open Set (Topology)|open set]] of $T$ which is [[Definition:Non-Empty Set|non-empty]]
+where $U^-$ denotes the [[Definition:Closure (Topology)|closure]] of $U$.
+By definition of [[Definition:Particular Point Topology|particular point space]], $U$ is [[Definition:Open Set (Topology)|open]] in $T$.
+By [[Closure of Open Set of Particular Point Space]], $U^- = S$.
+But $S$ is itself [[Definition:Open Set (Topology)|open]] in $T$ and [[Definition:Non-Empty Set|non-empty]], and so $U$ is not [[Definition:Nowhere Dense|nowhere dense]].
+From this [[Definition:Contradiction|contradiction]] it follows that $T$ is [[Definition:Non-Meager Space|non-meager]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Particular Point Space is not Countably Metacompact}
+Tags: Countably Metacompact Spaces, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be an [[Definition:Infinite Particular Point Topology|infinite particular point space]].
+Then $T$ is not [[Definition:Countably Metacompact Space|countably metacompact]].
+\end{theorem}
+
+\begin{proof}
+Suppose $T$ is a [[Definition:Countable Particular Point Topology|countable particular point space]].
+Let $\CC$ be the [[Definition:Open Cover|open cover]] of $T$ defined as:
+:$\CC = \set {\set {x, p}: x \in S}$
+$\CC$ is [[Definition:Countable Set|countable]] and has no [[Definition:Open Refinement|open refinement]] except $\CC$ itself.
+But $\CC$ is not [[Definition:Point Finite|point finite]] because $\forall U \in \CC: p \in U$, and $\CC$ is [[Definition:Countable|(countably) infinite]].
+Now suppose $T$ is an [[Definition:Uncountable Particular Point Topology|uncountable particular point space]].
+Let $S' \subseteq S$ be a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$.
+Let $\CC$ be the [[Definition:Open Cover|open cover]] of $T$ defined as:
+:$\CC = \set {\set {x, p}: x \in S'} \cup \set {S \setminus S' \cup \set p}$
+$\CC$ is also [[Definition:Countable Set|countable]] and has no [[Definition:Open Refinement|open refinement]] except $\CC$ itself.
+And similarly, $\CC$ is not [[Definition:Point Finite|point finite]] because $\forall U \in \CC: p \in U$, and $\CC$ is [[Definition:Countable|(countably) infinite]].
+Hence the result by definition of [[Definition:Countably Metacompact Space|countably metacompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exportation and Self-Conditional}
+Tags: Implication, Conjunction
+
+\begin{theorem}
+:$p \land q \implies r \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$
+\end{theorem}
+
+\begin{proof}
+From the [[Rule of Exportation]]:
+:$\paren {p \land q} \implies r \dashv \vdash p \implies \paren {q \implies r}$
+Then by [[Self-Distributive Law for Conditional]]:
+:$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$
+{{qed}}
+[[Category:Implication]]
+[[Category:Conjunction]]
+oc5bpk4w3hco5d9sobteg96jn3b38gs
+\end{proof}<|endoftext|>
+\section{Łoś-Vaught Test}
+Tags: Model Theory, Mathematical Logic
+
+\begin{theorem}
+Let $T$ be a satisfiable $\mathcal L$-theory with no finite models.
+Let $T$ be $\kappa$-[[Definition:Categorical (Model Theory)|categorical]] for some infinite cardinal $\kappa \ge \left|{\mathcal L}\right|$.
+Then $T$ is complete.
+\end{theorem}
+
+\begin{proof}
+We prove the contrapositive.
+The main idea is that if such a theory $T$ is incomplete, we can construct size $\kappa$ models which disagree on a sentence.
+Suppose $T$ is not complete.
+By the definition of complete, this means that there is some sentence $\phi$ such that both $T \not \models \phi$ and $T \not \models \neg \phi$.
+This in turn means that both $T \cup \left\{{\neg \phi}\right\}$ and $T \cup \left\{{\phi}\right\}$ have models.
+Since $T$ has no finite models, this means that $T \cup \left\{{\neg \phi}\right\}$ and $T \cup \left\{{\phi}\right\}$ both have infinite models.
+We have that $\kappa$ is infinite and greater than the cardinality of the language.
+We also have that these theories have infinite models.
+From the [[Upward Löwenheim-Skolem Theorem]] one can prove that there are size $\kappa$ models $\mathcal M_{\neg \phi}$ and $\mathcal M_{\phi}$ of $T \cup \left\{{\neg \phi}\right\}$ and $T \cup \left\{{\phi}\right\}$ respectively.
+In particular, $\mathcal M_{\neg\phi}$ and $\mathcal M_\phi$ are models of $T$ which disagree about the sentence $\phi$.
+Such models cannot be isomorphic since isomorphisms preserve the truth of sentences.
+Thus, $T$ is not $\kappa$-categorical.
+{{qed}}
+{{Namedfor|Jerzy Maria Michał Łoś|name2 = Robert Lawson Vaught|cat = Łoś|cat2 = Vaught}}
+This result is also known as [[Vaught's Test]].
+[[Category:Model Theory]]
+[[Category:Mathematical Logic]]
+0cehvfrmryvn8jtame1htqwwd9mm8o6
+\end{proof}<|endoftext|>
+\section{Sierpiński Space is Irreducible}
+Tags: Irreducible Spaces, Sierpiński Space
+
+\begin{theorem}
+Let $T = \struct {\set {0, 1}, \tau_0}$ be a [[Definition:Sierpiński Space|Sierpiński space]].
+Then $T$ is [[Definition:Irreducible Space|irreducible]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Sierpiński Space|Sierpiński space]] is a [[Definition:Particular Point Space|particular point space]] by definition.
+A [[Particular Point Space is Irreducible]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sierpiński Space is Ultraconnected}
+Tags: Ultraconnected Spaces, Sierpiński Space
+
+\begin{theorem}
+Let $T = \struct {\set {0, 1}, \tau_0}$ be a [[Definition:Sierpiński Space|Sierpiński space]].
+Then $T$ is [[Definition:Ultraconnected Space|ultraconnected]].
+\end{theorem}
+
+\begin{proof}
+The only [[Definition:Closed Set (Topology)|closed sets]] of $T$ are $\O, \set 1$ and $\set {0, 1}$.
+$\set 1$ and $\set {0, 1}$ are not [[Definition:Disjoint Sets|disjoint]].
+Hence the result by definition of [[Definition:Ultraconnected Space|ultraconnected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sierpiński Space is Path-Connected}
+Tags: Path-Connected Spaces, Sierpiński Space
+
+\begin{theorem}
+Let $T = \struct {\set {0, 1}, \tau_0}$ be a [[Definition:Sierpiński Space|Sierpiński space]].
+Then $T$ is [[Definition:Path-Connected Space|path-connected]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Sierpiński Space|Sierpiński space]] is a [[Definition:Particular Point Space|particular point space]] by definition.
+A [[Particular Point Space is Path-Connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sierpiński Space is not Arc-Connected}
+Tags: Arc-Connected Spaces, Sierpiński Space
+
+\begin{theorem}
+Let $T = \struct {\set {0, 1}, \tau_0}$ be a [[Definition:Sierpiński Space|Sierpiński space]].
+Then $T$ is not [[Definition:Arc-Connected Space|arc-connected]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Sierpiński Space|Sierpiński space]] is a [[Definition:Particular Point Space|particular point space]] by definition.
+A [[Particular Point Space is not Arc-Connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sierpiński Space is T5}
+Tags: T5 Spaces, Sierpiński Space
+
+\begin{theorem}
+Let $T = \struct {\set {0, 1}, \tau_0}$ be a [[Definition:Sierpiński Space|Sierpiński space]].
+Then $T$ is a [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+The only [[Definition:Closed Set (Topology)|closed sets]] in $T$ are $\O$, $\set 1$ and $\set {0, 1}$.
+So there are no two [[Definition:Separated Sets|separated sets]] $A, B \subseteq \set {0, 1}$.
+So $T$ is a [[Definition:T5 Space|$T_5$ space]] vacuously.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sierpiński Space is T4}
+Tags: T4 Spaces, Sierpiński Space
+
+\begin{theorem}
+Let $T = \struct {\set {0, 1}, \tau_0}$ be a [[Definition:Sierpiński Space|Sierpiński space]].
+Then $T$ is a [[Definition:T4 Space|$T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Sierpiński Space is T5|Sierpiński Space is $T_5$]].
+Then we have that a [[T5 Space is T4 Space|$T_5$ Space is $T_4$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upward Löwenheim-Skolem Theorem}
+Tags: Model Theory, Mathematical Logic
+
+\begin{theorem}
+{{Disambiguate|Definition:Model|I suspect model of a first-order theory $\LL$, which is more specific than what is linked to now}}
+Let $T$ be an $\LL$-theory with an infinite [[Definition:Model (Logic)|model]].
+Then for each infinite cardinal $\kappa \ge \card \LL$, there exists a [[Definition:Model (Logic)|model]] of $T$ with cardinality $\kappa$.
+\end{theorem}
+
+\begin{proof}
+The idea is:
+:to extend the language by adding $\kappa$ many new constants
+and:
+:to extend the theory by adding sentences asserting that these constants  are distinct.
+It is shown that this new theory is finitely satisfiable using an infinite model of $T$.
+Compactness then implies that the new theory has a model.
+Some care needs to be taken to ensure that we  construct a model of exactly size $\kappa$.
+Let $\LL^*$ be the language formed by adding new constants $\set {c_\alpha: \alpha < \kappa}$ to $\LL$.
+Let $T^*$ be the $\LL^*$-theory formed by adding the sentences $\set {c_\alpha \ne c_\beta: \alpha, \beta < \kappa, \ \alpha \ne \beta}$ to $T$.
+We show that $T^*$ is finitely satisfiable:
+Let $\Delta$ be a finite subset of $T^*$.
+Then $\Delta$ contains:
+:finitely  many sentences from $T$
+along with:
+:finitely many sentences of the form $c_\alpha \ne c_\beta$ for the new constant symbols.
+Since $T$ has an infinite model, it must have a model $\MM$ of cardinality at most $\card \LL + \aleph_0$.
+This model already satisfies everything in $T$.
+So, since we can find arbitrarily many distinct elements in it,  it can also be used as a model of $\Delta$ by interpreting the finitely many new constant symbols in $\Delta$ as distinct elements of $\MM$.
+Since $T^*$ is finitely satisfiable, it follows by the [[Compactness Theorem]] that $T^*$ itself is satisfiable.
+Since $T^*$ ensures the existence of $\kappa$ many distinct elements, this means it has models of size at least $\kappa$.
+It can be proved separately or observed from the ultraproduct proof of the compactness theorem that $T^*$ then has a model $\MM^*$ of exactly size $\kappa$.
+{{explain|That proof needs to be proved, and / or a link needs to be provided to that ultraproduct proof and its implications explained.}}
+Since $T^*$ contains $T$, $\MM^*$ is a model of $T$ of size $\kappa$.
+{{qed}}
+{{Namedfor|Leopold Löwenheim|name2 = Thoralf Albert Skolem|cat = Löwenheim|cat2 = Skolem}}
+\end{proof}<|endoftext|>
+\section{Compactness Theorem}
+Tags: Model Theory, Mathematical Logic, Named Theorems, Proofs by Contraposition
+
+\begin{theorem}
+Let $\LL$ be the [[Definition:Language of Predicate Logic|language of predicate logic]].
+Let $T$ be a set of $\LL$-[[Definition:Sentence|sentences]].
+Then $T$ is [[Definition:Satisfiable Set of Formulas|satisfiable]] {{iff}} $T$ is [[Definition:Finitely Satisfiable|finitely satisfiable]].
+\end{theorem}
+
+\begin{proof}
+By definition, $T$ is [[Definition:Finitely Satisfiable|finitely satisfiable]] means that every [[Definition:Finite Subset|finite subset]] of $T$ is [[Definition:Satisfiable|satisfiable]].
+Since the direction
+:$T$ [[Definition:Satisfiable Set of Formulas|satisfiable]] implies $T$ [[Definition:Finitely Satisfiable|finitely satisfiable]]
+is trivial, the proofs below all justify the converse
+:$T$ [[Definition:Finitely Satisfiable|finitely satisfiable]] implies $T$ [[Definition:Satisfiable Set of Formulas|satisfiable]].
+\end{proof}
+
+\begin{proof}
+Suppose $\mathbf H$ does '''not''' have a [[Definition:Model (Boolean Interpretations)|model]].
+By the [[Main Lemma of Propositional Tableaus]], $\mathbf H$ has a [[Definition:Tableau Confutation|tableau confutation]] $T$.
+By [[Tableau Confutation contains Finite Tableau Confutation]], $T$ may be assumed to be [[Definition:Finite Propositional Tableau|finite]].
+Hence the set $\mathbf H'$ of all [[Definition:WFF of Propositional Logic|WFFs]] in $\mathbf H$ used somewhere in $T$ is [[Definition:Finite Set|finite]].
+Now, let $T'$ be the [[Definition:Labeled Tree for Propositional Logic|labeled tree]] which is the same as $T$ but with root $\mathbf H'$ instead of $\mathbf H$.
+Then $T'$ is a [[Definition:Tableau Confutation|tableau confutation]] of $\mathbf H'$.
+By the [[Tableau Confutation implies Unsatisfiable]], $\mathbf H'$ has no [[Definition:Model (Boolean Interpretations)|models]].
+But this contradicts the assumption that all finite subsets of $\mathbf H$ have models.
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+If $\mathbf H$ is [[Definition:Finite Set|finite]], the result is trivial.
+So let $\mathbf H = \left\{{\mathbf A_n: n \in \N}\right\}$ be an [[Definition:Enumeration|enumeration]] of $\mathbf H$.
+Define $\mathbf H_m = \left\{{\mathbf A_n: n \le m}\right\}$.
+Let $T_1$ be the [[Definition:Propositional Tableau|propositional tableau]] consisting of only a [[Definition:Root Node|root node]] with [[Definition:Hypothesis Set|hypothesis set]] $\mathbf H_1$.
+For each $m \in \N$, construct $T_{m+1}$ from $T_m$ by applying the [[Tableau Extension Lemma]] to $\mathbf H_{m+1}$.
+This provides us with a [[Definition:Sequence|sequence]] $\left({T_m}\right)_{m \in \N}$ of [[Definition:Finished Propositional Tableau|finished propositional tableaus]].
+The $T_m$ form an [[Definition:Exhausting Sequence of Sets|exhausting sequence of sets]] for the [[Definition:Infinite Propositional Tableau|infinite propositional tableau]] $T = \displaystyle \bigcup_{m \mathop = 1}^\infty T_m$.
+If $T$ has a [[Definition:Finished Branch of Propositional Tableau|finished branch]] $\Gamma$, then by [[Finished Branch Lemma/Corollary|Finished Branch Lemma: Corollary]]:
+:$\Phi \left[{\Gamma}\right]$ is [[Definition:Satisfiable|satisfiable]]
+and hence $\mathbf H$ is [[Definition:Satisfiable|satisfiable]] (since $\mathbf H \subseteq \Phi \left[{\Gamma}\right]$).
+Suppose $T$ has no [[Definition:Finite Branch|finite]] [[Definition:Finished Branch of Propositional Tableau|finished branches]].
+Let $T'$ be the [[Definition:Subtree|subtree]] of $T$ given by:
+:$t \in T'$ {{iff}} $t$ is on a [[Definition:Finished Branch of Propositional Tableau|finished branch]] $\Gamma_m$ of every $T_m$ such that $t \in T_m$
+Suppose the [[Definition:Root Node|root node]] $r_T$ of $T$ were not in $T'$.
+Then for some $T_m$, $r_T$ would not be on a [[Definition:Finished Branch of Propositional Tableau|finished branch]] of $T_m$.
+But since $T_m$ is [[Definition:Finished Propositional Tableau|finished]], every [[Definition:Branch (Graph Theory)|branch]] of $T_m$ is [[Definition:Finished Branch of Propositional Tableau|finished]] or [[Definition:Contradictory Branch|contradictory]].
+Hence every [[Definition:Branch (Graph Theory)|branch]] of $T_m$ is [[Definition:Contradictory Branch|contradictory]].
+But then $T_m$ is a [[Definition:Tableau Confutation|tableau confutation]] of $\mathbf H_m$.
+By [[Tableau Confutation implies Unsatisfiable]], this contradicts the assumption that $\mathbf H_m$ is [[Definition:Satisfiable|satisfiable]].
+Therefore, $r_T$ is in $T'$.
+Suppose $T'$ were [[Definition:Finite Tree|finite]].
+Let $t$ be a [[Definition:Leaf Node|leaf]] of $T'$, which exists by [[Finite Tree has Leaf Nodes]].
+Suppose $t$ were a [[Definition:Leaf Node|leaf]] of $T$.
+Then $\Gamma_t$, the [[Definition:Branch (Graph Theory)|branch]] of $T$ identified by [[Leaf of Rooted Tree is on One Branch]], is [[Definition:Finished Branch of Propositional Tableau|finished]].
+For, as any $T_m$ is a [[Definition:Subtree|subtree]] of $T$, $\Gamma_t$ is the only [[Definition:Branch (Graph Theory)|branch]] of $T_m$ such that $t \in \Gamma_t$.
+The conclusion follows from the definition of $T'$.
+But then $\Gamma_t$ would be a [[Definition:Finite Branch|finite]] [[Definition:Finished Branch of Propositional Tableau|finished branch]] of $T$, a contradiction.
+Therefore, $T'$ cannot be [[Definition:Finite Tree|finite]].
+Hence, $T'$ is a [[Definition:Finitely Branching|finitely branching tree]].
+By [[König's Tree Lemma]], $T'$ has an [[Definition:Infinite Branch|infinite branch]] $\Gamma$.
+By definition of $T'$, the [[Definition:Branch (Graph Theory)|branch]] $\Gamma_m := \Gamma \cap T_m$ of $T_m$ is [[Definition:Finished Branch of Propositional Tableau|finished]] for each $m \in \N$.
+If $\Gamma$ were [[Definition:Contradictory Branch|contradictory]], then $\mathbf A \in \Gamma$ and $\neg\mathbf A \in \Gamma$ for some [[Definition:WFF of Propositional Logic|WFF]] $\mathbf A$.
+But then $\mathbf A, \neg \mathbf A \in \Gamma_m$ for some $m \in \N$, contradicting that $\Gamma_m$ is [[Definition:Finished Branch of Propositional Tableau|finished]].
+Also, if $\mathbf A$ [[Definition:Occurrence along Branch|occurs]] on $\Gamma$, then it occurs on some $\Gamma_m$.
+Since $\Gamma_m$ is [[Definition:Finished Branch of Propositional Tableau|finished]], it follows that $\mathbf A$ is [[Definition:Used WFF|used]] on $\Gamma_m$, and hence on $\Gamma$.
+In conclusion, $\Gamma$ is [[Definition:Finished Branch of Propositional Tableau|finished]].
+As established above, it follows that $\mathbf H$ is [[Definition:Satisfiable|satisfiable]] for [[Definition:Boolean Interpretation|boolean interpretations]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Łoś's Theorem}
+Tags: Model Theory, Mathematical Logic
+
+\begin{theorem}
+Let $\LL$ be a [[Definition:Language of Predicate Logic|language]].
+Let $I$ be an infinite set.
+Let $\UU$ be an [[Definition:Ultrafilter on Set|ultrafilter]] on $I$.
+Let $\map \phi {v_1, \ldots, v_n}$ be an $\LL$-[[Definition:Language of Predicate Logic/Formal Grammar|formula]].
+Let $\MM$ be the [[Definition:Ultraproduct|ultraproduct]]:
+:$\displaystyle \paren {\prod_{i \mathop \in I} \MM_i} / \UU$
+where each $\MM_i$ is an $\LL$-[[Definition:Formal Semantics/Structure|structure]].
+Then, for all $m_1 = \paren {m_{1, i} }_\UU, \dots, m_n = \paren {m_{n, i} }_\UU$ in $\MM$:
+:$\MM \models \map \phi {m_1, \ldots, m_n}$
+{{iff}}:
+:the set $\set {i \in I: \MM_i \models \map \phi {m_{1, i}, \ldots, m_{n, i} } }$ is in $\UU$.
+In particular, for all $\LL$-[[Definition:Classes of WFFs/Sentence|sentences]] $\phi$, we have that:
+:$\MM \models \phi$ {{iff}} $\set {i \in I: \MM_i \models \phi}$ is in $\UU$.
+\end{theorem}
+
+\begin{proof}
+We prove the $\LL$-[[Definition:Classes of WFFs/Sentence|sentences]] case by induction on the complexity of formulas. The general case trivially follows this proof. 
+We appeal to the interpretations of language symbols in the ultraproduct when viewed as an $\LL$-structure, the properties of ultrafilters, and make use of the [[Axiom:Axiom of Choice|Axiom of Choice]].
+The theorem holds trivially for statements of equality of terms and for relations, by definition of how to interpret language symbols for the ultraproduct.
+Suppose the theorem holds for $\psi_0$ and $\psi_1$.
+If $\phi$ is $\neg \psi_0$:
+We are assuming that $\MM \models \psi_0$ {{iff}}:
+:$\set {i: \MM_i \models \psi_0} \in \UU$.
+Thus:
+:$\MM \models \phi$ {{iff}} $\set {i: \MM_i \models \psi_0} \notin \UU$
+follows by negating both sides of this statement.
+Since $\UU$ is an ultrafilter, a set is absent from $\UU$ {{iff}} the set's complement is present in $\UU$.
+So, we may again rewrite the above statement equivalently as:
+:$\MM \models \phi \iff I \setminus \set {i: \MM_i \models \psi_0} \in \UU$
+Finally, we can further rewrite this set difference to see that:
+:$\MM \models \phi \iff \set {i: \MM_i \models \phi} \in \UU$
+which is the statement that the theorem holds for $\phi$.
+Let $\phi$ be $\psi_0 \wedge \psi_1$:
+For both $k \in \left\{{0, 1}\right\}$, we are assuming that:
+:$\MM \models \psi_k \iff \set {i: \MM_i \models \psi_k} \in \UU$
+By choice of $\phi$, we have $\MM \models \phi$ {{iff}} $\MM \models \psi_0 \wedge \psi_1$.
+The right side of this {{iff}} statement can be rewritten as $\MM \models \psi_0$ and $\MM \models \psi_1$.
+Thus, using the inductive hypothesis stated above for each $\psi_k$:
+:$\MM \models \phi \iff \set {i: \MM_i \models \psi_0} \in \UU$ and $\set {i: \MM_i \models \psi_1} \in \UU$
+Since $\UU$ is a filter, it is closed under intersections, and hence the right side of this statement can be written as:
+:$\set {i: \MM_i \models \psi_0 \text{ and } \MM_i \models \psi_1} \in \UU$
+Thus:
+:$\MM \models \phi \iff \set {i: \MM_i \models \phi} \in \UU$
+which is the statement that the theorem holds for $\phi$.
+Let $\phi$ be $\exists x \psi_0 \left({x}\right)$:
+If $x$ is not free in $\psi_0$ then earlier cases cover this, so we may assume $x$ is free in $\psi_0$.
+We are assuming then that for all $m = \left\langle{m_i}\right\rangle_\UU$ in $\MM$:
+: $\MM \models \psi_0 \left({m}\right) \iff \left\{{i \in I: \MM_i \models \psi_0 \left({m_i}\right)}\right\}\in \UU$
+Thus:
+: $\MM \models \phi \iff \exists m = \left\langle{m_i}\right\rangle_\UU \in \MM$
+for which:
+: $\left\{{i \in I: \MM_i \models \psi_0 \left({m_i}\right)}\right\}\in \UU$
+One direction of the theorem follows easily, since this above statement gives us the witnesses $m_i$:
+:$\MM \models \phi \implies \left\{{i \in I: \MM_i \models \psi_0 \left({m_i}\right)}\right\} \in \UU$
+And this above set is included in the set we're looking for, so that is an element of the ultrafilter as well:
+:$\left\{{i \in I: \MM_i \models \psi_0 \left({m_i}\right)}\right\} \subseteq \left\{{i \in I: \MM_i \models \exists x \psi_0 \left({x}\right)}\right\}\in \UU$
+For the converse, we need to find some appropriate $\left\langle{m_i}\right\rangle_\UU$ in order to apply the above [[Definition:Biconditional|biconditional statement]].
+To this end, let $\left\{{i \in I: \MM_i \models \exists x \psi_0 \left({x}\right)}\right\} \in \UU$, and apply the [[Axiom:Axiom of Choice|Axiom of Choice]] as follows:
+Select for each $i \in \left\{{i \in I: \MM_i \models \exists x \psi_0 \left({x}\right)}\right\}$ a witness $m_i \in \MM_i$ such that $\MM_i \models \psi_0 \left({m_i}\right)$
+Select for each $i$ not in this set an arbitrary element $m_i$ of $\MM_i$.
+Taking $\left\langle{m_i}\right\rangle_\UU$ as our element of $\MM$ then allows us to apply the above [[Definition:Biconditional|biconditional statement]] and complete the proof.
+{{qed}}
+{{Namedfor|Jerzy Maria Michał Łoś|cat = Łoś}}
+[[Category:Model Theory]]
+[[Category:Mathematical Logic]]
+97y8pu12rwidx7gtprot9k9zlpin4m6
+\end{proof}<|endoftext|>
+\section{Closed Extension Topology is Topology}
+Tags: Closed Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\tau^*_p$ be the [[Definition:Closed Extension Topology|closed extension topology]] of $\tau$.
+Then $\tau^*_p$ is a [[Definition:Topology|topology]] on $S^*_p = S \cup \set p$.
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \O$
+We have that $\varnothing \in \tau^*_p$ by definition.
+We also have that $S \in \tau$ so $S \cup \set p \in \tau^*_p$.
+Now let $U_1, U_2 \in \tau^*_p$.
+Then $U_1^* = U_1 \setminus \set p \in \tau, U_2^* = U_2 \setminus \set p \in \tau$.
+Then $p \in U_1$ and $p \in U_2$ and so $p \in U_1 \cap U_2$.
+So $U_1 \cap U_2 = \paren {U_1^* \cap U_2^*} \cup \set p \in \tau^*_p$. 
+Finally consider $\UU \subseteq \tau^*_p$.
+:$\displaystyle \bigcup \UU = \bigcup_{U \mathop \in \UU} \paren {U \setminus \set p} \cup \set p$
+So $\displaystyle \bigcup \UU \in \tau^*_p$.
+So $\tau^*_p$ is a [[Definition:Topology|topology]] on $S \cup \set p$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tarski-Vaught Test}
+Tags: Model Theory, Mathematical Logic
+
+\begin{theorem}
+Let $\mathcal M, \mathcal N$ be [[Definition:Structure|$\mathcal L$-structures]] such that $\mathcal M$ is a [[Definition:Substructure|substructure]] of $\mathcal N$.
+{{wtd|The page [[Definition:Structure]] is a disambiguation page, in which the form in which it is used here may not be included. The level of clarity in this page generally needs improving. Hence the invocation of the Disambiguate template.}}
+{{Disambiguate|Definition:Structure}}
+Then $\mathcal M$ is an [[Definition:Elementary Substructure|elementary substructure]] of $\mathcal N$ {{iff}}:
+:for every [[Definition:Logical Formula|$\mathcal L$-formula]] $\phi \left({x, \bar v}\right)$ and for every $\bar a$ in $\mathcal M$:
+::if there exists an $n$ in $\mathcal N$ such that $\mathcal N \models \phi \left({n, \bar a}\right)$
+::then there exists an $m$ in $\mathcal M$ such that $\mathcal N \models \phi \left({m, \bar a}\right)$.
+{{wtd|Before sense can be made of this page, [[Definition:Substructure]] and  [[Definition:Elementary Substructure]] need to be written.}}
+{{Disambiguate|Definition:Logical Formula}}
+The condition on the right side of the {{iff}} statement above can be rephrased as:
+:Every existential statement with parameters from $\mathcal M$ which is satisfied in $\mathcal N$ can be witnessed by an element from the substructure $\mathcal M$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\mathcal M$ be an elementary substructure of $\mathcal N$.
+Then $\mathcal N \models \exists x \phi \left({x, \bar a}\right)$ implies that $\mathcal M \models \exists x \phi \left({x, \bar a}\right)$.
+Hence there exists some $m$ in $\mathcal M$ such that:
+:$\mathcal M \models\phi \left({m, \bar a}\right)$.
+Passing back up to $\mathcal N$ yields the result.
+{{explain|Worth making it explicit exactly what that result is, so as to clarify what is being sought here.}}
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $\mathcal M$ be such that:
+:For every $\mathcal L$-formula $\phi \left({x, \bar v}\right)$ and for every $\bar a$ in $\mathcal M$:
+::if there exists an $n$ in $\mathcal N$ such that $\mathcal N \models \phi \left({n, \bar a}\right)$
+::then there exists an $m$ in $\mathcal M$ such that $\mathcal N \models \phi \left({m, \bar a}\right)$.
+It is to be shown that $\mathcal M$ is an elementary substructure of $\mathcal N$.
+That is:
+:for every $\mathcal L$-formula $\psi \left({\bar v}\right)$
+and:
+:for every $\bar a$ in $\mathcal M$:
+::$\left({\mathcal M \models \phi \left({\bar a}\right)}\right) \iff \left({\mathcal N \models\phi \left({\bar a}\right)}\right)$
+The proof proceeds by induction on complexity of formulas.
+Let $\psi$ be quantifier free.
+From [[Quantifier Free Formula is Preserved by Superstructure]], quantifier free formulas with parameters from $\mathcal M$ are preserved when passing to and from superstructures.
+Hence the result holds.
+Let the result holds for $\psi$.
+Consider $\neg \psi$:
+We have that:
+:$\left({\mathcal M \models \neg \psi \left({\bar a}\right)}\right) \iff \left({\mathcal M \not \models \psi \left({\bar a}\right)}\right)$
+By the inductive hypothesis:
+:$\left({\mathcal M \not\models\psi \left({\bar a}\right)}\right) \iff \left({\mathcal N \not\models\psi \left({\bar a}\right)}\right)$
+Thus the result follows for $\neg \psi$.
+Let the result hold for $\psi_0$ and $\psi_1$.
+Consider $\psi_0 \wedge \psi_1$:
+We have:
+:$\mathcal M \models \psi_0 \left({\bar a}\right) \wedge \psi_1 \left({\bar a}\right)$
+{{iff}}:
+:$\mathcal M \models\psi_0 \left({\bar a}\right)$
+and:
+:$\mathcal M \models \psi_1 \left({\bar a}\right)$
+By the inductive hypothesis:
+:$\mathcal M \models\psi_0 \left({\bar a}\right)$ and $\mathcal M \models \psi_1 \left({\bar a}\right)$
+{{iff}}:
+:$\mathcal N \models\psi_0 \left({\bar a}\right)$ and $\mathcal N \models \psi_1 \left({\bar a}\right)$
+Thus the result follows for $\psi_0 \wedge \psi_1$.
+Let the result holds for $\psi$.
+Consider:
+:$\exists x: \psi \left({x}\right)$
+We prove the two directions of this case separately.
+First let:
+:$\mathcal M \models \exists x: \psi \left({x, \bar a}\right)$
+Then there exists $m \in \mathcal M$ such that:
+:$\mathcal M \models \psi \left({m, \bar a}\right)$
+By the inductive hypothesis:
+:$\mathcal N \models \psi \left({m, \bar a}\right)$
+and so:
+:$\mathcal N \models \exists x: \psi \left({x, \bar a}\right)$
+Conversely, let:
+:$\mathcal N \models \exists x: \psi \left({x, \bar a}\right)$
+By assumption, there exists $m \in \mathcal M$ such that:
+:$\mathcal N \models \psi \left({m, \bar a}\right)$
+By the inductive hypothesis:
+:$\mathcal M \models \psi \left({m, \bar a}\right)$
+and hence:
+:$\mathcal M \models \exists x: \psi \left({x, \bar a}\right)$
+completing the proof.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quantifier Free Formula is Preserved by Superstructure}
+Tags: Model Theory, Mathematical Logic
+
+\begin{theorem}
+Let $\MM, \NN$ be $\LL$-[[Definition:Structure|structures]] such that  $\MM$ is a [[Definition:Substructure|substructure]] of $\NN$.
+{{Disambiguate|Definition:Structure}}
+Let $\map \phi {\bar x}$ be a [[Definition:Quantifier-Free Formula|quantifier-free $\LL$-formula]], and let $\bar a \in\MM$.
+Then $\MM \models \map \phi {\bar a}$ {{iff}} $\NN \models \map \phi {\bar a}$.
+\end{theorem}
+
+\begin{proof}
+The proof is done by induction on complexity of formulas.
+Note that since [[Interpretations of Terms are Preserved by Superstructures|interpretations of terms with parameters from $\MM$ are preserved when passing to superstructures]], we have that $\map {t^\MM} {\bar a} = \map {t^\NN} {\bar a}$ whenever $t$ is an $\LL$-[[Definition:Term|term]] and $\bar a$ is in $\MM$.
+First, we verify the theorem for atomic formulas.  This is the base for the inductive proof.
+Suppose $\phi$ is $t_1 = t_2$ for terms $t_1$ and $t_2$:
+We have $\MM \models \map \phi {\bar a}$ {{iff}} $\MM \models \map {t_1} {\bar a} = \map {t_2} {\bar a}$.
+The {{RHS}} of this holds {{iff}} $\map {t_1^\MM} {\bar a} = \map {t_2^\MM} {\bar a}$.
+By the note above, this in turn holds {{iff}} $\map {t_1^\NN} {\bar a} = \map {t_2^\NN} {\bar a}$, which is equivalent to $\NN \models \map \phi {\bar a}$.
+Suppose $\phi$ is $R(t_1, \dots, t_n)$ where $R$ is a relation symbol and each $t_i$ is a term:
+We have:
+:$\MM \models \map \phi {\bar a}$ iff $\paren {\map {t_1^\NN} {\bar a}, \dots, \map {t_n^\NN} {\bar a} } \in R^\MM$
+Since $\MM$ is a substructure of $\NN$, the interpretation of $R$ in $\NN$ must extend $R^\MM$, so the right side of this statement is equiavlent to:
+:$\paren {\map {t_1^\NN} {\bar a}, \dots, \map {t_n^\NN} {\bar a} } \in R^\NN$
+By the note above, this in turn is equivalent to:
+:$\paren {\map {t_1^\NN} {\bar a}, \dots, \map {t_n^\NN} {\bar a} } \in R^\NN$
+which is the same as:
+:$\NN \models \map \phi {\bar a}$
+Having verified the theorem for the atomic formulas, we proceed with the inductive step of the proof.
+Suppose the result holds for $\psi$, and consider $\neg \psi$:
+We have $\MM \models \neg \map \psi {\bar a}$ {{iff}} $\MM \not \models \map \psi {\bar a}$.
+By the inductive hypothesis, the right side of this statement is equivalent to $\NN \not\models \map \psi {\bar a}$, and so the result follows.
+Suppose the result holds for $\psi_1$ and $\psi_2$.#
+Consider $\psi_1 \wedge \psi_2$:
+We have $\MM \models \map {\psi_1} {\bar a} \wedge \map {\psi_2} {\bar a}$ {{iff}} $\MM \models \map {\psi_1} {\bar a}$ and $\MM \models \map {\psi_2} {\bar a}$.
+By the inductive hypothesis, the {{RHS}} of this is equivalent to $\NN \models \map {\psi_1} {\bar a}$ and $\NN \models \map {\psi_2} {\bar a}$.
+The result follows, completing the proof.
+{{qed}}
+{{MissingLinks}}
+[[Category:Model Theory]]
+[[Category:Mathematical Logic]]
+k9mvibx55zw5o2p5iik6700k6s5yzmr
+\end{proof}<|endoftext|>
+\section{Algebraically Closed Field is Infinite}
+Tags: Field Theory
+
+\begin{theorem}
+Let $F$ be an [[Definition:Algebraically Closed Field|algebraically closed field]].
+Then $F$ is [[Definition:Infinite Field|infinite]].
+\end{theorem}
+
+\begin{proof}
+We prove the [[Definition:Contrapositive Statement|contrapositive]]: that a [[Definition:Galois Field|Galois field]] cannot be [[Definition:Algebraically Closed Field|algebraically closed]].
+Let $F$ be [[Definition:Galois Field|Galois]].
+Define the [[Definition:Polynomial (Abstract Algebra)|polynomial]]:
+:$\displaystyle \map f x = 1 + \prod_{a \mathop \in F} \paren {x - a}$
+By definition, a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Ring (Abstract Algebra)|ring]].
+Thus by [[Ring Product with Zero]]:
+:$\displaystyle \forall x \in F: \prod_{a \mathop \in F} \paren {x - a} = 0$
+But:
+:$\forall a \in F: \map f a = 1 + 0 \ne 0$
+Therefore $\map f x$ cannot have a [[Definition:Root of Polynomial|root]] in $F$.
+Thus $F$ is not [[Definition:Algebraically Closed Field|algebraically closed]].
+The result follows by the [[Rule of Transposition]].
+{{qed}}
+[[Category:Field Theory]]
+dxwe85ej6e1gfs3xfgm4kzqok4lswql
+\end{proof}<|endoftext|>
+\section{Field of Uncountable Cardinality K has Transcendence Degree K}
+Tags: Field Theory, Uncountable Sets
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]] of [[Definition:Uncountable Set|uncountable]] [[Definition:Cardinality|cardinality]] $\kappa$.
+Then $F$ has [[Definition:Transcendence Degree|transcendence degree]] $\kappa$ over its [[Definition:Prime Subfield|prime field]].
+\end{theorem}
+
+\begin{proof}
+We prove the theorem for fields with characteristic $p = 0$.
+In the case where $p$ is a prime, the proof is similar, but instead we view the fields as extensions of $\Z / \Z_p$.
+The main idea is to exploit the lower cardinality of sets of polynomials in order to keep finding algebraically independent elements of $F$.
+Since each [[Definition:Characteristic of Ring|characteristic]] $0$ field contains a copy of $\Q$ as its prime field, we can view $F$ as a [[Definition:Field Extension|field extension]] over $\Q$.
+We will show that $F$ has a subset of cardinality $\kappa$ which is [[Definition:Algebraically Independent|algebraically independent]] over $\Q$.
+Since $\kappa$ is the largest possible cardinality for a subset of $F$, this will establish the theorem.
+We build the claimed subset of $F$ by [[Definition:Transfinite Induction|transfinite induction]] and implicit use of the [[Axiom:Axiom of Choice|axiom of choice]].
+For each [[Definition:Ordinal|ordinal]] $\alpha < \kappa$ we define a [[Definition:Set|set]] $S_\alpha$.
+We will build the sets so that each $S_\alpha$ has cardinality equal to that of $\alpha$ and is algebraically independent over $\Q$.
+Let $S_0 = \varnothing$.
+Let $S_1$ be a [[Definition:Singleton|singleton]] containing some element of $F$ which is not algebraic over $\Q$.
+This is possible from [[Algebraic Numbers are Countable]].
+Define $S_\beta$ for [[Definition:Successor Ordinal|successor ordinals]] $\beta = \alpha + 1 < \kappa$ to be $S_\alpha$ together with an element of $F$ which is not a [[Definition:Root of Polynomial|root]] of any non-[[Definition:Null Polynomial over Ring|trivial]] [[Definition:Polynomial|polynomial]] with [[Definition:Polynomial Coefficient|coefficients]] in $\Q \cup S_\alpha$.
+{{explain|Determine whether "trivial" polynomial actually means the [[Definition:Null Polynomial over Ring]] or maybe one of [[Definition:Polynomial of Degree Zero]] or something -- whichever is meant, replace with what we have here, and make a note on that page to indicate this alternative terminology.}}
+From [[Set of Polynomials over Infinite Set has Same Cardinality]] there are only $\left\vert{\Q \cup S_\alpha}\right\vert = \aleph_0 + \left\vert{\alpha}\right\vert < \kappa$ many such polynomials.
+hence the above construction is possible.
+From [[Polynomial over Field has Finitely Many Roots]] and the [[Cardinality of Infinite Union of Infinite Sets]], the union of $\kappa$ many finite sets is at most size $\kappa$.
+Thus there are are this many roots of such polynomials
+Define $S_\beta$ for [[Definition:Limit Ordinal|limit ordinals]] by $\displaystyle {S_\beta = \bigcup_{\alpha \mathop < \beta} S_\alpha}$.
+We can then define $S_\kappa$ to be $\displaystyle{\bigcup_{\alpha \mathop < \kappa} S_\alpha}$.
+This is a set of size $\kappa$ since its elements can be paired with the ordinals less than $\kappa$.  
+It is now demonstrated that it is algebraically independent over $\Q$.
+{{explain|At this point it is easy to lose track of what "it" is. I have indeed done so. So it is worth restating here what which particular object we are demonstrating to be algebraically independent.}}
+Let $P \left({x_1, \ldots, x_n}\right)$ be a non-trivial polynomial with coefficients in $\Q$ and elements $a_1, \ldots, a_n$ in $F$.  
+{{WLOG}}, it is assumed that $a_n$ was added at an ordinal $\alpha + 1$ later than the other elements.  
+Then $P \left({a_1, \ldots, a_{n-1}, x_n}\right)$ is a polynomial with coefficients in $\Q \cup S_\alpha$.  
+$a_n$ was selected at stage $\alpha + 1$ so as not to be root of such a polynomial.
+So $P \left({a_1, \ldots, a_{n-1}, x_n}\right) \ne 0$.
+Thus it is algebraically independent over $\Q$.
+{{qed}}
+{{wtd|... implicit use of the [[Axiom:Axiom of Choice|axiom of choice]]. Determine if its necessary}}
+[[Category:Field Theory]]
+[[Category:Uncountable Sets]]
+h0pxhtc8ixdqu51pz8zi9efmgtibyuy
+\end{proof}<|endoftext|>
+\section{Theory of Algebraically Closed Fields of Characteristic p is Complete}
+Tags: Model Theory, Mathematical Logic
+
+\begin{theorem}
+Let $p$ be either $0$ or a [[Definition:Prime Number|prime number]].
+Let $ACF_p$ be the [[Definition:Theory of Elementary Class|theory]] of [[Definition:Algebraically Closed Field|algebraically closed fields]] of [[Definition:Characteristic of Field|characteristic]] $p$ in the language $\mathcal L_r = \left\{ {0, 1, +, -, \cdot}\right\}$ for [[Definition:Ring (Abstract Algebra)|rings]], where:
+:$0, 1$ are [[Definition:Constant|constants]]
+and:
+:$+, -, \cdot$ are [[Definition:Binary Operation|binary functions]]. 
+Then $ACF_p$ is [[Definition:Complete Theory|complete]].
+\end{theorem}
+
+\begin{proof}
+By the [[Łoś-Vaught Test]], it suffices to show that $ACF_p$ is satisfiable, has no finite models, and is [[Definition:Categorical (Model Theory)|$\kappa$-categorical]] for some [[Definition:Uncountable Set|uncountable]] $\kappa$.
+=== Satisfiability ===
+$\C$ is an [[Definition:Algebraically Closed Field|algebraically closed field]] of [[Definition:Characteristic of Field|characteristic]] $0$.
+If $p$ is a [[Definition:Prime Number|prime]], then the [[Definition:Algebraic Closure|algebraic closure]] of $\Z / \Z_p$ is an [[Definition:Algebraically Closed Field|algebraically closed field]] of [[Definition:Characteristic of Field|characteristic]] $p$.
+Thus $ACF_p$ is satisfiable.
+{{explain|Needs to be made clear why what is above leads to that statement.}}
+{{qed|lemma}}
+=== No Finite Models ===
+From [[Algebraically Closed Field is Infinite]]:
+$ACF_p$ has no finite models
+{{qed|lemma}}
+=== $\kappa$-Categorical ===
+From:
+: [[Field of Uncountable Cardinality K has Transcendence Degree K|Field of Uncountable Cardinality $\kappa$ has Transcendence Degree $\kappa$]]
+and:
+: [[Algebraically Closed Fields are Isomorphic iff they have the same Characteristic and Transcendence Degree]]
+it follows that:
+: $ACF_p$ is $\kappa$-categorical for all uncountable $\kappa$
+Hence:
+: $ACF_p$ is $\kappa$-categorical for some uncountable $\kappa$
+as was to be proved.
+{{qed}}
+[[Category:Model Theory]]
+[[Category:Mathematical Logic]]
+q2tolc8ek8jowx6j72tpmdgmhgs5lmj
+\end{proof}<|endoftext|>
+\section{Lefschetz Principle (First-Order)}
+Tags: Algebraic Geometry, Field Theory, Model Theory, Mathematical Logic, Proofs by Contraposition
+
+\begin{theorem}
+Let $\phi$ be a sentence in the language $\LL_r = \set {0, 1, +, -, \cdot}$ for rings, where $0, 1$ are constants and $+, -, \cdot$ are binary functions.
+{{TFAE}}
+:$(1): \quad \phi$ is true in every algebraically closed field of characteristic $0$.
+:$(2): \quad \phi$ is true in some algebraically closed field of characteristic $0$.
+:$(3): \quad \phi$ is true in algebraically closed fields of characteristic $p$ for arbitrarily large primes $p$.
+:$(4): \quad \phi$ is true in algebraically closed fields of characteristic $p$ for sufficiently large primes $p$.
+Note in particular that since $\C$ is an algebraically closed field of characteristic $0$, these are equivalent to $\phi$ being true in the field $\C$.
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ iff $(2)$ ===
+From [[Theory of Algebraically Closed Fields of Characteristic p is Complete|Theory of Algebraically Closed Fields of Characteristic $p$ is Complete]]:
+the theory $ACF_p$ of algebraically closed fields of characteristic $p$ is complete.
+That is, all such fields satisfy the exact same $\LL_r$ sentences.
+{{qed|lemma}}
+=== $(2)$ implies $(4)$ ===
+Let $\phi$ be true in some such field.
+Then:
+:$ACF_0 \models \phi$
+By [[Gödel's Completeness Theorem]] and the finiteness of proofs, it follows that there is a finite subset $\Delta$ of $ACF_0$ such that $\Delta \models \phi$.
+Such a $\Delta$ can only make finitely many assertions about the character of its models.
+Hence, as long as $p$ is selected sufficiently large, an algebraically closed field of characteristic $p$ will satisfy $\phi$.
+{{qed|lemma}}
+=== $(4)$ implies $(3)$ ===
+We have that all sufficiently large $p$ work.
+Hence it follows that it is always possible to find arbitrarily large $p$ that work.
+{{qed|lemma}}
+=== $(3)$ implies $(2)$ ===
+We prove this by [[Proof by Contraposition|contraposition]].
+Suppose there is no algebraically closed field of characteristic $0$ where $\phi$ is true.
+Then $\phi$ is false in algebraically closed fields of characteristic $0$
+Since $ACF_0$ is complete, this means that $ACF_0 \models \neg \phi$.
+Similarly to the case of $(2)$ implies $(4)$, there must then be a finite subset $\Delta$ of $ACF_0$ such that $\Delta \models \neg \phi$.
+But then, for all sufficiently large $p$, we have that $\phi$ is false in the algebraically closed fields of characteristic $p$.
+Hence, it cannot be true for arbitrarily large $p$.
+By [[Rule of Transposition]], $\phi$ is true in algebraically closed fields for arbitrarily large primes $p$ implies there exists an algebraically closed field of characteristic $0$ where $\phi$ is true.
+{{qed}}
+{{Namedfor|Solomon Lefschetz|cat = Lefschetz}}
+[[Category:Algebraic Geometry]]
+[[Category:Field Theory]]
+[[Category:Model Theory]]
+[[Category:Mathematical Logic]]
+[[Category:Proofs by Contraposition]]
+2ztr54llb9zkvlktqc9933kmxbwp44x
+\end{proof}<|endoftext|>
+\section{CNF Satisfiability Problem is NP-Complete}
+Tags: Mathematical Logic
+
+\begin{theorem}
+The [[Definition:CNF Satisfiability Problem|conjunctive normal form boolean satisfiability problem]] (CNF SAT) is [[Definition:NP-Complete|NP-complete]].
+\end{theorem}
+
+\begin{proof}
+Let $P$ be a [[Definition:CNF Satisfiability Problem|CNF SAT problem]].
+=== CNF SAT is NP ===
+A potential solution to $P$ can be verified in polynomial time by checking every clause in $L$ to see if they all have at least one true un-negated variable or one false negated variable.
+Because a solution can be verified or rejected in polynomial time the problem is [[Definition:NP Complexity Class|NP]]
+=== CNF SAT is NP-hard ===
+We will show this by reducing the [[Definition:Boolean Satisfiability Problem|boolean satisfiability (SAT) problem]] to [[Definition:CNF Satisfiability Problem|CNF SAT]].
+The algorithm to convert the [[Definition:Boolean Satisfiability Problem|SAT) problem]] to [[Definition:CNF Satisfiability Problem|CNF SAT]] is recursive.
+Wherever  $A$, $B$,and $C$ are seen in the output it is understood that the  algorithm would call itself on those formulas and convert them into [[Definition:Conjunctive Normal Form|CNF]].
+==== iff and xor ====
+The first step will be to remove every instance of mutual implication and exclusive or from the problem.
+In every clause with one of these functions there must be at least one that does not have any instance of mutual implication or exclusive or in its arguments.
+Introduce a new variable $x$ and replace either:
+:$A \implies B$ with $x$
+or:
+:$\neg \left({A \implies B}\right)$ with $\neg x$
+as appropriate.
+Then add the following four clauses to $L$:
+: $A \lor B \lor x$
+: $A \lor \neg B \lor \neg x$
+: $\neg A \lor B \lor \neg x$
+: $\neg A \lor \neg B \lor x$
+Repeat this procedure until there are no more mutual implications or exclusive ors in the list of clauses.
+It is important to note that the amount this procedure adds to the size of the final problem is bounded.
+Specifically, every instance of mutual implication or exclusive or increases the number of variables by one, increases the number of "new" symbols in $L$ by a fixed number and reproduces every "old" symbol in $L$ at most four times.
+In total, this step increases the size of the problem at most linearly.
+The remaining functions that need to be considered can be converted to [[Definition:Conjunctive Normal Form|CNF]] with only a constant amount of increase to the size by a constant amount per production.
+Given a logical expression it will either be a variable, a unary function, or a binary function.
+==== Single Variable ====
+If $A$ is a single variable then it is in [[Definition:Conjunctive Normal Form|CNF]] and the task is complete.
+==== Unary functions ====
+With the two constant functions, one can simply replace $f_F \left({A}\right)$ and $f_T \left({A}\right)$ with $F$ and $T$ respectively.
+These can then be removed from the final expression as appropriate.
+{{explain|Link to the results linking $F$ and $T$ with disjunctions / conjunctions etc.}}
+For the identity function, replace $f_I \left({A}\right)$ with the [[Definition:Conjunctive Normal Form|CNF]] conversion of $A$.
+With the $\neg$ function the replacement depends on the argument.
+If  the argument is a single variable $x$ then the output $\neg x$ is in [[Definition:Conjunctive Normal Form|CNF]].
+If the argument is one of the constant functions then replace $\neg f_F \left({A}\right)$ and $\neg f_T \left({A}\right)$ with $T$  and $F$ respectively.
+If the argument is a binary function of the form  $f \left({A, B}\right)$ then replace the function with the appropriate one, following the list below:
+: $\neg f_F \left({A, B}\right) \to f_T \left({A, B}\right)$
+: $\neg f_T \left({A, B}\right) \to f_F \left({A, B}\right)$
+: $\neg A \land B \to A \uparrow B$
+: $\neg A \uparrow B \to A \land B$
+: $\neg \left({A \implies B}\right) \to \neg \left({A \implies B}\right)$
+: $\neg \neg \left({A \implies B}\right) \to \left({A \implies B}\right)$
+: $\neg \operatorname{pr}_1 \left({A, B}\right) \to \overline {\operatorname{pr}_1} \left({A, B}\right)$
+: $\neg \overline {\operatorname{pr}_1} \left({A, B}\right) \to \operatorname{pr}_1 \left({A, B}\right)$
+: $\neg \left({B \implies A}\right) \to \neg \left({B \implies A}\right)$
+: $\neg \neg \left({B \implies A}\right) \to \left({B \implies A}\right)$
+: $\neg \operatorname{pr}_2 \to \overline {\operatorname{pr}_2}$
+: $\neg \overline {\operatorname{pr}_2} \to \operatorname{pr}_2$
+: $\neg A \lor B \to A \downarrow B$
+: $\neg A \downarrow B \to A \lor B$
+==== Binary Functions ====
+Replace $f_T \left({A, B}\right)$ with $T$.
+$A \lor B$ is already in [[Definition:Conjunctive Normal Form|CNF]].
+Replace $A \implies B$ with $\neg A \lor B$.
+Replace $\operatorname{pr}_1 \left({A, B}\right)$ with $A$.
+Replace $B \implies A$ with $A \lor \neg B$.
+Replace $\operatorname{pr}_2 \left({A, B}\right)$ with $B$.
+In  the case of $A \land B$ the clause takes either the form:
+: $A \land B$
+or:
+:$\left({A \land B}\right) \lor C$
+In the first case simply replace the clause with  the two clauses $A$ and $B$.
+In the second case introduce the new variable $x$ and replace the clause with the three clauses:
+: $x \lor C$
+: $\neg x \lor A$
+: $\neg x \lor B$
+A simple case analysis will show that the above clauses have a solution {{iff}} $\left({A \land B}\right) \lor C$.
+Replace $A \uparrow B$ with $\neg A \lor \neg B$
+ 
+Replace $\overline {\operatorname{pr}_2} \left({A, B}\right)$ with $\neg B$.
+Replace $\neg \left({A \implies B}\right)$ with $A \land \neg B$ and use the rule for $A \land B$.
+Replace $\overline {\operatorname{pr}_2} \left({A, B}\right)$ with $\neg A$.
+Replace $\neg \left({B \implies A}\right)$ with $\neg A \land B$ and use the rule for $A \land B$.
+Replace $A \downarrow B$ with $\neg A \land \neg B$ and use the rule for $A \land B$.
+Replace $f_F \left({A, B}\right)$ with $F$.
+At this point there may be some instances of $T$ and $F$ in the clauses.
+They can be removed by:
+:deleting any clause from $L$ that contains $T$ in any of its conjunctions
+and:
+:removing $F$ from any clause that contains it.
+Note that if any clause simplifies to $F$ the problem is trivially unsatisfiable and the task is complete.
+In the end the size and number of variables in the CNF problem is $O(n)$ where $n$ is the number of symbols required to write down the original [[Definition:Boolean Satisfiability Problem|SAT problem]].
+Hence the conversion can be done in polynomial time.
+We have that [[Boolean Satisfiability Problem is NP-hard]].
+But [[Definition:Boolean Satisfiability Problem|SAT]] is reducible to [[Definition:CNF Satisfiability Problem|CNF SAT]].
+Therefore [[Definition:CNF Satisfiability Problem|CNF SAT]] is [[Definition:NP-Hard|NP-hard]].
+[[Definition:CNF Satisfiability Problem|CNF SAT]] is [[Definition:NP-Complete|NP-complete]].
+{{explain|The logical steps need to be clarified in the above.}}
+{{qed}}
+[[Category:Mathematical Logic]]
+8fhj0sypbhdkrrdnp5fv8a0zfherp05
+\end{proof}<|endoftext|>
+\section{Closed Sets of Closed Extension Topology}
+Tags: Closed Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the [[Definition:Closed Extension Space|closed extension space]] of $T$.
+Then the [[Definition:Closed Set (Topology)|closed sets]] of $T^*_p$ (apart from $S^*_p$) are the [[Definition:Closed Set (Topology)|closed sets]] of $T$.
+This explains why $\tau^*_p$ is called the [[Definition:Closed Extension Topology|closed extension topology]] of $\tau$.
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \O$
+Let $V \subseteq S^*_p$ be [[Definition:Closed Set (Topology)|closed]] in $T^*_p$.
+Then $S^*_p \setminus V$ is [[Definition:Open Set (Topology)|open]] in $T^*_p$.
+Then $\struct {S^*_p \setminus V} \setminus \set p$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Set Difference with Union]] we have:
+:$\struct {S^*_p \setminus V} \setminus \set p = S^*_p \setminus \struct {V \cup \set p}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Point Topology is Closed Extension Topology of Discrete Topology}
+Tags: Closed Extension Topology, Particular Point Topology, Discrete Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] and let $p \in S$.
+Let $\tau_p$ be the [[Definition:Particular Point Topology|particular point topology]] on $S$.
+Let $T = \struct {S \setminus \set p, \vartheta}$ be the [[Definition:Discrete Space|discrete topological space]] on $S \setminus \set p$.
+Then $T^* = \struct {S, \tau_p}$ is a [[Definition:Closed Extension Space|closed extension space]] of $T$.
+\end{theorem}
+
+\begin{proof}
+Directly apparent from the definitions of [[Definition:Particular Point Topology|particular point topology]], [[Definition:Discrete Space|discrete topological space]] and [[Definition:Closed Extension Space|closed extension space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Extension Topology is not T1}
+Tags: T1 Spaces, Closed Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the [[Definition:Closed Extension Space|closed extension space]] of $T$.
+Then $T^*_p$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \O$
+Let $x \in S^*_p, x \ne p$.
+Let $U = \set p$.
+Then $U \in \tau^*_p$ such that $p \in U, x \notin U$.
+But there is no $V \in \tau^*_p$ such that $x \in V, p \notin V$, by definition of the [[Definition:Closed Extension Topology|closed extension topology]].
+Hence $T^*_p$ can not be a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Algebraic Numbers are Countable}
+Tags: Countable Sets, Polynomial Theory, Algebraic Numbers
+
+\begin{theorem}
+The [[Definition:Set|set]] $\Bbb A$ of [[Definition:Algebraic Number|algebraic numbers]] is [[Definition:Countable Set|countable]].
+\end{theorem}
+
+\begin{proof}
+By definition, $\Bbb A$ is the [[Definition:Subset|subset]] of the [[Definition:Complex Number|complex numbers]] which consists of [[Definition:Root of Polynomial|roots of polynomials]] with coefficients in $\Q$.
+We can prove the theorem by a cardinality argument, counting the number of such polynomials and roots.
+By [[Set of Polynomials over Infinite Set has Same Cardinality]], the [[Definition:Set|set]] $\Q \sqbrk x$ of polynomials over $\Q$ is [[Definition:Countable Set|countable]].
+Next, note that $\Bbb A$ can be written as the [[Definition:Set Union|union]] of the set of roots of each polynomial.
+That is:
+:$\Bbb A = \displaystyle \bigcup_{\map p x \mathop \in \Q \sqbrk x} \set {x \in \C: \map p x = 0}$
+From [[Polynomial over Field has Finitely Many Roots]], $\Bbb A$ is a [[Definition:Countable Union|union of countably many finite sets]].
+From [[Countable Union of Countable Sets is Countable]] it follows that $\Bbb A$ is [[Definition:Countable|countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Points in Closed Extension Space}
+Tags: Limit Points in Closed Extension Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the [[Definition:Closed Extension Space|closed extension space]] of $T$.
+Let $x \in S$.
+Then $x$ is a [[Definition:Limit Point of Point|limit point]] of $p$.
+\end{theorem}
+
+\begin{proof}
+Every [[Definition:Open Set (Topology)|open set]] of $T^*_p = \struct {S^*_p, \tau^*_p}$ except $\O$ contains the point $p$ by [[Definition:Closed Extension Topology|definition]].
+So every [[Definition:Open Set (Topology)|open set]] $U \in \tau^*_p$ such that $x \in U$ contains $p$.
+So by definition of the [[Definition:Limit Point of Point|limit point of a point]], $x$ is a [[Definition:Limit Point of Point|limit point]] of $p$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Open Set of Closed Extension Space}
+Tags: Open Sets, Set Closures, Closed Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the [[Definition:Closed Extension Space|closed extension space]] of $T$.
+Then $U^- = S^*_p$ where $U^-$ denotes the [[Definition:Closure (Topology)|closure]] of $U$ in $T^*_p$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\forall U \in \tau^*_p, u \ne \O: p \in U$.
+From [[Limit Points in Closed Extension Space]], every point in $S^*_p$ is a [[Definition:Limit Point of Point|limit point of $p$]].
+So by definition of [[Definition:Closure (Topology)|closure]], every point in $S^*_p$ is in $U^-$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Extension Space is Irreducible}
+Tags: Irreducible Spaces, Closed Extension Topology, Closed Extension Space is Irreducible
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the [[Definition:Closed Extension Space|closed extension space]] of $T$.
+Then $T^*_p$ is [[Definition:Irreducible Space|irreducible]].
+\end{theorem}
+
+\begin{proof}
+Trivially, by definition, every [[Definition:Open Set (Topology)|open set]] in $T^*_p$ contains $p$.
+So:
+:$\forall U_1, U_2 \in \tau^*_p: p \in U_1 \cap U_2$
+for $U_1, U_2 \ne \O$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Closure of Open Set of Closed Extension Space]] we have that:
+:$\forall U \in \tau^*_p: U \ne \O \implies U^- = S$
+where $U^-$ is the [[Definition:Closure (Topology)|closure]] of $U$.
+The result then follows by definition of [[Definition:Irreducible Space/Definition 6|irreducible space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomial over Field has Finitely Many Roots}
+Tags: Polynomial Theory
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $F \left[{x}\right]$ be the [[Definition:Ring of Polynomial Functions|ring of polynomial functions]] in the indeterminate $x$.
+If $p \in F \left[{x}\right]$ be non-[[Definition:Null Polynomial over Ring|null]], then $p$ has finitely many roots in $F$.
+\end{theorem}
+
+\begin{proof}
+Let $n \ge 1$ be the [[Definition:Degree (Polynomial)|degree]] of $p$.
+We argue that $p$ has at most $n$ [[Definition:Root of Polynomial|roots]] in $F$.
+Let $A$ be the [[Definition:Set|set]] of roots of $p$.
+Let $a \in A$.
+By the [[Polynomial Factor Theorem]]:
+:$p \left({x}\right) = q_1 \left({x}\right) \cdot \left({x - a}\right)$
+:where $\deg q_1 = n - 1$.
+Let $a' \in A$ such that $a' \ne a$.
+Then since:
+:$p \left({a'}\right) = 0$
+but:
+:$\left({a' - a}\right) \ne 0$
+it follows that:
+:$q_1 \left({a'}\right) = 0$
+Again by the [[Polynomial Factor Theorem]]:
+:$q_1 \left({x}\right) = q_2 \left({x}\right) \cdot \left({x - a'}\right)$
+Therefore:
+:$p \left({x}\right) = q_2 \left({x}\right) \cdot \left({x - a'}\right) \cdot \left({x - a}\right)$
+where $\deg q_2 = n-2$.
+We can repeat this procedure as long as there are still distinct roots in $A$.
+After the $i$th iteration we obtain a polynomial $q_i$ of degree $n-i$.
+Now the degree of $q_i$ decreases each time we factor out a root of $p$, so $q_n$ is necessarily a constant term.
+Therefore $q_n$ can share no roots with $p$.
+So this procedure must stop by at most the $n$th step.
+That is, we can pick at most $n$ distinct elements from the set $A$, and:
+:$\left\vert{A}\right\vert \le n$
+{{qed}}
+[[Category:Polynomial Theory]]
+6s0xud8pptm377rznzyfrdvezyq5w0x
+\end{proof}<|endoftext|>
+\section{Set of Polynomials over Infinite Set has Same Cardinality}
+Tags: Polynomial Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] of [[Definition:Infinite Set|infinite]] [[Definition:Cardinality|cardinality]] $\kappa$.
+Let $S \left[{x}\right]$ be the set of [[Definition:Polynomial Form|polynomial forms]] over $S$ in the [[Definition:Indeterminate|indeterminate]] $x$.
+Then $S \left[{x}\right]$ has [[Definition:Cardinality|cardinality]] $\kappa$.
+\end{theorem}
+
+\begin{proof}
+Since $S \left[{x}\right]$ contains a copy of $S$ as [[Definition:Constant Polynomial|constant polynomials]], we have an [[Definition:Injection|injection]] $S \to S \left[{x}\right]$.
+We define an [[Definition:Injection|injection]] from $S \left[{x}\right]$ to the [[Definition:Set|set]] $\mathcal F$ of [[Definition:Finite Sequence|finite sequences]] over $S$ as follows:
+Each [[Definition:Polynomial|polynomial]] in $f \in S \left[{x}\right]$ is of the form:
+:$f = a_0 + a_1 x + a_2 x^2 + \dotsb + a_n x^n$
+where $a_n$ is non-zero and each $a_i$ is in $S$.
+We send each [[Definition:Polynomial|polynomial]] $f$ to the [[Definition:Finite Sequence|sequence]] of its [[Definition:Polynomial Coefficient|coefficients]] $\left({a_0, \dotsc, a_n}\right)$.
+By the definition of [[Equality of Polynomials|equality of polynomials]], this is [[Definition:Injection|injective]].
+Now the set of [[Definition:Finite Sequence|finite sequences]] over $S$ is a [[Definition:Countable Union|countable union]] of [[Definition:Set|sets]] of [[Definition:Cardinality|cardinality]] $\kappa$.
+From [[Cardinality of Infinite Union of Infinite Sets]], $\mathcal F$ has [[Definition:Cardinality|cardinality]] $\kappa$.
+Therefore there is a [[Definition:Bijection|bijection]] $\mathcal F \leftrightarrow S$.
+Composing this with the [[Definition:Injection|injectio]]n $S \left[{x}\right] \to \mathcal F$, we have an [[Definition:Injection|injection]] $S \left[{x}\right] \to S$.
+So by the [[Cantor-Bernstein-Schröder Theorem]] there is a [[Definition:Bijection|bijection]] $S \left[{x}\right] \leftrightarrow S$.
+Hence, we have:
+:$\left\vert{S \left[{x}\right]}\right\vert = \left\vert{S}\right\vert = \kappa$
+{{qed}}
+[[Category:Polynomial Theory]]
+p5prg956jy5h9i1jlhr1rm5qmxo5gqb
+\end{proof}<|endoftext|>
+\section{Excluded Point Topology is Topology}
+Tags: Excluded Point Topology
+
+\begin{theorem}
+Let $T = \left({S, \tau_{\bar p}}\right)$ be an [[Definition:Excluded Point Space|excluded point space]].
+Then $\tau_{\bar p}$ is a [[Definition:Topology|topology]] on $S$, and $T$ is a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+We have [[Definition:Excluded Point Space|by definition]] that $S \in \tau_{\bar p}$, and as $p \notin \varnothing$ we have that $\varnothing \in \tau_{\bar p}$.
+Now let $U_1, U_2 \in \tau_{\bar p}$.
+By definition $p \notin U_1$ and $p \notin U_2$, and so $p \notin U_1 \cap U_2$ by definition of [[Definition:Set Intersection|set intersection]].
+So $U_1 \cap U_2 \in \tau_{\bar p}$.
+Now let $\mathcal U \subseteq \tau_{\bar p}$.
+We have that $\forall U \in \mathcal U: p \notin U$.
+Hence from [[Subset of Union]] $p \notin \bigcup \mathcal U$.
+So all the properties are fulfilled for $\tau_{\bar p}$ to be a [[Definition:Topology|topology]] on $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Extension Topology is Topology}
+Tags: Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\tau^*_{\bar p}$ be the [[Definition:Open Extension Topology|open extension topology]] of $\tau$.
+Then $\tau^*_{\bar p}$ is a [[Definition:Topology|topology]] on $S^*_p = S \cup \set p$.
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$
+We have that $S^*_p \in \tau^*_{\bar p}$ by definition.
+We also have that $\O \in \tau$ so $\O \in \tau^*_{\bar p}$.
+Now let $U_1, U_2 \in \tau^*_{\bar p}$.
+Then $U_1, U_2 \in \tau \implies U_1 \cap U_2 \in \tau \implies U_1 \cap U_2 \in \tau^*_{\bar p}$.
+Finally consider $\UU \subseteq \tau^*_{\bar p}$.
+Assuming $S^*_p \notin \UU$ we have that $\UU \subseteq \tau$.
+So:
+:$\displaystyle \bigcup \UU \in \tau$ and so $\displaystyle \bigcup \UU \in \tau^*_p$.
+If $S^*_p \in \UU$ then $\displaystyle \bigcup \UU = S^*_p \in \tau^*_p$.
+So $\tau^*_p$ is a [[Definition:Topology|topology]] on $S \cup \set p$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Topology is Open Extension Topology of Discrete Topology}
+Tags: Open Extension Topology, Excluded Point Topology, Discrete Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] and let $p \in S$.
+Let $\tau_{\bar p}$ be the [[Definition:Excluded Point Topology|excluded point topology]] on $S$.
+Let $T = \struct {S \setminus \set p, \tau_D}$ be the [[Definition:Discrete Space|discrete topological space]] on $S \setminus \set p$.
+Then $T^* = \struct {S, \tau_{\bar p} }$ is an [[Definition:Open Extension Space|open extension space]] of $T$.
+\end{theorem}
+
+\begin{proof}
+Directly apparent from the definitions of [[Definition:Excluded Point Topology|excluded point topology]], [[Definition:Discrete Space|discrete topological space]] and [[Definition:Open Extension Space|open extension space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Countable Discrete Space with Sierpiński Space is Paracompact}
+Tags: Sierpiński Space, Discrete Topology, Paracompact Spaces
+
+\begin{theorem}
+Let $T_X = \struct {S, \tau}$ be a [[Definition:Countable Discrete Topology|countable discrete space]].
+Let $T_Y = \struct {\set {0, 1}, \tau_0}$ be a [[Definition:Sierpiński Space|Sierpiński space]].
+Let $T_X \times T_Y$ be the [[Definition:Product Space|product space]] of $T_X$ and $T_Y$.
+Then $T_X \times T_Y$ is [[Definition:Paracompact Space|paracompact]].
+\end{theorem}
+
+\begin{proof}
+From [[Discrete Space is Paracompact]], $T_X$ is [[Definition:Paracompact Space|paracompact]].
+We have that the [[Definition:Sierpiński Space|Sierpiński space]] $T_Y$ is a [[Definition:Finite Topological Space|finite topological space]].
+From [[Finite Topological Space is Compact]], $T_Y$ is a [[Definition:Compact Space|compact space]].
+{{finish|Steen and Seebach in Part $\text I$ chapter $3$ Compactness: Invariance Properties offer "If $X$ is compact, then in general $X \times Y$ has the compactness properties of $Y$." Whether this holds for $Y$ paracompact needs to be investigated.}}
+\end{proof}<|endoftext|>
+\section{Reduction of Explicit ODE to First Order System}
+Tags: Ordinary Differential Equations
+
+\begin{theorem}
+Let $\map {x^{\paren n} } t = \map F {t, x, x', \ldots, x^{\paren {n - 1} } }$, $\map x {t_0} = x_0$ be an [[Definition:Explicit ODE|explicit ODE]] with $x \in \R^m$.
+Let there exist $I \subseteq \R$ such that there exists a unique [[Definition:Particular Solution of Differential Equation|particular solution]]:
+:$x: I \to \R^m$
+to this [[Definition:Explicit ODE|ODE]].
+Then there exists a [[Definition:System of Differential Equations|system]] of [[Definition:First Order ODE|first order ODEs]]:
+:$y' = \map {\tilde F} {t, y}$
+with $y = \tuple {y_1, \ldots, y_{m n} }^T \in \R^{m n}$ such that:
+:$\tuple {\map {y_1} t, \ldots, \map {y_m} t} = \map x t$
+for all $t \in I$ and $\map y {t_0} = x_0$.
+\end{theorem}
+
+\begin{proof}
+Define the [[Definition:Mapping|mappings]]:
+:$z_1, \ldots, z_n: I \to \R^m$
+by:
+:$z_j = x^{\paren {j - 1} }$, $j = 1, \ldots, n$
+Then:
+{{begin-eqn}}
+{{eqn | l = z_1'
+      | r = z_2
+}}
+{{eqn | o = \vdots
+}}
+{{eqn | l = z_{n - 1}'
+      | r = z_n
+}}
+{{eqn | l = z_n'
+      | r = \map F {t, z_1, \ldots, z_n}
+}}
+{{end-eqn}}
+This is a [[Definition:System of Differential Equations|system]] of $m n$ [[Definition:First Order ODE|first order ODEs]].
+By construction:
+:$\map {z_1} t = \map x t$
+for all $t \in I$ and $\map {z_1} {t_0} = x_0$.
+Therefore we can take:
+:$y = \begin {pmatrix} z_1 \\ \vdots \\ z_{n - 1} \\ z_n \end {pmatrix}, \quad \tilde F: \begin {pmatrix} z_1 \\ \vdots \\ z_n \end{pmatrix} \mapsto \begin {pmatrix} z_2 \\ \vdots \\ z_n \\ \map F {t, z_1, \ldots, z_n} \end {pmatrix}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Continuous Image of Paracompact Space is not always Countably Metacompact}
+Tags: Paracompact Spaces, Countably Metacompact Spaces, Continuous Mappings, Open Mappings
+
+\begin{theorem}
+Let $T_A = \struct {X_A, \tau_A}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Paracompact Space|paracompact]].
+Let $T_B = \struct {X_B, \tau_B}$ be another [[Definition:Topological Space|topological space]].
+Let $\phi: T_A \to T_B$ be a [[Definition:Mapping|mapping]] which is both [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] and [[Definition:Open Mapping|open]].
+Then it is not necessarily even the case that $T_B$ is [[Definition:Countably Metacompact Space|countably metacompact]], let alone [[Definition:Paracompact Space|paracompact]].
+\end{theorem}
+
+\begin{proof}
+Let $T_X = \struct {X, \tau}$ be a [[Definition:Countable Discrete Topology|countable discrete space]].
+Let $T_Y = \struct {\set {0, 1}, \tau_0}$ be a [[Definition:Sierpiński Space|Sierpiński space]].
+Let $T_A = T_X \times T_Y$ be the [[Definition:Product Space (Topology)|product space]] of $T_X$ and $T_Y$ whose [[Definition:Product Topology|product topology]] is $\tau_A$
+Then $T_A$ is [[Definition:Paracompact Space|paracompact]] from [[Product of Countable Discrete Space with Sierpiński Space is Paracompact]].
+Let $T_B = \struct {S, \tau_p}$ be a [[Definition:Countable Particular Point Topology|countable particular point space]].
+Then $T_B$ is not [[Definition:Countably Metacompact Space|countably metacompact]] from [[Infinite Particular Point Space is not Countably Metacompact]]
+Let $\phi: T_A \to T_B$ be a [[Definition:Mapping|mapping]] defined as:
+:$\forall \tuple {x, y} \in T_A: \map \phi {x, y} = \begin {cases}
+p & : y = 0 \\
+x & : y = 1 \end {cases}$
+Then $\phi$ is [[Definition:Open Mapping|open]] and [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] as follows:
+Let $U \in \tau_A$ such that $U \ne \O$.
+By construction of $T_A$, $U$ is of the form $\tuple {U_A, U_B}$ where $U_A \in \tau$ and $U_B \in \tau_0$.
+By definition of $\tau_0$, $U_B \in \set {\set 0, \set {0, 1} }$.
+It follows that $U$ contains [[Definition:Element|elements]] of the form $\tuple {x, 0}$.
+Hence by definition of $\phi$ we have that:
+:$\exists \tuple {x, y} \in U: \map \phi {x, y} = p$
+and so:
+:$p \in \phi \sqbrk U$ 
+By definition of $T_B$ it follows that $\phi \sqbrk U$ is an [[Definition:Open Set (Topology)|open set]] of $T_B$.
+Thus $\phi$ maps from [[Definition:Open Set (Topology)|open sets]] to [[Definition:Open Set (Topology)|open sets]] and so $\phi$ is an [[Definition:Open Mapping|open mapping]].
+{{qed|lemma}}
+Let $V \in T_B$ be an [[Definition:Open Set (Topology)|open set]] such that $V \ne \O$.
+Then by definition:
+:$p \in V$
+Let $F = \phi^{-1} \sqbrk V$.
+As $p \in V$ we have that $F$ is of the form:
+:$F = A \times B$
+where:
+:$A \subseteq X$
+and:
+:$B \in \set {\set 0, \set {0, 1} }$
+Because $T_X$ is a [[Definition:Countable Discrete Topology|countable discrete space]], $A$ is [[Definition:Open Set (Topology)|open]] in $T_X$.
+By definition of $T_Y$, $B$ is [[Definition:Open Set (Topology)|open]] in $T_Y$.
+Hence by definition of [[Definition:Product Space (Topology)|product space]], $A \times B = F$ is [[Definition:Open Set (Topology)|open]] in $T_A$.
+Thus the [[Definition:Preimage of Subset under Mapping|preimage]] of $V$ under $\phi$ is an [[Definition:Open Set (Topology)|open set]].
+Hence $\phi$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+{{qed|lemma}}
+So we have constructed a [[Definition:Mapping|mapping]] $\phi$ which is both [[Definition:Open Mapping|open]] and [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] from a [[Definition:Paracompact Space|paracompact space]] to a [[Definition:Topological Space|space]] which is not [[Definition:Countably Metacompact Space|countably metacompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fixed Point Formulation of Explicit ODE}
+Tags: Ordinary Differential Equations
+
+\begin{theorem}
+Let $x' = \map f {t, x}$ with $\map x {t_0} = x_0$ be an [[Definition:Explicit ODE|explicit ODE]] of [[Definition:Dimension of Differential Equation|dimension]] $n$.
+For $a, b \in \R$, let $\mathcal X = \map {\mathcal C} {\closedint a b; \R^n}$ be the [[Continuous Functions on Interval Form Banach Space|space of continuous functions]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $T: \mathcal X \to \mathcal X$ be the [[Definition:Mapping|map]] defined by:
+:$\displaystyle \map {\paren {T x} } t = x_0 + \int_{t_0}^t \map f {s, \map x s} \rd s$
+{{explain|Definition of $\paren {T x}$}}
+Then a [[Definition:Fixed Point|fixed point]] of $T$ in $\mathcal X$ is a [[Definition:Solution of Differential Equation|solution]] to the above [[Definition:ODE|ODE]].
+\end{theorem}
+
+\begin{proof}
+Let $\map y t$ be a fixed point of the map $T$.
+That is:
+:$\displaystyle \map y t = x_0 + \int_{t_0}^t \map f {s, \map y s} \rd s$
+Then:
+:$\displaystyle \map y {t_0} = x_0 + \int_{t_0}^{t_0} \map f {s, \map y s} \rd s = x_0$
+By the [[Fundamental Theorem of Calculus/First Part|fundamental theorem of calculus]] we have that $y$ is [[Definition:Differentiable Mapping|differentiable]], and for $t \in \closedint a b$:
+{{begin-eqn}}
+{{eqn | l = \map {y'} t 
+      | r = \frac {\d} {\d t} \int_{t_0}^t \map f {s, \map y s} \rd s
+      | c = [[Derivative of Constant]]
+}}
+{{eqn | r = \map f {t, \map y t}
+      | c = [[Fundamental Theorem of Calculus/First Part|Fundamental Theorem of Calculus]]
+}}
+{{end-eqn}}
+This shows that $y$ is a solution to the [[Definition:ODE|ODE]] as claimed.
+{{Qed}}
+[[Category:Ordinary Differential Equations]]
+483130d2qqb3grax2ihsd466suob03y
+\end{proof}<|endoftext|>
+\section{Cardinality of Infinite Union of Infinite Sets}
+Tags: Infinite Sets
+
+\begin{theorem}
+Let $\kappa$ be an [[Definition:Infinite|infinite]] [[Definition:Cardinal|cardinal]].
+Let $X_i$ be [[Definition:Set|sets]] of [[Definition:Cardinality|cardinality]] at most $\kappa$ [[Definition:Indexing Set|indexed]] by a set $I$ of [[Definition:Cardinality|cardinality]] at most $\kappa$.
+Then [[Definition:Set Union|their union]] $\displaystyle \bigcup_{i \mathop \in I} X_i$ has [[Definition:Cardinality|cardinality]] at most $\kappa$.
+Furthermore, if at least one of the $X_i$ is size $\kappa$, then the [[Definition:Set Union|union]] has [[Definition:Cardinality|cardinality]] $\kappa$.
+\end{theorem}
+
+\begin{proof}
+We can assume that $I$ has [[Definition:Cardinality|cardinality]] $\kappa$ and that the sets $X_i$ are disjoint and all of size $\kappa$.
+The more general case follows since these other possibilities can only decrease the [[Definition:Cardinality|cardinality]] of the union.
+When at least one of the $X_i$ is size $\kappa$, then the union must be at least size $\kappa$ since it contains a copy of $X_i$.
+The idea is to prove by transfinite induction that for [[Definition:Limit Ordinal|limit ordinals]] $\omega_\alpha$, the order type of $\omega_\alpha \times \omega_\alpha$ is at most $\omega_\alpha$.
+From this we can conclude that the [[Definition:Cardinality|cardinality]] of $\aleph_\alpha \times \aleph_\alpha$ is still $\aleph_\alpha$.
+We can then identify $\displaystyle \bigcup_{i \mathop \in I} X_i$ with some $\aleph_\alpha \times \aleph_\alpha$ to yield the theorem.
+We begin by defining a well-ordering $\prec$ for each $\omega_\alpha \times \omega_\alpha$.
+Let $\left({\beta_1, \beta_2}\right) \prec \left({\gamma_1, \gamma_2}\right)$ if and only if one of the following holds:
+:$(1): \quad \max \beta_i < \max \gamma_i$
+or:
+:$(2): \quad \max \beta_i = \max \gamma_i$ and $\beta_1 < \gamma_1$
+or:
+:$(3): \quad \max \beta_i = \max \gamma_i$ and $\beta_1 = \gamma_1$ and $\beta_2 < \gamma_2$.
+Informally rephrased, this says that one pair is less than another pair if its max is smaller or if their max is the same but it is [[Definition:Lexicographic Order|lexicographically]] smaller.
+=== Basis for the Induction ===
+We have that $\left({\omega_0 \times \omega_0, \prec}\right)$ has order type at most $\left({\omega_0, <}\right)$
+The elements of $\omega_0 \times \omega_0$ can be arranged in an array as follows.
+Arrows are drawn so that $\left({a, b}\right) \to \left({c, d}\right)$ {{iff}} $\left({a, b}\right)$ is the predecessor of $\left({c, d}\right)$ in the ordering $\prec$.
+This diagram may be familiar from the proofs of theorems such as [[Countable Union of Countable Sets is Countable]] and [[Cartesian Product of Countable Sets is Countable]].
+We include it here to suggest motivation for the definition of $\prec$ by illuminating its behavior.
+$\begin{array} {*{4}c}
+ (0,0) & \rightarrow & (0,1) &             & (0,2) & \cdots \\
+       & \swarrow    &       & \nearrow    & \downarrow     \\
+ (1,0) & \rightarrow & (1,1) &             & (1,2) & \cdots \\
+       &             &       & \swarrow    &                \\
+       &             &\leftarrow                           \\
+       & \swarrow    &                                     \\
+ (2,0) & \rightarrow & (2,1) & \rightarrow & (2,2) & \cdots \\
+ \vdots  &           & \vdots & & \vdots & \ddots \\
+\end{array}$
+It is not hard to see the order-isomorphism suggested by this diagram between $\left({\omega_0 \times \omega_0, \prec}\right)$ and $\left({\omega_0, <}\right)$.
+The path traced by the arrows works its way out, hitting each element along the perimeter of the square formed by the first $n$ rows and $n$ columns where $\max \left({i, j}\right) = n$ before moving to the next layer, where $\max \left({i, j}\right) = n+1$.
+Within the $n$-th such perimeter, it starts at the element $\left({0, n}\right)$ moving down to $\left({n-1, n}\right)$.
+Then it jumps to $\left({n, 0}\right)$ and works rightward to $\left({n, n}\right)$.
+After that it jumps to the next perimeter at $\left({0, n+1}\right)$.
+The existence of such an order-isomorphism verifies the base case of our inductive proof.
+{{Handwaving|It is not hard to see the order-isomorphism suggested by this diagram...}}
+=== Induction Step ===
+We assume that $\left({\omega_\beta \times \omega_\beta, \prec}\right)$ has order type at most $\left({\omega_\beta, <}\right)$ for all $\beta < \alpha$.
+Then we prove that $\left({\omega_\alpha \times \omega_\alpha, \prec}\right)$ has order type at most $\left({\omega_\alpha, <}\right)$.
+First, note that if the order type of $\omega_\alpha \times \omega_\alpha$ exceeds that of $\omega_\alpha$, then $\left({\omega_\alpha + 1, <}\right)$ embeds into $\omega_\alpha \times \omega_\alpha$.
+The image of $\omega_\alpha$ (viewed as an element of $\omega_\alpha + 1$) under this embedding must be a pair $\left({\beta_1, \beta_2}\right)$ in $\omega_\alpha \times \omega_\alpha$ such that the set:
+: $\downarrow \left({\beta_1, \beta_2}\right) = \left\{ {\left({\gamma_1, \gamma_2}\right) \in \omega_\alpha \times \omega_\alpha : \left({\gamma_1, \gamma_2}\right) \prec \left({\beta_1, \beta_2}\right)}\right\}$
+has cardinality $\left\vert{\omega_\alpha}\right\vert = \aleph_\alpha$.
+{{explain|Establish by reading it carefully whether $\downarrow$ denotes the [[Definition:Lower Closure]], and exactly what f.}}
+Thus, it will be sufficient to show that for every $\left({\beta_1, \beta_2}\right) \in \omega_\alpha \times \omega_\alpha$, the set $\downarrow \left({\beta_1, \beta_2}\right)$ has cardinality less than $\aleph_\alpha$.
+We show this by containing $\downarrow \left({\beta_1, \beta_2}\right)$ in a set whose cardinality we can bound using the inductive hypothesis.
+Let $\beta = \max{\beta_i} + 1$.
+We have $\beta \in \omega_\alpha$ since $\omega_\alpha$ is a limit ordinal.
+Moreover, by the definition of $\prec$ and the equivalence of $<$ and $\in$ for ordinals, for each $\left({\gamma_1, \gamma_2}\right) \in \downarrow \left({\beta_1, \beta_2}\right)$ we have:
+:$\gamma_1 \in \beta$
+and:
+:$\gamma_2 \in \beta$.
+Thus $\downarrow \left({\beta_1, \beta_2}\right)$ is a subset of $\beta \times \beta$.
+However, since $\beta \in \omega_\alpha$ and $\omega_\alpha$ is a limit ordinal, we must have $\left\vert{\beta}\right\vert = \aleph_\gamma$ for some $\gamma < \alpha$ (unless it is finite, in which case the proof is easy).
+It follows that the cardinality of $\downarrow \left({\beta_1, \beta_2}\right)$ is at most:
+:$\left\vert{\aleph_\gamma \times \aleph_\gamma}\right\vert = \left\vert{\omega_\gamma \times \omega_\gamma}\right\vert$
+But, by the inductive hypothesis, the order-type of $\left({\omega_\gamma \times \omega_\gamma, \prec}\right)$ is at most $\left({\omega_\gamma, <}\right)$.
+So $\left\vert{\omega_\gamma \times \omega_\gamma}\right\vert$ is at most $\left\vert{\omega_\gamma}\right\vert = \aleph_\gamma < \aleph_\alpha$.
+{{qed|lemma}}
+The inductive proof demonstrates that $\left({\omega_\alpha \times \omega_\alpha, \prec}\right)$ has order type at most $\left({\omega_\alpha, <}\right)$ for all limit ordinals $\omega_\alpha$.
+We have that order-isomorphisms are bijections and hence preserve cardinality.
+We also have that $\omega_\alpha$ has cardinality $\aleph_\alpha$.
+This means that the cardinality of $\aleph_\alpha \times \aleph_\alpha$ is at most $\aleph_\alpha$.
+We have that $\aleph_\alpha$ clearly injects into $\aleph_\alpha \times \aleph_\alpha$
+From the [[Cantor-Bernstein-Schröder Theorem]]:
+:$\left\vert{\aleph_\alpha \times \aleph_\alpha}\right\vert = \aleph_\alpha$
+{{qed|lemma}}
+As a corollary of the [[Well-Ordering Theorem]], the cardinal $\kappa$ is the cardinality of some initial ordinal $\omega_\alpha$.
+Hence $\kappa$ is $\aleph_\alpha$ for some $\alpha$.
+We have that $\displaystyle{\bigcup_{i \mathop \in I} X_i}$ embeds in $I\times \aleph_\alpha$
+This in turn embeds in $\aleph_\alpha \times \aleph_\alpha$
+Hence the above result yields the theorem.
+{{qed}}
+[[Category:Infinite Sets]]
+1fi94bo6tv6yiypmsmvs69kan8cgj7h
+\end{proof}<|endoftext|>
+\section{Well-Ordering Theorem}
+Tags: Ordinals, Set Theory, Named Theorems
+
+\begin{theorem}
+Every [[Definition:Set|set]] is [[Definition:Well-Orderable Set|well-orderable]].
+\end{theorem}
+
+\begin{proof}
+Let $S$ be a [[Definition:Set|set]].
+Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
+By the [[Axiom:Axiom of Choice|Axiom of Choice]], there is a [[Definition:Choice Function|choice function]] $c$ defined on $\powerset S \setminus \set \O$.
+We will use $c$ and the [[Principle of Transfinite Induction]] to define a [[Definition:Bijection|bijection]] between $S$ and some [[Definition:Ordinal|ordinal]].
+Intuitively, we start by pairing $\map c S$ with $0$, and then keep extending the [[Definition:Bijection|bijection]] by pairing $\map c {S \setminus X}$ with $\alpha$, where $X$ is the set of elements already dealt with.
+=== Basis for the Induction ===
+$\alpha = 0$
+Let $s_0 = \map c S$.
+=== Inductive Step ===
+Suppose $s_\beta$ has been defined for all $\beta < \alpha$.
+If $S \setminus \set {s_\beta: \beta < \alpha}$ is [[Definition:Empty Set|empty]], we stop.
+Otherwise, define:
+:$s_\alpha := \map c {S \setminus \set {s_\beta: \beta < \alpha} }$
+The process eventually stops, else we have defined bijections between subsets of $S$ and arbitrarily large ordinals.
+{{handwaving}}
+Now, we can impose a [[Definition:Well-Ordering|well-ordering]] on $S$ by embedding it via $s_\alpha \to \alpha$ into the ordinal $\beta = \displaystyle {\bigcup_{s_\alpha \mathop \in S} \alpha}$ and using the [[Definition:Well-Ordering|well-ordering]] of $\beta$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Well-Ordering Theorem is Equivalent to Axiom of Choice}
+Tags: Set Theory, Axiom of Choice
+
+\begin{theorem}
+The [[Well-Ordering Theorem]] holds {{iff}} the [[Axiom:Axiom of Choice|Axiom of Choice]] holds.
+That is, every [[Definition:Set|set]] can be [[Definition:Well-Orderable Set|well-ordered]] {{iff}} every [[Definition:Set of Sets|collection of sets]] has a [[Definition:Choice Function|choice function]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose the [[Axiom:Axiom of Choice|Axiom of Choice]] holds.
+Then the [[Well-Ordering Theorem]] holds by the [[Well-Ordering Theorem]] itself.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Assume the [[Well-Ordering Theorem]] holds.
+Let $\mathcal F$ be an arbitrary [[Definition:Set of Sets|collection of sets]].
+By assumption all [[Definition:Set|sets]] can be [[Definition:Well-Ordered Set|well-ordered]].
+Hence the [[Definition:Set|set]] $\bigcup \mathcal F$ of all [[Definition:Element|elements]] of [[Definition:Set|sets]] contained in $\mathcal F$ is [[Definition:Well-Ordered Set|well-ordered]] by some [[Definition:Ordering|ordering]] $<$.
+By definition, in a [[Definition:Well-Ordered Set|well-ordered set]], every [[Definition:Subset|subset]] has a unique [[Definition:Smallest Element|least element]].
+Also, note that each [[Definition:Set|set]] in $\mathcal F$ is a [[Definition:Subset|subset]] of $\bigcup \mathcal F$.
+Thus, we may define the [[Definition:Choice Function|choice function]] $c$:
+:$\forall X \in \mathcal F: c: \mathcal F \to \bigcup \mathcal F$
+by letting $c \left({X}\right)$ be the [[Definition:Smallest Element|least element]] of $X$ under $<$.
+{{qed}}
+[[Category:Set Theory]]
+[[Category:Axiom of Choice]]
+85sp3pc1j6a9vxdjqwosz1qf6ftobab
+\end{proof}<|endoftext|>
+\section{Tychonoff Topology is Coarsest Topology such that Projections are Continuous}
+Tags: Product Spaces, Projections, Tychonoff Topology
+
+\begin{theorem}
+Let $\mathbb X = \family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
+Let $X$ be the [[Definition:Cartesian Product|cartesian product]] of $\mathbb X$:
+:$\displaystyle X := \prod_{i \mathop \in I} X_i$
+Let $\tau$ be the [[Definition:Tychonoff Topology|Tychonoff topology]] on $X$.
+For each $i \in I$, let $\pr_i : X \to X_i$ be the corresponding [[Definition:Projection (Mapping Theory)|projection]] which maps each [[Definition:Ordered Tuple|ordered tuple]] in $X$ to the corresponding element in $X_i$:
+:$\forall x \in X: \map {\pr_i} x = x_i$
+Then $\tau$ is the [[Definition:Coarser Topology|coarsest]] topology on $X$ such that all the $\pr_i$ are [[Definition:Continuous Mapping (Topology)|continuous]].
+\end{theorem}
+
+\begin{proof}
+The result follows from the definition of the [[Definition:Tychonoff Topology|Tychonoff topology]] and [[Equivalence of Definitions of Initial Topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Directed Hamilton Cycle Problem is NP-complete}
+Tags: Hamiltonian Graphs, Computer Science, Open Questions
+
+\begin{theorem}
+Both versions of the [[Definition:Directed Hamilton Cycle Problem|Directed Hamilton Cycle problem]] are [[Definition:NP-Complete|NP-complete]].
+\end{theorem}
+
+\begin{proof}
+=== Function Version is Reducible to Decision Version ===
+The algorithm described below solves the [[Definition:Directed Hamilton Cycle Problem/Function Version|function version]] of the problem with $O \left({n^2}\right)$ calls of the [[Definition:Directed Hamilton Cycle Problem/Decision Version|decision version]] of the problem.
+;Input: The [[Definition:Digraph|directed graph]] $G$
+;Output: Either:
+:A [[Definition:Hamilton Cycle|Hamilton cycle]] in $G$ if one exists
+or:
+:''no solution'' if not.
+;Auxiliary Function: $f \left({G}\right) \to \left\{{0, 1}\right\}$ defined as:
+:$f \left({G}\right) = 1$ {{iff}} $G$ has a [[Definition:Hamilton Cycle|Hamilton cycle]]
+:$f \left({G}\right) = 0$ {{iff}} $G$ has no [[Definition:Hamilton Cycle|Hamilton cycle]]
+If $f \left({G}\right) = 0$ output ''no solution''.
+Pick a starting [[Definition:Vertex of Graph|vertex]].
+Call it $v_0$.
+The solution path starts at $v_0$.
+While $G$ has more then one vertex:
+: Pick one of the edges going out of $v_0$.
+: Call that edge $\left({v_0, u}\right)$.
+: If $f \left({G - \left({v_0, u} \right)}\right) = 1$ remove $\left({v_0, u}\right)$ from $G$.
+: Otherwise:
+::add $u$ to the end of the solution set
+::remove the vertices $v_0$ and $u$ from $G$
+::replace them with a new $v_0$ where:
+:::$\left({v_0, w}\right)$ is in the new graph if $\left({u, w}\right)$ is in the old graph
+::and:
+:::$\left({w, v_0}\right)$ is in the new graph if $\left({w, v_0}\right)$ is in the old graph.
+Finally add $v_0$ to the end of the solution path and we have our solution.
+Every time $f \left({G}\right)$ is calculated at least one edge is removed.
+Thus $f \left({G}\right)$ is called $O \left({n^2}\right)$ times.
+This algorithm shows the [[Definition:Directed Hamilton Cycle Problem/Function Version|functional problem]] polynomially reduces to the [[Definition:Directed Hamilton Cycle Problem/Decision Version|decision problem]].
+If the output of the [[Definition:Directed Hamilton Cycle Problem/Function Version|functional problem]] is anything other than ''no solution'' then $G$ has a [[Definition:Hamilton Cycle|Hamilton cycle]].
+Thus the [[Definition:Directed Hamilton Cycle Problem/Function Version|functional problem]] reduces to the the [[Definition:Directed Hamilton Cycle Problem/Decision Version|decision problem]].
+{{explain|The below does not follow from the above -- where is the "mutually polynomially reducible"? We have two statements that FP reduces to DP but no indication that DP reduces to FP. Is the second of the above statements backwards?}}
+Because they are mutually polynomially reducible we can show that both are NP or NP-hard by showing that either of them are NP or NP-hard.
+{{qed|lemma}}
+=== Directed Hamilton Cycle Problem is NP ===
+Given a potential solution to the decision problem in the form of a sequence of vertices it is possible to determine if that sequence is a Hamilton Cycle by:
+: making sure every vertex appears exactly once
+and:
+: verifying that each vertex in the sequence follows is adjacent to the previous vertex.
+From [[NP Problem iff Solution Verifiable in Polynomial Time]], a  potential solution can be verified or rejected in polynomial time.
+Thus the Hamilton Cycle Problem is NP.
+{{qed|lemma}}
+=== Directed Hamilton Cycle Problem is NP-hard ===
+The objective here is to polynomially reduce the Conjunctive Normal Form Satisfiability problem with $m$ variables and $l$ clauses to the decision version of the Directed Hamilton Cycle problem.
+Because [[CNF Satisfiability Problem is NP-Complete]] that is enough to show the Hamilton Cycle is NP-hard.
+Consider the following diagram of a part of a graph:
+:[[File:CNF_to_HAMILTONIAN_graph.jpg]]
+It is clear that either $\left({A, B}\right)$ or $\left({B, A}\right)$ must be included in any Hamilton Cycle.
+Likewise either $\left({C, D}\right)$ or $\left({D, C}\right)$ must be included.
+The three cases for these four vertices are:
+: $\left({A, B, C, D}\right)$
+: $\left({A, B, E, C, D}\right)$
+: $\left({D, C, B, A}\right)$
+In short, the four vertices $A, B, C, D$ must be visited in order or in reverse order.
+Also, they can only visit the $E$ node if they are in the right order.
+These pieces of the graph can be concatenated by letting the $D$ node of one piece be the same as the $A$ node of the next.
+The final graph to be constructed takes a form similar to the following diagram:
+:[[File:CNF_to_HAMILTONIAN_graph2.jpg]]
+The boxes represent concatenations of the pieces we have just studied.
+In the final graph there will be one of these rows for every variable in the CNF SAT problem, each one linked to another row as shown here.
+This creates a circuit through the rows.
+A left to right path corresponds to the variable being true and a right to left path corresponds to the variable being false.
+In addition to these vertices there will also be one vertex for every clause in the original problem.
+These will take the $E$ vertex position in the first diagram.
+A vertex will have be attached to a pair of vertices in a row if the variable the row corresponds to appears in the clause, going:
+:left to right if the variable is not negated
+and:
+:right to left if it is negated.
+In the image the $E$ type node would contain $x_1 \lor \neg x_2$ if the first row corresponded to $x_1$, and the second to $x_2$.
+Clearly all the vertices could only be visited in a cycle if there was some choice of direction for each of the rows that allowed all the $E$ type vertices to be visited.
+That would only happen if there was some way of deciding values for the variables in the CNF SAT problem that gave each clause at least one true variable in its conjunction.
+And so the graph has a Hamilton Cycle iff the CNF SAT problem has a solution.
+The number of vertices in a given row is at most $3 l + 3$ because:
+:Each variable can only appear in each clause once, otherwise that clause is either internally redundant or trivially satisfied
+:Each additional instance of a variable only requires three additional vertices
+and:
+:there are only three nodes of overhead.
+The number of $E$ type vertices is $l$.
+Therefore the total number of vertices in the constructed graph is at most $3 l m + 3 m + l$ vertices.
+Because the size of the graph is bounded by a polynomial of the size of the CNF SAT problem this scheme is a polynomial reduction from CNF SAT to the directed Hamilton Cycle problem.
+The Directed Hamilton Cycle problem is NP-hard.
+Because the Directed Hamilton Cycle problem is NP and NP-hard it is NP-complete.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Closed Extension Space to be T0 Space}
+Tags: T0 Spaces, Closed Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the [[Definition:Closed Extension Space|closed extension space]] of $T$.
+Then $T^*_p$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]] {{iff}} $T$ is.
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \varnothing$
+Let $T = \struct {S, \tau}$ be a [[Definition:Kolmogorov Space|$T_0$ space]].
+Then:
+:$\forall x, y \in S$ such that $x \ne y$, either:
+::$\exists U \in \tau: x \in U, y \notin U$
+:or:
+::$\exists U \in \tau: y \in U, x \notin U$
+Let $x, y \in S^*_p$ such that $x \ne y, x \ne p, y \ne p$.
+Then:
+:$\exists U \in \tau: x \in U, y \notin U$
+or:
+:$\exists U \in \tau: y \in U, x \notin U$
+Now let $U^* = U \cup \set p$.
+It follows that:
+:$x \in U^*, y \notin U^*$
+or:
+:$y \in U^*, x \notin U^*$
+Now consider $x, p \in S^*_p$ such that $x \ne p$.
+We have that $\set p \in \tau^*_p: p \in \set p, x \notin \set p$
+So we have shown that:
+:$\forall x, y \in S^*_p$ such that $x \ne y$, either:
+::$\exists U \in \tau^*_p: x \in U, y \notin U$
+:or:
+::$\exists U \in \tau^*_p: y \in U, x \notin U$
+and so $T^*_p$ is a [[Definition:Kolmogorov Space|$T_0$ space]].
+Now suppose that $T = \struct {S, \tau}$ is '''not''' a [[Definition:Kolmogorov Space|$T_0$ space]].
+Let $x, y \in S$ such that:
+:$\forall U \in \tau: x \in U \implies y \in U$
+:$\forall U \in \tau: y \in U \implies x \in U$
+Now consider $U^* = U \cup \set p$ for any $U \in \tau$.
+Then:
+:As $x \in U \implies y \in U$ it follows that $x \in U^* \implies y \in U^*$.
+:As $y \in U \implies x \in U$ it follows that $y \in U^* \implies x \in U^*$.
+Hence $T^*_p$ is not a [[Definition:Kolmogorov Space|$T_0$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is T0}
+Tags: T0 Spaces, Excluded Point Topology, Excluded Point Space is T0
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Space|excluded point space]].
+Then $T$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T$ be a [[Definition:Trivial Topological Space|trivial space]].
+That is, let $S = \set p$.
+Then the result holds [[Definition:Vacuous Truth|vacuously]], as there are no two [[Definition:Distinct Elements|distinct points]] in $T$.
+Now suppose $T$ is not [[Definition:Trivial Topological Space|trivial]].
+Then $\exists x \in S: x \ne p$.
+Now we have that $\set x \subseteq T$ is [[Definition:Open Set (Topology)|open]] in $T$ such that $p \notin \set x$ but $x \in \set x$.
+Finally, suppose that $x, y \in S: x \ne y, x \ne p, y \ne p$.
+Then we have that (for example) $\set x \subseteq T$ is [[Definition:Open Set (Topology)|open]] in $T$ such that $x \in \set x$ but $y \notin \set x$
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have:
+: [[Excluded Point Topology is Open Extension Topology of Discrete Topology]]
+: [[Discrete Space satisfies all Separation Properties]] (including being a [[Definition:Kolmogorov Space|$T_0$ space]])
+Then by [[Condition for Open Extension Space to be T0 Space|Condition for Open Extension Space to be $T_0$ Space]], as a [[Definition:Discrete Space|discrete space]] is $T_0$ then so is its [[Definition:Open Extension Topology|open extension]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Extension Topology is not T1}
+Tags: T1 Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$
+Let $x \in S^*_p, x \ne p$.
+Let $U = \set x$.
+Then $U \in \tau^*_p$ such that $x \in U, p \notin U$.
+But the only $v \in \tau^*_p$ such that $p \in V$ is the set $S^*_p$, and we have that $x \in S^*_p$.
+So there is no $V \in \tau^*_p$ such that $x \notin V, p \in V$, by definition of the [[Definition:Open Extension Topology|open extension topology]].
+Hence $T^*_{\bar p}$ can not be a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Extreme Value Theorem}
+Tags: Metric Spaces, Continuous Functions
+
+\begin{theorem}
+Let $X$ be a [[Definition:Compact Metric Space|compact metric space]] and $Y$ a [[Definition:Normed Vector Space|normed vector space]].
+Let $f: X \to Y$ be a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
+Then $f$ is [[Definition:Bounded Mapping to Metric Space|bounded]], and there exist $x, y \in X$ such that:
+:$\forall z \in X: \norm {\map f x} \le \norm {\map f z} \le \norm {\map f y}$
+where $\norm {\map f x}$ denotes the [[Definition:Norm on Vector Space|norm]] of $\map f x$.
+Moreover, $\norm f$ attains its [[Definition:Minimum Value|minimum]] and [[Definition:Maximum Value|maximum]].
+\end{theorem}
+
+\begin{proof}
+{{AxiomReview|[[Definition:Zermelo-Fraenkel Set Theory|ZF]] version of this proof (use [[Continuous Image of Compact Space is Compact/Corollary 3]])}}
+By [[Continuous Image of Compact Space is Compact]], $f \sqbrk X \subseteq Y$ is [[Definition:Compact Metric Space|compact]].
+Therefore, by [[Compact Subspace of Metric Space is Bounded]], $f$ is [[Definition:Bounded Mapping to Metric Space|bounded]].
+Let $\displaystyle A = \inf_{x \mathop \in X} \norm {\map f x}$.
+It follows from the definition of [[Definition:Infimum of Mapping|infimum]] that there exists a [[Definition:Sequence|sequence]] $\sequence {y_n}$ in $X$ such that:
+:$\displaystyle \lim_{n \mathop \to \infty} \norm {\map f {y_n} } = A$
+By [[Sequence of Implications of Metric Space Compactness Properties]], $X$ is [[Definition:Sequentially Compact Space|sequentially compact]].
+So there exists a [[Definition:Convergent Sequence|convergent]] [[Definition:Subsequence|subsequence]] $\sequence {x_n}$ of $\sequence {y_n}$.
+Let $\displaystyle x = \lim_{n \mathop \to \infty} x_n$.
+Since $f$ is [[Definition:Continuous Mapping (Metric Spaces)|continuous]] and a [[Norm is Continuous|norm is continuous]], it follows by [[Composite of Continuous Mappings is Continuous/Point|Composite of Continuous Mappings is Continuous]] that:
+:$\displaystyle \norm {\map f x} = \norm {\map f {\lim_{n \mathop \to \infty} x_n} } = \norm {\lim_{n \mathop \to \infty} \map f {x_n} } = \lim_{n \mathop \to \infty} \norm {\map f {x_n} } = A$
+So $\norm f$ attains its [[Definition:Minimum Value|minimum]] at $x$.
+By replacing the [[Definition:Infimum of Mapping|infimum]] with the [[Definition:Supremum of Mapping|supremum]] in the definition of $A$, we also see that $\norm f$ attains its [[Definition:Maximum Value|maximum]] by the same reasoning.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Functions on Compact Space form Banach Space}
+Tags: Functional Analysis
+
+\begin{theorem}
+Let $X$ be a [[Definition:Compact Topological Subspace|compact]] [[Definition:Hausdorff Space|Hausdorff space]].
+Let $Y$ be a [[Definition:Banach Space|Banach space]].
+Let $\CC = \CC \struct {X; Y}$ be the [[Definition:Set|set]] of all [[Definition:Everywhere Continuous Mapping (Topology)|continuous mappings]] $X \to Y$.
+{{explain|Work out what convention ProofWiki has already evolved for $\CC \struct {X; Y}$.}}
+Let $\norm {\,\cdot\,}_\infty$ be the [[Definition:Supremum Norm|supremum norm]] on $\CC$.
+Then $\struct {\CC, \norm {\,\cdot\,} }$ is a [[Definition:Banach Space|Banach space]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Definition:Set|set]] of [[Definition:Everywhere Continuous Mapping (Topology)|continuous mappings]] $X \to Y$ is a [[Definition:Subset|subset]] of the [[Definition:Set|set]] $Y^X$ of ''all'' [[Definition:Mapping|mappings]] $X \to Y$.
+Therefore by [[Vector Space of All Mappings is Vector Space]], we need only show that $\CC$ is a [[Definition:Vector Subspace|subspace]] of $Y^X$.
+By the [[One-Step Vector Subspace Test]] we need only show that $\CC$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Linear Combination|linear combinations]] (clearly $\CC$ contains $0$).
+But this is shown by the [[Combined Sum Rule for Continuous Functions]].
+{{questionable|The above theorem has been proven only for $\Q, \R, \C$, not the general objects it is being asked to be applied to here.}}
+We have [[Supremum Norm is Norm]].
+It remains to be shown that $\CC$ is [[Definition:Complete Metric Space|complete]].
+But this is precisely the statement of the [[Uniform Limit Theorem]].
+{{questionable|The above statement says nothing about completeness -- just continuity.}}
+{{qed}}
+[[Category:Functional Analysis]]
+2caq0ms1yyf6jpkbjdy6ztf89r7sayw
+\end{proof}<|endoftext|>
+\section{Supremum Norm is Norm}
+Tags: Examples of Norms
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\struct {X, \norm {\, \cdot \,} }$ be a [[Definition:Normed Vector Space|normed vector space]] over $K \in \set {\R, \C}$.
+Let $\BB$ be the set of [[Definition:Bounded Mapping|bounded mappings]] $S \to X$.
+Let $\norm {\, \cdot \,}_\infty$ be the [[Definition:Supremum Norm|supremum norm]] on $\BB$.
+Then $\norm {\, \cdot \,}_\infty$ is a [[Definition:Norm on Vector Space|norm]] on $\BB$.
+{{MissingLinks|Add a link that establishes that $\BB$ is a vector space}}
+\end{theorem}
+
+\begin{proof}
+First:
+{{begin-eqn}}
+{{eqn | l = \norm f_\infty
+      | r = 0
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \sup_{x \mathop \in S} \norm {\map f x}
+      | r = 0
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \forall x \in S:
+      | l = \norm {\map f x}
+      | r = 0
+      | c = since $\norm {\, \cdot \,}$ is a [[Definition:Norm on Vector Space|norm]], and hence non-negative
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \forall x \in S:
+      | l = \map f x
+      | r = 0
+      | c = since $\norm x = 0 \iff x = 0$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = f
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Now let $\lambda \in K, f \in \BB$
+We have:
+{{begin-eqn}}
+{{eqn | l = \norm {\lambda f}_\infty
+      | r = \sup_{x \mathop \in S} \norm {\lambda \map f x}
+      | c = 
+}}
+{{eqn | r = \size \lambda_K \sup_{x \mathop \in S} \norm {\map f x}
+      | c = [[Multiple of Supremum]], and because $\norm {\, \cdot \,}$ is a [[Definition:Norm on Vector Space|norm]]
+}}
+{{eqn | r = \size \lambda_K \, \norm f_\infty
+      | c = 
+}}
+{{end-eqn}}
+Finally let $f, g \in \BB$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \norm {f + g}_\infty
+      | r = \sup_{x \mathop \in S} \norm {\map f x + \map g x}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \sup_{x \mathop \in S} \paren {\norm {\lambda \map f x} + \norm {\lambda \map g x} }
+      | c = because $\norm {\, \cdot \,}$ is a [[Definition:Norm on Vector Space|norm]]
+}}
+{{eqn | o = \le
+      | r = \sup_{x \mathop \in S} \norm {\lambda \map f x} + \sup_{x \mathop \in S} \norm {\lambda \map g x}
+      | c = [[Supremum of Sum]]
+}}
+{{eqn | r = \norm f_\infty + \norm g_\infty
+      | c = 
+}}
+{{end-eqn}}
+Thus $\norm {\, \cdot \,}_\infty$ has the defining properties of a [[Definition:Norm on Vector Space|norm]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{One-Step Vector Subspace Test}
+Tags: Linear Algebra, Vector Subspaces
+
+\begin{theorem}
+Let $V$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Division Ring|division ring]] $K$.
+Let $U \subseteq V$ be a non-[[Definition:Empty Set|empty]] [[Definition:Subset|subset]] of $V$ such that:
+:$\forall u, v \in U: \forall \lambda \in K: u + \lambda v \in U$
+Then $U$ is a [[Definition:Vector Subspace|subspace]] of $V$.
+\end{theorem}
+
+\begin{proof}
+We need to verify the [[Definition:Vector Space Axioms|vector space axioms]] for $U$.
+We start with observing that the properties for a [[Definition:Unitary Module|unitary module]] are true for all [[Definition:Element|elements]] of $V$.
+Hence, since $U \subseteq V$, they hold for all [[Definition:Element|elements]] of $U$ as well.
+The same holds for the axioms $(G1)$ and $(C)$.
+From [[Vector Inverse is Negative Vector]], we have for all $u \in U$:
+:$u + \paren {- 1_K} u = 0_V$
+which by assumption is an [[Definition:Element|element]] of $U$.
+Since $U$ is [[Definition:Non-Empty Set|non-empty]], this means $0_V \in U$.
+Hence it is seen that axioms $(G2)$ and $(G3)$ are satisfied.
+The last axiom that remains is $(G0)$.
+To this end we employ the knowledge that for all $v \in U$, we have:
+:$1_K v = v$
+and hence:
+:$u + 1_K v = u + v \in U$
+Having verified all the [[Definition:Vector Space Axioms|vector space axioms]], we conclude that $U$ is a [[Definition:Vector Subspace|subspace]] of $V$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ax-Grothendieck Theorem}
+Tags: Model Theory, Complex Analysis
+
+\begin{theorem}
+Let $f: \C^n \to \C^n$ be a [[Definition:Polynomial Map|polynomial map]].
+Let $f$ be [[Definition:Injection|injective]].
+Then $f$ is [[Definition:Surjection|surjective]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds as follows:
+:$(1): \quad$ showing that the theorem can be captured using a first-order sentences in the language of rings
+:$(2): \quad$ showing that the theorem is true for at least one field of every characteristic $p > 0$
+:$(3): \quad$ applying the [[Lefschetz Principle (First-Order)]].
+=== First-Order Sentences in Language of Rings ===
+Since $n$ is fixed, we can quantify over polynomials.
+There are only finitely many coefficients to quantify over.
+Therefore each variable occurs with degree at most $d$, say.
+Thus, for each $n$ and $d$, we can build an $\mathcal L_r$ sentence which holds in a field $F$ {{iff}} every injective polynomial map $F^n \to F^n$ where each variable occurs with at most degree $d$ is surjective.
+First, we write a formula:
+: $\phi_{\left({i_1, \dots, i_n}\right)}$
+which says that the $n$-variable polynomial map with coefficients $a_{\left({i_1, \dots, i_n}\right)}$ where $(\left({i_1, \dots, i_n}\right) \le \left({d, \dots, d}\right)$ is injective.
+Note that the polynomial map takes on $n$-tuple images, so is of the form:
+: $f = \left({f_1, \dots, f_n}\right)$
+where each $f_k$ is a polynomial in $n$ variables.
+The variables $a_{k, \left({i_1, \dots, i_n}\right)}$ in the sentence below are intended to be interpreted as the coefficients of $f_k$:
+:$\displaystyle \forall x_1 \cdots \forall x_n \forall y_1 \cdots \forall y_n \left({\left({\bigwedge_{k \mathop \le n} \ \sum_{\left({i_1, \ldots, i_n}\right)} a_{k, \left({i_1, \ldots, i_n}\right)} {x_1}^{i_1} \cdots  {x_n}^{i_n} = \sum_{ \left({i_1, \ldots, i_n}\right)} a_{k, \left({i_1, \ldots, i_n}\right)} {y_1}^{i_1} \cdots {y_n}^{i_n} }\right) \to \bigwedge_{i \mathop = 1, \ldots, n} x_i = y_i}\right)$
+We also write a formula $\psi_{(i_1,\dots,i_n)}$ which says that such a polynomial is surjective:
+:$\displaystyle \forall z_1 \cdots \forall z_n \exists x_1 \cdots \exists x_n \left(\bigwedge_{k \mathop \le n} \ \sum_{(i_1, \ldots, i_n)} a_{k, \left({i_1, \ldots, i_n}\right)} {x_1}^{i_1} \cdots {x_n}^{i_n} = z_k \right)$
+Finally, we combine these into the required implication, quantifying over all coefficients:
+:$\displaystyle \underset{k, \left({i_1, \ldots, i_n}\right)} {\huge \forall} a_{k, \left({i_1, \ldots, i_n}\right)} \left({ \phi_{\left({i_1, \ldots,i_n}\right)} \to \psi_{\left({i_1, \ldots, i_n}\right)} }\right)$
+Note that we have one of these sentences for every maximum degree $d$ of the variables in a polynomial map.
+Thus it has been demonstrated that the theorem can be captured using a first-order sentences in the language of rings.
+{{explain|Is that "a first-order sentence" or "first-order sentences"? It's one or the other. Also in the exposition in the first line.}}
+{{qed|lemma}}
+=== Field of Characteristic $p > 0$ ===
+Since injections on finite sets are necessarily surjective, every injective polynomial map $k^n \to k^n$ is surjective when $k$ is a finite field.
+We extend this to the algebraic closure of $k$.
+This will demonstrate that the sentence above is satisfied by at least one model of the theory of algebraically closed fields of characteristic $p$ for every $p > 0$.
+Let $\mathbb F^{\operatorname{alg} }_p$ be the algebraic closure of the finite field with $p$ elements.
+Suppose there is an injective polynomial map:
+:$f: \left({\mathbb F^{\operatorname{alg} }_p}\right)^n \to \left({\mathbb F^{\operatorname{alg} }_p}\right)^n$
+which is not surjective.
+Let $A$ be the set of coefficients appearing in $f$.
+Let $\left({z_1, \ldots, z_n}\right) \in \left({\mathbb F^{\operatorname{alg} }_p}\right)^n$ be an element not in the range of $f$.
+Consider the subfield $k$ of $\mathbb F^{\operatorname{alg}}_p$ generated by the elements of $A$ and the elements $z_1, \ldots, z_n$.
+We have that:
+:$\displaystyle \mathbb F^{\operatorname{alg} }_p = \bigcup_{n \mathop = 1, 2, \ldots} \mathbb F_{p^n}$
+Therefore any finitely generated subfield is contained in some finite sub-union:
+:$\displaystyle \bigcup_{n \mathop = 1, 2, \ldots, N} \mathbb F_{p^n}$
+Hence $k$ is finite.
+Therefore, $f {\restriction_{k^n} }$ is an injective polynomial map on a finite field which is not surjective.
+This is a contradiction.
+So it must be the case that every injective polynomial map on $\left({\mathbb F^{\operatorname{alg} }_p}\right)^n$ is surjective.
+That is:
+:$\left({\mathbb F^{\operatorname{alg} }_p}\right)^n$ satisfies the sentences above for each characteristic $p > 0$.
+Thus it has been demonstrated that the theorem is true for at least one field of every characteristic $p > 0$.
+{{qed|lemma}}
+=== Application of Lefschetz Principle ===
+It has been shown that the sentences above are true in some algebraically closed field of characteristic $p$ for all $p > 0$.
+By the [[Lefschetz Principle (First-Order)|Lefschetz Principle]], it follows that they are true in $\C$.
+{{qed}}
+{{Namedfor|James Burton Ax|name2=Alexander Grothendieck|cat=Ax|cat2=Grothendieck}}
+[[Category:Model Theory]]
+[[Category:Complex Analysis]]
+7e9ntfvhtkr9fox900mtt6h25z9jch0
+\end{proof}<|endoftext|>
+\section{Condition for Open Extension Space to be T0 Space}
+Tags: T0 Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]] {{iff}} $T$ is.
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$
+Let $T = \struct {S, \tau}$ be a [[Definition:Kolmogorov Space|$T_0$ space]].
+Let $\forall x, y \in S$ such that $x \ne y$.
+First, suppose $x \ne p \ne y$.
+Then by definition of $T$ as a [[Definition:Kolmogorov Space|$T_0$ space]]:
+:$\exists U \in \tau: x \in U, y \notin U$
+or:
+:$\exists U \in \tau: y \in U, x \notin U$
+Then as $U \in \tau^*_{\bar p}$, it follows that:
+:$\exists U \in \tau^*_{\bar p}: x \in U, y \notin U$
+or:
+:$\exists U \in \tau^*_{\bar p}: y \in U, x \notin U$
+Now suppose that $y = p$.
+Then by definition:
+:$\forall U \in \tau^*_{\bar p}: x \in U, p \notin U$
+unless $U = S^*_p$.
+So we have shown that:
+:$\forall x, y \in S^*_p$ such that $x \ne y$, either:
+::$\exists U \in \tau^*_{\bar p}: x \in U, y \notin U$
+:or:
+::$\exists U \in \tau^*_{\bar p}: y \in U, x \notin U$
+and so $T^*_{\bar p}$ is a [[Definition:Kolmogorov Space|$T_0$ space]].
+Now suppose that $T = \struct {S, \tau}$ is '''not''' a [[Definition:Kolmogorov Space|$T_0$ space]].
+Let $x, y \in S$ such that:
+:$\forall U \in \tau: x \in U \implies y \in U$
+:$\forall U \in \tau: y \in U \implies x \in U$
+As $U \in \tau^*_{\bar p}$ it follows that:
+:$\forall U \in \tau^*_{\bar p}: x \in U \implies y \in U$
+:$\forall U \in \tau^*_{\bar p}: y \in U \implies x \in U$
+and if $U = S^*_p$ the same applies.
+Hence $T^*_{\bar p}$ is not a [[Definition:Kolmogorov Space|$T_0$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Trivial Excluded Point Topology is not T1}
+Tags: T1 Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be a [[Definition:Excluded Point Topology|excluded point space]] such that $S$ is not a [[Definition:Singleton|singleton]].
+Then $T$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+:[[Excluded Point Topology is Open Extension Topology of Discrete Topology]]
+:[[Open Extension Topology is not T1|Open Extension Topology is not $T_1$]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Points in Open Extension Space}
+Tags: Limit Points, Open Extension Topology, Limit Points in Open Extension Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Let $x \in S$.
+Then $p$ is a [[Definition:Limit Point of Point|limit point]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Every [[Definition:Open Set (Topology)|open set]] of $T^*_p = \struct {S^*_p, \tau^*_{\bar p} }$ except $S^*_p$ does not contain the point $p$ by [[Definition:Open Extension Topology|definition]].
+So every [[Definition:Open Set (Topology)|open set]] $U \in \tau^*_{\bar p}$ such that $p \in U$ (there is only the one such [[Definition:Open Set (Topology)|open set]]) contains $x$.
+So by definition of the [[Definition:Limit Point of Point|limit point of a point]], $p$ is a [[Definition:Limit Point of Point|limit point]] of $x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Points in Excluded Point Space}
+Tags: Limit Points, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Space|excluded point space]].
+Let $x \in S$ such that $x \ne p$.
+Then $p$ is the only [[Definition:Limit Point of Point|limit point]] of $x$.
+Similarly, let $U \subseteq S$.
+Then $p$ is the only [[Definition:Limit Point of Set|limit point]] of $U$.
+\end{theorem}
+
+\begin{proof}
+Let $U \subseteq S$.
+Let $x \in S$ such that $x \ne p$.
+From:
+:[[Excluded Point Topology is Open Extension Topology of Discrete Topology]]
+:[[Limit Points in Open Extension Space]]
+it follows that:
+:$p$ is a [[Definition:Limit Point of Set|limit point of $U$]]
+:$p$ is a [[Definition:Limit Point of Point|limit point of $x$]].
+Now suppose $y \in S$ such that $y \ne p$.
+We have by definition of [[Definition:Excluded Point Space|excluded point space]] that $\set y$ is [[Definition:Open Set (Topology)|open]] in $T$.
+So there is no $z \in \set y: z \ne y, z \in U$. 
+Hence $y$ can not be a [[Definition:Limit Point of Set|limit point of $U$]].
+Similarly $x \notin \set y$.
+So $y$ can not be a [[Definition:Limit Point of Point|limit point of $x$]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is T5}
+Tags: T5 Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $T$ is a [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $A, B \subseteq S$ such that:
+:$A^- \cap B = A \cap B^- = \O$
+where $A^-$ denotes the [[Definition:Closure (Topology)|closure]] of $A$.
+Every [[Definition:Closed Set (Topology)|closed set]] of $T$ contains $\set p$.
+From [[Limit Points in Excluded Point Space]] and the definition of [[Definition:Closure (Topology)|closure]]:
+:$A^- = A \cup \set p, B^- = B \cup \set p$
+So if $A^- \cap B = A \cap B^- = \O$ then $p \in A^-$ and $p \in B^-$.
+Thus, $p \notin A$ and $p \notin B$ and so by definition $A$ and $B$ are [[Definition:Open Set (Topology)|open sets]] in $T$.
+It follows that $A \cap B = \O$.
+Hence the result, as we have shown that:
+:$\forall A, B \subseteq X, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau_{\bar p}: A \subseteq U, B \subseteq V, U \cap V = \O$
+where $U = A$ and $V = B$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is Compact}
+Tags: Compact Spaces, Excluded Point Topology, Excluded Point Space is Compact
+
+\begin{theorem}
+Let $T = \left({S, \tau_{\bar p}}\right)$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $T$ is a [[Definition:Compact Topological Space|compact space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+: [[Excluded Point Topology is Open Extension Topology of Discrete Topology]]
+: [[Open Extension Space is Compact]]
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of [[Definition:Excluded Point Space|excluded point space]], the only [[Definition:Open Set (Topology)|open set]] of $T$ which contains $p$ is $S$.
+So any [[Definition:Open Cover|open cover]] $\CC$ of $T$ must have $S$ in it.
+So $\set S$ will be a [[Definition:Subcover|subcover]] of $\CC$, whatever $\CC$ may be.
+And $\set S$ (having only one set in it) is trivially a [[Definition:Finite Cover|finite cover]] of $T$.
+Hence the result, by definition of [[Definition:Compact Topological Space|compact topological space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Extension Space is Compact}
+Tags: Compact Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is a [[Definition:Compact Topological Space|compact space]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Open Extension Space|open extension space]], the only [[Definition:Open Set (Topology)|open set]] of $T^*_{\bar p}$ containing $p$ is $S^*_p$.
+So any [[Definition:Open Cover|open cover]] $\CC$ of $T^*_{\bar p}$ must have $S^*_p$ in it.
+So $\set {S^*_p}$ will be a [[Definition:Subcover|subcover]] of $\CC$, whatever $\CC$ may be.
+And $\set {S^*_p}$ (having only one set in it) is trivially a [[Definition:Finite Cover|finite cover]] of $T^*_{\bar p}$.
+Hence the result, by definition of [[Definition:Compact Topological Space|compact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Traveling Salesman Problem is NP-Complete}
+Tags: Complexity Theory
+
+\begin{theorem}
+Both versions of the Traveling Salesman Problem are NP-complete.
+\end{theorem}<|endoftext|>
+\section{Open Extension Space is Connected}
+Tags: Connected Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is a [[Definition:Connected Topological Space|connected space]].
+\end{theorem}
+
+\begin{proof}
+The only [[Definition:Open Set (Topology)|open set]] of $T$ which contains $p$ is $S^*_p$.
+Therefore it is impossible to set up a [[Definition:Separation (Topology)|separation]] of $T$, as $S^*_p$ will always need to be an [[Definition:Element|element]] of such a [[Definition:Separation (Topology)|separation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is Connected}
+Tags: Connected Spaces, Excluded Point Topology, Excluded Point Space is Connected
+
+\begin{theorem}
+Let $T = \left({S, \tau_{\bar p}}\right)$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $T^*_{\bar p}$ is a [[Definition:Connected Topological Space|connected space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+: [[Excluded Point Topology is Open Extension Topology of Discrete Topology]]
+: [[Open Extension Space is Connected]]
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The only [[Definition:Open Set (Topology)|open set]] of $T$ which contains $p$ is $S$.
+Therefore it is impossible to set up a [[Definition:Separation (Topology)|separation]] of $T$, as $S$ will always need to be an [[Definition:Element|element]] of such a [[Definition:Separation (Topology)|separation]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+: [[Excluded Point Space is Ultraconnected]]
+: [[Ultraconnected Space is Path-Connected]]
+: [[Path-Connected Space is Connected]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Extension Space is Ultraconnected}
+Tags: Ultraconnected Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is [[Definition:Ultraconnected Space|ultraconnected]].
+\end{theorem}
+
+\begin{proof}
+Apart from $S^*_p$, every [[Definition:Open Set (Topology)|open set]] of $T^*_{\bar p}$ does not contain $p$, by definition of [[Definition:Open Extension Space|open extension space]].
+So, apart from $\O$, every [[Definition:Closed Set (Topology)|closed set]] of $T$ does contain $p$, by definition of [[Definition:Closed Set (Topology)|closed set]].
+So every pair of [[Definition:Closed Set (Topology)|closed sets]] of $T$ has an [[Definition:Set Intersection|intersection]] which contains at least $p$.
+So there are no [[Definition:Non-Empty Set|non-empty]] [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed sets]] of $T$.
+Hence the result, by definition of [[Definition:Ultraconnected Space|ultraconnected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is Ultraconnected}
+Tags: Ultraconnected Spaces, Excluded Point Topology, Excluded Point Space is Ultraconnected
+
+\begin{theorem}
+Let $T = \left({S, \tau_{\bar p}}\right)$ be an [[Definition:Excluded Point Space|excluded point space]].
+Then $T$ is [[Definition:Ultraconnected Space|ultraconnected]].
+\end{theorem}
+
+\begin{proof}
+: [[Excluded Point Topology is Open Extension Topology of Discrete Topology]]
+: [[Open Extension Space is Ultraconnected]]
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Apart from $S$, every [[Definition:Open Set (Topology)|open set]] of $T$ does not contain $p$, by definition of [[Definition:Excluded Point Space|excluded point space]].
+So, apart from $\O$, every [[Definition:Closed Set (Topology)|closed set]] of $T$ does contain $p$, by definition of [[Definition:Closed Set (Topology)|closed set]].
+So every pair of [[Definition:Closed Set (Topology)|closed sets]] of $T$ has an [[Definition:Set Intersection|intersection]] which contains at least $p$.
+So there are no [[Definition:Non-Empty Set|non-empty]] [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed sets]] of $T$.
+Hence the result, by definition of [[Definition:Ultraconnected Space|ultraconnected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is not Irreducible}
+Tags: Irreducible Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Topology|excluded point space]] with at least three points.
+Then $T^*_{\bar p}$ is not [[Definition:Irreducible Space|irreducible]].
+\end{theorem}
+
+\begin{proof}
+By definition, [[Definition:Open Set (Topology)|open sets]] of $S$ are precisely the [[Definition:Open Set (Topology)|open sets]] of $S \setminus \set p$ under the [[Definition:Discrete Topology|discrete topology]].
+Let $x, y \in S \setminus \set p: x \ne y$.
+Then $\set x$ and $\set y$ are both [[Definition:Open Set (Topology)|open sets]] of $T$ such that $\set x \cap \set y = \O$.
+Hence the result, by definition of [[Definition:Irreducible Space|irreducible]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is Path-Connected}
+Tags: Path-Connected Spaces, Excluded Point Topology, Excluded Point Space is Path-Connected
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $T^*_{\bar p}$ is [[Definition:Path-Connected Space|path-connected]].
+\end{theorem}
+
+\begin{proof}
+: [[Excluded Point Topology is Open Extension Topology of Discrete Topology]]
+: [[Open Extension Space is Path-Connected]]
+{{qed}}
+\end{proof}
+
+\begin{proof}
+:[[Excluded Point Space is Ultraconnected]]
+:[[Ultraconnected Space is Path-Connected]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Extension Space is Path-Connected}
+Tags: Path-Connected Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is [[Definition:Path-Connected Space|path-connected]].
+\end{theorem}
+
+\begin{proof}
+:[[Open Extension Space is Ultraconnected]]
+:[[Ultraconnected Space is Path-Connected]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is not Arc-Connected}
+Tags: Arc-Connected Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] with at least two [[Definition:Distinct Elements|distinct elements]].
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $T$ is not [[Definition:Arc-Connected Space|arc-connected]].
+\end{theorem}
+
+\begin{proof}
+Let $q \in S$ be such that $q \ne p$.
+Let $f: \closedint 0 1 \to T$ be an [[Definition:Injection|injection]] such that:
+:$\map f 0 = p$
+:$\map f 1 = q$
+Because $f$ is an [[Definition:Injection|injection]], it must be that:
+:$f^{-1} \sqbrk {\set q} = \set 1$
+where $f^{-1}$ denotes the [[Definition:Preimage of Subset under Mapping|preimage]] under $f$.
+Now since $p \notin \set q$, $\set q$ is [[Definition:Open Set (Topology)|open]] in $\tau_{\bar p}$, by definition of the [[Definition:Excluded Point Topology|excluded point topology]].
+By [[Closed Real Interval is not Open Set]], $\set 1 = \closedint 1 1$ is not [[Definition:Open Set (Topology)|open]].
+Thus we have exhibited an [[Definition:Open Set (Topology)|open set]] whose [[Definition:Preimage of Subset under Mapping|preimage]] under $f$ is not [[Definition:Open Set (Topology)|open]].
+Hence $f$ is not [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Since $f$ was an arbitrary [[Definition:Injection|injection]], no [[Definition:Arc (Topology)|arc]] between $p$ and $q$ can exist.
+Hence $T$ is not [[Definition:Arc-Connected Space|arc-connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis for Excluded Point Space}
+Tags: Excluded Point Topology, Topological Bases
+
+\begin{theorem}
+Let $T = \left({S, \tau_{\bar p}}\right)$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Consider the set $\mathcal B$ defined as:
+:$\mathcal B = \left\{{\left\{{x}\right\}: x \in S \setminus \left\{{p}\right\}}\right\} \cup \left\{{S}\right\}$
+Then $B$ is a [[Definition:Basis (Topology)|basis]] for $S$.
+\end{theorem}
+
+\begin{proof}
+Let $H \in \tau_{\bar p}$ be [[Definition:Open Set (Topology)|open]] in $T$.
+If $H = S$ then trivially $H$ is the [[Definition:Set Union|union]] of elements of $\mathcal B$.
+So suppose $H \ne S$.
+Then by definition $p \notin H$ and so:
+: $\forall y \in H: \exists \left\{{y}\right\} \in \mathcal B$
+Thus:
+:$\displaystyle H = \bigcup_{y \in H} \left\{{y}\right\}$
+So $\mathcal B$ is an [[Definition:Analytic Basis|analytic basis]] for $T$.
+{{qed}}
+It could equally well be shown that $\mathcal B$ is also a [[Definition:Synthetic Basis|synthetic basis]] for $T$.
+[[Category:Excluded Point Topology]]
+[[Category:Topological Bases]]
+mz0v244fyb4gjyq4r16iac42au531re
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is Locally Path-Connected}
+Tags: Locally Path-Connected Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $T$ is [[Definition:Locally Path-Connected Space|locally path-connected]].
+\end{theorem}
+
+\begin{proof}
+Consider the set $\BB$ defined as:
+:$\BB = \set {\set x: x \in S \setminus \set p} \cup \set S$
+Then $\BB$ is a [[Basis for Excluded Point Space|basis for $T$]].
+Let $H \in \BB$.
+Then $\exists x \in S: H = \set x$.
+From [[Point is Path-Connected to Itself]] we have that $H$ is [[Definition:Path-Connected Set (Topology)|path-connected]].
+Now consider the [[Definition:Open Set (Topology)|open set]] $S \in \BB$.
+Let $a, b \in S$.
+Let $\mathbb I$ be the [[Definition:Closed Unit Interval|closed unit interval]] in $\R$.
+Let $f: \mathbb I \to S$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in \mathbb I: \map f x = \begin{cases}
+a & : x \in \hointr 0 1 \\
+b & : x = 1
+\end{cases}$
+Then $f^{-1} \sqbrk S = \closedint 0 1$ which is [[Definition:Open Set (Topology)|open]] in $\mathbb I$ because $\closedint 0 1 = \mathbb I$.
+As $a$ and $b$ are any two points of $S$ it follows that $S$ is [[Definition:Path-Connected Set (Topology)|path-connected]].
+So we have shown that all elements of a [[Definition:Basis (Topology)|basis]] of $T$ are [[Definition:Path-Connected Set (Topology)|path-connected]].
+Hence by definition $T$ is [[Definition:Locally Path-Connected Space|locally path-connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is not Locally Arc-Connected}
+Tags: Locally Arc-Connected Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $T$ is not [[Definition:Locally Arc-Connected Space|locally arc-connected]].
+\end{theorem}
+
+\begin{proof}
+Let $\BB \subseteq \tau_{\bar p}$ be a [[Definition:Basis (Topology)|basis]] for $\tau_{\bar p}$.
+Since $\BB$ [[Definition:Cover of Set|covers]] $S$, there must be an [[Definition:Open Set (Topology)|open set]] $B \in \BB$ such that $p \in B$.
+By definition of the [[Definition:Excluded Point Topology|excluded point topology]], the only [[Definition:Open Set (Topology)|open set]] containing $p$ is $S$ itself.
+Hence necessarily $S \in \BB$.
+But by [[Excluded Point Space is not Arc-Connected]], $S$ is not [[Definition:Arc-Connected Space|arc-connected]].
+Hence $\BB$ does not consist only of [[Definition:Arc-Connected Set|arc-connected sets]].
+Because $\BB$ was arbitrary, there cannot exist a [[Definition:Basis (Topology)|basis]] for $\tau_{\bar p}$ comprising only [[Definition:Arc-Connected Set|arc-connected sets]].
+Hence, by definition, $T$ is not [[Definition:Locally Arc-Connected Space|locally arc-connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partition Topology is Zero Dimensional}
+Tags: Zero Dimensional Spaces, Partition Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Partition Space|partition space]].
+Then $T$ is [[Definition:Zero Dimensional Space|zero dimensional]].
+\end{theorem}
+
+\begin{proof}
+Let $\PP$ be the [[Definition:Partition (Set Theory)|partition]] which is the [[Basis for Partition Topology|basis for $T$]].
+From [[Open Set in Partition Topology is also Closed]], all the elements of $\PP$ are [[Definition:Clopen Set|both closed and open]].
+Hence the result, by definition of [[Definition:Zero Dimensional Space|zero dimensional space]]
+{{qed}}
+[[Category:Zero Dimensional Spaces]]
+[[Category:Partition Topology]]
+cipbm5n8qxfxks85cia91feioap6cyt
+\end{proof}<|endoftext|>
+\section{Discrete Space is Zero Dimensional}
+Tags: Zero Dimensional Spaces, Discrete Topology, Discrete Space is Zero Dimensional
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Space|discrete space]].
+Then $T$ is [[Definition:Zero Dimensional Space|zero dimensional]].
+\end{theorem}
+
+\begin{proof}
+We have from [[Partition of Singletons yields Discrete Topology]] that a [[Definition:Discrete Space|discrete space]] is a [[Definition:Partition Space|partition space]].
+The result follows from [[Partition Topology is Zero Dimensional]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\BB$ be the set:
+:$\BB := \set {\set x: x \in S}$
+From [[Basis for Discrete Topology]], $\BB$ is a [[Definition:Basis (Topology)|basis]] for $T$.
+From [[Set in Discrete Topology is Clopen]], all the elements of $\mathcal B$ are [[Definition:Clopen Set|both closed and open]].
+Hence the result, by definition of [[Definition:Zero Dimensional Space|zero dimensional space]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space is Scattered}
+Tags: Scattered Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] where $\tau$ is the [[Definition:Discrete Topology|discrete topology]] on $S$.
+Then $T$ is a [[Definition:Scattered Space|scattered space]].
+\end{theorem}
+
+\begin{proof}
+We have that [[Topological Space is Discrete iff All Points are Isolated]].
+So, by definition, no [[Definition:Subset|subset]] $H \subseteq S$ of $T$ such that $H \ne \varnothing$ is [[Definition:Dense-in-itself|dense-in-itself]].
+So, again, by definition, $T$ is [[Definition:Scattered Space|scattered]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space is Extremally Disconnected}
+Tags: Extremally Disconnected Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] where $\tau$ is the [[Definition:Discrete Topology|discrete topology]] on $S$.
+Then $T$ is [[Definition:Extremally Disconnected Space|extremally disconnected]].
+\end{theorem}
+
+\begin{proof}
+First we note that as [[Discrete Space satisfies all Separation Properties]], $T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+Then from [[Interior Equals Closure of Subset of Discrete Space]], it follows directly that the [[Definition:Closure (Topology)|closure]] of every [[Definition:Open Set (Topology)|open set]] of $T$ is [[Definition:Open Set (Topology)|open]].
+Hence by definition $T$ is [[Definition:Extremally Disconnected Space|extremally disconnected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Totally Separated Space is Totally Disconnected}
+Tags: Totally Separated Spaces, Totally Disconnected Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Totally Separated Space|totally separated]].
+Then $T$ is [[Definition:Totally Disconnected Space|totally disconnected]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Totally Separated Space|totally separated space]].
+Then for every $x, y \in S: x \ne y$ there exists a [[Definition:Partition (Topology)|partition]] $U \mid V$ of $T$ such that $x \in U, y \in V$.
+Suppose $T$ were not [[Definition:Totally Disconnected Space|totally disconnected]].
+Then $\exists H \subseteq S: x, y \in U$ such that $H$ is [[Definition:Connected Set (Topology)|connected]].
+But by definition of [[Definition:Connected Set (Topology)|connected]], there can be no [[Definition:Partition (Topology)|partition]] $U \mid V$ of $T$ such that $x \in U, y \in V$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Extension Topology is not Perfectly T4}
+Tags: Perfectly T4 Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is not a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$
+We have that $S$ is an [[Definition:Open Set (Topology)|open set]] in $T$ and so [[Definition:Open Set (Topology)|open set]] in $T^*_{\bar p}$.
+So $\set p = S^*_p \setminus S$ is [[Definition:Closed Set (Topology)|closed]] in $T^*_{\bar p}$.
+The only [[Definition:Open Set (Topology)|open set]] in $T^*_{\bar p}$ which contains $p$ is $S^*_p$.
+So $\set p$ can not be the [[Definition:Set Intersection|intersection]] of [[Definition:Open Set (Topology)|open sets]] of $T^*_{\bar p}$, whether that [[Definition:Set Intersection|intersection]] be [[Definition:Countable Intersection|countable]] or not.
+So by definition $T^*_{\bar p}$ is not a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is not Perfectly T4}
+Tags: Perfectly T4 Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $T$ is not a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+:[[Excluded Point Topology is Open Extension Topology of Discrete Topology]]
+:[[Open Extension Topology is not Perfectly T4]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Type is Realized in some Elementary Extension}
+Tags: Model Theory
+
+\begin{theorem}
+Let $\mathcal M$ be an [[Definition:First-Order Structure|$\mathcal L$-structure]].
+Let $A$ be a subset of the universe of $\mathcal M$.
+Let $p$ be an [[Definition:Type|$n$-type]] over $A$.
+There exists an [[Definition:Elementary Extension|elementary extension]] of $\mathcal M$ which [[Definition:Realization|realizes]] $p$.
+\end{theorem}
+
+\begin{proof}
+The idea is to work in a language with constant symbols for all elements of $\mathcal M$ and show that the union of $p$ and the [[Definition:Elementary Diagram|elementary diagram]] of $\mathcal M$ is satisfiable.
+Since $\mathcal M$ naturally embeds into any model of such a theory, this will prove the theorem.
+Let $\mathcal L_\mathcal M$ be the language obtained by adding to $\mathcal L$ constant symbols for each element of $\mathcal M$.
+Denote by $\operatorname {Diag}_{\mathrm {el}} \left({\mathcal M}\right)$ the elementary diagram of $\mathcal M$.
+Let $T$ be $p \cup \operatorname {Diag}_{\mathrm{el}} \left({\mathcal M}\right)$.
+We will show that $T$ is finitely satisfiable.  It will follow by the [[Compactness Theorem]] that $T$ is satisfiable.
+To this end, let $\Delta$ be a finite subset of $T$.
+We have that $\Delta$ is finite.
+Thus it consists of:
+:finitely many $\mathcal L_A$-sentences $\phi_0, \dotsc, \phi_n$ from $p$ (which are $\mathcal L_\mathcal M$ sentences since $A \subseteq \mathcal M$)
+along with:
+:finitely many $\mathcal L_\mathcal M$-sentences $\psi_0,\dots,\psi_k$ from $\operatorname{Diag}_{\mathrm{el}} \left({\mathcal M}\right)$.
+By definition, $p$ is satisfiable by some $\mathcal L_A$-structure $\mathcal N$ such that:
+:$\mathcal N \models p \cup \operatorname{Th}_A \left({\mathcal M}\right)$
+Thus, since $\phi_0, \dotsc, \phi_n \in p$:
+:$\mathcal N$ satisfies $\phi_0, \dotsc, \phi_n$.
+We will show that the same $\mathcal N$ also satisfies $\psi_0, \dotsc, \psi_k$.
+The obstacle to overcome is that the $\psi_i$ are $\mathcal L_\mathcal M$-formulas, and we only know $\mathcal N$ as an $\mathcal L_A$-structure which satisfies sentences with parameters from $A$.
+The $\psi_i$ may have parameters from $\mathcal M$ outside of $A$.
+The idea is to quantify away the excess parameters and appropriately select the interpretation of new symbols so that $\mathcal N$ is a good $\mathcal L_\mathcal M$-structure.
+Explicitly:
+Let $\psi$ be the conjunction $\psi_0 \wedge \cdots \wedge \psi_k$.
+Note that since $\psi$ is an $\mathcal L_\mathcal M$-sentence, it can be written as an $\mathcal L_A$-formula $\psi \left({\bar b}\right)$, where $\bar b$ is a tuple of parameters from $\mathcal M$ not in $A$.
+By existentially quantifying away the tuple $\bar b$, we obtain an $\mathcal L_A$-sentence $\exists \bar x \psi \left({\bar x}\right)$.
+Now, since $\mathcal M \models \psi \left({\bar b}\right)$, we have:
+: $\mathcal M \models \exists \bar x: \psi \left({\bar x}\right)$
+Hence $\exists \bar x: \psi \left({\bar x}\right)$ is in $\operatorname{Th}_\mathcal A \left({\mathcal M}\right)$.
+By choice of $\mathcal N$, it follows that:
+:$\mathcal N \models \exists \bar x: \psi \left({\bar x}\right)$
+and thus there must be some tuple $\bar c$ of elements from $\mathcal N$ such that:
+:$\mathcal N \models \psi \left({\bar c}\right)$
+Now, by interpreting the $\mathcal L_\mathcal M$-symbols $\bar b$ as the elements $\bar c$, we can view $\mathcal N$ as an $\mathcal L_\mathcal M$-structure which satisfies:
+:$\phi_0 \wedge \cdots \wedge \phi_n \wedge \psi_0 \wedge \cdots \wedge \psi_k$. 
+Thus $\mathcal N$ satisfies all of $\Delta$.
+This demonstrates that $T$ is finitely satisfiable and hence satisfiable by the [[Compactness Theorem]].
+This means that there is an $\mathcal L_\mathcal M$-structure $\mathcal M^*$ which satisfies:
+:$p \cup \operatorname{Diag}_{\mathrm {el} } \left({\mathcal M}\right)$
+Since $\mathcal M^*$ interprets a symbol for each element of $\mathcal M$, there is an obvious embedding of $\mathcal M$ into $\mathcal M^*$.
+This embedding is [[Definition:Elementary Map (Model Theory)|elementary]] since $\mathcal M^*$ satisfies the elementary diagram of $\mathcal M$.
+Thus $\mathcal M^*$ is an elementary extension of $\mathcal M$.
+Finally, since $\mathcal M^*$ satisfies $p$, there must be a tuple of elements $\bar{d}$ such that $\mathcal M^* \models \phi \left({d}\right)$ for each $\phi \left({\bar v}\right) \in p$.
+Thus $\mathcal M^*$ realizes $p$.
+{{qed}}
+[[Category:Model Theory]]
+i06sa9qyaa7xhjubk1uegxek51ty53e
+\end{proof}<|endoftext|>
+\section{Subset of Excluded Point Space is not Dense-in-itself}
+Tags: Denseness, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be a [[Definition:Excluded Point Topology|excluded point space]] such that $S$ is not a [[Definition:Singleton|singleton]].
+Let $H \subseteq S$.
+Then $H$ is not [[Definition:Dense-in-itself|dense-in-itself]].
+\end{theorem}
+
+\begin{proof}
+From [[Limit Points in Excluded Point Space]], the only [[Definition:Limit Point of Set|limit point]] of $H$ is $p$.
+So by definition, all points of $H$ are [[Definition:Isolated Point of Subset|isolated in $H$]] except $p$.
+So if  $H \ne \set p$, $H$ contains at least one point which is [[Definition:Isolated Point of Subset|isolated in $H$]].
+As for $p$ itself, from [[Singleton Point is Isolated]] we have that $p$ is itself [[Definition:Isolated Point of Subset|isolated]] in $\set p$.
+So if  $H = \set p$, $H$ also contains one point which is [[Definition:Isolated Point of Subset|isolated in $H$]].
+Hence the result, by definition of [[Definition:Dense-in-itself|dense-in-itself]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is Scattered}
+Tags: Scattered Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $T$ is a [[Definition:Scattered Space|scattered space]].
+\end{theorem}
+
+\begin{proof}
+We have that [[Subset of Excluded Point Space is not Dense-in-itself]].
+So, by definition, $T$ is [[Definition:Scattered Space|scattered]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dispersion Point of Excluded Point Space}
+Tags: Dispersion Points, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $p$ is a [[Definition:Dispersion Point|dispersion point]] of $T$.
+\end{theorem}
+
+\begin{proof}
+We have that the [[Excluded Point Topology is Open Extension Topology of Discrete Topology]].
+So $S \setminus \set p$ is a [[Definition:Discrete Space|discrete space]].
+Then a [[Totally Disconnected and Locally Connected Space is Discrete|discrete space is totally disconnected]].
+The result follows from definition of [[Definition:Dispersion Point|dispersion point]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is First-Countable}
+Tags: First-Countable Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $T$ is [[Definition:First-Countable Space|first-countable]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$ such that $x \ne p$.
+By definition, the set $\set x$ is [[Definition:Open Set (Topology)|open]] in $T$.
+Let $U \in \tau_{\bar p}: x \in U$.
+Then $\set x \subseteq U$ and so $\set {\set x}$ is a [[Definition:Local Basis|local basis]] at $x$ which is trivially [[Definition:Countable Set|countable]].
+Now if $x = p$ there is only one $U \in \tau_{\bar p}: p \in U$, and that is $S$.
+So $\set S$ is a [[Definition:Local Basis|local basis]] at $x$ which is trivially [[Definition:Countable Set|countable]].
+Hence the result, by definition of [[Definition:First-Countable Space|first-countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Space is Sequentially Compact}
+Tags: Sequentially Compact Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $T$ is a [[Definition:Sequentially Compact Space|sequentially compact space]].
+\end{theorem}
+
+\begin{proof}
+:[[Excluded Point Space is Compact]]
+:[[Compact Space is Countably Compact]]
+:[[Excluded Point Space is First-Countable]]
+:[[First-Countable Space is Sequentially Compact iff Countably Compact]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Excluded Point Space is Second-Countable}
+Tags: Countable Excluded Point Space is Second-Countable
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be a [[Definition:Countable Excluded Point Topology|countable excluded point space]].
+Then $T$ is a [[Definition:Second-Countable Space|second-countable space]].
+\end{theorem}
+
+\begin{proof}
+Consider the set $\mathcal B$ defined as:
+:$\mathcal B = \left\{{\left\{{x}\right\}: x \in S \setminus \left\{{p}\right\}}\right\} \cup \left\{{S}\right\}$
+From [[Basis for Excluded Point Space]], $\mathcal B$ is a [[Definition:Basis (Topology)|basis]] for $T$, and trivially has the same [[Definition:Cardinality|cardinality]] as $S$.
+So by definition, if $S$ is [[Definition:Countable Set|countable]], then $T$ is [[Definition:Second-Countable Space|second-countable]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have:
+: [[Countable Discrete Space is Second-Countable]]
+: [[Excluded Point Topology is Open Extension Topology of Discrete Topology]]
+The result follows from [[Condition for Open Extension Space to be Second-Countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Excluded Point Space is not Second-Countable}
+Tags: Second-Countable Spaces, Excluded Point Topology, Uncountable Excluded Point Space is not Second-Countable
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Uncountable Excluded Point Topology|uncountable excluded point space]].
+Then $T$ is not [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+Let $H = S \setminus \left\{{p}\right\}$ where $\setminus$ denotes [[Definition:Set Difference|set difference]].
+By definition, $H$ is an [[Definition:Uncountable Discrete Topology|uncountable discrete space]].
+The result follows from [[Uncountable Discrete Space is not Second-Countable]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have:
+: [[Uncountable Discrete Space is not Second-Countable]]
+: [[Excluded Point Topology is Open Extension Topology of Discrete Topology]]
+The result follows from [[Condition for Open Extension Space to be Second-Countable]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fractional Sobolev Embedding Theorem}
+Tags: Fourier Analysis
+
+\begin{theorem}
+Let $S'$ denote the space of [[Definition:Tempered Distribution|tempered distributions]].
+Let $\mathcal F : S' \to S'$ denote the [[Definition:Fourier Transform|Fourier transform]].
+For each $s \in \R$ and $p \in [1, \infty]$ let $W^{s,p} \left({\R^n}\right) = \left\{{u \in S': \langle \xi \rangle^s \hat{u} \in L^p \left({\R^n}\right)}\right\}$ where $\displaystyle \left \langle{\xi}\right \rangle = \left({1 + \left|{\xi}\right|^2}\right)^\frac 1 2$.
+Then
+: 1. If $s > t$ then $W^{s,p} \left({\R^n}\right)$ [[Definition:Continuously Embedded|embeds continuously]] into $W^{t,q}\left({\R^n}\right)$ where $q$ is given by $\displaystyle \frac 1 q = \frac 1 p - \frac {s-t} n$.
+{{stub}}
+{{namedfor|Sergei Lvovich Sobolev}}
+[[Category:Fourier Analysis]]
+33i3o3ouq6duaqzehxwbx7ff7dp5v5c
+\end{theorem}<|endoftext|>
+\section{Countable Excluded Point Space is Separable}
+Tags: Separable Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be a [[Definition:Countable Excluded Point Topology|countable excluded point space]].
+Then $T$ is a [[Definition:Separable Space|separable space]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Closure (Topology)|closure]] of the set $S$ (trivially) equals $S$.
+That is, $S$ is [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+But as $S$ is [[Definition:Countable Set|countable]] it follows by definition that $T$ is [[Definition:Separable Space|separable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Excluded Point Space is not Separable}
+Tags: Separable Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be an [[Definition:Uncountable Excluded Point Topology|uncountable excluded point space]].
+Then $T$ is not [[Definition:Separable Space|separable]].
+\end{theorem}
+
+\begin{proof}
+Let $H \subseteq S$ such that $H$ is [[Definition:Countable Set|countable]].
+Then $H \ne S$ as $S$ is [[Definition:Uncountable Set|uncountable]] by hypothesis.
+From [[Limit Points in Excluded Point Space]], the only [[Definition:Limit Point of Set|limit point]] of $H$ is $p$.
+So, by definition, the [[Definition:Closure (Topology)|closure]] of $H$ is $H \cup \set p$.
+From [[Countable Union of Countable Sets is Countable]] we have that $H \cup \set p$ is [[Definition:Countable Set|countable]].
+So $H \cup \set p \ne S$.
+So $H$ is not [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+Thus $T$ can have no [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$ which is [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+Hence the result by definition of [[Definition:Separable Space|separable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Set Topology is Topology}
+Tags: Excluded Set Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar H} }$ be an [[Definition:Excluded Set Space|excluded set space]].
+Then $\tau_{\bar H}$ is a [[Definition:Topology|topology]] on $S$, and $T$ is a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+We have [[Definition:Excluded Set Space|by definition]] that $S \in \tau_{\bar H}$.
+Also, as $H \cap \O = \O$, we have that $\O \in \tau_{\bar H}$.
+Now let $U_1, U_2 \in \tau_{\bar H}$.
+By definition:
+:$H \cap U_1 = \O$
+and:
+:$H \cap U_2 = \O$
+and so by definition of [[Definition:Set Intersection|set intersection]]:
+:$H \cap \paren {U_1 \cap U_2} = \O$
+So:
+:$U_1 \cap U_2 \in \tau_{\bar H}$
+Now let $\UU \subseteq \tau_{\bar H}$.
+We have that:
+:$\forall U \in \UU: H \cap U = \O$
+Hence from [[Subset of Union]]:
+:$H \cap \bigcup \UU = \O$
+So all the conditions are fulfilled for $\tau_{\bar H}$ to be a [[Definition:Topology|topology]] on $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Set Topology is not T0}
+Tags: T0 Spaces, Excluded Set Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar H} }$ be an [[Definition:Excluded Set Topology|excluded set space]] where $H$ has at least two [[Definition:Distinct Elements|distinct points]].
+Then $T$ is not a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in H$ such that $x \ne y$.
+Then $x, y \in S$, but by definition $x$ and $y$ are in no other [[Definition:Open Set (Topology)|open sets]] of $T$.
+Hence there is no $U \in \tau_{\bar H}$ such that $x \in U, y \notin U$ or $y \in U, x \notin U$.
+Hence the result, by definition of [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is Topology}
+Tags: Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $\tau$ is a [[Definition:Topology|topology]] on $T$.
+\end{theorem}
+
+\begin{proof}
+From the definition:
+:$H \in \tau \iff \paren {\set 0 \nsubseteq H \lor \openint {-1} 1 \subseteq H}$
+for any $H \subseteq S$.
+First note that:
+:$\set 0 \nsubseteq \O$ and so $\O \in \tau$
+:$\openint {-1} 1 \subseteq S$ and so $S \in \tau$
+Now suppose $H_1, H_2 \in \tau$.
+If either $\set 0 \nsubseteq H_1$ or $\set 0 \nsubseteq H_2$ then from [[De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection|De Morgan's Laws: Difference with Intersection]] (indirectly):
+:$\set 0 \nsubseteq H_1 \cap H_2$
+Otherwise:
+:$\openint {-1} 1 \subseteq H_1$ and $\openint {-1} 1 \subseteq H_2$
+and so from [[Intersection is Largest Subset]]:
+:$\openint {-1} 1 \subseteq H_1 \cap H_2$
+So:
+:$H_1, H_2 \in \tau \implies H_1 \cap H_2 \in \tau$
+Now let $\HH \subseteq \tau$ be a [[Definition:Set of Sets|set of elements]] of $\tau$.
+Then either:
+:$\forall H \in \HH: \set 0 \nsubseteq H$ in which case $\set 0 \nsubseteq \bigcup \HH$
+or:
+:$\exists H \in \HH: \openint {-1} 1 \subseteq H$ in which case $\openint {-1} 1 \subseteq \bigcup \HH$
+In both cases $\displaystyle \bigcup \HH \in \tau$.
+Note that $\set 0 \nsubseteq H \implies \openint {-1} 1 \nsubseteq H$ so there does not exist the confusion of what happens if the conditions are contradictory.
+Hence the result, from the definition of [[Definition:Topology|topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Sets of Either-Or Topology}
+Tags: Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Space|either-or space]].
+Then the [[Definition:Closed Set (Topology)|closed sets]] of $T$ are:
+:$\O$
+:$S$
+:$\set {-1}$
+:$\set 1$
+:$\set {-1, 1}$
+:Any subset of $\closedint {-1} 1$ containing $\set 0$ as a [[Definition:Subset|subset]].
+\end{theorem}
+
+\begin{proof}
+From [[Open and Closed Sets in Topological Space]] we have that $\O$ and $S$ are [[Definition:Closed Set (Topology)|closed sets]] trivially.
+From the definition of [[Definition:Closed Set (Topology)|closed set]], we have:
+:$U$ is [[Definition:Open Set (Topology)|open]] in $T$ {{iff}} $S \setminus U$ is [[Definition:Closed Set (Topology)|closed]] in $T$
+:$U$ is [[Definition:Closed Set (Topology)|closed]] in $T$ {{iff}} $S \setminus U$ is [[Definition:Open Set (Topology)|open]] in $T$
+where $S \setminus U$ denotes the [[Definition:Relative Complement|relative complement]] of $U$ in $S$.
+Now we have that:
+:$\openint {-1} 1$ is [[Definition:Open Set (Topology)|open]] in $T$ so $S \setminus \openint {-1} 1 = \set {-1, 1}$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+:Both $\hointr {-1} 1 \supseteq \openint {-1} 1$ and $\hointr {-1} 1 \supseteq \openint {-1} 1$ therefore are [[Definition:Open Set (Topology)|open]] in $T$, so $\set {-1}$ and $\set 1$ are [[Definition:Closed Set (Topology)|closed]] in $T$.
+Apart from $S$ itself, there are no other subsets of $S = \closedint {-1} 1$ which have $\openint {-1} 1$ as a subset.
+Finally, let $U \in \tau: \set 0 \nsubseteq U$.
+Then by the definition of [[Definition:Relative Complement|relative complement]] $\set 0 \subseteq S \setminus U$.
+Now suppose $\set 0 \subseteq V$.
+Then again by the definition of [[Definition:Relative Complement|relative complement]] $\set 0 \nsubseteq S \setminus V$.
+So $S \setminus V$ is [[Definition:Open Set (Topology)|open]] in $T$ and so $V$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boubaker's Theorem}
+Tags: Boubaker Polynomials
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Commutative Ring|commutative ring]].
+Let $\left({D, +, \circ}\right)$ be an [[Definition:Subdomain|integral subdomain]] of $R$ whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
+Let $X \in R$ be [[Definition:Transcendental over Integral Domain|transcendental]] over $D$.
+Let $D \left[{X}\right]$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] in $X$ over $D$.
+Finally, consider the following properties:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \sum_{k \mathop = 1}^N {p_n \left({0}\right)}
+      | r = -2N
+}}
+{{eqn | n = 2
+      | l = \sum_{k \mathop = 1}^N {p_n \left({\alpha_k}\right)}
+      | r = 0
+}}
+{{eqn | n = 3
+      | l = \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_x \left({x}\right)} {\mathrm d x} }\right\vert_{x \mathop = 0}
+      | r = 0
+}}
+{{eqn | n = 4
+      | l = \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d {p_n}^2 \left({x}\right)} {\mathrm d x^2} }\right\vert_{x \mathop = 0}
+      | r = \frac 8 3 N \left({N^2 - 1}\right)
+}}
+{{end-eqn}}
+where, for a given [[Definition:Positive Integer|positive integer]] $n$, $p_n \in D \left[{X}\right]$ is a non-[[Definition:Null Polynomial over Ring|null polynomial]] such that $p_n$ has $N$ roots $\alpha_k$ in $F$.
+Then the [[Definition:Subsequence|subsequence]] $\left \langle {B_{4 n} \left({x}\right)}\right \rangle$ of the [[Definition:Boubaker Polynomials|Boubaker polynomials]] is the unique polynomial sequence of $D \left[{X}\right]$ which verifies simultaneously the four properties $(1) - (4)$.
+\end{theorem}
+
+\begin{proof}
+=== Proof of validity ===
+We first prove that the [[Definition:Boubaker Polynomials|Boubaker Polynomials]] sub-sequence $ B_{4n}(x)$, defined in $D \left[{X}\right]$ verifies properties $(1)$, $(2)$, $(3)$ and $(4)$.
+Let:
+: $\left({R, +, \circ}\right)$ be a [[Definition:Commutative Ring|commutative ring]]
+: $\left({D, +, \circ}\right)$ be an [[Definition:Subdomain|integral subdomain]] of $R$ whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$
+: $X \in R$ be [[Definition:Transcendental over Integral Domain|transcendental]] over $D$.
+;Property $(1)$:
+We have the [[Definition:Closed Form|closed form]] of the the [[Definition:Boubaker Polynomials|Boubaker Polynomials]]:
+:$\displaystyle B_n \left({x}\right) = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p x^{n - 2 p}$
+which gives:
+:$B_{4 n} \left({0}\right) = \dfrac {n - 4 n / 2} {n - n / 2} \dbinom {n - n / 2} {n / 2} = -2$
+and finally:
+: $(1): \quad \displaystyle \sum_{k \mathop = 1}^N B_{4 n} \left({0}\right) = \sum_{k \mathop = 1}^N -2 = - 2 N$
+;Property $(2)$:
+We have, for given integer $n$, $ B_{4n} \in D \left[{X}\right]$ is a non-[[Definition:Null Polynomial over Ring|null]] polynomial with $N$ roots $\alpha_k$ in $F$.  
+Since:
+:$B_{4 n} \left({\alpha_k}\right) = 0$
+then the equality:
+: $(2): \quad \displaystyle \sum_{k \mathop = 1}^N B_{4 n} \left({\alpha_k}\right) = 0$
+holds.
+;Property $(3)$:
+According to the [[Definition:Closed Form|closed form]] of the the [[Definition:Boubaker Polynomials|Boubaker Polynomials]]:
+:$\displaystyle B_n \left({x}\right) = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p x^{n - 2 p}$
+We have:
+:$\displaystyle \frac {\mathrm d B_{4 n} \left({x}\right)} {\mathrm d x} = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor - 1} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p \left({n - 2 p}\right) x^{n - 2 p - 1}$  
+       
+The minimal power in this expansion is obtained for $p = {\left\lfloor{n / 2}\right\rfloor - 1}$, hence:
+:$\dfrac {\mathrm d B_{4 n} } {\mathrm d x} \left({0}\right) = 0$
+and the equality:
+: $(3): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d B_{4 n} \left({x}\right)} {\mathrm d x} } \right\vert_{x \mathop = 0} = 0$
+holds.
+ 
+;Property $(4)$:
+Starting from the [[Definition:Closed Form|closed form]] of the the [[Definition:Boubaker Polynomials|Boubaker Polynomials]]:
+:$\displaystyle B_n \left({x}\right) = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p x^{n - 2 p}$
+we have consequently:
+:$\displaystyle \frac {\mathrm d^2 B_{4n} \left({x}\right)} {\mathrm d x^2} = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor - 2} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p \left({n - 2 p}\right) \left({n - 2 p - 1}\right) x^{n - 2 p - 2}$      
+   
+The minimal power in this expansion is obtained for $p = \left\lfloor{n / 2}\right\rfloor - 2$, hence:
+:$\displaystyle \frac {\mathrm d^2 B_{4 n} } {\mathrm d x^2} \left({0}\right) = \left({-1}\right)^p \left({n - 2 p}\right) \left({n - 2 p - 1}\right) 0$
+and the equality:
+: $(4): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d^2 B_{4 n} \left({x}\right)} {\mathrm d x^2}}\right\vert_{x \mathop = 0} = \frac 8 3 N \left({N^2 - 1}\right)$
+holds.
+{{qed}}
+{{Proofread}}
+=== [[Boubaker's Theorem/Proof of Uniqueness|Proof of Uniqueness]] ===
+{{:Boubaker's Theorem/Proof of Uniqueness}}
+{{finish}}
+{{Namedfor|Boubaker Boubaker|cat = Boubaker}}
+[[Category:Boubaker Polynomials]]
+3symgdip87gorka3pmkubxjyvqfnlc2
+\end{proof}
+
+\begin{proof}
+Let:
+: $\left({R, +, \circ}\right)$ be a [[Definition:Commutative Ring|commutative ring]]
+: $\left({D, +, \circ}\right)$ be an [[Definition:Subdomain|integral subdomain]] of $R$ whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$
+: $X \in R$ be [[Definition:Transcendental over Integral Domain|transcendental]] over $D$.
+It has been demonstrated that the [[Definition:Boubaker Polynomials|Boubaker Polynomials]] sub-sequence $B_{4 n} \left({x}\right)$, defined in $D \left[{X}\right]$ as:
+:$\displaystyle B_{4 n} \left({x}\right) = 4 \sum_{p \mathop = 0}^{2 n} \frac {n - p} {4 n - p} \binom {4 n - p} p \left({-1}\right)^p x^{2 \left({2n - p}\right)}$
+satisfies the properties:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \sum_{k \mathop = 1}^N {p_n \left({0}\right)}
+      | r = -2N
+}}
+{{eqn | n = 2
+      | l = \sum_{k \mathop = 1}^N {p_n \left({\alpha_k}\right)}
+      | r = 0
+}}
+{{eqn | n = 3
+      | l = \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_x \left({x}\right)} {\mathrm d x} }\right\vert_{x \mathop = 0}
+      | r = 0
+}}
+{{eqn | n = 4
+      | l = \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d {p_n}^2 \left({x}\right)} {\mathrm d x^2} }\right\vert_{x \mathop = 0}
+      | r = \frac 8 3 N \left({N^2 - 1}\right)
+}}
+{{end-eqn}}
+with $\left. {\alpha_k}\right\vert_{k \mathop = 1 \,.\,.\, N}$ roots of $B_{4 n}$.
+Suppose there exists another $4 n$-indexed polynomial $q_{4 n} \left({x}\right)$, with $N$ roots $\left.{\beta_k}\right\vert_{k \mathop = 1 \,.\,.\, N}$ in $F$ and which also satisfies simultaneously properties $(1)$ to $(4)$.
+Let:
+:$\displaystyle B_{4 n} \left({x}\right) = \sum_{p \mathop = 0}^{2 n} a_{4 n, p} x^{2 \left({2 n - p}\right)}$
+and:
+:$\displaystyle q_{4 n} \left({x}\right) = \sum_{p \mathop = 0}^{2 n} b_{4 n, p} x^{2 \left({2 n - p}\right)}$
+and:
+:$\displaystyle \mathrm d_{4 n, p} = a_{4 n, p} - b_{4 n, p}$ for $p = 0 \,.\,.\, 2 n$
+then, simultaneous expressions of conditions $(1)$ and $(3)$ give:
+:$ \quad \displaystyle \sum_{k \mathop = 1}^N \mathrm d_{4 n, 2 n} = 0$
+:$ \quad \displaystyle \sum_{k \mathop = 1}^N \mathrm d_{4 n, 2 n - 2} = 0$
+It has also been demonstrated that $ B_{4 n}$ has exactly $4 n - 2$ real roots inside the domain $\left[{-2 \,.\,.\, 2}\right]$.
+So application of conditions $(3)$ and $(4)$ give $4n-2$ linear equation with variables $\left.{ d_{4n,p}}\right|_{p \mathop = 0 \,.\,.\, 2n-3}$.
+Finally, since $B_{4 n}$ contains $2 n$ monomial terms (see definition), we obtain a Cramer system in variables $\left.{ d_{4 n, p}}\right\vert_{p \mathop = 0 \,.\,.\, 2 n}$, with evident solution:
+:$\left.{\mathrm d_{4 n, p}}\right\vert_{p \mathop = 0 \,.\,.\, 2 n} = 0 $
+and consequently:
+:$\left.{a_{4 n, p} }\right\vert_{p \mathop = 0 \,.\,.\, 2 n} = \left.{b_{4 n, p} }\right\vert_{p \mathop = 0 \,.\,.\, 2 n}$
+which means:
+:$q_{4 n} \left({x}\right) = B_{4 n} \left({x}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Type Space is Compact}
+Tags: Model Theory
+
+\begin{theorem}
+Let $\mathcal M$ be an $\mathcal L$-[[Definition:First-Order Structure|structure]], and let $A$ be a subset of the universe of  $\mathcal M$.
+The [[Definition:Type Space|type space]] $S_n^{\mathcal M}(A)$ of $n$-types over $A$ is [[Definition:Compact Topological Space|compact]].
+\end{theorem}
+
+\begin{proof}
+It will suffice to show that every [[Definition:Open Cover|open cover]] of $S_n^{\mathcal M}(A)$ by the [[Definition:Synthetic Basis|basic open sets]] $[\phi]$ of the topology has a [[Definition:Subcover|finite subcover]].
+Let $\mathcal U = \{ [\phi_i] : i\in I \}$ be a cover of $S_n^{\mathcal M}(A)$ by basic open sets.
+This means that every complete $n$-type over $A$ contains some $\phi_i$.
+We will find a finite subcover of $\mathcal U$.
+Let $\Gamma = \{ \neg\phi_i : i\in I \}$.
+Then $\operatorname{Th}_\mathcal A(M) \cup \Gamma$ cannot be [[Definition:Satisfiable|satisfied]], since if $\mathcal N \models \operatorname{Th}_\mathcal A(M) \cup \Gamma (\bar{b})$, then the [[Definition:Type|type]] $\operatorname{tp}_\mathcal N (\bar{b}/A)$ is a [[Definition:Complete Type|complete $n$-type]] in $S_n^{\mathcal M}(A)$ which does not contain any $\phi_i$.
+By the [[Compactness Theorem]], $\operatorname{Th}_\mathcal A(M) \cup \Gamma$ must have a finite subset $\Delta$ which is not satisfiable.
+Since $\Delta$ is not satisfiable but $\operatorname{Th}_\mathcal A(M)$ is, $\Delta$ must contain some of the $\neg\phi_i$ from $\Gamma$.
+Furthermore, we must have that every model of $\operatorname{Th}_\mathcal A(M)$ fails to satisfy at least one of these finitely many $\neg\phi_i$ in $\Delta$.
+We claim that the finite set $\mathcal F = \{ [\phi_i] : \neg\phi_i \in \Delta \}$ is a finite subcover of $\mathcal U$.
+Since $\mathcal F$ is clearly a subset of $\mathcal U$, we need only show that every $p\in S_n^{\mathcal M}(A)$ is contained in some $[\phi_i]\in\mathcal F$.
+That is, we must show that each $p\in S_n^{\mathcal M}(A)$ contains one of these $\phi_i$.
+Let $p\in S_n^{\mathcal M}(A)$.
+By definition, this means that $p\cup \operatorname{Th}_\mathcal A(M)$ is satisfiable by some $\mathcal N$.
+But, as mentioned above, since $\mathcal N\models \operatorname{Th}_\mathcal A(M)$, we have that $\mathcal N\not\models \neg\phi_i$ for some $\neg\phi_i \in \Delta$.
+Since $p$ is complete, it must contain either $\phi_i$ or $\neg\phi_i$.
+However, if $p$ contained $\neg\phi_i$, then $\mathcal N\models \neg\phi_i$, which contradicts $\mathcal N\not\models \neg\phi_i$.
+Thus, $p$ contains $\phi_i$.
+This demonstrates that $\mathcal F$ is a finite subcover for $\mathcal U$.
+Thus, $S_n^{\mathcal M}(A)$ is compact.
+{{qed}}
+[[Category:Model Theory]]
+nn84c5wfi332meprrm3tew7ig41k53n
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is T0}
+Tags: T0 Spaces, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $T$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$ such that $x \ne y$.
+{{WLOG}}, let $x \ne 0$.
+Then $U = \set x$ is [[Definition:Open Set (Topology)|open]] in $T$ from the definition of the [[Definition:Either-Or Topology|either-or topology]].
+We have that $x$ and $y$ are arbitrary.
+So:
+:$\forall x, y \in S: \exists U \in \tau: x \in U, y \notin U$
+and the result follows by definition of [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is not T1}
+Tags: T1 Spaces, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $T$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+Let $x = \dfrac 1 2$.
+We have that $V = \set x$ such that $x \in V, 0 \notin V$.
+However, by definition of the [[Definition:Either-Or Topology|either-or topology]], the only [[Definition:Open Set (Topology)|open sets]] of $T$ containing $0$ also contain $\openint {-1} 1$, and so must also contain $x$.
+So we have that $\nexists U, V \in \tau: 0 \in U, x \notin U, x \in V, 0 \notin V$.
+Hence the result by definition of [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Omitting Types Theorem}
+Tags: Model Theory, Named Theorems
+
+\begin{theorem}
+Let $\mathcal L$ be the [[Definition:Language of Predicate Logic|language of predicate logic]] with a [[Definition:Countable Set|countable]] [[Definition:Signature for Predicate Logic|signature]].
+Let $T$ be an $\mathcal L$-[[Definition:Theory (Logic)|theory]].
+Let $\left\{{p_i: i \in \N}\right\}$ be a [[Definition:Countable Set|countable set]] of [[Definition:Isolated Type|non-isolated]] [[Definition:Type|$n$-types of $T$]].
+There is a [[Definition:Countable Set|countable]] $\mathcal L$-[[Definition:First Order Structure|structure]] $\mathcal M$ such that [[Definition:Model (Logic)|$\mathcal M \models T$]] and $\mathcal M$ [[Definition:Omission|omits]] each $p_i$.
+\end{theorem}
+
+\begin{proof}
+This proof is done by constructing a model using a method known as a Henkin construction.
+This results in a model all of whose elements are the interpretations of constant symbols from some language.
+The construction for this proof in particular is done so that the theory this model satisfies asserts that each tuple of constants fails to satisfy at least one $\phi$ from each type $p_i$.
+As a result, the model we construct will not realize any of the types.
+We will prove the theorem for the special case where we only omit one type.
+That is, assume $\left\{{p_i: i \in \N}\right\}$ is $\left\{{p}\right\}$ for some $p$.
+Comments on the general case will be made after the proof.
+Our goal is to eventually be able to apply [[Maximal Finitely Satisfiable Theory with Witness Property is Satisfiable]] to a suitable $\mathcal L^*$-theory $T^*$, where $\mathcal L^*$ contains $\mathcal L$ and $T^*$ contains $T$.
+Let $\mathcal L^*$ be $\mathcal L \cup \left\{{c_i : i \in \N}\right\}$ where each $c_i$ is a new constant symbol.
+Note that $\mathcal L^*$ is still countable.
+We will recursively construct $T^*$ by adding a sentence $\theta_k$ at each stage.
+In order to ensure that it has the necessary properties, we will construct it in stages, each of which has three steps.
+The goal of the first step of each stage is ensure that $T^*$ will be maximal upon completion of the definition.
+The goal of the second step of each stage is to ensure that $T^*$ will satisfy the [[Definition:Witness Property|witness property]].
+The goal of the third step of each stage is to ensure that $p$ is not realized by any tuple of constants.
+Since $\mathcal L^*$ is countable, the set of $\mathcal L$-formulas is countable, and hence in particular, the $\mathcal L^*$-sentences can be listed as $\phi_0, \phi_1, \ldots$
+Similarly, since $\mathcal L^*$ is countable, the set of $n$-tuples of its constant symbols can be listed as $\bar d_0, \bar d_1, \dots$
+=== Recursive definition of $T^*$ ===
+We now recursively define sets $T_k$ whose union will be $T^*$.
+==== Stage $k = 0$ ====
+Let $\theta_0$ be any $\mathcal L$-sentence which is a [[Definition:Tautology|tautology]]; for example, $\forall x : x = x$.
+Let $T_0 = T \cup \left\{{\theta_0}\right\}$.
+Note that $T_0$ is satisfiable since $T$ is satisfiable, and $T \models \theta_0$ trivially.
+Suppose that $\theta_k$ and $T_k$ have been defined and $T_k$ is satisfiable.
+We now handle the three steps of the $k$-th stage as mentioned above.
+==== Stage $k + 1 = 3 i + 1$ ====
+In this step we ensure that $T^*$ will be maximal.
+If $T_k \cup \left\{{\phi_i}\right\}$ is satisfiable, let $\theta_{k+1}$ be $\phi_i$.
+Otherwise, since $T_k$ is satisfiable, $T_k \cup \left\{{\neg \phi_i}\right\}$ must be satisfiable.
+In this case we let $\theta_{k+1}$ be $\neg \phi_i$.
+Let $T_{k+1} = T_k \cup \left\{{\theta_{k+1} }\right\}$.
+==== Stage $k + 1 = 3 i + 2$ ====
+In this step we ensure that $T^*$ will have the witness property.
+Suppose that:
+:$T_k \models \phi_j$ for some set of $j \le i$
+and that:
+:$\phi_j$ are each of the form $\exists v \psi_j \left({v}\right)$.
+Let $\theta_{k+1}$ be $\bigwedge \psi \left({c}\right)$ where $c$ is a new constant which doesn't appear in $T_k$.
+This is possible since $T_k$ only has finitely many $\mathcal L^*$-sentences which are not $\mathcal L$-sentences.
+Let $T_{k+1} = T_k \cup \{\theta_{k+1}\}$.
+Since $T_k$ is satisfiable and $T_k \models \bigwedge \exists v \psi_j \left({v}\right)$, any model $\mathcal M$ of $T_k$ has an element $a_j$ such that $\mathcal M \models \psi_j (a_j)$.
+By interpreting $c_j$ as $a_j$, we can view $\mathcal M$ as a model of $T_{k+1}$.
+Thus $T_{k+1}$ is satisfiable.
+If $T_k$ doesn't satisfy any existential statements $\phi_j$ for $j \le i$, let $\theta_{k+1}$ be $\theta_k$ and let $T_{k+1}$ be $T_{k}$.
+==== Stage $k + 1 = 3 i + 3$ ====
+In this step we ensure that $p$ will be omitted by any model of $T^*$.
+Let $(e_1,\dots,e_n)$ be the $i$-th $n$-tuple of constants $\bar d_i$.
+We will add a sentence to ensure that $\bar d_i$ fails to realize $p$.
+Let $\psi$ be the $\mathcal L$-formula obtained from $\theta_0 \wedge \cdots \theta_k$ by:
+:$(1): \quad$ replacing [[Definition:Occurrence (Predicate Logic)|occurrences]] of the constants $e_1, \dotsc, e_n$ with variables $v_1, \dotsc, v_n$
+:$(2): \quad$ replacing occurrences of $c_i$ besides $e_1, \dotsc, e_n$ by variables $v_{c_i}$ other than $v_1, \dotsc, v_n$
+and finally:
+:$(3): \quad$ adding an existential quantifier $\exists v_{c_i}$ for each of the $c_i$ that were replaced other than $e_1, \dotsc, e_n$.
+This results in $\psi$ being an $\mathcal L$-formula with $n$ free variables.
+The free variables are those that correspond to the $e_i$.
+Since $p$ is not isolated, there must be some $\phi\in p$ such that:
+:$T \not \models \forall \bar v \left({\psi \left({\bar v}\right) \to \phi \left({\bar v}\right)}\right)$
+Thus:
+:$T \models \exists \bar v \left({\psi \left({\bar v}\right) \wedge \neg \phi \left({\bar v}\right)}\right)$
+Let $\theta_{k+1}$ be $\neg \phi \left({\bar d_i}\right)$.
+Let $T_{k+1} = T_k \cup \left\{ {\theta_{k+1} }\right\}$.
+Since $T$ is satisfiable, it has a model $\mathcal M$.
+By the above:
+:$\mathcal M \models \exists \bar v \left({\psi \left({\bar v}\right) \wedge \neg \phi \left({\bar v}\right)}\right)$
+Hence there is some tuple $\bar a$ in $\mathcal M$ such that:
+:$\mathcal M \models \psi \left({\bar a}\right) \wedge \neg \phi \left({\bar a}\right)$
+Let the constants $e_1, \dotsc, e_n$ be interpreted as the constants in the tuple $\bar a$.
+Let the the constants $c_i$ other than $e_1, \dotsc, e_n$ be interpreted as the elements from $\mathcal M$ which satisfy the existential quantifiers in $\psi \left({\bar a}\right)$.
+Then we have that $\mathcal M$ can be viewed as a model of $\theta_0 \wedge \dotsb \wedge \theta_k \wedge \neg \phi \left({\bar d_i}\right)$ as well.
+Thus $T_{k+1}$ is satisfiable.
+=== Verification of properties of $T^*$ ===
+Now, let $\displaystyle T^* = \bigcup_{k \mathop \in \N} T_k$, or equivalently:
+:$T^* = T \cup \left\{ {\theta_k: k \in \N}\right\}$
+We verify that $T^*$ is a maximal finitely satisfiable theory with the witness property.
+Thus we may apply [[Maximal Finitely Satisfiable Theory with Witness Property is Satisfiable]] which is the theorem that will construct our desired model.
+==== $T^*$ is finitely satisfiable ====
+Let $\Delta$ be a finite subset of $T^*$.
+Then $\Delta$ is contained in one of the $T_k$.
+But by construction each $T_k$ was satisfiable.
+Heence by the [[Compactness Theorem]] $\Delta$ is satisfiable.
+==== $T^*$ is a maximal theory ====
+Let $\phi$ is an $\mathcal L^*$-sentence.
+Then it is some $\phi_i$.
+Hence either $\phi$ or $\neg \phi$ is contained in $T_{3i + 1}$.
+==== $T^*$ has the witness property ====
+Let $\psi \left({v}\right)$ be an $\mathcal L^*$-formula with one free variable.
+Then $\exists v \psi \left({v}\right)$ is an $\mathcal L^*$-sentence and hence is some $\phi_j$.
+Suppose $T^*\models \exists v \psi \left({v}\right)$.
+There must be a finite subset $\Delta$ of $T^*$ such that $\Delta \models \exists v \psi \left({v}\right)$.
+Otherwise, by the [[Compactness Theorem]], since $\Delta \cup \left\{ {\neg \exists v \psi \left({v}\right)}\right\}$ is always satisfiable, $T^* \cup \left\{ {\neg \exists v \psi \left({v}\right)}\right\}$ is satisfiable.
+Since such a $\Delta$ must be contained in one of the $T_k$, there is some $T_k$ such that $T_k \models \exists v \psi \left({v}\right)$.
+Since each $T_k$ is contained in $T_{k+1}$, we can assume that $k = 3 i + 2$ for some $i \le j$.
+But, $T_{3 i + 2}$ contains $\psi \left({c}\right)$ for some constant $c$ since $j \le i$ and $T_{3i+2}\models \phi_j$.
+=== Verification that the constructed model omits $p$ ===
+Recall that $\mathcal L$ was assumed to be countable.
+Thus $\mathcal L^*$ is countable as well.
+Thus, by [[Maximal Finitely Satisfiable Theory with Witness Property is Satisfiable]] applied to $T^*$:
+:there is an $\mathcal L^*$-structure $\mathcal M$ such that $\mathcal M \models T^*$
+:every element of $\mathcal M$ is the interpretation of one of the constant symbols from $\mathcal L^*$
+and consequently:
+:$\mathcal M$ is at most countable.
+We verify that $\mathcal M$ omits $p$.
+That is, every $n$-tuple of elements in $\mathcal M$ fails some sentence in $p$.
+Let $\left({a_1, \dotsc, a_n}\right)$ be an $n$-tuple of elements from $\mathcal M$.
+Then each of $a_1, \dotsc, a_n$ is the interpretation of constant symbol $e_1, \dotsc, e_n$ respectively from $\mathcal L^*$.
+In turn, $\left({e_1, \dotsc, e_n}\right)$ is one of the $\bar d_i$.
+But $T_{3 i + 3} \models \neg \phi \left({\bar d_i}\right)$ for some $\phi \in p$.
+Thus $\mathcal M \models \neg \phi \left({a_1, \dots, a_n}\right)$ for some $\phi \in p$.
+Hence, $\mathcal M$ omits $p$.
+=== Generalization to countable set of non-isolated types ===
+The above proof can be generalized by extending stage $k + 1 = 3i + 3$.
+As it is written above, we force $\bar d_i$ to fail to satisfy some $\phi$ in a single type $p$.
+But suppose we list the pairs $\left\{ {\left({\bar d_\alpha, p_\beta}\right) : \alpha, \beta \in \N}\right\}$ as a set $\left\{ {\pi_i: i \in \N}\right\}$.
+This is possible since the set of such pairs is countable.
+Then at stage $3 i + 3$, where $\pi_i = \left({\bar d_\alpha, p_\beta}\right)$, we can force $\bar d_\alpha$ to fail to satisfy some $\phi$ in $p_\beta$.
+This is achieved analogously to how it is done in the above proof.
+This ensures that for each $\bar d_\alpha$ with $\alpha \in \N$, and for each $p_\beta$, eventually some $T_k$ includes a sentence $\neg \phi \left({\bar d_\alpha}\right)$ for some $\phi \in p_\beta$.
+That is, eventually some $T_k$ ensures that $\bar d_\alpha$ does not realize $p_\beta$.
+{{qed}}
+{{finish}}
+[[Category:Model Theory]]
+[[Category:Named Theorems]]
+g07qco0li1v0r7459nvaquev67uvukv
+\end{proof}<|endoftext|>
+\section{Vinogradov's Theorem}
+Tags: Number Theory, Analytic Number Theory, Vinogradov's Theorem
+
+\begin{theorem}
+Let $\Lambda$ be the [[Definition:Von Mangoldt Function|von Mangoldt function]].
+For $N \in \Z$, let:
+:$\displaystyle \map R N = \sum_{n_1 + n_2 + n_3 \mathop = N} \map \Lambda {n_1} \, \map \Lambda {n_2} \, \map \Lambda {n_3}$
+be a [[Definition:Weighted Count|weighted count]] of the number of [[Definition:Representation|representations]] of $N$ as a sum of three [[Definition:Prime Power|prime powers]].
+Let $\SS$ be the [[Definition:Arithmetic Function|arithmetic function]]:
+:$\displaystyle \map \SS N = \prod_{p \mathop \nmid N} \paren {1 + \frac 1 {\paren {p - 1}^3} } \prod_{p \mathop \divides N} \paren {1 - \frac 1 {\paren {p - 1}^2} }$
+where:
+:$p$ ranges over the [[Definition:Prime Number|primes]]
+:$p \nmid N$ denotes that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $N$
+:$p \divides N$ denotes that $p$ is a [[Definition:Divisor of Integer|divisor]] of $N$.
+Then for any $A > 0$ and [[Definition:Sufficiently Large|sufficiently large]] [[Definition:Odd Integer|odd integers]] $N$:
+:$\map R N = \dfrac 1 2 \map \SS N N^2 + \map \OO {\dfrac {N^2} {\paren {\log N}^A} }$
+where $\OO$ denotes [[Definition:Big-O Notation|big-O notation]].
+\end{theorem}<|endoftext|>
+\section{Vinogradov's Theorem/Minor Arcs}
+Tags: Vinogradov's Theorem
+
+\begin{theorem}
+For any $B > 0$:
+:$\displaystyle \int_\mathcal M \map F \alpha^3 \map e {-\alpha N} \rd \alpha \ll \frac {N^2} {\paren {\ln N}^{B/2 - 5} }$
+\end{theorem}
+
+\begin{proof}
+=== [[Vinogradov's Theorem/Minor Arcs/Lemma 1|Lemma 1]] ===
+{{:Vinogradov's Theorem/Minor Arcs/Lemma 1}}
+{{qed|lemma}}
+{{finish|The below need proofs}}
+=== Lemma 2 ===
+Let $a, q \in \Z$ such that:
+:$(1): \quad \displaystyle \size {\alpha - \frac a q} \le \frac 1 {q^2} \quad 1 \le q \le N, \quad \gcd \set {a, q} = 1$
+Let $m, n \in \Z$ be [[Definition:Integer|integers]].
+Then:
+:$\displaystyle\sum_{k mathop = 1}^m \min \set {\frac {m n} k, \frac 1 {\norm {\alpha k} } } \ll \paren {m + \frac {m n} q + q} \map \ln {2 q m}$
+{{qed|lemma}}
+=== Lemma 3 ===
+Let $\alpha$ satisfy the condition $(1)$ of Lemma 2.
+Let $x, y \in \N$.
+Let $\beta_k, \gamma_k$ be any [[Definition:Complex Number|complex numbers]] such that $\size {\beta_k}, \size {\gamma_k} \le 1$.
+Then:
+:$\displaystyle \sum_{k \mathop = y}^{x/y} \sum_{\ell \mathop = y}^{x/k} \alpha_k \beta_\ell \map e {\alpha k \ell} \ll x^{1/2} \paren {\ln x}^2 \paren {\frac x y + y + \frac x q + q}^{1/2}$
+{{qed|lemma}}
+{{qed}}
+[[Category:Vinogradov's Theorem]]
+dofrtafwri6ze1f5k2ftmsxii6qjyh7
+\end{proof}<|endoftext|>
+\section{Vinogradov's Theorem/Major Arcs}
+Tags: Vinogradov's Theorem
+
+\begin{theorem}
+Let $B \in \R_{>0}$.
+Then:
+:$\displaystyle \int_\mathcal M \map F \alpha^3 \map e {-N \alpha} \rd \alpha = \frac {N^2} 2 \map {\mathcal S} N + \map {\mathcal O} {\frac {N^2} {\paren {\ln N}^{B/2} } }$
+where the [[Definition:Implied Constant|implied constant]] depends only on $B$.
+\end{theorem}
+
+\begin{proof}
+=== [[Vinogradov's Theorem/Major Arcs/Lemma 1|Lemma 1]] ===
+{{:Vinogradov's Theorem/Major Arcs/Lemma 1}}
+{{qed|lemma}}
+{{finish|These need proof.}}
+=== Lemma 2 ===
+Let $N \in \N_{\ge 1}$.
+Let $\beta \in \R$.
+Let:
+:$\displaystyle \map u \beta = \sum_{n \mathop \le N} \map e {n \beta}$
+For $P \ge 1$, define:
+:$\displaystyle \map {J_P} N = \int_{-P/N}^{P/N} \map u \beta^3 \map e {-N \beta} \rd \beta$
+:$\map J N = \map {J_{N/2} } N$
+{{explain|What is $P$? Real, integer ...?}}
+Then with $Q = \paren {\ln N}^B$ as above:
+{{explain|Above where? This page needs to be made standalone.}}
+:$\map {J_Q} N = \map J N + \map {\mathcal O} {\dfrac {N^2} {Q^2} }$
+and:
+:$\map J N = \dfrac {N^2} 2 + \map {\mathcal O} N$
+{{qed|lemma}}
+=== Lemma 3 ===
+Let $\alpha \in \map {\mathcal M} {q, a}$ for some $q, a$ such that:
+:$\map {\mathcal M} {q, a} \subseteq \mathcal M$
+Let $\beta = \alpha - \dfrac a q$.
+Then:
+:$\map F \alpha^3 = \dfrac {\map \mu q} {\map \phi q^3} \map u \beta^3 + \map {\mathcal O} {N^3 \map \exp {-C \sqrt {\ln N} } }$
+where $C$ is a constant that depends only on $B$.
+{{qed|lemma}}
+{{qed}}
+[[Category:Vinogradov's Theorem]]
+ejkfvj0y2jdoat0l4ygfprf6eag1rt1
+\end{proof}<|endoftext|>
+\section{Exponential Dominates Polynomial}
+Tags: Exponential Function, Limits of Functions, Polynomial Theory
+
+\begin{theorem}
+Let $\exp$ denote the [[Definition:Real Exponential Function|real exponential function]].
+For any fixed $k \in \N$ and $\alpha > 0$ there exists $N \in \N$ such that $x^k < \exp \left({\alpha x}\right)$ for all [[Definition:Real Number|real]] $x > N$.
+{{expand|it seems that the proof would apply to $x^k, x \in \R$ without any changes to the proof. Changes made as suggested Nov. 2016.}}
+\end{theorem}
+
+\begin{proof}
+Choose any $N > \dfrac{\left({k + 1}\right)!} {\alpha^{k+1}}$, where $!$ denotes the [[Definition:Factorial|factorial]].
+By [[Taylor Series Expansion for Exponential Function]] we have for any $x \in \R_{\ge 0}$:
+:$\displaystyle \exp \left({\alpha x}\right) = \sum_{m \mathop \ge 0} \frac{\left({\alpha x}\right)^m}{m!} > \frac{\left({\alpha x}\right)^{k+1}}{\left({k + 1}\right)!}$
+Therefore, for any $x > N$ we have:
+{{begin-eqn}}
+{{eqn | l = \exp \left({\alpha x}\right)
+      | o = >
+      | r = \frac{\left({\alpha x}\right)^{k+1} } {\left({k + 1}\right)!}
+}}
+{{eqn | o = >
+      | r = \frac{N \alpha^{k+1} } {\left({k + 1}\right)!} x^k
+      | c = because $x > N$
+}}
+{{eqn | o = >
+      | r = x^k
+      | c = by definition of $N$
+}}
+{{end-eqn}}
+This establishes the result.
+{{qed}}
+[[Category:Exponential Function]]
+[[Category:Limits of Functions]]
+[[Category:Polynomial Theory]]
+toxnjiz0eq6qyr8zc8m7xcmvpyfys0g
+\end{proof}<|endoftext|>
+\section{Power Dominates Logarithm}
+Tags: Limits of Functions, Logarithms, Powers
+
+\begin{theorem}
+Let $\epsilon \in \R_{>0}$.
+Let $B \in \N$ be arbitrary.
+Then there exists $N \in \N$ such that:
+:$\forall n > N: \left({\ln n}\right)^B < n^\epsilon$
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+First we show that there exists $N$ such that $\ln n < n^\epsilon$ for all $n > N$,.
+By [[Exponential is Strictly Increasing]], the [[Definition:Real Exponential Function|real exponential function]] is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+Therefore:
+:$\ln n < n^\epsilon \iff n < \exp \left({n^\epsilon}\right)$
+Choose $k \in \N$ such that $k \epsilon > 2$ and $N = k!$.
+By [[Taylor Series Expansion for Exponential Function]]:
+:$\displaystyle \exp \left({n^\epsilon}\right) = \sum_{m \mathop \ge 0} \frac{n^{m \epsilon}}{m!} > \frac{n^{k \epsilon}}{k!}$
+Then for all $n > N$:
+{{begin-eqn}}
+{{eqn | l = \exp \left({n^\epsilon}\right)
+      | r = \frac {n^{k \epsilon} } {k!}
+      | o = >
+      | c = 
+}}
+{{eqn | r = \frac N {k!} n^{k \epsilon - 1}
+      | o = >
+      | c =
+}}
+{{eqn | r = n
+      | o = >
+      | c = because $k \epsilon > 2$ and $N = k!$
+}}
+{{end-eqn}}
+This completes the proof when $B = 1$.
+Now let $B \in \N$ be arbitrary.
+By the above we can find $N \in \N$ such that:
+:$\forall n > N: \ln n < n^{\epsilon / B}$
+Then:
+:$\forall n > N: \left({\ln n}\right)^B < \left({n^{\epsilon / B} }\right)^B = n^\epsilon$
+Hence the result.
+{{qed}}
+[[Category:Limits of Functions]]
+[[Category:Logarithms]]
+[[Category:Powers]]
+j8f0ldxcjxzqhv7zu4zcj5058qmpp24
+\end{proof}<|endoftext|>
+\section{Cover of Interval By Closed Intervals is not Pairwise Disjoint}
+Tags: Measure Theory
+
+\begin{theorem}
+Let $\closedint a b$ be a [[Definition:Closed Interval|closed interval]] in $\R$.
+{{explain|Title mentions only "interval"; this does not affect truth of statement so may "closed" above line be removed as superfluous?}}
+Let $\JJ$ be a [[Definition:Set|set]] of two or more [[Definition:Closed Interval|closed intervals]] contained in $\closedint a b$ such that $\displaystyle \bigcup \JJ = \closedint a b$.
+Then the intervals in $\JJ$ are not [[Definition:Pairwise Disjoint|pairwise disjoint]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that the intervals of $\JJ$ are [[Definition:Pairwise Disjoint|pairwise disjoint]].
+Let $I = \closedint p q$ be the [[Definition:Unique|unique]] [[Definition:Closed Interval|interval]] of $\JJ$ such that $a \in I$.
+Let $J = \closedint r s$ be the [[Definition:Unique|unique]] [[Definition:Closed Interval|interval]] of $\JJ$ containing the least [[Definition:Real Number|real number]] not in $I$.
+These choices are possible since $\JJ$ has at least two elements, and they are supposed [[Definition:Pairwise Disjoint|disjoint]].
+If $q = r$ then $I \cap J \ne \O$, a contradiction.
+If $r < q$ then $I \cap J \ne \O$, a contradiction.
+If $r > q$ then there is some [[Definition:Real Number|real number]] with $q < \alpha < r$.
+Therefore $\alpha \notin I$.
+Since $J$ contains the least [[Definition:Real Number|real number]] not in $I$ it follows that there is $\beta \in J$ with $\beta < \alpha$.
+But we also have that $\alpha < r \le \beta$ for all $\beta \in J$, a contradiction.
+This exhausts all the possibilities, and we conclude that the intervals of $\JJ$ are not [[Definition:Pairwise Disjoint|pairwise disjoint]].
+{{qed}}
+[[Category:Measure Theory]]
+9wda0ujr5f54g8ssj2kpco4umgx8h92
+\end{proof}<|endoftext|>
+\section{Condition for Open Extension Space to be T5 Space}
+Tags: T5 Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is a [[Definition:T5 Space|$T_5$ space]] {{iff}} $T$ is a [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}<|endoftext|>
+\section{Open Extension Topology is T4}
+Tags: T4 Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is a [[Definition:T4 Space|$T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+We have that an [[Open Extension Space is Ultraconnected]].
+That means none of its [[Definition:Closed Set (Topology)|closed sets]] are [[Definition:Disjoint Sets|disjoint]].
+Hence, [[Definition:Vacuous Truth|vacuously]], any two of its [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed subsets]] are [[Definition:Separated by Neighborhoods|separated by neighborhoods]].
+The result follows by definition of [[Definition:T4 Space|$T_4$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Topology is T4}
+Tags: T4 Spaces, Excluded Point Topology, Excluded Point Topology is T4
+
+\begin{theorem}
+Let $T = \left({S, \tau_{\bar p}}\right)$ be an [[Definition:Excluded Point Topology|excluded point space]].
+Then $T$ is a [[Definition:T4 Space|$T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+We have that an [[Excluded Point Space is Ultraconnected]].
+That means none of its [[Definition:Closed Set (Topology)|closed sets]] are [[Definition:Disjoint Sets|disjount]].
+Hence, [[Definition:Vacuous Truth|vacuously]], any two of its [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed subsets]] of $S$ are [[Definition:Separated by Neighborhoods|separated by neighborhoods]].
+The result follows by definition of [[Definition:T4 Space|$T_4$ space]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have:
+: [[Excluded Point Topology is Open Extension Topology of Discrete Topology]]
+: [[Open Extension Topology is T4|Open Extension Topology is $T_4$]]
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have:
+: [[Excluded Point Topology is T5|Excluded Point Topology is $T_5$]]
+: [[T5 Space is T4 Space|$T_5$ Space is $T_4$]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Extension Topology is not T3}
+Tags: T3 Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is not a [[Definition:T3 Space|$T_3$ space]].
+\end{theorem}
+
+\begin{proof}
+As $S$ is (trivially) [[Definition:Open Set (Topology)|open]] in $T$, it is also open in $T^*_{\bar p}$.
+As $S^*_p = S \cup \set p$, it follows that $\set p$ is [[Definition:Closed Set (Topology)|closed]] in $T^*_{\bar p}$.
+But by definition, the only [[Definition:Open Set (Topology)|open set]] of $T^*_{\bar p}$ that contains $\set p$ is $S^*_p$ itself.
+So there can be no $x \in S^*_p: x \notin \set p$ contained in an [[Definition:Open Set (Topology)|open set]] of $T^*_{\bar p}$ which is [[Definition:Disjoint Sets|disjoint]] from $S^*_p$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Excluded Point Topology is not T3}
+Tags: T3 Spaces, Excluded Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{\bar p} }$ be a [[Definition:Excluded Point Topology|excluded point space]].
+Then $T$ is not a [[Definition:T3 Space|$T_3$ space]].
+\end{theorem}
+
+\begin{proof}
+:[[Excluded Point Topology is Open Extension Topology of Discrete Topology]]
+:[[Open Extension Topology is not T3]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Open Extension Space to be Separable}
+Tags: Separable Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is a [[Definition:Separable Space|separable space]] {{iff}} $T$ is.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Separable Space|separable space]].
+Then there exists a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] $H \subseteq S$ which is [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+That is, $H^- = S$ where $H^-$ is the [[Definition:Closure (Topology)|closure]] of $H$ in $S$.
+So in $T_{\bar p}^*$, $S\subseteq H^-=S^-\subseteq S^*_p$.
+Hence $H^- = S$ or $H^- = S \cup \set p = S^*_p$.
+From [[Topological Closure is Closed]] we deduce that $H^-$ is [[Definition:Closed Set (Topology)|closed]] in $T_{\bar p}^*$.
+From the definition of [[Definition:Open Extension Topology|open extension topology]], the [[Definition:Closed Set (Topology)|closed sets]] of $T_{\bar p}^*$ are $S^*_p \setminus U$ where $U \in \tau$.
+Because $\set p \notin \tau$, $S$ is not [[Definition:Closed Set (Topology)|closed]].
+Thus it must be that $H^- = S^*_p$.
+Hence $H$ is a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] $H \subseteq S^*_p$ which is [[Definition:Everywhere Dense|everywhere dense]] in $T^*_{\bar p}$.
+{{qed|lemma}}
+{{AimForCont}} $T^*_{\bar p} = \struct {S^*_p, \tau}$ is [[Definition:Separable Space|separable]], but that $T = \struct {S, \tau}$ is not [[Definition:Separable Space|separable]].
+Let $H \subseteq S^*_p$ be a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] which is [[Definition:Everywhere Dense|dense]] in $T^*_{\bar p}$.
+If $H \subseteq S$, then $H^- = S$ in $T$, but this cannot be since $T$ is not [[Definition:Separable Space|separable]].
+Then $p \in H$.
+Consider the subset $V = H \setminus \set p$.
+From [[Topological Closure is Closed]] we have that $V^-$ is [[Definition:Closed Set (Topology)|closed]] in $T_{\bar p}^*$.
+From the definition of [[Definition:Open Extension Space|open extension space]], all [[Definition:Non-Empty Set|non-empty]] [[Definition:Closed Set (Topology)|closed sets]] contain the point $p$.
+So $V \cup \set p = H \subseteq V^-$. 
+From [[Set Closure is Smallest Closed Set in Topological Space]], $S^*_p = H^- \subseteq V^- \implies V$ is dense in $T_{\bar p}^*$.
+But because $V \subseteq S$, $V$ is [[Definition:Everywhere Dense|dense]] in $S$ also.
+From [[Subset of Countably Infinite Set is Countable]], $V$ is [[Definition:Countable Set|countable]] as it is a [[Definition:Subset|subset]] of a [[Definition:Countable Set|countable set]].
+Thus $T$ is [[Definition:Separable Space|separable]], which is a [[Definition:Contradiction|contradiction]].
+Finally, our assumption is false: $T$ is indeed [[Definition:Separable Space|separable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension}
+Tags: Model Theory
+
+\begin{theorem}
+Let $T$ be a [[Definition:Finitely Satisfiable|finitely satisfiable]] [[Definition:Mathematical Theory|$\mathcal L$-theory]].
+Then there exists a [[Definition:Finitely Satisfiable|finitely satisfiable]] $\mathcal L$-theory $T'$ which contains $T$ as a [[Definition:Subset|subset]] such that:
+:for all $\mathcal L$-sentences $\phi$, either $\phi \in T'$ or $\neg \phi \in T'$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathcal A$ be the [[Definition:Set|set]] of finitely satisfiable extensions of $T$.
+By the [[Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Lemma|lemma]], for each element $S$ of $\mathcal A$ and each $\mathcal L$-sentence $\phi$, either $S \cup \left\{ {\phi}\right\} \in \mathcal A$ or $S \cup \left\{ {\neg \phi}\right\} \in \mathcal A$.
+$\mathcal A$ has [[Definition:Finite Character|finite character]], by the following argument:
+Let $S \in \mathcal A$.
+Let $F$ be a [[Definition:Finite Subset|finite subset]] of $S$.
+Then $S$ is satisfiable and hence finitely satisfiable.
+Thus in $\mathcal A$.
+Let $S$ be a theory on $\mathcal L$.
+Let every [[Definition:Finite Subset|finite subset]] of $S$ be finitely satisfiable.
+Then every [[Definition:Finite Subset|finite subset]] of $S$ is satisfiable.
+Therefore $S$ is finitely satisfiable.
+Thus $\mathcal A$ has finite character.
+By the [[Restricted Tukey-Teichmüller Theorem]], $\mathcal A$ has an element $T'$ such that:
+:for each $\mathcal L$-sentence $\phi$, either $\phi \in T'$ or $\neg \phi \in T'$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The set of all [[Definition:Finitely Satisfiable|finitely satisfiable]] [[Definition:Mathematical Theory|$\mathcal L$-theories]] containing $T$ forms an [[Definition:Ordered Set|ordered set]] using [[Definition:Subset|subset inclusion]] as the [[Definition:Ordering|ordering]].
+{{explain|Link to "subset is ordering"}}
+Let $C$ be a [[Definition:Non-Empty Set|nonempty]] [[Definition:Chain (Set Theory)|chain]] in this ordered set.
+Let $\displaystyle T_C = \bigcup_{\Sigma \mathop \in C} \Sigma$.
+Let $\Delta$ be a [[Definition:Finite Subset|finite subset]] of $T_C$.
+Then there exists a single $\Sigma$ in $C$ which contains $\Delta$.
+Since this $\Sigma$ is finitely satisfiable by definition, this means that $\Delta$ is satisfiable.
+Hence $T_C$ is finitely satisfiable.
+Since each $\Sigma \in C$ is contained in $T_C$, this means that $T_C$ is an upper bound for $C$ in the ordered set.
+Thus, by [[Zorn's Lemma]], there is a finitely satisfiable $\mathcal L$-theory $T'$ containing $T$ such that $T'$ contains all other such theories.
+Let $\phi$ be an $\mathcal L$-sentence.
+Let $\phi \nsubseteq T'$.
+Aiming for a contradiction, suppose $T' \cup \left\{ {\phi}\right\}$ were finitely satisfiable.
+Then by definition of $T'$, $T'$ would contain $T' \cup \left\{ {\phi}\right\}$ as a subset.
+This would mean that $T'$ contains $\phi$, which contradicts the assumption.
+Thus $T' \cup \left\{ {\phi}\right\}$ is not finitely satisfiable.
+By the [[Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Lemma|lemma]], $T' \cup \left\{ {\neg \phi}\right\}$ is finitely satisfiable.
+Thus, by definition of $T'$, $T'$ contains $T' \cup \left\{{\neg \phi}\right\}$ as a [[Definition:Subset|subset]].
+Hence $T'$ contains $\neg \phi$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Open Extension Space to be First-Countable}
+Tags: First-Countable Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is a [[Definition:First-Countable Space|first-countable space]] {{iff}} $T$ is.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:First-Countable Space|first-countable space]].
+Then every point in $S$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]].
+Every [[Definition:Open Set (Topology)|open set]] of $T$ is an [[Definition:Open Set (Topology)|open set]] of $T^*_{\bar p}$ by definition.
+So if $x$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] in $T$, then it has one in $T^*_{\bar p}$ as well.
+Finally we note that $p$ is by definition in exactly one [[Definition:Open Set (Topology)|open set]] of $T^*_{\bar p}$, that is, $S^*_p$, and thus (trivially) has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] in $T^*_{\bar p}$.
+Then every point in $S^*_p$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] in $T^*_{\bar p}$.
+So if $T$ is a [[Definition:First-Countable Space|first-countable space]], then $T^*_{\bar p}$ is also a [[Definition:First-Countable Space|first-countable space]].
+Now suppose $T = \struct {S, \tau}$ is not a [[Definition:First-Countable Space|first-countable space]].
+Then $\exists x \in S$ such that $x$ has no [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] in $T$.
+As every [[Definition:Open Set (Topology)|open set]] of $T$ is an open set of $T^*_{\bar p}$ by definition, it follows that $x$ has no [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] in $T^*_{\bar p}$ either.
+So if $T $ is not a [[Definition:First-Countable Space|first-countable space]], then neither is $T^*_{\bar p}$ a [[Definition:First-Countable Space|first-countable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Open Extension Space to be Second-Countable}
+Tags: Second-Countable Spaces, Open Extension Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the [[Definition:Open Extension Space|open extension space]] of $T$.
+Then $T^*_{\bar p}$ is a [[Definition:Second-Countable Space|second-countable space]] {{iff}} $T$ is.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Second-Countable Space|second-countable space]].
+Then $\tau$ has a [[Definition:Countable Basis|countable basis]].
+Every [[Definition:Open Set (Topology)|open set]] of $T$ is an [[Definition:Open Set (Topology)|open set]] of $T^*_{\bar p}$ by definition.
+So if $\tau$ has a [[Definition:Countable Basis|countable basis]] in $T$, then $\tau^*_{\bar p} = \tau \cup \set {S^*_p}$ has one as well.
+So if $T$ is a [[Definition:Second-Countable Space|second-countable space]], then $T^*_{\bar p}$ is also a [[Definition:Second-Countable Space|second-countable space]].
+Now suppose $T = \struct {S, \tau}$ is not a [[Definition:Second-Countable Space|second-countable space]].
+Then $\tau$ has no [[Definition:Countable Basis|countable basis]] in $T$.
+As every [[Definition:Open Set (Topology)|open set]] of $T$ is an [[Definition:Open Set (Topology)|open set]] of $T^*_{\bar p}$ by definition, it follows that $\tau^*_{\bar p} = \tau \cup \set {S^*_p}$ has no [[Definition:Countable Basis|countable basis]] either.
+So if $T$ is not a [[Definition:Second-Countable Space|second-countable space]], then neither is $T^*_{\bar p}$ a [[Definition:Second-Countable Space|second-countable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is not T3}
+Tags: T3 Spaces, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $T$ is not a [[Definition:T3 Space|$T_3$ space]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Set|set]] $\set 0$ which is [[Definition:Closed Set (Topology)|closed]] from [[Closed Sets of Either-Or Topology]].
+Let $x = \dfrac 1 2$.
+Let $U \in \tau$ such that $\set 0 \subseteq U$.
+Then by definition $\openint {-1} 1 \subseteq U$ and so $x \in U$.
+So we have found a [[Definition:Closed Set (Topology)|closed set]] of $T$ and a point in $S$ which do not satisfy the conditions for $T$ to be a [[Definition:T3 Space|$T_3$ space]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Separation Properties in Open Extension of Particular Point Topology}
+Tags: Separation Axioms, Open Extension Topology, Particular Point Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Space|particular point space]] such that $S$ is not a [[Definition:Singleton|singleton]] or a [[Definition:Doubleton|doubleton]].
+Let $T^*_{\bar q} = \struct {S^*_q, \tau^*_{\bar q} }$ be an [[Definition:Open Extension Space|open extension space]] of $T$.
+Then:
+:$T^*_{\bar q}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+:$T^*_{\bar q}$ is a [[Definition:T4 Space|$T_4$ (space]].
+:$T^*_{\bar q}$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+:$T^*_{\bar q}$ is not a [[Definition:T5 Space|$T_5$ (space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Particular Point Space is T0|Particular Point Space is $T_0$]].
+Then from [[Condition for Open Extension Space to be T0 Space|Condition for Open Extension Space to be $T_0$ Space]], it follows that $T^*_{\bar q}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+We have directly that:
+:An [[Open Extension Topology is not T1|Open Extension Topology is not $T_1$]].
+:An [[Open Extension Topology is T4|Open Extension Topology is $T_4$]].
+Finally, we have that a [[Particular Point Topology with Three Points is not T4|Particular Point Topology with three points or more is not $T_4$]].
+From [[T5 Space is T4 Space|$T_5$ Space is $T_4$ Space]], it follows that $T$ is not a [[Definition:T5 Space|$T_5$ space]].
+It follows from [[Condition for Open Extension Space to be T5 Space|Condition for Open Extension Space to be $T_5$ Space]] that $T^*_{\bar q}$ is not a [[Definition:T5 Space|$T_5$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Points of Either-Or Topology}
+Tags: Limit Points, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Space|either-or space]].
+Let $H \subseteq S$ be any [[Definition:Subset|subset]] of $S$.
+Then no element of $S$ can be a [[Definition:Limit Point of Set|limit point]] of $H$ except $0$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$ such that $x \ne 0$.
+Then, as $0 \notin \set x$, we have by definition of the [[Definition:Either-Or Topology|either-or topology]] that $x$ is [[Definition:Open Set (Topology)|open]] in $T$.
+So whatever $H$ is, $\set x$ never contains any points of $H$ which are different from $x$.
+So $x$ can not be a [[Definition:Limit Point of Set|limit point]] of $H$.
+However, every [[Definition:Open Set (Topology)|open set]] of $T$ which contains $0$ also contains the [[Definition:Open Real Interval|interval]] $\openint {-1} 1$ and so can contain points of $H$.
+It follows that the only [[Definition:Non-Empty Set|non-empty sets]] of $T$ for which $0$ is ''not'' a [[Definition:Limit Point of Set|limit point]] are $\set {-1}$, $\set 0$ and $\set 1$ and their [[Definition:Set Union|unions]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is T5}
+Tags: T5 Spaces, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Space|either-or space]].
+Then $T$ is a [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $A, B \subseteq S$ such that $A \cap B \ne \O$.
+Suppose neither $A$ nor $B$ contain $0$.
+Then $A$ and $B$ are both [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Limit Points of Either-Or Topology]], the only [[Definition:Limit Point of Set|limit point]] that either $A$ or $B$ may have is $0$.
+So either $A^- = A \cup \set 0$ or $A^- = A$, where $A^-$ is the [[Definition:Closure (Topology)|closure]] of $A$
+Similarly for $B$.
+So $A^- \cap B = \O$, and so are [[Definition:Separated Sets|separated]], and such that:
+:$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
+where $U = A$ and $V = B$, fulfilling the condition for a [[Definition:T5 Space|$T_5$ space]].
+Suppose one of $A$ and $B$ contains $0$.
+{{WLOG}}, suppose $0 \in A$.
+Then for $A$ and $B$ to be [[Definition:Separated Sets|separated]], $0 \notin B^-$ and so $B^-$ has to be $\set {-1}$ or $\set 1$ or their [[Definition:Set Union|union]].
+In any of these cases we have that $B$ and $S \setminus B$ are [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] containing $B$ and $A$ respectively.
+So in all cases we see that if $A$ and $B$ are [[Definition:Separated Sets|separated]], then:
+:$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
+where $U = A$ and $V = B$.
+and so by definition $T$ is a [[Definition:T5 Space|$T_5$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is Lindelöf}
+Tags: Lindelöf Spaces, Either-Or Topology, Either-Or Topology is Lindelöf
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $T$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Either-Or Topology is Compact]]
+:[[Compact Space is Lindelöf]]
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subspace of Either-Or Space less Zero is not Lindelöf}
+Tags: Lindelöf Spaces, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Let $H = S \setminus \set 0$ be the [[Definition:Set|set]] $S$ without [[Definition:Zero (Number)|zero]].
+Then the [[Definition:Topological Subspace|topological subspace]] $T_H = \struct {H, \tau_H}$ is not a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Topological Subspace|topological subspace]], $U \subseteq H$ is [[Definition:Open Set (Topology)|open]] in $T_H$ {{iff}}:
+:$(1): \quad \set 0 \nsubseteq U$
+or:
+:$(2): \quad \openint {-1} 1 \subseteq U$
+But for all $U \subseteq H$, condition $(1)$ holds as $0 \notin H$.
+So $T_H$ is by definition a [[Definition:Discrete Space|discrete space]].
+As $T_H$ is [[Definition:Uncountable Set|uncountable]], we have that [[Uncountable Discrete Space is not Lindelöf]] holds.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is First-Countable}
+Tags: First-Countable Spaces, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $T$ is a [[Definition:First-Countable Space|first-countable space]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$ such that $x \ne 0$.
+Then $\set x$ is [[Definition:Open Set (Topology)|open]] in $T$ and so on its own forms a [[Definition:Local Basis|local basis]] of $x$ which is (trivially) [[Definition:Countable Set|countable]].
+Let $x = 0$.
+Let $U \in \tau$ be [[Definition:Open Set (Topology)|open]] in $T$ such that $x \in U$.
+Then by definition of the [[Definition:Either-Or Topology|either-or space]], $U$ contains the [[Definition:Open Set (Topology)|open set]] $\openint {-1} 1$.
+So $\openint {-1} 1$ forms a [[Definition:Local Basis|local basis]] of $0$ which is (trivially) [[Definition:Countable Set|countable]].
+Hence the result by definition of [[Definition:First-Countable Space|first-countable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is not Separable}
+Tags: Separable Spaces, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $T$ is not a [[Definition:Separable Space|separable space]].
+\end{theorem}
+
+\begin{proof}
+From [[Limit Points of Either-Or Topology]], the only [[Definition:Limit Point of Set|limit point]] of any set of $S$ is $0$.
+So the only set whose [[Definition:Closure (Topology)|closure]] is $S$ are $S \setminus \set 0$ and $S$ itself.
+So these two are the only [[Definition:Subset|subsets]] of $S$ which are [[Definition:Everywhere Dense|everywhere dense]] in $S$.
+Both of these are [[Definition:Uncountable Set|uncountable]].
+Hence the result, by definition of [[Definition:Separable Space|separable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is Non-Meager}
+Tags: Non-Meager Spaces, Either-Or Topology, Either-Or Topology is Non-Meager
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $T$ is [[Definition:Non-Meager Space|non-meager]].
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Either-Or Topology|either-or space]], we have that every point $x$ in $T$ (apart from $0$) forms an [[Definition:Open Set (Topology)|open set]] of $T$.
+So every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $T$ (apart from $\left\{{0}\right\})$ contains at least one [[Definition:Open Set (Topology)|open set]] of $T$.
+So no [[Definition:Subset|subset]] of $T$ is [[Definition:Nowhere Dense|nowhere dense]] in $T$.
+So $T$ is not a [[Definition:Countable Union|countable union]] of [[Definition:Subset|subsets]] of $S$ which are [[Definition:Nowhere Dense|nowhere dense]] in $T$.
+Hence the result by definition of [[Definition:Non-Meager Space|non-meager]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From the definition of the [[Definition:Either-Or Topology|either-or space]], we have that every point $x$ in $T$ (apart from $0$) forms an [[Definition:Open Set (Topology)|open set]] of $T$.
+The result follows directly from [[Space with Open Point is Non-Meager]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis for Either-Or Topology}
+Tags: Either-Or Topology, Topological Bases
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be the [[Definition:Either-Or Topology|either-or space]].
+Let $\mathcal B$ be the set:
+:$\mathcal B := \left\{{\left\{{x}\right\}: x \in S, x \ne 0}\right\} \cup \left\{{ \left({-1 \,.\,.\, 1}\right) }\right\}$
+... that is, the set of all [[Definition:Singleton|singleton]] subsets of $S$ less $\left\{{0}\right\}$ and including the [[Definition:Open Real Interval|open real interval]] $\left({-1 \,.\,.\, 1}\right)$.
+Then $\mathcal B$ is a [[Definition:Basis (Topology)|basis]] for $T$.
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$ such that $0 \notin U$.
+Then:
+:$\displaystyle U = \bigcup_{x \mathop \in U} \left\{{x}\right\}$
+where $x \ne 0$.
+Hence for all $x \in U$, we have $\left\{{x}\right\} \in \mathcal B$.
+Thus $U$ is the [[Definition:Set Union|union]] of elements of $\mathcal B$.
+Now suppose $U \in \tau$ such that $0 \in U$.
+Then $\left({-1 \,.\,.\, 1}\right) \subseteq U$ by definition.
+So one of four cases holds:
+:$U = \left({-1 \,.\,.\, 1}\right)$
+:$U = \left[{-1 \,.\,.\, 1}\right) = \left({-1 \,.\,.\, 1}\right) \cup \left\{{-1}\right\}$
+:$U = \left({-1 \,.\,.\, 1}\right] = \left({-1 \,.\,.\, 1}\right) \cup \left\{{1}\right\}$
+:$U = \left[{-1 \,.\,.\, 1}\right] = \left({-1 \,.\,.\, 1}\right) \cup \left\{{-1}\right\} \cup \left\{{1}\right\}$
+All of these sets are in $\mathcal B$, so that $U$ is the [[Definition:Set Union|union]] of elements of $\mathcal B$.
+Hence, by definition, $\mathcal B$ is a [[Definition:Basis (Topology)|basis]] for $T$.
+{{qed}}
+[[Category:Either-Or Topology]]
+[[Category:Topological Bases]]
+hu2ftr2gslktr0yrxrxkre01xmye643
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is Locally Path-Connected}
+Tags: Locally Path-Connected Spaces, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $T$ is a [[Definition:Locally Path-Connected Space|locally path-connected space]].
+\end{theorem}
+
+\begin{proof}
+Consider the set:
+:$\BB := \set {\set x: x \in S, x \ne 0} \cup \openint {-1} 1$
+Then by [[Basis for Either-Or Topology]], $\BB$ is a [[Definition:Basis (Topology)|basis]] for $T$.
+From [[Point is Path-Connected to Itself]], all $x \in S, x \ne 0$ are [[Definition:Path-Connected Set (Topology)|path-connected]] elements of $\BB$.
+Finally we consider $\openint {-1} 1 \in \BB$.
+Let $p \in \openint {-1} 1$ and consider the [[Definition:Mapping|mapping]] $f: \closedint 0 1 \to \openint {-1} 1$:
+:$\forall x \in \closedint 0 1: \map f x = \begin{cases}
+0 & : x = 0 \\
+p & : x \in \hointl 0 1 \end{cases}$
+Then $f$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] and so a [[Definition:Path (Topology)|path]] from $0$ to $p$.
+So $\openint {-1} 1$ is a [[Definition:Path-Connected Set (Topology)|path-connected]] [[Definition:Element|element]] of $\BB$.
+Hence the result by definition of [[Definition:Locally Path-Connected Space|locally path-connected space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is Scattered}
+Tags: Scattered Spaces, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $T$ is a [[Definition:Scattered Space|scattered space]].
+\end{theorem}
+
+\begin{proof}
+By definition, $T$ is '''[[Definition:Scattered Space|scattered]]''' {{iff}} every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] $H$ of $S$ contains at least one point which is [[Definition:Isolated Point of Subset|isolated in $H$]].
+Let $H$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
+Let $x \in H$.
+Let $x \ne 0$.
+By definition of [[Definition:Either-Or Topology|either-or space]], $\set x$ is [[Definition:Open Set (Topology)|open]] in $T$.
+So $\set x$ is an [[Definition:Open Set (Topology)|open set]] of $x$ containing only $x$.
+Thus by definition $x$ is [[Definition:Isolated Point (Topology)|isolated]].
+Thus if $H$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$ which contains any [[Definition:Element|element]] other than $0$, $H$ contains at least one [[Definition:Isolated Point (Topology)|isolated point]].
+The other case is that $H$ contains no $x$ such that $x \ne 0$.
+That is: $H = \set 0$.
+But from [[Singleton Point is Isolated]], $0$ is an [[Definition:Isolated Point (Topology)|isolated point]] in $H$.
+Thus in all cases $H$ contains at least one point which is [[Definition:Isolated Point of Subset|isolated in $H$]].
+So by definition $T$ is a [[Definition:Scattered Space|scattered space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Stability implies Stability for All Infinite Cardinalities}
+Tags: Model Theory
+
+\begin{theorem}
+Let $T$ be a [[Definition:Complete Theory|complete $\mathcal L$-theory]] whose [[Definition:Logical Language|language]] $\mathcal L$ is [[Definition:Countable Language|countable]].
+If $T$ is [[Definition:Kappa-Stable Theory|$\omega$-stable]], then $T$ is [[Definition:Kappa-Stable Theory|$\kappa$-stable]] for all [[Definition:Infinite Cardinal|infinite $\kappa$]].
+\end{theorem}
+
+\begin{proof}
+We prove the [[Definition:Contrapositive Statement|contrapositive]].
+Let $\kappa$ be an [[Definition:Infinite Cardinal|infinite cardinal]].
+Suppose that $T$ is not [[Definition:Kappa-Stable Theory|$\kappa$-stable]].
+Then there exists some $\mathcal M \models T$ and $A \subseteq \mathcal M$ with $\left\vert{A}\right\vert = \kappa$ such that:
+:$\left\vert{ {S_n}^\mathcal M \left({A}\right)}\right\vert > \kappa$
+Let $\mathcal L_A$ denote $\mathcal L \cup \left\{{a: a \in A}\right\}$, the [[Definition:Logical Language|language]] obtained from $\mathcal L$ by adding new [[Definition:Constant Symbol|constant symbols]] for each $a \in A$.
+For each [[Definition:Well-Formed Formula|$\mathcal L_A$-formula]] $\phi$, let $\left[{\phi}\right] = \left\{{p \in {S_n}^\mathcal M \left({A}\right): \phi \in p}\right\}$, the [[Definition:Set|set]] of [[Definition:Complete Type|complete $n$-types over $A$]] which contain $\phi$.
+Our goal will be to find a [[Definition:Countable Set|countable set]] $B$ in $\mathcal M$ such that:
+:$\left\vert{ {S_n}^\mathcal M \left({B}\right)}\right\vert = 2^{\aleph_0} \ne \aleph_0$
+which will demonstrate non-[[Definition:Kappa-Stable Theory|$\omega$-stability]] of $T$.
+We will do this by constructing a [[Definition:Countable Set|countable]] [[Definition:Binary Tree|binary tree]] of [[Definition:Well-Formed Formula|formulas]] such that each of the $2^{\aleph_0}$ distinct [[Definition:Simple Path|simple paths]] from the [[Definition:Root of Tree|root of the tree]] out to infinity correspond to distinct [[Definition:Type|types]].
+Before we can build the tree, we need the following lemma.
+=== Lemma ===
+Suppose $\left\vert{ \left[{\phi}\right] }\right\vert > \kappa$.
+We argue that we can select some $\mathcal L_A$-formula $\psi$ such that both:
+: $\left\vert{\left[{\phi \land \psi}\right]}\right\vert > \kappa$
+and:
+: $\left\vert{\left[{\phi \land \neg \psi}\right]}\right\vert > \kappa$
+=== Proof of Lemma ===
+The argument is a [[Proof by Contradiction]].
+Suppose the proposition is not true.
+Let $p$ be the [[Definition:Subset|subset]]:
+:$\left\{ {\psi: \left\vert{\left[{\phi \land \psi}\right]}\right\vert > \kappa }\right\}$
+where each $\psi$ is an $\mathcal L_A$-formula in $n$ [[Definition:Free Variable|free variables]].
+We will eventually write $\left[{\phi}\right]$ as a union of $\left\{{p}\right\}$ and other sets which are "too small", so that we contradict the cardinality of $\left[{\phi}\right]$.
+In order to do this, we first need to show that $p$ is a type.
+First, note that if both $\left\vert{\left[{\phi \land \neg \psi}\right]}\right\vert \le \kappa$ and $\left\vert{\left[{\phi \land \neg \psi}\right]}\right\vert \le \kappa$, then $\left\vert{\left[{\phi}\right]}\right\vert \le \kappa$, which is not the case.
+{{explain|The above makes no sense -- both expressions being anded are the same. Presumably one of these should be $\left\vert{\left[{\phi \land \psi}\right]}\right\vert > \kappa$.}}
+Hence, for each $\psi$, either $\psi \in p$ or $\neg \psi \in p$.
+Next, note that by assumption, it cannot be the case that both $\psi$ and $\neg \psi$ in $p$.
+Let $\Delta = \left\{{\psi_1, \ldots, \psi_m}\right\} \cup \Delta'$ be a finite subset of $p \cup \operatorname{Th}_A \left({\mathcal M}\right)$, where $\Delta$ is written so that any sentences from $\operatorname{Th}_A \left({\mathcal M}\right)$ are in $\Delta'$.
+Suppose $\psi_1 \land \cdots \land \psi_m$ is not in $p$.
+Then by the above comment, $\neg \left({\psi_1 \land \cdots \land \psi_m}\right)$ is in $p$.
+But this means that:
+{{begin-eqn}}
+{{eqn | l = \left\vert{\left[{\phi \land \neg \left({\psi_1 \land \cdots \land \psi_m}\right)}\right]}\right\vert
+      | r = \left\vert{\left[{\left({\phi \land \neg \psi_1}\right) \lor \cdots \lor \left({\phi \land \neg \psi_m}\right)}\right]}\right\vert
+      | c = [[De Morgan's Laws (Logic)]]
+}}
+{{eqn | r = \left\vert{\left[{\phi \land \neg \psi_1}\right] \cup \cdots \cup \left[{\phi \land \neg \psi_m}\right]}\right\vert
+      | c = 
+}}
+{{eqn | o = >
+      | r = \kappa
+      | c = 
+}}
+{{end-eqn}}
+Thus, by [[Cardinality of Infinite Union of Infinite Sets]], at least one of the $\psi_i$ must satisfy $\left\vert{\left[{\phi \land \neg \psi_i}\right]}\right\vert > \kappa$.
+This is impossible, since $\psi_i \in p$.
+So, $\psi_1 \land \cdots \land \psi_m$ is in $p$.
+By definition of $p$ this means:
+: $\left\vert{\left[{\phi \land \psi_1 \land \cdots \land \psi_m}\right]}\right\vert > \kappa$
+Hence there are types containing $\psi_1 \land \cdots \land \psi_m$.
+So $\Delta$ is satisfiable.
+By the [[Compactness Theorem]], this means that $p \cup \operatorname{Th}_A \left({\mathcal M}\right)$ is satisfiable.
+Hence:
+: $p \in S_n^\mathcal M \left({A}\right)$
+We have that $p$ is a type.
+So we can write:
+:$\displaystyle \left[{\phi}\right] = \left\{{p}\right\} \cup \bigcup_{\psi \notin p} \left[{\phi \land \psi}\right]$
+since every type besides $p$ which contains $\phi$ must contain some $\psi \notin p$.
+Note the cardinalities involved in this union:
+Clearly, $\left\{{p}\right\}$ has cardinality $1 < \kappa$.
+By definition of $p$ each $\left[{\phi \land \psi}\right]$ for $\psi \notin p$ has cardinality at most $\kappa$.
+We have noted earlier in the main proof that there are only $\kappa$-many $\mathcal L_A$-formulas.
+Thus, by [[Cardinality of Infinite Union of Infinite Sets]], it is to be concluded that:
+:$\left\vert{\left[{\phi}\right]}\right\vert \le \kappa$
+However, this contradicts our supposition.
+The lemma follows by [[Proof by Contradiction]].
+{{qed|lemma}}
+Now the tree is to be built.
+This amounts to recursively defining formulas $\phi_\sigma$ for each finite sequence $\sigma$ over $\left\{{0, 1}\right\}$.
+First, the root of the tree $\phi_{()}$ is defined where the subscript is the empty sequence.
+The assumption is:
+:$\displaystyle \left\vert{\bigcup \left[{\phi}\right]}\right\vert = \left\vert{S_n^\mathcal M \left({A}\right)}\right\vert > \kappa$
+where the union is taken over all  $\mathcal L_A$-formulas $\phi$
+But there are only $\kappa$ many such formulas.
+Thus by [[Cardinality of Infinite Union of Infinite Sets]] there must be some $\mathcal L_A$-formula $\phi_{()}$ such  that the cardinality of $\left[{\phi_{()} }\right]$ is strictly larger than $\kappa$.
+Suppose $\phi_\sigma$ has been defined and that:
+:$\left\vert{\left[{\phi_\sigma}\right]}\right\vert > \kappa$
+Let $\sigma = \left({\sigma_0, \dots, \sigma_k}\right)$.
+By the lemma above, we can choose an $\mathcal L_A$ formula $\psi$ such that both:
+:$\left\vert{\left[{\phi \land \psi}\right]}\right\vert > \kappa$
+and:
+:$\left\vert{\left[{\phi \land \neg \psi}\right]}\right\vert > \kappa$
+Define $\phi_{\left({\sigma_0, \dots, \sigma_k, 0}\right)}$ to be $\phi_\sigma \land \psi$.
+Define $\phi_{\left({\sigma_0, \dots, \sigma_k, 1}\right)}$ to be $\phi_\sigma \land \neg \psi$.
+Now, let $B$ be the set of elements of $A$ which occur as constant symbols in any of the $\phi_\sigma$.
+Since only countably many $\phi_\sigma$ have been defined, $B$ is countable.
+We will define an [[Definition:Injection|injection]] from the set of infinite sequences over $\left\{{0, 1}\right\}$ to $S_n^\mathcal M \left({B}\right)$ using our tree.
+This will demonstrate that our theory $T$ is not $\omega$-stable.
+From [[Type Space is Compact]], $S_n^\mathcal M \left({A}\right)$ is compact (when viewed as a [[Definition:Type Space|type space]]).
+Thus it satisfies the [[Definition:Finite Intersection Axiom|finite intersection axiom]] by [[Equivalent Definitions of Compactness]].
+We have that each $\left[{\phi_\sigma}\right]$ is closed, essentially by definition of the type space topology.
+Also, any finite intersection $\left[{\phi_{()} }\right] \cap \left[{\phi_{(\sigma_0)} }\right] \cap \cdots \cap \left[{\phi_{\left({\sigma_0, \dots, \sigma_k}\right)} }\right]$ is equal to $\left[{\phi_{\left({\sigma_0, \dots, \sigma_k}\right)} }\right]$ by construction.
+Hence it is nonempty (by its cardinality)
+Thus, by the [[Definition:Finite Intersection Axiom|finite intersection axiom]], for each infinite sequence $\Sigma = \left({\Sigma_0, \Sigma_1, \Sigma_2, \ldots}\right)$ over $\left\{{0, 1}\right\}$, the intersection $\displaystyle \bigcap_{k \mathop \in \N} \left[{\phi_{\left({\Sigma_0, \Sigma_1, \ldots, \Sigma_k}\right)} }\right]$ is nonempty.
+Moreover, let $\Sigma = \left({\Sigma_0, \Sigma_1, \Sigma_2, \ldots}\right)$ and $\Sigma' = \left({\Sigma'_0, \Sigma'_1, \Sigma'_2, \ldots}\right)$ be two distinct infinite sequences over $\left\{{0, 1}\right\}$.
+Then there is some $k$ for which $\Sigma_i = \Sigma'_i$ for $i \le k$ and $\Sigma_{k+1} \ne \Sigma_{k+1}$.
+But $\phi_{\left({\Sigma_1, \ldots, \Sigma_k, 0}\right)}$ and $\phi_{\left({\Sigma_1, \ldots, \Sigma_k, 1}\right)}$ were defined to imply $\psi$ and $\neg \psi$ respectively for some $\psi$.
+So no type can satisfy both of them simultaneously.
+Thus $\displaystyle \bigcap_{k \mathop \in \N} \left[{\phi_{\left({\Sigma_0, \Sigma_1, \ldots, \Sigma_k}\right)} }\right]$ and $\displaystyle \bigcap_{k \mathop \in \N} \left[{\phi_{\left({\Sigma'_0, \Sigma'_1, \ldots, \Sigma'_k}\right)} }\right]$ cannot both contain the same type.
+Thus, we can define our injection by sending each infinite sequence $\Sigma$ over $\left\{{0, 1}\right\}$ to a type chosen from $\displaystyle \bigcap_{k \mathop \in \N} \left[{\phi_{\left({\Sigma_0, \Sigma_1, \ldots, \Sigma_k}\right)} }\right]$.
+The existence of this injection implies that the cardinality of $S_n^\mathcal M \left({B}\right)$ is at least $2^{\aleph_0}$, as this is the cardinality of the set of infinite sequences over $\left\{{0, 1}\right\}$.
+Hence, $T$ is not $\omega$-stable.
+The theorem now follows by the [[Rule of Transposition]].
+{{qed}}
+[[Category:Model Theory]]
+1peli88zcxfhadt88up612lnahav6x4
+\end{proof}<|endoftext|>
+\section{Finite Complement Topology is Topology}
+Tags: Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Space|finite complement space]].
+Then $\tau$ is a [[Definition:Topology|topology]] on $T$.
+\end{theorem}
+
+\begin{proof}
+By definition, we have that $\O \in \tau$.
+We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially [[Definition:Finite Set|finite]].
+{{qed|lemma}}
+Let $A, B \in \tau$.
+Let $H = A \cap B$.
+Then:
+{{begin-eqn}}
+{{eqn | l = H
+      | r = A \cap B
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp S H
+      | r = \relcomp S {A \cap B}
+      | c =
+}}
+{{eqn | r = \relcomp S A \cup \relcomp S B
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection|De Morgan's laws: Complement of Intersection]]
+}}
+{{end-eqn}}
+But as $A, B \in \tau$ it follows that $\relcomp S A$ and $\relcomp S B$ are both [[Definition:Finite Set|finite]].
+Hence their [[Definition:Set Union|union]] is also [[Definition:Finite Set|finite]].
+Thus $\relcomp S H$ is [[Definition:Finite Set|finite]].
+So $H = A \cap B \in \tau$ as its [[Definition:Relative Complement|complement]] is [[Definition:Finite Set|finite]].
+{{qed|lemma}}
+Let $\UU \subseteq \tau$.
+From [[De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Union|De Morgan's laws: Complement of Union]]:
+:$\displaystyle \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
+But as:
+:$\forall U \in \UU: \relcomp S U \in \tau$
+each of the $\relcomp S U$ is [[Definition:Finite Set|finite]].
+Hence so is their [[Definition:Set Intersection|intersection]].
+So $\displaystyle \relcomp S {\bigcup \UU}$ is [[Definition:Finite Set|finite]].
+So by definition:
+:$\displaystyle \bigcup \UU \in \tau$
+So $\tau$ is a [[Definition:Topology|topology]] on $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Points of Infinite Subset of Finite Complement Space}
+Tags: Limit Points, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Space|finite complement space]].
+Let $H \subseteq S$ be an [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$.
+Then every point of $S$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$ be any [[Definition:Open Set (Topology)|open set]] of $T$.
+From [[Infinite Subset of Finite Complement Space Intersects Open Sets]], we have that $U \cap H$ is [[Definition:Infinite Set|infinite]] {{iff}} $H$ is [[Definition:Infinite Set|infinite]].
+Let $x \in S$.
+Then every [[Definition:Open Set (Topology)|open set]] $U$ in $T$ such that $x \in U$ also contains an [[Definition:Infinite Set|infinite number]] of points of $H$ other than $x$.
+Thus, by definition, $x$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Complement Topology is Separable}
+Tags: Separable Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite]] set $S$.
+Then $T$ is a [[Definition:Separable Space|separable space]].
+\end{theorem}
+
+\begin{proof}
+Let $H$ be a [[Definition:Countably Infinite Set|countably infinite subset]] of $S$.
+From [[Closure of Infinite Subset of Finite Complement Space]], the [[Definition:Closure (Topology)|closure]] of $H$ is $S$.
+So by definition $H$ is [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+Hence the result by definition of [[Definition:Separable Space|separable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Infinite Subset of Finite Complement Space}
+Tags: Set Closures, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Space|finite complement space]].
+Let $H \subseteq S$ be an [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$.
+Then $H^- = S$ where $H^-$ is the [[Definition:Closure (Topology)|closure]] of $S$.
+\end{theorem}
+
+\begin{proof}
+Let $H$ be an [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$.
+From [[Limit Points of Infinite Subset of Finite Complement Space]], every point of $S$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+Hence the result from the definition of [[Definition:Closure (Topology)|closure]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subspace of Finite Complement Topology is Compact}
+Tags: Compact Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite]] set $S$.
+Then every [[Definition:Topological Subspace|topological subspace]] of $T$, including $T$ itself, is a [[Definition:Compact Topological Space|compact space]].
+\end{theorem}
+
+\begin{proof}
+Let $T_H = \struct {H, \tau_H}$ be a [[Definition:Topological Subspace|subspace]] of $T$.
+Let $\CC$ be an [[Definition:Open Cover|open cover]] of $T_H$.
+Let $U \in \CC$ be any [[Definition:Set|set]] in $C$.
+$U$ [[Definition:Cover of Set|covers]] all but a [[Definition:Finite Set|finite number]] of points of $T_H$.
+So for each of those points we pick an [[Definition:Element|element]] of $\CC$ which [[Definition:Cover of Set|covers]] each of those points.
+Hence we have a [[Definition:Finite Subcover|finite subcover]] of $T_H$.
+So by definition $T_H$ is a [[Definition:Compact Topological Space|compact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Finite Complement Topology is not Perfectly T4}
+Tags: Perfectly T4 Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable]] set $S$.
+Then $T$ is not a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+Recall the definition of a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]]
+:Every [[Definition:Closed Set (Topology)|closed set]] in $T$ can be written as a [[Definition:Countable Intersection|countable intersection]] of [[Definition:Open Set (Topology)|open sets]] of $T$.
+Let $V$ be a [[Definition:Closed Set (Topology)|closed set]] in $T$.
+From [[Closed Set of Uncountable Finite Complement Topology is not G-Delta|Closed Set of Uncountable Finite Complement Topology is not $G_\delta$]]:
+:$V$ is not a [[Definition:G-Delta Set|$G_\delta$ set]].
+The result follows by definition of [[Definition:Perfectly T4 Space|perfectly $T_4$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kluyver's Formula for Ramanujan's Sum}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $q \in \N_{>0}$.
+Let $n \in \N$.
+Let $\map {c_q} n$ be [[Definition:Ramanujan Sum|Ramanujan's sum]].
+Let $\mu$ denote the [[Definition:Möbius Function|Möbius function]].
+Then:
+:$\displaystyle \map {c_q} n = \sum_{d \mathop \divides \gcd \set {q, n} } d \map \mu {\frac q d}$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+\end{theorem}
+
+\begin{proof}
+Let $\alpha \in \R$.
+Let $e: \R \to \C$ be the [[Definition:Mapping|mapping]] defined as:
+:$\map e \alpha := \map \exp {2 \pi i \alpha}$
+Let $\zeta_q$ be a [[Definition:Primitive Root of Unity|primitive $q$th root of unity]].
+Let:
+:$\displaystyle \map {\eta_q} n := \sum_{1 \mathop \le a \mathop \le q} \map e {\frac {a n} q}$
+By [[Complex Roots of Unity in Exponential Form]] this is the sum of all $q$th [[Definition:Complex Roots of Unity|roots of unity]].
+Therefore:
+:$\displaystyle \map {\eta_q} n = \sum_{d \mathop \divides q} \map {c_d} n$
+By the [[Möbius Inversion Formula]], this gives:
+:$\displaystyle \map {c_q} n = \sum_{d \mathop \divides q} \map {\eta_d} n \map \mu {\frac q d}$
+Now by [[Sum of Roots of Unity]], we have:
+:$\displaystyle \map {c_q} n = \sum_{d \mathop \divides q} d \map \mu {\frac q d}$
+as required.
+{{questionable|This demonstrates only the sum over $d \divides q$, not over $d \divides \gcd \set {q, n}$.}}
+{{qed}}
+{{Namedfor|Jan Cornelis Kluyver|cat = Kluyver}}
+[[Category:Analytic Number Theory]]
+kcc0luy1raznt4yqkvk73ebjz1alvnx
+\end{proof}<|endoftext|>
+\section{Condition for Complex Root of Unity to be Primitive}
+Tags: Complex Roots of Unity
+
+\begin{theorem}
+Let $n, k \in \N$.
+Then $\alpha_k = \map \exp {\dfrac {2 \pi i k} n}$ is a [[Definition:Primitive Complex Root of Unity|primitive $n$th root of unity]] {{iff}} $\gcd \set {n, k} = 1$.
+\end{theorem}
+
+\begin{proof}
+Let $U_n = \set {\map \exp {\dfrac {2 \pi i k} n}: 0 \le k \le n - 1}$.
+Let $V = \set {1, \dotsc, {\alpha_k}^{n - 1} }$.
+By [[Complex Roots of Unity in Exponential Form]] it is sufficient to show that $U_n = V$ {{iff}} $\gcd \set {n, k} = 1$.
+Let $\gcd \set {n, k} = d > 1$.
+Then there are $n', k' \in \N$ such that:
+:$n' = d n$
+and:
+:$k' = d k$
+Then we have:
+:$\alpha_k = \map \exp {\dfrac {2 \pi i k'} {n'} }$
+and:
+:$\alpha_k^{n'} = \map \exp {2 \pi i k'} = 1$
+Therefore:
+:$V = \set {1, \dotsc, \alpha^{n' - 1} }$
+such that $n' < n$.
+So:
+:$\cmod V = n' < n = \cmod {U_n}$
+and $U_n \ne V$.
+Let $\gcd \set {n, k} = 1$.
+Let:
+:$\map \exp {\dfrac {2 \pi i k} n}^d = \map \exp {\dfrac {2 \pi i k} n} = 1$
+Then it must be the case that $\dfrac {k d} n \in \Z$.
+Since $\gcd \set {n, k}\ = 1$ it follows that:
+:$n \divides d$
+and so:
+:$d \ge n$
+Therefore $\set {1, \dotsc, \alpha^{n - 1} }$ are [[Definition:Distinct|distinct]]
+Hence $\card V = \card {U_n}$.
+Moreover each element of $V$ can be written in the form:
+:$\map \exp {\dfrac {2 \pi i k} n}$
+with $0 \le k \le n - 1$.
+It follows that $V = U_n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation of Boubaker Polynomials to Chebyshev Polynomials}
+Tags: Boubaker Polynomials
+
+\begin{theorem}
+The [[Definition:Boubaker Polynomials|Boubaker polynomials]] are related to [[Definition:Chebyshev Polynomials|Chebyshev polynomials]] through the equations:
+: $B_n \left({2x}\right) = \dfrac {4x}n \dfrac{\mathrm d}{\mathrm d x} T_n \left({x}\right) - 2 T_n \left({x}\right)$
+: $B_n \left({2x}\right) = -2 T_n \left({x}\right) + 4 x U_{n-1} \left({x}\right)$
+where:
+: $T_n$ denotes the [[Definition:Chebyshev Polynomials of the First Kind|Chebyshev polynomials of the first kind]]
+: $U_n$ denotes the [[Definition:Chebyshev Polynomials of the Second Kind|Chebyshev polynomials of the second kind]].
+\end{theorem}
+
+\begin{proof}
+('''Using Riordan Matrix''')
+The ordinary generating function of the Boubaker Polynomials can be expressed in terms of [[Definition:Riordan Matrix|Riordan matrices]]:
+: $\displaystyle \sum_{n \geqslant 0} B_n \left({t}\right)x^n = \frac {1+3x^2 }{1-xt+x^2}= \left (1+3x^2 \biggl\lvert 1+x^2 \right )\left({\frac 1 {1-xt}}\right)$       
+    
+By considering the [[Definition:Riordan Matrix|Riordan matrices]] of Chebyshev Polynomials of the first and second kind $T_n$ and $U_n$, respectively:
+:$\displaystyle T \left({\frac 1 4 - \frac {x^2} 4 \biggl \lvert \frac 1 2 + \frac 1 2 x^2}\right)$
+and:
+:$\displaystyle T \left({\frac 1 2 \biggl\lvert \frac 1 2 + \frac 1 2 x^2}\right)$
+and by setting:
+:$\displaystyle T \left({1+3x^2 \biggl \lvert 1+x^2}\right) = T \left({1+3x^2 \biggl\lvert 1}\right) T \left({\frac 1 2 \biggl\lvert \frac 1 2 + \frac 1 2 x^2}\right)$
+we obtain finally:
+:$\displaystyle B_n \left({x}\right) = U_n \left({\frac x 2}\right) + 3U_{n-2} \left({\frac x 2}\right)$
+which gives, due the relation between Chebyshev Polynomials of the first and second kind $T_n$ and $U_n$ and their derivatives:
+ 
+: $B_n \left({2x}\right) = -2 T_n \left({x}\right) + 4 x U_{n-1} \left({x}\right)$
+: $B_n \left({2x}\right) = \dfrac {4x}n \dfrac{\mathrm d}{\mathrm d x} T_n \left({x}\right) - 2 T_n \left({x}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation of Boubaker Polynomials to Dickson Polynomials}
+Tags: Boubaker Polynomials
+
+\begin{theorem}
+The [[Definition:Boubaker Polynomials|Boubaker polynomials]] $B_n$ are linked to the  [[Definition:Dickson Polynomials|Dickson polynomials]] by the relations:
+: $B_{n+1} \left({x}\right) B_{n+j} \left({x}\right) - B_{n+j+1} \left({x}\right) B_n \left({x}\right) = \left({3 x^2 + 4}\right) D_{n+1} \left({x, \dfrac 1 4}\right)$
+: $B_n \left({x}\right) = D_n \left({2x, \dfrac 1 4}\right) + 4 D_{n-1} \left({2 x, \dfrac 1 4}\right)$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|The material on this page was originally raised by [[User:Yohji kinomoto]].}}
+\end{proof}<|endoftext|>
+\section{Relation of Boubaker Polynomials to Fermat Polynomials}
+Tags: Boubaker Polynomials
+
+\begin{theorem}
+The [[Definition:Boubaker Polynomials|Boubaker polynomials]] are related to [[Definition:Fermat Polynomials|Fermat polynomials]] by:
+: $B_n \left({x}\right) = \dfrac 1 {\left({\sqrt 2}\right)^n} F_n \left({\dfrac {2 \sqrt 2 x} 3}\right) + \dfrac 3 {\left({\sqrt 2}\right)^{n-2}} F_{n-2} \left({\dfrac {2 \sqrt 2 x} 3}\right): \quad n = 0, 1, 2, \ldots$
+\end{theorem}
+
+\begin{proof}
+('''Using Riordan Matrix''')
+Since the ordinary generating function of the Boubaker Polynomials can be expressed in terms of [[Definition:Riordan Matrix|Riordan matrices]]:
+: $\displaystyle \sum_{n \geqslant 0}^{} B_n \left({t}\right)x^n= \frac {1+3x^2 }{1-xt+x^2}= \left (1+3x^2 \biggl\lvert 1+x^2 \right )\left(\frac {1}{1-xt} \right )$     
+then, by writing differently:
+:$\displaystyle T \left(1+3x^2 \biggl\lvert 1+x^2\right )= T \left(1+3x^2 \biggl\lvert 1\right ) T \left( \frac {1}{2}\biggl\lvert \frac {1}{2}+\frac {1}{2}x^2\right ) T \left( 2\biggl\lvert 2 \right )$
+and:
+:$\displaystyle T \left(1+3x^2 \biggl\lvert 1+x^2\right )= T \left(1+3x^2 \biggl\lvert 1\right ) T \left( 1\biggl\lvert \sqrt{2}
+ \right ) T \left( \frac {1}{3}\biggl\lvert \frac {1}{3}+\frac {2}{3}x^2\right ) T \left( 3\biggl\lvert \frac {3}{\sqrt{2}} \right )$
+and taking into account the Riordan matrix of Fermat polynomials:
+:$\displaystyle T \left( \frac {1}{3}\biggl\lvert \frac {1}{3}+\frac {1}{3}x^2\right )$
+we obtain finally:
+: $B_n \left({x}\right) = \dfrac 1 {\left({\sqrt 2}\right)^n} F_n \left({\dfrac {2 \sqrt 2 x} 3}\right) + \dfrac 3 {\left({\sqrt 2}\right)^{n-2}} F_{n-2} \left({\dfrac {2 \sqrt 2 x} 3}\right): \quad n = 0, 1, 2, \ldots$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{F-Sigma and G-Delta Subsets of Uncountable Finite Complement Space}
+Tags: F-Sigma Sets, G-Delta Sets, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable]] set $S$.
+Then [[Definition:Countably Infinite Set|countably infinite]] subsets of $S$ are [[Definition:F-Sigma Set|$F_\sigma$ sets]] and are neither [[Definition:Open Set (Topology)|open]] nor [[Definition:Closed Set (Topology)|closed sets]].
+Their [[Definition:Relative Complement|relative complements in $S$]] are [[Definition:G-Delta Set|$G_\delta$ sets]], and are also neither [[Definition:Open Set (Topology)|open]] nor [[Definition:Closed Set (Topology)|closed sets]].
+\end{theorem}
+
+\begin{proof}
+Let $U$ be a [[Definition:Countably Infinite Set|countably infinite]] subset of $S$.
+Then it can be written as:
+:$\displaystyle U = \bigcup_{x \mathop \in U} \set x$
+But as $\set x$ is a [[Definition:Finite Set|finite]] subset of $S$, it is by definition the [[Definition:Relative Complement|complement]] of an [[Definition:Open Set (Topology)|open set]] in $T$.
+So by definition $\set x$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+So:
+:$\displaystyle U = \bigcup_{x \mathop \in U} \left\{{x}\right\}$
+is a [[Definition:Countable Union|countable union]] of [[Definition:Closed Set (Topology)|closed sets]] of $T$.
+That is, by definition, $U$ is an [[Definition:F-Sigma Set|$F_\sigma$ set]].
+As $U$ is [[Definition:Countably Infinite Set|countably infinite]], its [[Definition:Relative Complement|complement]] is [[Definition:Uncountable Set|uncountable]], not [[Definition:Finite Set|finite]], and so $U$ is not [[Definition:Open Set (Topology)|open]] in $T$.
+But as $U$ is also not [[Definition:Finite Set|finite]], it is not the [[Definition:Relative Complement|complement]] of an [[Definition:Open Set (Topology)|open set]] and so is not [[Definition:Closed Set (Topology)|closed]] in $T$ either.
+From [[Complement of F-Sigma Set is G-Delta Set|Complement of $F_\sigma$ Set is $G_\delta$ Set]] it follows that the [[Definition:Relative Complement|complement]] of $U$ is a [[Definition:G-Delta Set|$G_\delta$ set]], and similarly neither [[Definition:Open Set (Topology)|open]] nor [[Definition:Closed Set (Topology)|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Skolem's Paradox}
+Tags: Mathematical Logic, Set Theory
+
+\begin{theorem}
+Let $\mathcal L$ be a [[Definition:Countable|countable]] [[Definition:First-Order (Logic)|first-order]] [[Definition:Logical Language|language]].
+Let $T$ be an $\mathcal L$-[[Definition:Theory (Logic)|theory]] which [[Definition:Axiomatize (Logic)|axiomatizes]] some version of [[Definition:Axiomatic Set Theory|set theory]] (for example, [[Definition:ZFC|ZFC]]).
+There is a countable model of $T$.
+\end{theorem}
+
+\begin{proof}
+This is a straightforward application of the downward [[Löwenheim-Skolem Theorem]].
+{{qed}}
+{{refactor|From here on down, decide where it belongs and how it is to be presented. At the moment it is too much like an encyclopedia article to be compatible with {{ProofWiki}}'s dictionary style.}}
+\end{proof}<|endoftext|>
+\section{Uncountable Finite Complement Space is not First-Countable}
+Tags: First-Countable Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is not [[Definition:First-Countable Space|first-countable]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} some $x \in S$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]].
+That means:
+:there exists a [[Definition:Countable Set|countable]] [[Definition:Set of Sets|set]] $\BB_x \subseteq \tau$
+such that:
+:$\forall B \in \BB_x: x \in B$
+and such that:
+:every [[Definition:Open Neighborhood|open neighborhood]] of $x$ contains some $B \in \BB_x$.
+Let $y \in S$ with $y \ne x$.
+Because its [[Definition:Relative Complement|complement relative to $S$]] is [[Definition:Finite Set|finite]], we have that $S \setminus \set y$ is an [[Definition:Open Set (Topology)|open set]] containing $x$.
+Because  $\BB_x$ is a [[Definition:Neighborhood Basis|neighborhood basis]] at $x$, there exists some $B \in \BB_x$ such that:
+:$B \subseteq S \setminus \set y$
+whence $y \notin B$.
+So:
+:$y \notin \bigcap \BB_x$
+So:
+{{begin-eqn}}
+{{eqn | l = \bigcap \BB_x
+      | r = \set x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = S \setminus \set x
+      | r = S \setminus \bigcap \BB_x
+      | c = 
+}}
+{{eqn | r = \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{end-eqn}}
+By definition, each of $S \setminus B$ is [[Definition:Finite Set|finite]].
+From [[Countable Union of Countable Sets is Countable]] it follows that $\displaystyle \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}$ is [[Definition:Countable Set|countable]].
+So $S \setminus \set x$ and therefore $S$ is also [[Definition:Countable Set|countable]].
+From this [[Definition:Contradiction|contradiction]] (as we have specified that $S$ is [[Definition:Uncountable Set|uncountable]]) it follows that our assumption that $x \in S$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] must be false.
+Hence by definition $T$ can not be [[Definition:First-Countable Space|first-countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Separable Space need not be First-Countable}
+Tags: First-Countable Spaces, Separable Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Separable Space|separable]].
+Then $T$ does not necessarily have to be [[Definition:First-Countable Space|first-countable]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Uncountable Finite Complement Topology|finite complement topology]] on an [[Definition:Uncountable Set|uncountable]] set $S$.
+We have that a [[Finite Complement Topology is Separable]].
+But we also have that an [[Uncountable Finite Complement Space is not First-Countable]].
+Hence the result, by [[Proof by Counterexample]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ramanujan Sum is Multiplicative}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $q \in \N_{>0}$, $n \in \N$.
+Let $\map {c_q} n$ be the [[Definition:Ramanujan Sum|Ramanujan sum]].
+Then $\map {c_q} n$ is [[Definition:Multiplicative Arithmetic Function|multiplicative]] in $q$.
+\end{theorem}
+
+\begin{proof}
+Let $q, r \in \N$ such that:
+:$\gcd \set {q, r} = 1$
+where $\gcd \set {q, r}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $q$ and $r$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {c_q} n \, \map {c_r} n
+      | r = \sum_{d_1 \mathop \divides \gcd \set {n, q} } d_1 \, \map \mu {\frac q {d_1} } \sum_{d_2 \mathop \divides \gcd \set {n, r} } d_2 \, \map \mu {\frac r {d_2} }
+      | c = [[Kluyver's Formula for Ramanujan's Sum]]
+}}
+{{eqn | r = \sum_{\substack {d_1 \mathop \divides \gcd \set {n, q} \\ d_2 \mathop \divides \gcd \set {n, r} } } d_1 d_2 \, \map \mu {\frac q {d_1} } \, \map \mu {\frac r {d_2} }
+      | c = 
+}}
+{{eqn | r = \sum_{d \mathop \divides \gcd \set {n, q} \gcd \set {n, r} } d \, \map \mu {\frac {q r} d}
+      | c = [[Möbius Function is Multiplicative]], and setting $d = d_1 d_2$
+}}
+{{eqn | r = \sum_{d \mathop \divides \gcd \set {n, q r} } d \, \map \mu {\frac {q r} d}
+      | c = [[GCD with One Fixed Argument is Multiplicative Function]]
+}}
+{{eqn | r = \map {c_{q r} } n
+      | c = [[Kluyver's Formula for Ramanujan's Sum]]
+}}
+{{end-eqn}}
+This completes the proof.
+{{qed}}
+[[Category:Analytic Number Theory]]
+92a5u5ragspb52bp1qa1wgrtfjbbata
+\end{proof}<|endoftext|>
+\section{Asymptotic Growth of Euler Phi Function}
+Tags: Analytic Number Theory, Euler Phi Function
+
+\begin{theorem}
+Let $\phi$ be the [[Definition:Euler Phi Function|Euler $\phi$ function]].
+For any $\epsilon > 0$ and [[Definition:Sufficiently Large|sufficiently large]] $n$:
+:$n^{1 - \epsilon} < \phi \left({n}\right) < n$
+\end{theorem}
+
+\begin{proof}
+It is clear that $\phi \left({n}\right) < n$ for all $n$, so it is sufficient to prove that:
+:$\displaystyle \lim_{n \to \infty} \frac {n^{1 - \epsilon}}{\phi \left({n}\right)} = 0$
+By [[Multiplicative Function that Converges to Zero on Prime Powers]] it is sufficient to prove that:
+:$\displaystyle \lim_{p^k \to \infty} \frac{p^{k(1-\epsilon)}}{\phi(p^k)} = 0$
+as $p^k$ ranges through all prime powers.
+By [[Euler Phi Function of Prime Power]] we have:
+: $\phi \left({p^k}\right) = p^k - p^{k-1}$
+for a prime power $p^k$.
+Therefore:
+{{begin-eqn}}
+{{eqn|l=\frac{p^{k \left({1 - \epsilon}\right)} } {\phi \left({p^k}\right)}
+     |r=\frac{p^{k \left({1 - \epsilon}\right)} } {p^k - p^{k-1} }
+     |c=
+}}
+{{eqn|l=
+     |r=\frac p {p-1} \frac{p^{k \left({1 - \epsilon}\right)} }{p^k}
+     |c=
+}}
+{{eqn|l=
+     |r=\frac p {p-1} \frac 1 {p^{k \epsilon} }
+     |c=
+}}
+{{eqn|l=
+     |o=\le
+     |r=\frac 2 {p^{k \epsilon} }
+     |c=since $p / \left({p - 1}\right) \le 2$ for all primes $p$
+}}
+{{end-eqn}}
+Therefore:
+:$\displaystyle \lim_{p^k \to \infty} \frac{p^{k \left({1 - \epsilon}\right)}} {\phi \left({p^k}\right)} \le \lim_{p^k \to \infty} \frac 2 {p^{k \epsilon} } = 0$
+{{qed}}
+[[Category:Analytic Number Theory]]
+[[Category:Euler Phi Function]]
+f8cck1beowzt7p4l1eapw1am9ggntt1
+\end{proof}<|endoftext|>
+\section{Finite Complement Space is T1}
+Tags: T1 Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on any [[Definition:Set|set]] $S$ which contains at least two points.
+Then $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$.
+Then:
+:$S \setminus \set x$ is an [[Definition:Open Set (Topology)|open set]] of $T$ containing $y$ but not $x$.
+:$S \setminus \set y$ is an [[Definition:Open Set (Topology)|open set]] of $T$ containing $x$ but not $y$.
+Hence the result from definition of [[Definition:Fréchet Space (Topology)|$T_1$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Complement Space is Irreducible}
+Tags: Irreducible Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite]] set $S$.
+Then $T$ is [[Definition:Irreducible Space|irreducible]].
+\end{theorem}
+
+\begin{proof}
+Let $U_1, U_2 \in \tau$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open sets]] of $T$.
+We have that $U_1$ and $U_2$ are both [[Definition:Infinite Set|infinite]] if $S$ is.
+From [[Infinite Subset of Finite Complement Space Intersects Open Sets]], they [[Definition:Set Intersection|intersect]] each other.
+Hence the result from definition of [[Definition:Irreducible Space|irreducible space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Subset of Finite Complement Space Intersects Open Sets}
+Tags: Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Let $H \subseteq S$ be an [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$.
+Then the [[Definition:Set Intersection|intersection]] of $H$ with any [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open set]] of $T$ is [[Definition:Infinite Set|infinite]].
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$ be any [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open set]] of $T$.
+Then $\relcomp S U$ is [[Definition:Finite Set|finite]].
+We have that:
+:$H = H \cap \paren {U \cup \relcomp S U} = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$
+{{AimForCont}} $H \cap U$ is [[Definition:Finite Set|finite]].
+Since $H \cap \relcomp S U \subseteq \relcomp S U$, $H \cap \relcomp S U$ is also [[Definition:Finite Set|finite]].
+$H = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$ is the [[Definition:Set Union|union]] of two [[Definition:Finite Set|finite]] sets, and so it is [[Definition:Finite Set|finite]].
+It is a [[Definition:Contradiction|contradiction]] that $H$ is [[Definition:Infinite Set|infinite]] and [[Definition:Finite Set|finite]] at the same time, so $H \cap U$ must be [[Definition:Infinite Set|infinite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Irreducible Hausdorff Space is Singleton}
+Tags: Irreducible Spaces, Hausdorff Spaces
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Irreducible Space|irreducible]] and [[Definition:Hausdorff Space|Hausdorff]].
+Then $T$ contains only one point.
+\end{theorem}
+
+\begin{proof}
+Suppose $T$ has at least two points $x, y \in S$.
+Because $T$ is [[Definition:Irreducible Space|irreducible]], there are no two [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] such that $x$ is in one and $y$ is in the other.
+This contradicts the fact that $T$ is [[Definition:Hausdorff Space|Hausdorff]].
+Thus $T$ has only one point.
+{{qed}}
+[[Category:Irreducible Spaces]]
+[[Category:Hausdorff Spaces]]
+77maej8b2acclyq5geemvmi4h4g2vae
+\end{proof}<|endoftext|>
+\section{Finite Complement Space is not T2}
+Tags: Hausdorff Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite]] set $S$.
+Then $T$ is not a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Finite Complement Space is Irreducible]]
+:[[Irreducible Hausdorff Space is Singleton]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Complement Space is not T3, T4 or T5}
+Tags: T3 Spaces, T4 Spaces, T5 Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite]] set $S$.
+Then $T$ is not a [[Definition:T3 Space|$T_3$ space]], [[Definition:T4 Space|$T_4$ space]] or [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Finite Complement Space is T1|Finite Complement Space is a $T_1$ space]].
+From [[T1 Space is T0 Space|$T_1$ Space is $T_0$ Space]], $T$ is a [[Definition:Kolmogorov Space|$T_0$ space]].
+We then have that a [[Finite Complement Space is not T2|Finite Complement Space is not $T_2$]].
+From [[Regular Space is T2 Space|Regular Space is $T_2$ Space]], $T$ is not a [[Definition:Regular Space|regular space]].
+By definition, a [[Definition:Regular Space|regular space]] is a [[Definition:Topological Space|space]] that is both a [[Definition:Kolmogorov Space|$T_0$ space]] and a [[Definition:T3 Space|$T_3$ space]].
+But $T$ is a [[Definition:Kolmogorov Space|$T_0$ space]] and not a [[Definition:Regular Space|regular space]].
+So it follows that $T$ cannot be a [[Definition:T3 Space|$T_3$ space]].
+Next we have that a [[Normal Space is T3 Space|Normal Space is a $T_3$ Space]].
+But as $T$ is not a [[Definition:T3 Space|$T_3$ space]], $T$ cannot be a [[Definition:Normal Space|normal space]].
+By definition, a [[Definition:Normal Space|normal space]] is a space that is both a [[Definition:Fréchet Space (Topology)|$T_1$ space]] and a [[Definition:T4 Space|$T_4$ space]].
+But $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ space]] and not a [[Definition:Normal Space|normal space]].
+So it follows that $T$ cannot be a [[Definition:T4 Space|$T_4$ space]].
+Finally we have that a [[T5 Space is T4 Space|$T_5$ Space is a $T_4$ Space]].
+But as $T$ is not a [[Definition:T4 Space|$T_4$ space]], $T$ cannot be a [[Definition:T5 Space|$T_5$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Double Pointed Finite Complement Topology is Compact}
+Tags: Compact Spaces, Double Pointed Topologies, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Let $T \times D$ be the [[Definition:Double Pointed Topology|double pointed topology]] on $T$.
+Then $T \times D$ is [[Definition:Compact Topological Space|compact]].
+\end{theorem}
+
+\begin{proof}
+The [[Subspace of Finite Complement Topology is Compact|Finite Complement Topology is Compact]].
+Also, a [[Finite Topological Space is Compact]].
+So $D$, defined as the [[Definition:Indiscrete Space|indiscrete space]] on a [[Definition:Doubleton|doubleton]], is also [[Definition:Compact Topological Space|compact]].
+The result follows directly from [[Tychonoff's Theorem]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Double Pointed Finite Complement Topology fulfils no Separation Axioms}
+Tags: Separation Axioms, Double Pointed Topologies, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Let $T \times D$ be the [[Definition:Double Pointed Topology|double pointed topology]] on $T$.
+Then $T \times D$ is not a [[Definition:Kolmogorov Space|$T_0$ space]], [[Definition:Fréchet Space (Topology)|$T_1$ space]], [[Definition:Hausdorff Space|$T_2$ space]], [[Definition:T3 Space|$T_3$ space]], [[Definition:T4 Space|$T_4$ space]] or [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+From [[Separation Axioms on Double Pointed Topology]], we have that:
+:$T \times D$ is not a [[Definition:Kolmogorov Space|$T_0$ space]].
+:$T \times D$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ space]].
+:$T \times D$ is not a [[Definition:Hausdorff Space|$T_2$ space]].
+Also from [[Separation Axioms on Double Pointed Topology]], we have that:
+:$T \times D$ is a [[Definition:T3 Space|$T_3$ space]] {{iff}} $T$ is a [[Definition:T3 Space|$T_3$ space]]
+:$T \times D$ is a [[Definition:T4 Space|$T_4$ space]] {{iff}} $T$ is a [[Definition:T4 Space|$T_4$ space]]
+:$T \times D$ is a [[Definition:T5 Space|$T_5$ space]] {{iff}} $T$ is a [[Definition:T5 Space|$T_5$ space]]
+But from [[Finite Complement Space is not T3, T4 or T5|Finite Complement Space is not $T_3$, $T_4$ or $T_5$]]:
+:$T$ is not a [[Definition:T3 Space|$T_3$ space]]
+:$T$ is not a [[Definition:T4 Space|$T_4$ space]]
+:$T$ is not a [[Definition:T5 Space|$T_5$ space]]
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vaughan's Identity}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $\Lambda$ be [[Definition:Von Mangoldt Function|von Mangoldt's function]].
+Let $\mu$ be the [[Definition:Möbius Function|Möbius function]].
+Then for $y, z \ge 1$ and $n > z$:
+:$\displaystyle \map \Lambda n = \sum_{\substack {d \mathop \divides n \\ d \mathop \le y}} \map \mu d \, \map \ln {\frac n d} - \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop \le y, \, c \mathop \le z}} \map \mu d \, \map \Lambda c + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop > z}} \map \mu d \, \map \Lambda c$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+\end{theorem}
+
+\begin{proof}
+By [[Sum Over Divisors of von Mangoldt is Logarithm]]:
+:$\displaystyle \ln n = \sum_{d \mathop \divides n} \map \Lambda d$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map \Lambda n
+      | r = \sum_{d \mathop \divides n} \map \mu d \, \map \ln {\frac n d}
+      | c = [[Möbius Inversion Formula]]
+}}
+{{eqn | r = \sum_{\substack {d \mathop \divides n \\ d \mathop \le y} } \map \mu d \, \map \ln {\frac n d} + \sum_{\substack {d \mathop \divides n \\ d \mathop > y} } \map \mu d \, \map \ln {\frac n d}
+      | c = 
+}}
+{{end-eqn}}
+Taking the second [[Definition:Summation|summation]] in that last line:
+{{begin-eqn}}
+{{eqn | l = \sum_{\substack {d \mathop \divides n \\ d \mathop > y} } \map \mu d \, \map \ln {\frac n d}
+      | r = \sum_{\substack {d \mathop \divides n \\ d \mathop > y} } \map \mu d \sum_{c \mathop \divides n / d} \map \Lambda c
+      | c = 
+}}
+{{eqn | r = \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y} } \map \mu d \, \map \Lambda c
+      | c = 
+}}
+{{eqn | r = \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop > z} } \map \mu d \, \map \Lambda c + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop \le z} } \map \mu d \, \map \Lambda c
+      | c = 
+}}
+{{end-eqn}}
+Again, taking the second [[Definition:Summation|summation]] in that last line:
+:$\displaystyle \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop \le z} } \map \mu d \, \map \Lambda c = \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \, \map \Lambda c - \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop \le y, \, c \mathop \le z} } \map \mu d \, \map \Lambda c$
+Putting this together:
+:$\displaystyle \map \Lambda n = \sum_{\substack {d \mathop \divides n \\ d \mathop \le y} } \map \mu d \, \map \ln {\frac n d} + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, c \mathop > z} } \map \mu d \, \map \Lambda c + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \, \map \Lambda c - \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop \le y, c \mathop \le z} } \map \mu d \, \map \Lambda c$
+It remains to be shown that:
+:$\displaystyle \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \, \map \Lambda c = 0$
+The [[Definition:Summation|summation]] is expressed as:
+:$\displaystyle \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \, \map \Lambda c = \sum_{\substack {c \mathop \le z \\ c \mathop \divides n} } \map \Lambda c \sum_{d \mathop \divides \frac n c} \map \mu d$
+Now we have $c \le z < n$, so:
+:$\dfrac n c > 1$
+Therefore, by [[Sum of Möbius Function over Divisors/Lemma|the lemma to Sum of Möbius Function over Divisors]]:
+:for each $c$ the inner sum vanishes.
+This shows that:
+:$\displaystyle \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \, \map \Lambda c = 0$
+as required.
+{{qed}}
+{{Namedfor|Robert Charles Vaughan|cat = Vaughan}}
+[[Category:Analytic Number Theory]]
+6ci23vilevkgb4exsd9dt9uypq89ciz
+\end{proof}<|endoftext|>
+\section{Sum Over Divisors of von Mangoldt is Logarithm}
+Tags: Analytic Number Theory
+
+\begin{theorem}
+Let $\Lambda$ be [[Definition:Von Mangoldt Function|von Mangoldt's function]].
+Then for $n \geq 1$
+:$\displaystyle \sum_{d | n} \Lambda(d) = \log n$
+\end{theorem}
+
+\begin{proof}
+Let $n \geq 1$, and by the [[Fundamental Theorem of Arithmetic]] write $n = p_1^{e_1}\cdots p_k^{e_k}$ with $p_1,\ldots,p_k$ distinct [[Definition:Prime Number|primes]] and $e_1,\ldots,e_k > 0$.
+Now $d | n$ if any only if $d = p_1^{f_1}\cdots p_k^{f_k}$ with $0 \leq f_i \leq e_i$ for $i = 1,\ldots, k$.
+By the definition of $\Lambda$, for such $d$ we have $\Lambda(d) \neq 0$ if and only if there is exactly one $i \in \{1,\ldots, k\}$ such that $f_i > 0$.
+If this is the case, say $d = p_i^{f_i}$ then $\Lambda(d) = \log p_i$. Therefore,
+:$\displaystyle \sum_{d | n}\Lambda(d) = \sum_{i = 1}^k e_i \log p_i$
+Also, we have
+{{begin-eqn}}
+{{eqn|l=\log n
+     |r=\log(p_1^{e_1} \cdots p_k^{e_k})
+     |c=
+}}
+{{eqn|l=
+     |r= \sum_{i = 1}^k e_i \log p_i
+     |c=By [[Sum of Logarithms]]
+}}
+{{end-eqn}}
+Thus we indeed have
+:$\displaystyle \displaystyle \sum_{d | n}\Lambda(d) = \log n$
+{{qed}}
+[[Category:Analytic Number Theory]]
+48twvhkcbcsk2ja85z9y4r3rv9l2yyz
+\end{proof}<|endoftext|>
+\section{Multiplicative Function that Converges to Zero on Prime Powers}
+Tags: Analytic Number Theory, Multiplicative Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Multiplicative Arithmetic Function|multiplicative function]] such that:
+:$\displaystyle \lim_{p^k \mathop \to \infty} \map f {p^k} = 0$
+where $p^k$ runs though ''all'' [[Definition:Prime Power|prime powers]].
+Then:
+:$\displaystyle \lim_{n \mathop \to \infty} \map f n = 0$
+where $n$ runs through the [[Definition:Integer|integers]].
+\end{theorem}
+
+\begin{proof}
+By hypothesis, there exist only [[Definition:Finite Set|finitely many]] [[Definition:Prime Power|prime powers]] $p^k$ such that $\size {\map f {p^k} } > 1$.
+Let $\displaystyle A = \prod_{\size {\map f {p^k} } \mathop > 1} \size {\map f {p^k} }$.
+Thus $A \ge 1$.
+Let $0 < \dfrac \epsilon A$. 
+There exist only [[Definition:Finite Set|finitely many]] [[Definition:Prime Power|prime powers]] $p^k$ such that $\size {\map f {p^k} } > \dfrac \epsilon A$.
+Therefore there are only [[Definition:Finite Set|finitely many]] [[Definition:Integer|integers]] $n$ such that:
+:$\size {\map f {p^k} } > \dfrac \epsilon A$
+for every [[Definition:Prime Power|prime power]] $p^k$ that [[Definition:Divisor of Integer|divides]] $n$.
+Therefore if $n$ is [[Definition:Sufficiently Large|sufficiently large]] there exists a [[Definition:Prime Power|prime power]] $p^k$ that [[Definition:Divisor of Integer|divides]] $n$ such that:
+:$\size {\map f {p^k} } < \dfrac \epsilon A$
+Therefore $n$ can be written as:
+:$\displaystyle n = \prod_{i \mathop = 1}^ r p_i^{k_i} \prod_{i \mathop = r + 1}^{r + s} p_i^{k_i} \prod_{i \mathop = r + s + 1}^{r + s + t} p_i^{k_i}$
+where $t \ge 1$ and:
+{{begin-eqn}}
+{{eqn | l = 1
+      | o = \le
+      | m = \size {\map f {p_i^{k_i} } }
+      | mo= 
+      | r = 
+      | c = for $i = 1, \ldots, r$
+}}
+{{eqn | l = \frac \epsilon A
+      | o = \le
+      | m = \size {\map f {p_i^{k_i} } }
+      | mo= <
+      | r = 1
+      | c = for $i = r + 1, \ldots, r + s$
+}}
+{{eqn | o = 
+      | m = \size {\map f {p_i^{k_i} } }
+      | mo= <
+      | r = \frac \epsilon A
+      | c = for $i = r + s + 1, \ldots, r + s + t$
+}}
+{{end-eqn}}
+Therefore:
+{{begin-eqn}}
+{{eqn | l = \size {\map f n}
+      | r = \prod_{i \mathop = 1}^r \map f {p_i^{k_i} } \prod_{i \mathop = r + 1}^{r + s} \map f {p_i^{k_i} } \prod_{i \mathop = r + s + 1}^{r + s + t} \map f {p_i^{k_i} }
+      | c = because $f$ is [[Definition:Multiplicative Arithmetic Function|multiplicative]]
+}}
+{{eqn | o = <
+      | r = A \paren {\frac \epsilon A}^t
+      | c = 
+}}
+{{eqn | o = <
+      | r = \epsilon
+      | c = because $t \ge 1$
+}}
+{{end-eqn}}
+This shows that $\map f n$ can be made arbitrarily small for [[Definition:Sufficiently Large|sufficiently large]] $n$.
+{{qed}}
+[[Category:Analytic Number Theory]]
+[[Category:Multiplicative Functions]]
+sl3n43xpxm69xs440y1uacxtzcykkja
+\end{proof}<|endoftext|>
+\section{Finite T1 Space is Discrete}
+Tags: Finite Topological Spaces, T1 Spaces, Discrete Topology, Finite T1 Space is Discrete
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+Then $\tau$ is the [[Definition:Discrete Topology|discrete topology]] on $S$.
+\end{theorem}
+
+\begin{proof}
+We have from [[Finite Complement Topology is Minimal T1 Topology|Finite Complement Topology is Minimal $T_1$ Topology]] that $\tau$ space is an [[Definition:Expansion of Topology|expansion]] of a [[Definition:Finite Complement Space|finite complement space]].
+But from the definition, any such [[Definition:Finite Complement Space|finite complement space]] on a [[Definition:Finite Set|finite set]] is [[Definition:Discrete Space|discrete]].
+But as the [[Discrete Topology is Finest Topology]], $\tau$ must itself be the [[Definition:Discrete Topology|discrete topology]] on $S$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Fréchet Space (Topology)|$T_1$ space]] on a [[Definition:Finite Set|finite set]] $S$.
+Let $U \subseteq S$ be any [[Definition:Subset|subset]] of $S$. 
+Let $H = \complement_S \left({U}\right)$ be the [[Definition:Relative Complement|complement of $U$ relative to $S$]].
+Then by [[Relative Complement of Relative Complement]] we have that $U = \complement_S \left({H}\right)$.
+We can write $H$ as:
+:$\displaystyle H = \bigcup_{x \mathop \in H} \left\{{x}\right\}$
+From [[Equivalence of Definitions of T1 Space|Equivalence of Definitions of $T_1$ Space]], $\forall x \in H: \left\{{x}\right\}$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+As $S$ is a [[Definition:Finite Set|finite set]], it follows that $H$ is a [[Definition:Finite Union|finite union]] of [[Definition:Closed Set (Topology)|closed sets]] of $T$.
+By [[Topology Defined by Closed Sets]], $H$ is therefore [[Definition:Closed Set (Topology)|closed]] in $T$.
+By definition of [[Definition:Closed Set (Topology)|closed set]], $U = \complement_S \left({H}\right)$ is [[Definition:Open Set (Topology)|open]] in $T$.
+As $U$ is arbitrary, it follows that:
+:$\forall U \subseteq S: S \in \tau$
+by definition of a [[Definition:Topology|topology]].
+Hence the result, by definition of the [[Definition:Discrete Topology|discrete topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Complement Topology is Minimal T1 Topology}
+Tags: T1 Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Space|finite complement space]].
+Let $\tau'$ be a [[Definition:Topology|topology]] on $S$ such that $T' = \struct {S, \tau'}$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+Then $\tau$ is [[Definition:Comparable Topologies|comparable]] with $\tau'$ such that $\tau$ is [[Definition:Coarser Topology|coarser]] than $\tau'$.
+That is, of all the [[Definition:Topology|topologies]] on $S$ fulfilling the [[Definition:Fréchet Space (Topology)|$T_1$ separation axiom]], the [[Definition:Finite Complement Space|finite complement space]] is the [[Definition:Minimal Set|smallest]].
+Thus the [[Definition:Finite Complement Topology|finite complement topology]] is known as the '''minimal $T_1$ topology''' on any given set.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be the [[Definition:Finite Complement Space|finite complement space]] on $S$.
+Let $U \in \tau$ be any [[Definition:Open Set (Topology)|open set]] of $T$.
+Let $H = \relcomp S U$ be the [[Definition:Relative Complement|complement of $U$ relative to $S$]].
+By definition of [[Definition:Finite Complement Topology|finite complement topology]], $H \subseteq S$ is a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $S$.
+Let $T' = \struct {S, \tau'}$ be any arbitrary [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]] on $S$.
+From [[Equivalence of Definitions of T1 Space|Equivalence of Definitions of $T_1$ Space]]:
+:$\forall x \in S: \set x$ is [[Definition:Closed Set (Topology)|closed]] in $T'$.
+We can write $H$ as:
+:$\displaystyle H = \bigcup_{x \mathop \in H} \set x$
+As $H$ is [[Definition:Finite Set|finite]], it follows that $H$ is a [[Definition:Finite Union|finite union]] of [[Definition:Closed Set (Topology)|closed sets]] of $T'$.
+By [[Topology Defined by Closed Sets]], $H$ is therefore [[Definition:Closed Set (Topology)|closed]] in $T'$.
+By [[Relative Complement of Relative Complement]] we have that $U = \relcomp S H$.
+So, by definition of [[Definition:Closed Set (Topology)|closed set]], we have that $U$ is [[Definition:Open Set (Topology)|open]] in $T'$.
+So we have shown that for any arbitrary $U \in \tau$ it follows that $U \in \tau'$.
+So $\tau \subseteq \tau'$ and so by definition $\tau$ is [[Definition:Coarser Topology|coarser]] than $\tau'$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Complement Space is Connected}
+Tags: Connected Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $T$ is a [[Definition:Connected Topological Space|connected space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Finite Complement Space is Irreducible]].
+The result follows from [[Irreducible Space is Connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Complement Space is Locally Connected}
+Tags: Locally Connected Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $T$ is a [[Definition:Locally Connected Space|locally connected space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Finite Complement Space is Irreducible]].
+The result follows from [[Irreducible Space is Locally Connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping Induces Partition on Domain}
+Tags: Mapping Theory
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let $F$ be defined as:
+:$F = \set {\map {f^{-1} } x: x \in T}$
+where $\map {f^{-1} } x$ is the [[Definition:Preimage of Element under Mapping|preimage]] of $x$.
+Then $F$ is a [[Definition:Partition (Set Theory)|partition]] of $S$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathcal R_f \subseteq S \times S$ be the [[Definition:Equivalence Relation Induced by Mapping|relation induced by $f$]]:
+:$\tuple {s_1, s_2} \in \mathcal R_f \iff \map f {s_1} = \map f {s_2}$
+Then from [[Relation Induced by Mapping is Equivalence Relation]], $\mathcal R_f$ is an [[Definition:Equivalence Relation|equivalence relation]].
+The result follows from [[Relation Partitions Set iff Equivalence]].
+{{qed}}
+[[Category:Mapping Theory]]
+mn1p432i9cwga3h01701aau75sdxeus
+\end{proof}<|endoftext|>
+\section{Countable Finite Complement Space is not Path-Connected}
+Tags: Path-Connected Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Finite Complement Topology|countable finite complement topology]].
+Then $T$ is not [[Definition:Path-Connected Space|path-connected]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $f: \closedint 0 1 \to S$ is a [[Definition:Path (Topology)|path]] on $T$.
+Then $f$ is by definition [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+By definition of the [[Definition:Finite Complement Topology|finite complement topology]], for all $x \in S$, the set $\set x$ is [[Definition:Closed Set (Topology)|closed]].
+Now consider the set:
+:$F = \set {\map {f^{-1} } x: x \in S}$
+From [[Continuity Defined from Closed Sets]], each of the [[Definition:Element|elements]] of $F$ is [[Definition:Closed Set (Topology)|closed]].
+Also, from [[Mapping Induces Partition on Domain]], the [[Definition:Element|elements]] of $F$ are [[Definition:Pairwise Disjoint|pairwise disjoint]].
+As $T$ is a [[Definition:Countable Finite Complement Topology|countable finite complement topology]], its [[Definition:Underlying Set of Topological Space|underlying set]] $S$ is [[Definition:Countably Infinite Set|countably infinite]] by definition.
+Hence $F$ is also [[Definition:Countably Infinite Set|countably infinite]] by nature of $f$ being a [[Definition:Mapping|mapping]].
+Furthermore, we have $\displaystyle \bigcup F = \closedint 0 1$.
+Hence $F$ is a [[Definition:Countably Infinite Set|countably infinite set]] of [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Closed Set (Topology)|closed sets]] whose [[Definition:Set Union|union]] is $\closedint 0 1$.
+From [[Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets]], this is impossible.
+From this [[Definition:Contradiction|contradiction]], $f$ cannot be [[Definition:Everywhere Continuous Mapping (Topology)|continuous]], and so cannot be a [[Definition:Path (Topology)|path]] on $T$.
+So $T$ cannot be [[Definition:Path-Connected Space|path-connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Finite Complement Space is not Locally Path-Connected}
+Tags: Locally Path-Connected Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Finite Complement Topology|countable finite complement topology]].
+Then $T$ is not [[Definition:Locally Path-Connected|locally path-connected]].
+\end{theorem}
+
+\begin{proof}
+Let $\BB$ be a [[Definition:Basis (Topology)|basis]] for $T$.
+Let $B \in \BB$.
+{{AimForCont}} $f: \closedint 0 1 \to B$ is a [[Definition:Path (Topology)|path]] on $T$.
+Then $f$ is by definition [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Now consider the set:
+:$F = \set {\map {f^{-1} } x: x \in S}$
+From [[Continuity Defined from Closed Sets]], each of the [[Definition:Element|elements]] of $F$ is [[Definition:Closed Set (Topology)|closed]].
+Also, from [[Mapping Induces Partition on Domain]], the [[Definition:Element|elements]] of $F$ are [[Definition:Pairwise Disjoint|pairwise disjoint]].
+From [[Basis of Countable Finite Complement Topology consists of Countably Infinite Sets]], $B$ is a [[Definition:Countably Infinite Set|countably infinite set]].
+Hence $F$ is also [[Definition:Countably Infinite Set|countably infinite]] by nature of $f$ being a [[Definition:Mapping|mapping]].
+Furthermore, we have $\displaystyle \bigcup F = \closedint 0 1$.
+Hence $F$ is a [[Definition:Countably Infinite Set|countably infinite set]] of [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Closed Set (Topology)|closed sets]] whose [[Definition:Set Union|union]] is $\closedint 0 1$.
+From [[Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets]], this is impossible.
+From this [[Definition:Contradiction|contradiction]], $f$ cannot be [[Definition:Everywhere Continuous Mapping (Topology)|continuous]], and so cannot be a [[Definition:Path (Topology)|path]] on $B$.
+So $B$ cannot be [[Definition:Path-Connected Set (Topology)|path-connected]].
+So, by definition, $T$ is not [[Definition:Locally Path-Connected Space|locally path-connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arc-Connectedness in Uncountable Finite Complement Space}
+Tags: Continuum Hypothesis, Arc-Connected Spaces, Locally Arc-Connected Spaces, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on a [[Definition:Uncountable Set|uncountable]] set $S$.
+If the [[Continuum Hypothesis]] is accepted as [[Definition:True|true]], then:
+:$T$ is [[Definition:Arc-Connected Space|arc-connected]]
+:$T$ is [[Definition:Locally Arc-Connected Space|locally arc-connected]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in S$ such that $a \ne b$.
+Let us assume the truth of the [[Continuum Hypothesis]].
+Then $a$ and $b$ are contained in a [[Definition:Subset|subset]] $X \subseteq S$ whose [[Definition:Cardinality|cardinality]] is the same as that of $\closedint 0 1$.
+So by definition of [[Definition:Cardinality|cardinality]] we can set up a [[Definition:Bijection|bijection]] $f: \closedint 0 1 \leftrightarrow X$ such that $\map f 0 = a$ and $\map f 1 = b$.
+Each element of $S$ is [[Definition:Closed Point|closed]] from [[Finite Complement Space is T1|Finite Complement Space is $T_1$]].
+Hence $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] and therefore an [[Definition:Arc (Topology)|arc]] in $T$.
+As $a$ and $b$ are arbitrary, $T$ is [[Definition:Arc-Connected Space|arc-connected]].
+Now let $\BB$ be a [[Definition:Basis (Topology)|basis]] for $T$.
+Let $B \in \BB$ and let $a, b \in B$.
+By definition of a [[Definition:Finite Complement Topology|finite complement topology]] on a [[Definition:Uncountable Set|uncountable set]], $B$ is [[Definition:Uncountable Set|uncountable]].
+So, by the same argument as above, we can set up an [[Definition:Arc (Topology)|arc]] $f$ between any points of $B$.
+So $T$ is [[Definition:Locally Arc-Connected Space|locally arc-connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Topology is Topology}
+Tags: Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Space|countable complement space]].
+Then $\tau$ is a [[Definition:Topology|topology]] on $T$.
+\end{theorem}
+
+\begin{proof}
+By definition, we have that $\O \in \tau$.
+We also have that $S \in \tau$ as $\complement_S \left({S}\right) = \O$ which is trivially [[Definition:Countable Set|countable]].
+Now suppose $A, B \in \tau$.
+Let $H = A \cap B$.
+Then:
+{{begin-eqn}}
+{{eqn | l = H
+      | r = A \cap B
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp S H
+      | r = \relcomp S {A \cap B}
+      | c = 
+}}
+{{eqn | r = \relcomp S A \cup \relcomp S B
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection|De Morgan's Laws: Complement of Intersection]]
+}}
+{{end-eqn}}
+But as $A, B \in \tau$ it follows that $\relcomp S A$ and $\relcomp S B$ are both [[Definition:Countable Set|countable]].
+Hence from [[Countable Union of Countable Sets is Countable]] their [[Definition:Set Union|union]] is also [[Definition:Countable Set|countable]] and so $\relcomp S H$ is [[Definition:Countable Set|countable]].
+So $H = A \cap B \in \tau$ as its [[Definition:Relative Complement|complement]] is [[Definition:Countable Set|countable]].
+Now let $\UU \subseteq \tau$.
+Then:
+:$\displaystyle \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
+by [[De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Union|De Morgan's Laws: Complement of Union]].
+But as:
+:$\forall U \in \UU: \relcomp S U \in \tau$
+each of the $\relcomp S U$ is [[Definition:Countable Set|countable]].
+Hence so is their [[Definition:Set Intersection|intersection]].
+So $\displaystyle \relcomp S {\bigcup \UU}$ is [[Definition:Countable Set|countable]] which means:
+:$\displaystyle \bigcup \UU \in \tau$
+So $\tau$ is a [[Definition:Topology|topology]] on $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Topology is Expansion of Finite Complement Topology}
+Tags: Countable Complement Topology, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Let $T' = \struct {S, \tau'}$ be the [[Definition:Finite Complement Topology|finite complement topology]] on the same [[Definition:Infinite Set|infinite set]] $S$.
+Then $\tau$ is an [[Definition:Expansion of Topology|expansion]] of $\tau'$.
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau', U \ne \O$.
+Then $\relcomp S U$ is [[Definition:Finite Set|finite]] by definition of [[Definition:Finite Complement Topology|finite complement topology]].
+Then by definition of [[Definition:Countable Complement Topology|countable complement topology]], we have that $U \in \tau$.
+So $\tau' \subseteq \tau$ and hence the result by definition of [[Definition:Expansion of Topology|expansion]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is T1}
+Tags: T1 Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]].
+Then $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Countable Complement Topology is Expansion of Finite Complement Topology]]. 
+We also have that a [[Finite Complement Space is T1|Finite Complement Space is $T_1$]].
+Then from [[Separation Properties Preserved by Expansion]], we have that $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Subset of Countable Complement Space Intersects Open Sets}
+Tags: Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable]] set $S$.
+Let $H \subseteq S$ be an [[Definition:Uncountable Set|uncountable]] [[Definition:Subset|subset]] of $S$.
+Then the [[Definition:Set Intersection|intersection]] of $H$ with any [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open set]] of $T$ is [[Definition:Uncountable Set|uncountable]].
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$ be any [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open set]] of $T$.
+Then $\relcomp S U$ is [[Definition:Countable Set|countable]].
+Suppose $H \cap U = \O$.
+Then from [[Intersection with Complement is Empty iff Subset]] it follows that $H \subseteq \relcomp S U$ and so $H$ is [[Definition:Countable Set|countable]].
+So if $H$ is [[Definition:Uncountable Set|uncountable]] it is bound to have a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set Intersection|intersection]] with every [[Definition:Open Set (Topology)|open set]] in $T$.
+{{finish|It still needs to be shown that $U \cap H$ is [[Definition:Uncountable Set|uncountable]] {{iff}} $H$ is [[Definition:Uncountable Set|uncountable]].}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is not T2}
+Tags: Hausdorff Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is not a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Countable Complement Space is Irreducible]]
+:[[Irreducible Hausdorff Space is Singleton]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is Irreducible}
+Tags: Irreducible Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is an [[Definition:Irreducible Space|irreducible space]].
+\end{theorem}
+
+\begin{proof}
+Let $U_1, U_2 \in \tau$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open sets]] of $T$.
+We have that both $U_1$ and $U_2$ are both [[Definition:Uncountable Set|uncountable]] if $S$ is.
+From [[Uncountable Subset of Countable Complement Space Intersects Open Sets]], they [[Definition:Set Intersection|intersect]] each other.
+Hence the result from definition of [[Definition:Irreducible Space|irreducible space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is not T3, T4 or T5}
+Tags: T3 Spaces, T4 Spaces, T5 Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is not a [[Definition:T3 Space|$T_3$ space]], [[Definition:T4 Space|$T_4$ space]] or [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Countable Complement Space is T1|Countable Complement Space is a $T_1$ space]].
+From [[T1 Space is T0 Space|$T_1$ Space is $T_0$ Space]], $T$ is a [[Definition:Kolmogorov Space|$T_0$ space]].
+We then have that a [[Countable Complement Space is not T2|Countable Complement Space is not $T_2$]].
+From [[Regular Space is T2 Space|Regular Space is $T_2$ Space]], $T$ is not a [[Definition:Regular Space|regular space]].
+By definition, a [[Definition:Regular Space|regular space]] is a space that is both a [[Definition:Kolmogorov Space|$T_0$ space]] and a [[Definition:T3 Space|$T_3$ space]].
+But $T$ is a [[Definition:Kolmogorov Space|$T_0$ space]] and not a [[Definition:Regular Space|regular space]].
+So it follows that $T$ cannot be a [[Definition:T3 Space|$T_3$ space]].
+Next we have that a [[Normal Space is T3 Space|Normal Space is a $T_3$ Space]].
+But as $T$ is not a [[Definition:T3 Space|$T_3$ space]], $T$ cannot be a [[Definition:Normal Space|normal space]].
+By definition, a [[Definition:Normal Space|normal space]] is a space that is both a [[Definition:Fréchet Space (Topology)|$T_1$ space]] and a [[Definition:T4 Space|$T_4$ space]].
+But $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ space]] and not a [[Definition:Normal Space|normal space]].
+So it follows that $T$ cannot be a [[Definition:T4 Space|$T_4$ space]].
+Finally we have that a [[T5 Space is T4 Space|$T_5$ Space is a $T_4$ Space]].
+But as $T$ is not a [[Definition:T4 Space|$T_4$ space]], $T$ cannot be a [[Definition:T5 Space|$T_5$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Higher-Aleph Complement Topology is Topology}
+Tags: Higher-Aleph Complement Topology
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be an [[Definition:Higher-Aleph Complement Space|$\aleph_n$ complement space]].
+Then $\tau$ is a [[Definition:Topology|topology]] on $T$.
+\end{theorem}
+
+\begin{proof}
+By definition, we have that $\varnothing \in \tau$.
+We also have that $S \in \tau$ as $\complement_S \left({S}\right) = \varnothing$ which is trivially [[Definition:Finite|finite]].
+Now suppose $A, B \in \tau$.
+Let $H = A \cap B$.
+Then:
+{{begin-eqn}}
+{{eqn | l = H
+      | r = A \cap B
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \complement_S \left({H}\right)
+      | r = \complement_S \left({A \cap B}\right)
+      | c = 
+}}
+{{eqn | r = \complement_S \left({A}\right) \cup \complement_S \left({B}\right)
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection|De Morgan's Laws: Complement of Intersection]]
+}}
+{{end-eqn}}
+But as $A, B \in \tau$ it follows that $\complement_S \left({A}\right)$ and $\complement_S \left({B}\right)$ both have a [[Definition:Cardinality|cardinality]] strictly smaller than that of $S$.
+Hence their [[Definition:Set Union|union]] also has a [[Definition:Cardinality|cardinality]] strictly smaller than that of $S$.
+So $\complement_S \left({H}\right)$ has a [[Definition:Cardinality|cardinality]] strictly smaller than that of $S$.
+So $H = A \cap B \in \tau$ as its [[Definition:Relative Complement|complement]] has a [[Definition:Cardinality|cardinality]] strictly smaller than that of $S$.
+Now let $\mathcal U \subseteq \tau$.
+Then from [[De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Union|De Morgan's laws: Complement of Union]]:
+: $\displaystyle \complement_S \left({\bigcup \mathcal U}\right) = \bigcap_{U \in \mathcal U} \complement_S \left({U}\right)$
+But as:
+:$\forall U \in \mathcal U: \complement_S \left({U}\right) \in \tau$
+each of the $\complement_S \left({U}\right)$ has a [[Definition:Cardinality|cardinality]] strictly smaller than that of $S$.
+Hence so has their [[Definition:Set Intersection|intersection]].
+So $\displaystyle \complement_S \left({\bigcup \mathcal U}\right)$ has a [[Definition:Cardinality|cardinality]] strictly smaller than that of $S$, which means:
+: $\displaystyle \bigcup \mathcal U \in \tau$
+So $\tau$ is a [[Definition:Topology|topology]] on $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets}
+Tags: Closed Sets, Topology
+
+\begin{theorem}
+Let $\mathbb I = \closedint 0 1$ be the [[Definition:Closed Unit Interval|closed unit interval]].
+Then $\mathbb I$ cannot be expressed as the [[Definition:Set Union|union]] of a [[Definition:Countably Infinite Set|countably infinite set]] of [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Closed Set (Topology)|closed sets]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\displaystyle \mathbb I = \bigcup_{i \mathop = 1}^\infty C_i$ where $\set {C_i}$ is a [[Definition:Set|set]] of [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Closed Set (Topology)|closed sets]].
+Let:
+:$\displaystyle B = \bigcup \partial C_i = \mathbb I \setminus \bigcup C_i^\circ$
+where $\partial C_i$ is the [[Definition:Boundary (Topology)|boundary]] of $C_i$ and $C_i^\circ$ is the [[Definition:Interior (Topology)|interior]] of $C_i$.
+Let $J \subseteq \mathbb I$ be a [[Definition:Subinterval|subinterval]] of $\mathbb I$.
+Then $J$ is [[Definition:Non-Meager Space|non-meager]], and so some $C_k$ is [[Definition:Dense-in-itself|dense]] in some [[Definition:Open Real Interval|open interval]] $L \subseteq J$.
+Since $C_k$ is [[Definition:Closed Set (Topology)|closed]] we have that $L \subseteq C_k^\circ$ and so $L \cap B = \O$.
+So $J$ contains an [[Definition:Open Set (Topology)|open subset]] $L$ [[Definition:Disjoint Sets|disjoint]] from $B$.
+So $B$ is [[Definition:Nowhere Dense|nowhere dense]] in $\mathbb I$.
+So every [[Definition:Open Real Interval|open interval]] $U$ containing some $x \in \partial C_j$ must intersect $B \setminus \partial C_j$.
+This is because $U$ is an [[Definition:Open Neighborhood|open neighborhood]] of $x$ and therefore contains some point $u \in \mathbb I \setminus C_j$, say $u \in C_m$.
+Then if $U \cap B \cap \partial C_m = \varnothing$ it follows that $C_m^\circ \cap U$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open set]] in $U$. which is also [[Definition:Relatively Closed Set|relatively closed]].
+{{explain|Closed relative to what?}}
+Now $B$ is itself [[Definition:Non-Meager Space|non-meager]], since it is a [[Definition:Closed Set (Topology)|closed]] subset of $\mathbb I$.
+Thus some $\partial C_k$ is [[Definition:Dense-in-itself|dense]] in some [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open set]] $U \cap B$, for some [[Definition:Open Real Interval|open interval]] $U$ in $\mathbb I$.
+Since $\partial C_k$ is [[Definition:Closed Set (Topology)|closed]], this means $\partial C_k \cap U = B \cap U$.
+But this is impossible, since if $U \cap C_k \ne \O$ then $U \cap \paren {B \setminus \partial C_k} \ne \O$.
+This contradiction shows that $\mathbb I$ cannot be expressed as $\displaystyle \bigcup_{i \mathop = 1}^\infty C_i$.
+{{qed}}
+{{proofread}}
+\end{proof}<|endoftext|>
+\section{Compact Sets in Countable Complement Space}
+Tags: Compact Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then the [[Definition:Compact Topological Subspace|compact sets]] of $T$ are exactly the [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $S$.
+\end{theorem}
+
+\begin{proof}
+From [[Finite Topological Space is Compact]], any [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $S$ is [[Definition:Compact Topological Subspace|compact]].
+{{AimForCont}} $H \subseteq S$ is an [[Definition:Infinite Set|infinite]] [[Definition:Compact Topological Subspace|compact set]].
+Take a [[Definition:Infinite Sequence|(countably) infinite sequence]] $\sequence {a_n}_{n \mathop \ge 0}$ of [[Definition:Distinct Elements|distinct elements]] of $H$.
+Consider the [[Definition:Open Set (Topology)|open sets]]:
+:$V_m := S \setminus \set {a_{m + n} }_{n \mathop \ge 0}$
+for $m \ge 0$, which satisfy $V_{m_1} \subset V_{m_2}$ if $m_1 < m_2$.
+Then:
+:$\displaystyle H \subseteq \bigcup_{m \mathop \ge 0} V_m$
+is an [[Definition:Open Cover|open cover]] of $H$.
+Since $H$ is [[Definition:Compact Topological Subspace|compact]], it has a [[Definition:Finite Subcover|finite open subcover]], say:
+:$\displaystyle H \subseteq \bigcup_{i \mathop = 0}^N V_{m_i}$
+with $m_0 < m_1 < \cdots < m_N$.
+But then:
+:$\displaystyle \bigcup_{i \mathop = 0}^N V_{m_i} = V_{m_N}$ 
+and so:
+:$\displaystyle a_{m_N + 1}, a_{m_N + 2}, \ldots \notin V_{m_N} = \bigcup_{i \mathop = 0}^N V_{m_i}$
+which implies $a_{m_N + 1}, a_{m_N + 2}, \ldots \notin H$, a [[Definition:Contradiction|contradiction]].
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is not Sigma-Compact}
+Tags: Sigma-Compact Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is not a [[Definition:Sigma-Compact Space|$\sigma$-compact space]].
+\end{theorem}
+
+\begin{proof}
+From [[Compact Sets in Countable Complement Space]], the only [[Definition:Compact Topological Subspace|compact sets]] in $T$ are [[Definition:Finite Set|finite]].
+A [[Definition:Countable Union|countable union]] of [[Definition:Finite Set|finite sets]] can not be an [[Definition:Uncountable Set|uncountable set]].
+Hence the result by definition of [[Definition:Sigma-Compact Space|$\sigma$-compact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is not Countably Compact}
+Tags: Countably Compact Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is not a [[Definition:Countably Compact Space|countably compact space]].
+\end{theorem}
+
+\begin{proof}
+Consider $U \in \tau$.
+By definition of [[Definition:Countable Complement Topology|countable complement topology]], $\relcomp S U$ is [[Definition:Countably Infinite Set|countably infinite]].
+Then for each $x \in \relcomp S U$, $\relcomp S U \setminus \set x$ is [[Definition:Countably Infinite Set|countably infinite]].
+Then for each $x \in \relcomp S U$, $\relcomp S {\relcomp S U \setminus \set x} = U \cup \set x$ is [[Definition:Open Set (Topology)|open]].
+Hence $\set {U \cup \set x: x \in \relcomp S U}$ is an [[Definition:Open Cover|open cover]].
+This [[Definition:Cover of Set|cover]] is [[Definition:Countable Set|countable]] because it is [[Definition:Set Equivalence|equivalent]] to $\relcomp S U$.
+{{AimForCont}} $T$ is a [[Definition:Countably Compact Space|countably compact space]].
+Then by definition there exists a [[Definition:Finite Subcover|finite subcover]] $\set {U \cup \set {x_1}, \ldots, U \cup \set {x_n} }$ of $T$.
+Then:
+:$\paren {U \cup \set {x_1} } \cup \dotsb \cup \paren {U \cup \set {x_n} } = U \cup \set {x_1, \ldots, x_n}$
+Since $\relcomp S U$ is not [[Definition:Finite Set|finite]], it follows that $U \cup \set {x_1, \ldots, x_n} \ne S$.
+Thus there is no [[Definition:Finite Subcover|finite subcover]] of $T$.
+It follows by [[Proof by Contradiction]] that $T$ cannot be a [[Definition:Countably Compact Space|countably compact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is Lindelöf}
+Tags: Lindelöf Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+By definition, $T$ is a [[Definition:Lindelöf Space|Lindelöf space]] {{iff}} every [[Definition:Open Cover|open cover]] of $S$ has a [[Definition:Countable Subcover|countable subcover]].
+Let $\CC$ be an [[Definition:Open Cover|open cover]] of $T$.
+Let $U \in \CC$ be any [[Definition:Set|set]] in $C$.
+It [[Definition:Cover of Set|covers]] all but a [[Definition:Countable Set|countable number]] of points of $T$.
+So for each of those points we pick an element of $\CC$ which [[Definition:Cover of Set|covers]] each of those points.
+Hence we have a [[Definition:Countable Subcover|countable subcover]] of $T$.
+So by definition $T$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is not First-Countable}
+Tags: First-Countable Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is not a [[Definition:First-Countable Space|first-countable space]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} some $x \in S$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]].
+That means:
+:there exists a [[Definition:Countable Set|countable]] [[Definition:Set of Sets|set of sets]] $\BB_x \subseteq \tau$ 
+such that:
+:$\forall B \in \BB_x: x \in B$
+and such that:
+:every [[Definition:Open Neighborhood|open neighborhood]] of $x$ contains some $B \in \BB_x$.
+So:
+{{begin-eqn}}
+{{eqn | l = \bigcap \BB_x
+      | r = \set x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = S \setminus \set x
+      | r = S \setminus \bigcap \BB_x
+      | c = 
+}}
+{{eqn | r = \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{end-eqn}}
+By definition, each of $S \setminus B$ is [[Definition:Countable Set|countable]].
+From [[Countable Union of Countable Sets is Countable]] it follows that $\displaystyle \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}$ is also [[Definition:Countable Set|countable]].
+So $S \setminus \set x$ and therefore $S$ is also [[Definition:Countable Set|countable]].
+From this [[Definition:Contradiction|contradiction]] (as we have specified that $S$ is [[Definition:Uncountable Set|uncountable]]) it follows that our assumption that $x \in S$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] must be false.
+Hence by definition $T$ can not be [[Definition:First-Countable Space|first-countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Form for Number of Derangements on Finite Set}
+Tags: Combinatorics, Counting Arguments, Closed Forms, Number of Derangements on Finite Set
+
+\begin{theorem}
+The number of [[Definition:Derangement|derangements]] $D_n$ on a [[Definition:Finite Set|finite set]] $S$ of [[Definition:Cardinality|cardinality]] $n$ is:
+:$D_n = n! \paren {1 - \dfrac 1 {1!} + \dfrac 1 {2!} - \dfrac 1 {3!} + \cdots + \dfrac {\paren {-1}^n} {n!} }$
+\end{theorem}
+
+\begin{proof}
+Let $s_i$ be the $i$th element of set $S$.
+Begin by defining set $A_m$, which is all of the [[Definition:Permutation|permutations]] of $S$ which [[Definition:Fixed Element of Permutation|fixes]] $S_m$.
+Then the number of [[Definition:Permutation|permutations]], $W$, with ''at least'' one [[Definition:Element|element]] [[Definition:Fixed Element of Permutation|fixed]], $m$, is:
+:$\displaystyle W = \size {\bigcup_{m \mathop = 1}^n A_m}$
+Applying the [[Inclusion-Exclusion Principle]]:
+{{begin-eqn}}
+{{eqn | l = W
+      | r = \sum_{m_1 \mathop = 1}^n \size {A_{m_1} }
+      | c = 
+}}
+{{eqn | o = 
+      | ro= -
+      | r = \sum_{m_1, m_2 : 1 \mathop \le m_1 \mathop < m_2 \mathop \le n} \size {A_{m_1} \cap A_{m_2} }
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \sum_{m_1, m_2, m_3 : 1 \mathop \le m_1 \mathop < m_2 \mathop < m_3 \mathop \le n} \size {A_{m_1} \cap A_{m_2} \cap A_{m_3} }
+      | c = 
+}}
+{{eqn | o = 
+      | ro= -
+      | r = \cdots
+      | c = 
+}}
+{{end-eqn}}
+Each value $A_{m_1} \cap \cdots \cap A_{m_p}$ represents the set of [[Definition:Permutation|permutations]] which [[Definition:Fixed Element of Permutation|fix]] $p$ values $m_1, \ldots, m_p$.
+Note that the number of [[Definition:Permutation|permutations]] which [[Definition:Fixed Element of Permutation|fix]] $p$ values only depends on $p$, not on the particular values of $m$.
+Thus from [[Cardinality of Set of Subsets]] there are $\dbinom n p$ terms in each summation. 
+So:
+{{begin-eqn}}
+{{eqn | l = W
+      | r = \binom n 1 \size {A_1}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= -
+      | r = \binom n 2 \size {A_1 \cap A_2}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \binom n 3 \size {A_1 \cap A_2 \cap A_3}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= -
+      | r = \cdots
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \paren {-1}^{p - 1} \binom n p \size {A_1 \cap \cdots \cap A_p}
+      | c = 
+}}
+{{eqn | o = 
+      | r = \cdots
+      | c = 
+}}
+{{end-eqn}}
+$\size {A_1 \cap \cdots \cap A_p}$ is the number of [[Definition:Permutation|permutations]] fixing $p$ [[Definition:Element|elements]] in position.
+This is equal to the number of [[Definition:Permutation|permutations]] which rearrange the remaining $n - p$ [[Definition:Element|elements]], which is $\paren {n - p}!$.
+Thus we finally get:  
+:$W = \dbinom n 1 \paren {n - 1}! - \dbinom n 2 \paren {n - 2}! + \dbinom n 3 \paren {n - 3}! - \cdots + \paren {-1}^{p - 1} \dbinom n p \paren {n - p}! \cdots$
+That is:
+:$\displaystyle W = \sum_{p \mathop = 1}^n \paren {-1}^{p - 1} \binom n p \paren {n - p}!$
+Noting that $\dbinom n p = \dfrac {n!} {p! \paren {n - p}!}$, this reduces to:
+:$\displaystyle W = \sum_{p \mathop = 1}^n \paren {-1}^{p - 1} \dfrac {n!} {p!}$ 
+{{qed}}
+\end{proof}
+
+\begin{proof}
+A [[Definition:Derangement|derangement]] is a [[Definition:Permutation|permutation]] that [[Definition:Fixed Element|fixes no elements]].
+So, to compute the total number of [[Definition:Derangement|derangements]], we compute the number of [[Definition:Permutation|permutations]] that [[Definition:Fixed Element|fix elements]], that is leave [[Definition:Element|elements]] in the same place.
+Let $\pi$ be a permutation of $S$.
+Let $x, y \in S$.
+Let $i$ be a position in a given [[Definition:Permutation|permutation]] of [[Definition:Set|set]] $S$, where $1 \le i \le n$.
+Let $\map {P_i} \pi$ be the [[Definition:Property|property]] that $\map \pi x = y$ at position $i$.
+Let $\map N {P_i, \dots, P_j}$ be the number of [[Definition:Permutation|permutations]] where $\map {P_i} \pi, \dots, \map {P_j} \pi$ hold.
+If we consider all the [[Definition:Permutation|permutations]] of a set where the property $P_i$ holds, we have fixed a single [[Definition:Element|element]] of the [[Definition:Set|set]].
+So there are $\paren {n - 1}!$ such possible [[Definition:Permutation|permutations]].
+Furthermore, we could have chosen $i$ as any element in the set so there are $\dbinom n 1$ terms to choose from.
+So $\map N {P_i} = \paren {n - 1}!$ and there are $\dbinom n 1$ terms.
+Similarly, $\map N {P_i, P_j} = \paren {n - 2}!$ and there are $\dbinom n 2$ terms.
+So now we can compute the total number of [[Definition:Derangement|derangements]]:
+{{begin-eqn}}
+{{eqn | l = D_n
+      | r = n! - \dbinom n 1 \paren {n - 1}! + \dbinom n 2 \paren {n - 2}! - \ldots + \paren {-1}^n \dbinom n n \paren {n - n}!
+      | c = 
+}}
+{{eqn | r = n! - \frac {n!} {1! \paren {n - 1}!} \paren {n - 1}! + \frac {n!} {2! \paren {n - 2}!} \paren {n - 2}! - \cdots + \paren {-1}^n \frac {n!} {n! \paren {n - n}!} \paren {n - n}!
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{eqn | r = n! \paren {1 - \frac 1 {1!} + \frac 1 {2!} - \frac 1 {3!} + \cdots + \paren {-1}^n \frac 1 {n!} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Points of Countable Complement Space}
+Tags: Limit Points, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Space|countable complement space]].
+Let $H \subseteq S$ be an [[Definition:Uncountable Set|uncountable]] [[Definition:Subset|subset]] of $S$.
+Then every point of $S$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+Let $H \subseteq S$ be an [[Definition:Countable Set|countable]] or [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $S$.
+Then $H$ contains all its [[Definition:Limit Point of Set|limit points]].
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$ be any [[Definition:Open Set (Topology)|open set]] of $T$.
+Let $H \subseteq S$ be an [[Definition:Uncountable Set|uncountable]].
+From [[Uncountable Subset of Countable Complement Space Intersects Open Sets]], we have that $U \cap H$ is [[Definition:Uncountable Set|uncountable]] {{iff}} $H$ is [[Definition:Uncountable Set|uncountable]].
+Let $x \in S$.
+Then every [[Definition:Open Set (Topology)|open set]] $U$ in $T$ such that $x \in U$ also contains an [[Definition:Uncountable Set|uncountable number]] of points of $H$ other than $x$.
+Thus, by definition, $x$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+Now let $H$ be a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$.
+By definition $\relcomp S H$ is [[Definition:Open Set (Topology)|open]] in $T$ and so $H$ is [[Definition:Closed Set (Topology)|closed]].
+So by [[Closed Set Equals its Closure]] we have by definition that if $x$ is a [[Definition:Limit Point of Set|limit point]] of $H$, then $x \in H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is not Separable}
+Tags: Separable Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is not a [[Definition:Separable Space|separable space]].
+\end{theorem}
+
+\begin{proof}
+Let $U$ be a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$.
+By the definition of $T$, $U$ is [[Definition:Closed Set (Topology)|closed]].
+As $U$ is [[Definition:Countable Set|countable]] but $S$ is [[Definition:Uncountable Set|uncountable]]:
+:$U \subsetneq S$
+and so:
+:$U \ne S$
+From [[Closed Set Equals its Closure]]:
+:$U^- = U$
+where $U^-$ is the [[Definition:Closure (Topology)|closure]] of $U$.
+Thus:
+:$U^- \ne S$
+So by definition, $U$ is not [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+As $U$ is arbitrary, there is no [[Definition:Countable Set|countable set]] which is [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+The result follows by definition of [[Definition:Separable Space|separable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space Satisfies Countable Chain Condition}
+Tags: Countable Chain Condition, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ satisfies the [[Definition:Countable Chain Condition|countable chain condition]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Countable Complement Space is Irreducible]].
+Therefore there is no [[Definition:Disjoint Sets|disjoint]] set of [[Definition:Open Set (Topology)|open sets]] of $T$.
+So the result holds [[Definition:Vacuous Truth|vacuously]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is Connected}
+Tags: Connected Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is a [[Definition:Connected (Topology)|connected space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Countable Complement Space is Irreducible]].
+The result follows from [[Irreducible Space is Connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is Locally Connected}
+Tags: Locally Connected Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is a [[Definition:Locally Connected Space|locally connected space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Countable Complement Space is Irreducible]].
+The result follows from [[Irreducible Space is Locally Connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is Pseudocompact}
+Tags: Pseudocompact Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is [[Definition:Pseudocompact Space|pseudocompact]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Countable Complement Space is Irreducible]].
+The result follows from [[Irreducible Space is Pseudocompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is not Countably Metacompact}
+Tags: Countably Metacompact Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is not [[Definition:Countably Metacompact Space|countably metacompact]].
+\end{theorem}
+
+\begin{proof}
+From [[Uncountable Subset of Countable Complement Space Intersects Open Sets]], the [[Definition:Set Intersection|intersection]] of any [[Definition:Open Set (Topology)|open sets]] is [[Definition:Uncountable Set|uncountable]].
+So for any [[Definition:Open Cover|open cover]] $\CC$ of $T$, every point is in an [[Definition:Infinite Set|infinite number]] of [[Definition:Open Set (Topology)|open sets]] of $T$.
+So no [[Definition:Refinement of Cover|refinement]] of any [[Definition:Open Cover|open cover]] of $T$ can be [[Definition:Point Finite|point finite]].
+Hence the result by definition of [[Definition:Countably Metacompact Space|countably metacompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Complement Space is not Weakly Countably Compact}
+Tags: Weakly Countably Compact Spaces, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is not [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+\end{theorem}
+
+\begin{proof}
+By definition, $T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]] {{iff}} every [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$ has a [[Definition:Limit Point of Set|limit point]] in $S$.
+Let $H \subseteq S$ be a [[Definition:Countable Set|countable set]].
+From [[Limit Points of Countable Complement Space]], it contains all its [[Definition:Limit Point of Set|limit points]].
+Let $x \in H$.
+$\relcomp S {\relcomp S H \cup \set x} = H \setminus \set x$ is [[Definition:Countable Set|countable]].
+Then $\relcomp S H \cup \set x$ is an [[Definition:Open Neighborhood|open neighborhood]] of $x$. 
+However $H \cap \paren {\paren {\relcomp S H \cup \set x} \setminus \set x} = H \cap \relcomp S H = \O$.
+So $x$ is not a [[Definition:Limit Point of Set|limit point]] of $H$.
+Since $x$ is arbitrary, there are no [[Definition:Limit Point of Set|limit points]] in $H$. 
+This shows that $T$ is not [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Saturated Implies Universal}
+Tags: 
+
+\begin{theorem}
+Let $\kappa$ be an [[Definition:Infinite|infinite]] [[Definition:Cardinal|cardinal]].
+Let $\mathcal{M}$ be a [[Definition:Model (Logic)|model]] of the $\mathcal{L}$-[[Definition:Theory (Logic)|theory]] $T$.
+If $\mathcal{M}$ is $\kappa$-[[Definition:Saturated Model|saturated]], then it is $\kappa^+$-[[Definition:Universal Model|universal]], where $\kappa^+$ is the [[Definition:Successor Cardinal|successor cardinal]] of $\kappa$.
+\end{theorem}
+
+\begin{proof}
+The idea of the proof is that $\mathcal{M}$ being saturated means that when we want to define an elementary map $\mathcal{N}\to\mathcal{M}$, we can find an image $y\in \mathcal{M}$ for an element $x\in \mathcal{N}$ by realizing the type made up of the formulas that such a $y$ would need to satisfy.
+Let $\mathcal{N}$ be a model of $T$ with universe of cardinality strictly less than $\kappa$.
+We will construct an elementary embedding of $\mathcal{N}$ into $\mathcal{M}$ by transfinite recursion.
+Since $|\mathcal{N}|<\kappa$, we can write its elements as $n_\alpha$ for ordinals $\alpha < \kappa$.
+For each ordinal $\alpha < \kappa$, let $A_\alpha$ be the subset $\{n_\beta : \beta < \alpha\}$ of the universe of $\mathcal{N}$.
+Note for clarity that $n_\alpha \in A_{\alpha+1}$ but $n_\alpha \notin A_\alpha$.
+*Base case $\alpha = 0$:
+Define $f_0 = \emptyset$.
+Note that $f_0$ is trivially an [[Definition:Elementary Embedding|elementary embedding]] from $A_0 = \emptyset$ into $\mathcal{M}$
+*[[Definition:Limit Ordinal|Limit ordinals]] $\alpha$, assuming $f_\beta$ is defined and elementary $A_\beta \to \mathcal{M}$ for all $\beta < \alpha$:
+Let $\displaystyle f_\alpha = \bigcup_{\beta < \alpha} f_\beta$.
+If $\phi$ is an $\mathcal{L}$-sentence with parameters from $A_\alpha$, then since it involves only finitely many such parameters, they must all be contained in some $A_\beta$ for $\beta < \alpha$.  But $f_\alpha \restriction A_\beta = f_\beta$ is elementary, so $f_\alpha$ must be as well.
+* [[Definition:Successor Ordinal|Successor ordinals]] $\alpha = \beta + 1$, assuming $f_\beta$ is defined and elementary $A_\beta \to \mathcal{M}$:
+We need to extend $f_\beta$ to $A_\alpha = A_\beta \cup \{n_\beta\}$ so that truth of $\mathcal{L}$-sentences with parameters from $A_\alpha$ is preserved.
+Consider the subset $p = \{\phi(v, f_\beta(\bar{a})):\bar{a}\text{ is a tuple from }A_\beta \text{ and }\mathcal{N}\models\phi(n_\beta,\bar{a})\}$ of the set of $\mathcal{L}$-formulas with one free variable and parameters from the image $f_\beta (A_\beta)$ of $A_\beta$ under $f_\beta$.
+The set $p$ is a $1$-type over the image $f_\beta (A_\beta)$ in $\mathcal{M}$.
+Since $|A_\beta| < \kappa$ and by assumption $\mathcal{M}$ is $\kappa$-saturated, this means that $p$ is realized in $\mathcal{M}$ by some element $b$.
+Thus, defining $f_\alpha$ to be $f_\beta \cup \{(n_\beta, b)\}$ makes it an elementary embedding $A_\alpha \to \mathcal{M}$.
+Now, define $\displaystyle f = \bigcup_{\alpha < \kappa} f_\alpha$.
+Then $f$ is an elementary embedding from $\mathcal{N}$ to $\mathcal{M}$ since $\displaystyle \bigcup_{\alpha < \kappa} A_\alpha = \mathcal{N}$, any finite set of parameters from $\mathcal{N}$ must belong to one $A_\alpha$, and $f \restriction A_\alpha = f_\alpha$ is elementary.
+{{qed}}
+{{MissingLinks}}
+o0chgtswe94jl7ke5tvpt0un4m8c9fg
+\end{proof}<|endoftext|>
+\section{Double Pointed Countable Complement Topology fulfils no Separation Axioms}
+Tags: Separation Axioms, Double Pointed Topologies, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Let $T \times D$ be the [[Definition:Double Pointed Topology|double pointed topology]] on $T$.
+Then $T \times D$ is not a [[Definition:Kolmogorov Space|$T_0$ space]], [[Definition:Fréchet Space (Topology)|$T_1$ space]], [[Definition:Hausdorff Space|$T_2$ space]], [[Definition:T3 Space|$T_3$ space]], [[Definition:T4 Space|$T_4$ space]] or [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+From [[Separation Axioms on Double Pointed Topology]], we have that:
+:$T \times D$ is not a [[Definition:Kolmogorov Space|$T_0$ space]].
+:$T \times D$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ space]].
+:$T \times D$ is not a [[Definition:Hausdorff Space|$T_2$ space]].
+Also from [[Separation Axioms on Double Pointed Topology]], we have that:
+:$T \times D$ is a [[Definition:T3 Space|$T_3$ space]] {{iff}} $T$ is a [[Definition:T3 Space|$T_3$ space]]
+:$T \times D$ is a [[Definition:T4 Space|$T_4$ space]] {{iff}} $T$ is a [[Definition:T4 Space|$T_4$ space]]
+:$T \times D$ is a [[Definition:T5 Space|$T_5$ space]] {{iff}} $T$ is a [[Definition:T5 Space|$T_5$ space]]
+But from [[Countable Complement Space is not T3, T4 or T5|Countable Complement Space is not $T_3$, $T_4$ or $T_5$]]:
+:$T$ is not a [[Definition:T3 Space|$T_3$ space]]
+:$T$ is not a [[Definition:T4 Space|$T_4$ space]]
+:$T$ is not a [[Definition:T5 Space|$T_5$ space]]
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Double Pointed Countable Complement Topology is Weakly Countably Compact}
+Tags: Weakly Countably Compact Spaces, Double Pointed Topologies, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Let $T \times D$ be the [[Definition:Double Pointed Topology|double pointed topology]] on $T$.
+Then $T \times D$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+\end{theorem}
+
+\begin{proof}
+By definition, $T$ is [[Definition:Weakly Countably Compact Space|weakly countably compact]] {{iff}} every [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$ has a [[Definition:Limit Point of Set|limit point]] in $S$.
+Let $D = \set {0, 1}$.
+Let $\tuple {p, 0}$ belong to some [[Definition:Infinite Set|infinite]] $A \subseteq S$.
+Then its twin $\tuple {p, 1}$ is a [[Definition:Limit Point of Set|limit point]] of $A$.
+Hence the result by definition of [[Definition:Weakly Countably Compact Space|weakly countably compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Extension of Double Pointed Countable Complement Topology is T4 Space}
+Tags: T4 Spaces, Open Extension Topology, Double Pointed Topologies, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_S}$ be a [[Definition:Countable Complement Topology|countable complement topology]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Let $D = \struct {\set {0, 1}, \tau_D}$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on two points.
+Let $T \times D$ be the [[Definition:Double Pointed Topology|double pointed topology]] on $T$.
+Let $\paren {T \times D}^*_{\bar p}$ be the [[Definition:Open Extension Topology|open extension topology]] on $S \times \set {0, 1} \cup \set p$ where $p \notin S \times \set {0, 1}$.
+Then $\paren {T \times D}^*_{\bar p}$ is a [[Definition:T4 Space|$T_4$ space]], and no other [[Definition:Tychonoff Separation Axioms|separation axioms]] are fulfilled.
+That is, $\paren {T \times D}^*_{\bar p}$ is not a [[Definition:Kolmogorov Space|$T_0$ space]], [[Definition:Fréchet Space (Topology)|$T_1$ space]], [[Definition:Hausdorff Space|$T_2$ space]], [[Definition:T3 Space|$T_3$ space]] or [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+From [[Open Extension Topology is T4]] we have that $\paren {T \times D}^*_{\bar p}$ is a [[Definition:T4 Space|$T_4$ space]].
+From [[Double Pointed Countable Complement Topology fulfils no Separation Axioms]], we have that $T \times D$ is not a [[Definition:Kolmogorov Space|$T_0$ space]] or a [[Definition:T5 Space|$T_5$ space]].
+From [[Condition for Open Extension Space to be T0 Space|Condition for Open Extension Space to be $T_0$ Space]], it follows that $\paren {T \times D}^*_{\bar p}$ is not a [[Definition:Kolmogorov Space|$T_0$ space]].
+From [[Condition for Open Extension Space to be T5 Space|Condition for Open Extension Space to be $T_5$ Space]], it follows that $\paren {T \times D}^*_{\bar p}$ is not a [[Definition:Kolmogorov Space|$T_5$ space]].
+From [[Open Extension Topology is not T1|Open Extension Topology is not $T_1$]], we have that $\paren {T \times D}^*_{\bar p}$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ space]].
+From [[Open Extension Topology is not T3|Open Extension Topology is not $T_3$]] we have that $\paren {T \times D}^*_{\bar p}$ is not a [[Definition:T3 Space|$T_3$ space]].
+Finally, from [[T2 Space is T1 Space|$T_2$ Space is $T_1$ Space]], as $\paren {T \times D}^*_{\bar p}$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ space]] it is not a [[Definition:Hausdorff Space|$T_2$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Saturated Models of same Cardinality are Isomorphic}
+Tags: Model Theory
+
+\begin{theorem}
+Let $T$ be an $\LL$-[[Definition:Theory (Logic)|theory]].
+Let $\kappa$ be an [[Definition:Infinite|infinite]] [[Definition:Cardinal|cardinal]].
+If $\MM$ and $\NN$ are [[Definition:Saturated Model|saturated models]] of $T$ and the [[Definition:Cardinality|cardinality]] of the universes of $\MM$ and $\NN$ are both $\kappa$, then $\MM$ and $\NN$ are [[Definition:Isomorphism (Model Theory)|isomorphic]].
+\end{theorem}
+
+\begin{proof}
+=== Outline ===
+The idea of the proof is that since the models are saturated, we can define an isomorphism $f: \NN \to \MM$ by picking the image $\map f x$ of an element $x \in \NN$ as something which realizes the type that $\map f x$ would need to satisfy in $\MM$.
+This is done using a (transfinite) recursive construction which extends an elementary map at each step.
+The proof is very similar to that of [[Saturated Implies Universal]]; in both proofs we attempt to construct a formula-preserving map.
+In this case, however, the map must be bijective between the universes of the models.
+=== Body of Proof ===
+First, since $\MM$ and $\NN$ have cardinality $\kappa$, we can denote their elements by $m_\alpha$ and $n_\alpha$ respectively for ordinals $\alpha < \kappa$.
+For each $\alpha < \kappa$, let $A_\alpha$ denote the domain of the $f_\alpha$ we define below.  Note that we do not know in advance what the domains will be.
+Base case $\alpha = 0$:
+Define $f_0 = \O$.
+Note  that $f_0$ is trivially an [[Definition:Elementary Embedding|elementary embedding]] from $A_0 = \O$ into $\MM$.
+{{qed|lemma}}
+[[Definition:Limit  Ordinal|Limit ordinals]] $\alpha$, assuming $f_\beta: A_\beta \to \MM$ is defined and elementary, and that $\card {A_\beta} < \kappa$ for all $\beta < \alpha$:
+Let $\displaystyle f_\alpha = \bigcup_{\beta < \alpha} f_\beta$.
+If  $\phi$ is an $\LL$-sentence with parameters from $A_\alpha$, then since it involves only finitely many such parameters, they must all  be contained in some $A_\beta$ for $\beta < \alpha$.
+But $f_\alpha  \restriction A_\beta = f_\beta$ is elementary, so $f_\alpha$ must be as well.
+Note that $\card {A_\alpha} < \kappa$ by [[Cardinality of Infinite Union of Infinite Sets]].
+{{qed|lemma}}
+[[Definition:Successor  Ordinal|Successor ordinals]] $\alpha = \beta + 1$, assuming $f_\beta: A_\beta \to \MM$ is defined and elementary, and that $\card {A_\beta} < \kappa$:
+We need to extend $f_\beta$ to some $f_\alpha$ so that the domain includes $n_\beta$, the image includes $m_\beta$, and so that $\LL$-formulas with parameters from $A_\alpha$ are preserved by $f_\alpha$.
+First we add $n_\beta$ to the domain:
+Consider the subset:
+:$p = \set {\map \phi {v, \map {f_\beta} {\bar a} }: \bar a \text { is a tuple  from } A_\beta \text { and } \NN \models \map \phi {n_\beta, \bar a} }$
+of the set of $\LL$-formulas with one free variable and parameters from the image $\map {f_\beta} {A_\beta}$ of $A_\beta$ under $f_\beta$.
+The set $p$ is a $1$-type over the image $\map {f_\beta} {A_\beta}$ in $\MM$.
+Since $\card {A_\beta} < \kappa$ by the inductive hypothesis and since by assumption $\MM$ is $\kappa$-saturated, this means that $p$ is realized in $\MM$ by some element $b$.
+Thus $f'_\alpha = f_\beta \cup \set {\tuple {n_\beta, b} }$ is an elementary embedding $A_\beta \cup \set {n_\beta} \to \MM$.
+Now we add $m_\beta$ to the image:
+This is done similarly to the above.
+Consider the subset:
+:$q = \set {\map \phi {v, \bar a}: \bar a \text { is a tuple  from } A_\beta \cup \set {n_\beta} \text { and } \MM \models \map \phi {m_\beta, \map {f'_\alpha} {\bar a} } }$
+of the set of $\LL$-formulas with one free variable and parameters from $A_\beta \cup \{n_\beta\}$.
+The set $q$ is a $1$-type over $A_\beta \cup \{n_\beta\}$ in $\NN$.
+Since $\card {A_\beta} < \kappa$ by the inductive hypothesis and hence $\card {A_\beta \cup \set {n_\beta} } < \kappa$ as well, and since by assumption $\NN$ is $\kappa$-saturated, this means that $q$ is realized in $\NN$ by some element $c$.
+Thus $f_\alpha = f'_\alpha \cup \set {\tuple {c, m_\beta} } = f_\beta \cup \set {\tuple {n_\beta, b}, \tuple {c, m_\beta} }$ is an elementary embedding $A_\beta \cup \set {n_\beta} \to \MM$ which includes $m_\beta$ in its range.
+Finally define $\displaystyle f = \bigcup_{\alpha < \kappa} f_\alpha$.
+The map $f$ is an elementary embedding $\NN \to \MM$ since $\displaystyle \bigcup_{\alpha < \kappa} A_\alpha = \NN$, any finite set of parameters from $\NN$ must belong to some single $A_\alpha$, and $f\restriction A_\alpha$ is elementary.
+$f$ is onto $\MM$ since we have constructed it so that $m_\alpha$ is in the image of $f_{\alpha + 1}$.
+So, $f$ is a bijection.
+Since elementary embeddings are by definition $\LL$-embeddings, this means that $f$ is an isomorphism by definition.
+{{qed}}
+[[Category:Model Theory]]
+1mf8vzyw8ow9lnqwkgsdn87sjxzzb2a
+\end{proof}<|endoftext|>
+\section{Compact Complement Topology is Topology}
+Tags: Compact Complement Topology
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Space|compact complement space]].
+Then $\tau$ is a [[Definition:Topology|topology]] on $T$.
+\end{theorem}
+
+\begin{proof}
+By definition, we have that $\O \in \tau$.
+We also have that $\R \in \tau$ as $\relcomp \R \R = \O$ which is trivially [[Definition:Compact Topological Subspace|compact]].
+Now suppose $A, B \in \tau$.
+Let $H = A \cap B$.
+Then:
+{{begin-eqn}}
+{{eqn | l = H
+      | r = A \cap B
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp \R H
+      | r = \relcomp \R {A \cap B}
+      | c = 
+}}
+{{eqn | r = \relcomp \R A \cup \relcomp \R B
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection|De Morgan's Laws: Complement of Intersection]]
+}}
+{{end-eqn}}
+But as $A, B \in \tau$ it follows that $\relcomp \R A$ and $\relcomp \R B$ are both [[Definition:Compact Topological Subspace|compact]].
+Hence from [[Union of Compact Sets is Compact]] their [[Definition:Set Union|union]] is also [[Definition:Compact Topological Subspace|compact]] and so $\relcomp \R H$ is [[Definition:Compact Topological Subspace|compact]].
+So $H = A \cap B \in \tau$ as its [[Definition:Relative Complement|complement]] is [[Definition:Compact Topological Subspace|compact]].
+Now let $\UU \subseteq \tau$.
+Then from [[De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Union|De Morgan's Laws: Complement of Union]]:
+:$\displaystyle \relcomp \R {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp \R U$
+But as:
+:$\forall U \in \UU: \relcomp \R U \in \tau$
+each of the $\relcomp \R U$ is [[Definition:Compact Topological Subspace|compact]].
+Hence from [[Intersection of Compact Sets is Compact]], so is their [[Definition:Set Intersection|intersection]].
+So $\displaystyle \relcomp \R {\bigcup \UU}$ is [[Definition:Compact Topological Subspace|compact]] which means $\displaystyle \bigcup \UU \in \tau$.
+So $\tau$ is a [[Definition:Topology|topology]] on $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Ramsey's Theorem}
+Tags: Ramsey Theory
+
+\begin{theorem}
+Let $k, n \in \N$.
+For any set $S$, let $S^{\left({n}\right)}$ denote the set $\left\{ {\left\{ {s_1, \ldots, s_n}\right\}: \text{each } s_i \in S}\right\}$ of [[Definition:Cardinality|cardinality]] $n$ [[Definition:Subset|subsets]] of $S$.
+Let $X$ be an [[Definition:Infinite Set|infinite set]].
+Then:
+:for every [[Definition:Partition (Set Theory)|partition]] $P$ of $X^{\left({n}\right)}$ into $k$ many components
+:there is an infinite subset $Y \subseteq X$
+such that:
+:each member of $Y^{\left({n}\right)}$ is in the same component of $P$.
+\end{theorem}
+
+\begin{proof}
+We will prove the theorem for fixed $k$ by [[Proof by Mathematical Induction|induction]] on $n$.
+=== Basis for the Induction ===
+The case $n = 1$ ("Infinite Pigeonhole Principle"):
+In this case, $X^{\left({n}\right)}$ is the set of singletons of elements of $X$.
+So to ease exposition, we will identify $X^{\left({n}\right)}$ with $X$ and just speak of partitions of $X$.
+Let $P$ be a partition of $X$ into $k$ many components $X_1, \dots, X_k$.
+We have:
+:$X = X_1 \cup \cdots \cup X_k$
+If each $X_i$ were finite, then $X$ would be a finite union of finite sets, and hence finite.
+Thus, at least one of the $X_i$ must be infinite.
+But taking $Y$ to be an infinite $X_i$ gives us the required subset for the theorem.
+This is the [[Proof by Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+The [[Proof by Mathematical Induction#Induction Hypothesis|induction hypothesis]] is that:
+For every [[Definition:Partition (Set Theory)|partition]] $P$ of $X^{\left({n - 1}\right)}$ into $k$ many components
+:there is an infinite subset $Y \subseteq X$
+such that:
+:each member of $Y^{\left({n - 1}\right)}$ is in the same component of $P$.
+It is to be proved that:
+For every [[Definition:Partition (Set Theory)|partition]] $P$ of $X^{\left({n}\right)}$ into $k$ many components
+:there is an infinite subset $Y \subseteq X$
+such that:
+:each member of $Y^{\left({n}\right)}$ is in the same component of $P$.
+=== Induction Step ===
+This is the [[Proof by Mathematical Induction#Induction Step|induction step]]:
+Suppose the theorem holds for all $i < n$.
+It is to be proved that it holds for $n$.
+Since $X$ is infinite, there is an [[Definition:Injection|injection]] of $\N$ into $X$.
+All that is needed to be done to prove the theorem is to find an infinite subset of $X$
+So, to ease exposition, $\N$ will be viewed as a subset of $X$.
+The proof will proceed by finding an infinite subset of $\N$.
+Write the components of $P$ as $S_1, \ldots, S_k$.
+For each $a \in \N$, let $P_a$ be the partition on $\left({\N \setminus \left\{ {a}\right\} }\right)^{ \left({n - 1}\right)}$ into $k$ components:
+:$S_{a, 1}, \dots, S_{a, k}$
+defined by assigning:
+:$\left\{ {b_1, \ldots, b_{n - 1} }\right\} \in \left({\N \setminus \left\{ {a}\right\} }\right)^{\left({n - 1}\right)}$
+to $S_{a, j}$ {{iff}}:
+: $\left\{ {a, b_1, \ldots, b_{n - 1} }\right\} \in \N^{\left({n}\right)}$ is in $S_j$
+Note that the theorem holds for $P_a$ by the [[Infinite Ramsey's Theorem#Induction Hypothesis|induction hypothesis]].
+Now, we inductively define $a_0 < a_1 < \dots$ and infinite sets $X_0 \supseteq X_1 \supseteq \cdots$ such that $a_i$ is the smallest element of $X_i$.
+Let $a_0 = 0$, and let $X_0 = \N$.
+Once $X_i$ and $a_i$ are defined:
+Let $X_{i + 1}$ be an infinite subset of $X_i \setminus \left\{{a_i}\right\}$ satisfying the theorem for the partition $P_{a_i}$ of $n-1$ size subsets.
+This exists by the [[Infinite Ramsey's Theorem#Induction Hypothesis|induction hypothesis]].
+Let $a_{i + 1}$ be the smallest element of $X_{i+1}$.
+This exists by the [[Well-Ordering Principle]].
+Note that $a_{i + 1} > a_i$ since $X_{i + 1}$ is a subset of $X_i$ not containing the smallest element $a_i$ of $X_i$.
+We constructed each $X_{i + 1}$ so that each element of $X_{i + 1}^{\left({n - 1}\right)}$ is in the same component of $P_{a_i}$.
+So, we may consider the sequence $\left({k_i}\right)_{i \in \N}$ where each $k_i$ is such that all elements of $X_{i + 1}^{\left({n - 1}\right)}$ are in the component $S_{a_i, k_i}$ of $P_{a_i}$.
+Since each $k_i$ is one of $1, \ldots, k$, but there are infinitely many $i \in \N$, this forms a partition of $\N$ to which we can apply the base case of this proof.
+This means that there is some $c$ among $1, \ldots, k$, there is an infinite set $I = \left\{ {i \in \N : k_i = c}\right\}$.
+Define $Y = \left\{ {a_i: i \in I}\right\}$.
+$Y$ is infinite since $I$ is infinite and each $a_i$ is distinct by construction.
+We now verify that each element of $Y^{\left({n}\right)}$ is in the same component $S_c$ of $P$.
+Let $\left\{ {y_1, \ldots, y_n}\right\} \in Y^{\left({n}\right)}$.
+We may take $y_1 < \cdots < y_n$ by relabeling if necessary.
+There is some $i \in I$ for which $y_1 = a_i$.
+By construction:
+:$y_1 = a_i$ is the least element of $X_i$
+:each $y_2, \dots, y_n$ is in $X_i$
+and:
+:each element of $X_i^{\left({n - 1}\right)}$ is in the component $S_{a_i, c}$ of the partition $P_{a_i}$.
+But then, by definition of $P_{a_i}$, since $\left\{ {y_2, \ldots, y_n}\right\} \in S_{a_i , c}$, we have:
+:$\left\{{y_1, y_2, \ldots, y_n}\right\} = \left\{ {a_i, y_2, \ldots, y_n}\right\} \in S_c$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Complement Topology is T1}
+Tags: T1 Spaces, Compact Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Compact Complement Topology|compact complement space]].
+Then $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Finite Topological Space is Compact]].
+So any [[Definition:Finite Set|finite]] [[Definition:Topological Subspace|subspace]] of $T$ is [[Definition:Compact Topological Subspace|compact]].
+Let $\tau^*$ be the [[Definition:Set|set]] defined as:
+:$\tau^* = \leftset {S \subseteq \R: S = \O \text { or } \relcomp \R S}$ is [[Definition:Finite Set|finite]] $\rightset {}$
+where $\relcomp \R S$ denotes the [[Definition:Relative Complement|complement of $S$ in $\R$]].
+Then $\tau^*$ is a [[Definition:Subset|subset]] of $\tau$ by definition of the [[Definition:Compact Complement Topology|compact complement topology]].
+But $\tau^*$ is the [[Definition:Finite Complement Topology|finite complement topology]].
+The result follows from [[Finite Complement Topology is Minimal T1 Topology|Finite Complement Topology is Minimal $T_1$ Topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Complement Topology is Irreducible}
+Tags: Irreducible Spaces, Compact Complement Topology
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
+Then $T$ is an [[Definition:Irreducible Space|irreducible space]].
+\end{theorem}
+
+\begin{proof}
+Let $U_1, U_2 \in \tau$ be [[Definition:Open Set (Topology)|open]] in $T$.
+Let $\relcomp \R {U_1} = V_1$ and $\relcomp \R {U_2} = V_2$.
+By definition of [[Definition:Compact Complement Topology|compact complement topology]], $V_1, V_2 \subseteq \R$ such that $V_1$ and $V_2$ are both [[Definition:Compact Topological Subspace|compact]].
+$V_1$ and $V_2$ are both [[Definition:Bounded Subset of Real Numbers|bounded]] by definition of [[Definition:Compact Topological Subspace|compact]] in $\R$, so their [[Definition:Set Union|union]] is likewise [[Definition:Bounded Subset of Real Numbers|bounded]]: [[Definition:Bounded Above Set|above]] by $\max \set {\sup V_1, \sup V_2}$ and [[Definition:Bounded Below Set|below]] by $\min \set {\inf V_1, \inf V_2}$.
+So $V_1 \cup V_2$ can not equal $\R$ as $\R$ is not [[Definition:Bounded Subset of Real Numbers|bounded]].
+So:
+{{begin-eqn}}
+{{eqn | l = \relcomp \R {V_1 \cup V_2}
+      | o = \ne
+      | r = \O
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp \R {V_1} \cap \relcomp \R {V_2}
+      | o = \ne
+      | r = \O
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union|De Morgan's Laws: Complement of Union]]
+}}
+{{eqn | ll= \leadsto
+      | l = U_1 \cap U_2
+      | o = \ne
+      | r = \O
+      | c = 
+}}
+{{end-eqn}}
+So any two [[Definition:Open Set (Topology)|open sets]] in $T$ are not [[Definition:Disjoint Sets|disjoint]] and so $T$ is [[Definition:Irreducible Space|irreducible space]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Complement Space is not T2, T3, T4 or T5}
+Tags: Completely Hausdorff Spaces, Hausdorff Spaces, T3 Spaces, T4 Spaces, T5 Spaces, Compact Complement Topology
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
+Then $T$ is not a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]], [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]], [[Definition:T3 Space|$T_3$ space]], [[Definition:T4 Space|$T_4$ space]] or [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Compact Complement Topology is T1|Compact Complement Space is a $T_1$ space]].
+From [[T1 Space is T0 Space|$T_1$ Space is $T_0$ Space]], $T$ is a [[Definition:Kolmogorov Space|$T_0$ space]].
+We have:
+:[[Compact Complement Topology is Irreducible]]
+:[[Irreducible Hausdorff Space is Singleton]]
+From [[Completely Hausdorff Space is Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space is $T_2$ (Hausdorff) space]], $T$ is not a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+From [[Regular Space is T2 Space|Regular Space is $T_2$ Space]], $T$ is not a [[Definition:Regular Space|regular space]].
+By definition, a [[Definition:Regular Space|regular space]] is a space that is both a [[Definition:Kolmogorov Space|$T_0$ space]] and a [[Definition:T3 Space|$T_3$ space]].
+But $T$ is a [[Definition:Kolmogorov Space|$T_0$ space]] and not a [[Definition:Regular Space|regular space]].
+So it follows that $T$ can not be a [[Definition:T3 Space|$T_3$ space]].
+Next we have that a [[Normal Space is T3 Space|Normal Space is a $T_3$ Space]].
+But as $T$ is not a [[Definition:T3 Space|$T_3$ space]], $T$ can not be a [[Definition:Normal Space|normal space]].
+By definition, a [[Definition:Normal Space|normal space]] is a space that is both a [[Definition:Fréchet Space (Topology)|$T_1$ space]] and a [[Definition:T4 Space|$T_4$ space]].
+But $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ space]] and not a [[Definition:Normal Space|normal space]].
+So it follows that $T$ can not be a [[Definition:T4 Space|$T_4$ space]].
+Finally we have that a [[T5 Space is T4 Space|$T_5$ Space is a $T_4$ Space]].
+But as $T$ is not a [[Definition:T4 Space|$T_4$ space]], $T$ can not be a [[Definition:T5 Space|$T_5$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Weierstrass-Casorati Theorem}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $f$ be a [[Definition:Holomorphic Function|holomorphic function]] defined on the [[Definition:Open Ball|open ball]] $\map B {a, r} \setminus \set a$.
+Let $f$ have an [[Definition:Essential Singularity|essential singularity]] at $a$.
+Then:
+:$\forall s < r: f \sqbrk {\map B {a, s} \setminus \set a}$ is a [[Definition:Dense|dense]] [[Definition:Subset|subset]] of $\C$.
+{{Disambiguate|Definition:Dense}}
+\end{theorem}
+
+\begin{proof}
+{{WLOG}}, suppose $a = 0$ and $r = 1$.
+{{AimForCont}} $\exists s < 1$ such that:
+:$f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$.
+Then, by definition of [[Definition:Dense|dense subset]]:
+:$\exists z_0 \in \C: \exists r_0 > 0: \map B {z_0, r_0} \cap f \sqbrk {\map B {0, s} \setminus \set 0} = \O$
+{{explain|This definition of dense still needs to be added to {{ProofWiki}}. Whichever definition of denseness (either everywhere dense or dense-in-itself) will need to be expanded for the Complex case so as to make it clear that the above definition follows.}}
+Hence, the function $\varphi$ defined on $\map B {z_0, r_0}$ by:
+:$\map \varphi z = \dfrac 1 {\map f z - z_0}$
+is [[Definition:Analytic Function|analytic]] on $\map B {0, s} \setminus \set 0$ and [[Definition:Bounded Mapping|bounded]] near to $0$, because:
+:$\forall z \in \map B {0, s} \setminus \set 0: \cmod {\map f z - z_0} > r_0 \implies \cmod {\map \varphi z} < \dfrac 1 {r_0}$
+Therefore, we can extend the domain of $\varphi$ (using the [[Analytic Continuation Principle]]).
+Let $\map \varphi 0 \ne 0$.
+Then:
+:$\map f 0 = z_0 + \dfrac 1 {\map \varphi 0}$
+and the [[Definition:Singularity|singularity]] of $f$ was [[Definition:Removable Singularity|removable]].
+Otherwise, let the [[Definition:Power Series|power series]] of $\varphi$ be written:
+:$\displaystyle \map \varphi z = \sum_{n \mathop = 1}^{+\infty} a_n z^n$
+Then as $\varphi \ne 0$:
+:$E = \set {k \in \N: a_k \ne 0} \ne \O$
+Let $p = \min E$.
+Then $0$ is a [[Definition:Pole (Complex Analysis)|pole]] of order $p$ of $f$.
+In each case, the assumption that:
+:$\exists s < 1: f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$ contradicts the fact that $0$ is an essential singularity of $f$.
+Hence the result, by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Complement Topology is Connected}
+Tags: Connected Spaces, Compact Complement Topology
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]].
+Then $T$ is a [[Definition:Connected Topological Space|connected space]].
+\end{theorem}
+
+\begin{proof}
+Follows from:
+:[[Compact Complement Topology is Irreducible]]
+:[[Irreducible Space is Connected]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Complement Topology is Locally Connected}
+Tags: Locally Connected Spaces, Compact Complement Topology
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
+Then $T$ is a [[Definition:Locally Connected Space|locally connected space]].
+\end{theorem}
+
+\begin{proof}
+Follows from:
+:[[Compact Complement Topology is Irreducible]]
+:[[Irreducible Space is Locally Connected]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Complement Topology is not Ultraconnected}
+Tags: Ultraconnected Spaces, Compact Complement Topology, Compact Complement Topology is not Ultraconnected
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
+Then $T$ is not an [[Definition:Ultraconnected Space|ultraconnected space]].
+\end{theorem}
+
+\begin{proof}
+By definition, [[Definition:Closed Set (Topology)|closed sets]] of $T$ are [[Definition:Compact Topological Space|compact sets]] of $T$.
+So, for example, $\left[{0 \,.\,.\, 1}\right]$ and $\left[{2 \,.\,.\, 3}\right]$ are [[Definition:Disjoint Sets|disjoint]] [[Definition:Compact Topological Space|compact sets]] and therefore [[Definition:Closed Set (Topology)|closed sets]] of $T$.
+Hence the result by definition of [[Definition:Ultraconnected Space|ultraconnected]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have:
+: [[Compact Complement Topology is T1|Compact Complement Topology is $T_1$]]
+: [[Non-Trivial Ultraconnected Space is not T1|Non-Trivial Ultraconnected Space is not $T_1$]]
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Complement Topology is Compact}
+Tags: Compact Spaces, Compact Complement Topology
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
+Then $T$ is a [[Definition:Compact Topological Space|compact space]].
+\end{theorem}
+
+\begin{proof}
+Let $\CC$ be an [[Definition:Open Cover|open cover]] of $\R$.
+Let $U \in \CC$.
+Then by definition $\R \setminus U$ is [[Definition:Compact Topological Subspace|compact]] in the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
+Since each $U$ is [[Definition:Open Set (Topology)|open]] in the [[Definition:Euclidean Topology on Real Number Line|Euclidean topology]], a [[Definition:Finite Set|finite number]] must cover $\R \setminus U$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Local Basis in Compact Complement Topology}
+Tags: Compact Complement Topology
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
+Let $p \in \R$.
+Then [[Definition:Set|sets]] of the form:
+:$\openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty$
+form a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] for $p$.
+\end{theorem}
+
+\begin{proof}
+Let $p \in \R$.
+Let:
+:$\BB_p = \set {\openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty: n \in \N}$
+Let $n \in \N$ and $P_n \in \BB$, so that:
+:$P_n = \openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty$
+Then:
+:$\R \setminus P_n = \closedint {-n} {p - \dfrac 1 n} \cup \closedint {p + \dfrac 1 n} n$
+which is the [[Definition:Set Union|union]] of two [[Definition:Compact Topological Subspace|compact sets]] in $\R$ and therefore itself [[Definition:Compact Topological Subspace|compact]].
+Clearly $p \in P_n$.
+So $\BB$ is a set of [[Definition:Open Neighborhood|open neighborhoods]] of $p$ in $T$
+Now let $U \in \tau$ such that $p \in U$.
+Then $V := \R \setminus U$ is [[Definition:Compact Topological Subspace|compact]] and is therefore [[Definition:Bounded Subset of Real Numbers|bounded]].
+Suppose $\sup V = M, \inf V = m$.
+Then $\exists n_B \in \N: n_T > M, -n_B > m$.
+But as $p \notin V$ we have:
+:$\exists \epsilon \in \R: p + \epsilon \notin V, p - \epsilon \notin V$
+So $\exists n \in \N: \epsilon > \dfrac 1 n$
+and so by making $n$ large enough you can fix it so that:
+:$\openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty \subseteq U$
+so fulfilling the conditions for $\BB$ to be a [[Definition:Local Basis|local basis]] for $p$ which is [[Definition:Countable Set|countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Complement Topology is First-Countable}
+Tags: First-Countable Spaces, Compact Complement Topology
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
+Then $T$ is a [[Definition:First-Countable Space|first-countable space]].
+\end{theorem}
+
+\begin{proof}
+Let $\CC$ be an [[Definition:Open Cover|open cover]] of $\R$.
+Let $p \in \R$.
+Consider the set:
+:$\BB_p = \set {\openint {-\infty} {-n} \cup \openint {p - \dfrac 1 n} {p + \dfrac 1 n} \cup \openint n \infty: n \in \N}$
+From [[Countable Local Basis in Compact Complement Topology]], $\BB_p$ is a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] for $T$.
+Hence the result, by definition of [[Definition:First-Countable Space|first-countable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Complement Topology is Second-Countable}
+Tags: Second-Countable Spaces, Compact Complement Topology
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
+Then $T$ is a [[Definition:Second-Countable Space|second-countable space]].
+\end{theorem}<|endoftext|>
+\section{Compact Complement Topology is Sequentially Compact}
+Tags: Sequentially Compact Spaces, Compact Complement Topology
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
+Then $T$ is a [[Definition:Sequentially Compact Space|sequentially compact space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Compact Complement Topology is Compact]]
+:[[Compact Space is Countably Compact]]
+:[[Compact Complement Topology is First-Countable]]
+:[[First-Countable Space is Sequentially Compact iff Countably Compact]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Complement Topology is Coarser than Euclidean Topology}
+Tags: Compact Spaces, Compact Complement Topology
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
+Then $\tau$ is [[Definition:Coarser Topology|coarser]] than the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]] in $\R$.
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$.
+Then $V = \R \setminus U$ is [[Definition:Compact Topological Subspace|compact]].
+So by definition $V$ is [[Definition:Closed Set (Topology)|closed]] in the [[Definition:Euclidean Topology on Real Number Line|Euclidean topology]].
+That is, $U = \R \setminus V$ is [[Definition:Open Set (Topology)|open]] in the [[Definition:Euclidean Topology on Real Number Line|Euclidean topology]].
+That is, every [[Definition:Open Set (Topology)|open set]] in the [[Definition:Compact Complement Topology|compact complement topology]] is also [[Definition:Open Set (Topology)|open]] in the [[Definition:Euclidean Topology on Real Number Line|Euclidean topology]].
+Hence the result, by definition of [[Definition:Coarser Topology|coarser topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Theories with Infinite Models have Models with Order Indiscernibles}
+Tags: Model Theory
+
+\begin{theorem}
+Let $T$ be an $\LL$-[[Definition:Theory (Logic)|theory]] with [[Definition:Infinite|infinite]] [[Definition:Model (Logic)|models]].
+Let $\struct {I, <$ be an infinite [[Definition:Strict Total Ordering|strict linearly ordered set]].
+There is a model $\MM \models T$ containing an [[Definition:Order Indiscernible|order indiscernible]] set $\set {x_i : i \in I}$.
+\end{theorem}
+
+\begin{proof}
+We will construct the claimed model using the [[Compactness Theorem]].
+Let $LL^*$ be the language obtained by adding constant symbols $c_i$ to $\LL$ for each $i \in I$.
+Let $T^*$ be the $\LL^*$-theory obtained by adding to $T$ the $\LL^*$-sentences:
+:$c_i \ne c_j$ for each $i \ne j$ in $I$
+and:
+:$\map \phi {c_{i_1}, \dotsc, c_{i_n} } \leftrightarrow \map \phi {c_{j_1}, \dotsc, c_{j_n} }$ for each $n \in \N$, each $\LL$-formula $\phi$ with $n$ [[Definition:Free Variable|free variables]], and each pair of [[Definition:Chain (Set Theory)|chains]] $i_1 < \cdots < i_n$ and $j_1 < \cdots < j_n$ in $I$.
+For future reference, we will refer to these last sentences using as those which '''assert indiscernibility with respect to $\phi$'''.
+If we can find a model of $T^*$, then its interpretations of the constants $c_i$ for $i\in I$ will be order indiscernibles.
+Suppose $\Delta$ is a finite subset of $T^*$.
+{{refactor|Extract this section as a lemma}}
+:We will show that there is a model of $\Delta$ using the [[Infinite Ramsey's Theorem]].
+:Let $I_\Delta$ be the finite subset of $I$ containing those $i$ for which $c_i$ occurs in some sentence in $\Delta$.
+:Let $\psi_1, \dots, \psi_k$ be the finitely many $\LL$-formulas in $\Delta$ which assert indiscernibility with respect to $\phi_1, \dots, \phi_k$.
+:Let $m$ be the maximum number of free variables in the formulas $\phi_1, \dots, \phi_k$. 
+:Let $\MM$ be an infinite model of $T$ (which exists by assumption).
+:Let $X$ be any infinite subset of the universe of $\MM$ such that $X$ is linearly ordered by some relation $\prec$.
+:Define a [[Definition:Set Partition|partition]] $P$ of $X^{\paren m} = \set {X' \subseteq X: \card {X'} = m}$ into $2^k$ components $S_A$ for each $A \subseteq \set {1, 2, \dotsc, k}$ by:
+::$\set {x_1, \dots, x_m} \in S_A \iff x_1 \prec \cdots \prec x_m \text { and } A = \set {i: \MM \models \map {\phi_i} {x_1, \dotsc, x_m} }$
+:By [[Infinite Ramsey's Theorem]], there is an infinite subset $Y\subseteq X$ such that each element of $Y^{\paren m} = \set {Y' \subseteq Y: \card {Y'} = m}$ is in the same component $S_A$ of $P$.
+:Note that $Y$ is indexed by some infinite subset $J_Y$ of $J$ and hence still linearly ordered by $\prec$.
+:We now show that the constants $c_i$ for $i \in I_\Delta$ can be interpreted as elements of $Y$ in $\MM$, and that the sentences $\psi_1, \dots, \psi_k$ will be satisfied using this interpretation.
+:Since $I_\Delta$ is finite and linearly ordered, and $J_Y$ is infinite and linearly ordered, there is clearly an [[Definition:Increasing|increasing]] function $f:I_\Delta \to J_Y$.
+:For each $h = 1, \dotsc, k$ and any pair of chains $i_1 < \cdots < i_n$ and $j_1 < \cdots < j_n$ in $I_\Delta$, we thus have:
+::$\MM \models \map {\phi_h} {y_{\map f {i_1} }, \dotsc, y_{\map f {i_n} } }$ {{iff}} $h \in A$ {{iff}} $\MM \models \map {\phi_h} {y_{\map f {j_1} }, \dotsc, y_{\map f {j_n} } }$
+:and hence:
+::$\MM \models \map {\phi_h} {y_{\map f {i_1} }, \dotsc, y_{\map f {i_n} } } \iff \map {\phi_h} {y_{\map f {j_1} }, \dotsc, y_{\map f {j_n} } }$
+:So, if we interpret each $c_i$ for $i \in I_\Delta$ as $y_{\map f i}$, we will have $\MM \models \psi_h$ for each $h = 1, \dotsc, k$.
+:Thus, we have shown that $\MM$ can be extended to an $\LL^*$-structure which models $\Delta$.
+Hence all finite subsets $\Delta$ of $T^*$ are satisfiable.
+By the [[Compactness Theorem]], we have that $T^*$ is satisfiable.
+Let $\NN$ be any model of $T^*$.
+Let $n_i$ be the interpretation of $c_i$ in $\NN$ for each $i \in I$.
+Then $\set {n_i: i \in I}$ is easily seen to be an order indiscernible set, since $T^*$ was defined to include sentences guaranteeing such.
+{{qed}}
+[[Category:Model Theory]]
+desk4iksvybgz4e4huhqre1dhkqymuh
+\end{proof}<|endoftext|>
+\section{Fort Topology is Topology}
+Tags: Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]].
+Then $\tau_p$ is a [[Definition:Topology|topology]] on $T$.
+\end{theorem}
+
+\begin{proof}
+We have that $p \in \relcomp S \O = S$, so $\O \in \tau_p$.
+We have that $\relcomp S S = \O$ which is [[Definition:Finite Set|finite]], so $S \in \tau_p$.
+Now consider $A, B \in \tau_p$, and let $H = A \cap B$.
+If $p \notin A$ or $p \notin B$ then $p \notin A \cap B$ and so $H \in \tau_p$.
+Now suppose $p \in A$ and $p \in B$.
+Then:
+{{begin-eqn}}
+{{eqn | l = H
+      | r = A \cap B
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp S H
+      | r = \relcomp S {A \cap B}
+      | c = 
+}}
+{{eqn | r = \relcomp S A \cup \relcomp S B
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union|De Morgan's Laws: Complement of Union]]
+}}
+{{end-eqn}}
+In order for $A$ and $B$ to be [[Definition:Open Set (Topology)|open sets]] it follows that $\relcomp S A$ and $\relcomp S B$ are both [[Definition:Finite Set|finite]].
+Hence their [[Definition:Set Union|union]] is also [[Definition:Finite Set|finite]] and so $\relcomp S H$ is [[Definition:Finite Set|finite]].
+So $H = A \cap B \in \tau_p$ as its [[Definition:Relative Complement|complement]] is [[Definition:Finite Set|finite]].
+Now let $\UU \subseteq \tau_p$.
+Then from [[De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Union|De Morgan's Laws: Complement of Union]]:
+:$\displaystyle \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
+We have either of two options:
+:$(1): \quad \forall U \in \UU: p \in \relcomp S U$
+in which case:
+:$\displaystyle p \in \bigcap_{U \mathop \in \UU} \relcomp S U$
+Or:
+:$(2): \quad \exists U \in \UU: \relcomp S U$ is [[Definition:Finite Set|finite]]
+in which case:
+:$\displaystyle \bigcap_{U \mathop \in \UU} \relcomp S U$ is [[Definition:Finite Set|finite]], from [[Intersection is Subset]].
+So in either case $\displaystyle \bigcup \UU \in \tau_p$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fort Space is Excluded Point Space with Finite Complement Space}
+Tags: Fort Spaces, Excluded Point Topology, Finite Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $\tau_p$ is the [[Definition:Minimal Topology|minimal topology]] that is [[Definition:Topology Generated by Synthetic Sub-Basis|generated]] by the [[Definition:Excluded Point Topology|excluded point topology]] and the [[Definition:Finite Complement Topology|finite complement topology]].
+\end{theorem}
+
+\begin{proof}
+Let $T_1 = \struct {S, \tau_1}$ be the [[Definition:Excluded Point Space|excluded point space]] on $S$ from $p$.
+Let $T_2 = \struct {S, \tau_2}$ be the [[Definition:Finite Complement Space|finite complement space]] on $S$.
+By definition:
+:$\tau_1 = \set {H \subseteq S: p \in \relcomp S H} \cup \set S$ 
+:$\tau_2 = \leftset {H \subseteq S: \relcomp S H}$ is [[Definition:Finite Set|finite]]$\rightset {} \cup \set \O$
+By definition of [[Definition:Fort Space|Fort space]], we have:
+:$U \in \tau_1 \implies U \in \tau_p$
+:$U \in \tau_2 \implies U \in \tau_p$
+So $\tau_1 \cup \tau_2 \subseteq \tau_p$.
+Similarly:
+:$U \in \tau_p \implies U \in \tau_1 \lor U \in \tau_2$
+and so $\tau_p \subseteq \tau_1 \cup \tau_2$.
+So $\tau_p = \tau_1 \cup \tau_2$ and the result follows from [[Union is Smallest Superset]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fort Space is T1}
+Tags: T1 Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+From [[Fort Space is Excluded Point Space with Finite Complement Space]], $T$ is an [[Definition:Expansion of Topology|expansion]] of a [[Definition:Finite Complement Space|finite complement space]].
+Then we have that a [[Finite Complement Space is T1|Finite Complement Space is $T_1$]].
+Then from [[Separation Properties Preserved by Expansion]] we have that as a [[Definition:Finite Complement Space|finite complement space]] is a [[Definition:Fréchet Space (Topology)|$T_1$ space]], then so is $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fort Space is T5}
+Tags: T5 Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $T$ is a [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $A, B \in \tau_p$ such that $A$ and $B$ are [[Definition:Separated Sets|separated]].
+If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both [[Definition:Open Set (Topology)|open]].
+Otherwise $p$ must be in exactly one of them, because if $p$ were in both they could not be [[Definition:Separated Sets|separated]].
+{{WLOG}}, suppose that $p \in A$.
+Then $p \notin B$ so $B$ is [[Definition:Open Set (Topology)|open]].
+Now suppose $B$ were not [[Definition:Closed Set (Topology)|closed]].
+Then $B$ is [[Definition:Infinite Set|infinite]] by definition of [[Definition:Fort Space|Fort space]].
+But every [[Definition:Open Set (Topology)|open set]] containing $p$ is [[Definition:Cofinite Set|cofinite]] in $S$.
+So $p$ would be in $B^-$, where $B^-$ is the [[Definition:Closure (Topology)|closure]] of $B$.
+But then $p \in A$, [[Definition:By Hypothesis|by hypothesis]].
+So $A \cap B^- \ne \O$ and so $A$ and $B$ are not [[Definition:Separated Sets|separated]].
+So $B$ must be [[Definition:Closed Set (Topology)|closed]].
+Therefore $\relcomp S B$ is an [[Definition:Open Set (Topology)|open set]] such that $A \subseteq \relcomp S B$, and the [[Definition:T5 Space|$T_5$ separation axiom]] is satisfied.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fort Space is Completely Normal}
+Tags: Completely Normal Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $T$ is a [[Definition:Completely Normal Space|completely normal space]].
+Consequently, $T$ satisfies all weaker [[Definition:Separation Axioms|separation axioms]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Fort Space is T1|Fort Space is $T_1$]]
+:[[Fort Space is T5|Fort Space is $T_5$]]
+and so by definition $T$ is [[Definition:Completely Normal Space|completely normal]].
+{{qed}}
+See [[Sequence of Implications of Separation Axioms]] for confirmation of the statement about weaker [[Definition:Separation Axioms|separation axioms]].
+\end{proof}<|endoftext|>
+\section{Exchange Principle}
+Tags: Model Theory, Named Theorems
+
+\begin{theorem}
+Let $D$ be a [[Definition:Minimal (Model Theory)|strongly minimal]] set in $\mathcal M$.
+Let $A$ be a [[Definition:Subset|subset]] of $D$.
+Let $b, c \in D$.
+If $b$ is [[Definition:Algebraic (Model Theory)|algebraic]] over $A \cup \left\{ {c}\right\}$ but not over $A$, then $c$ is [[Definition:Algebraic (Model Theory)|algebraic]] over $A \cup \left\{ {b}\right\}$.
+\end{theorem}
+
+\begin{proof}
+Let $D \left({x}\right)$ be a formula defining $D$, which exists since by definition, strongly minimal sets are definable.
+To simplify exposition, we will assume below that all further mentioned formulas are  $\mathcal L$-formulas with parameters from $A$, in addition to whatever other parameters are supplied.
+Suppose $b$ is algebraic over $A \cup \left\{ {c}\right\}$ but not over $A$.
+By definition of being algebraic, there is some $\phi \left({x, c}\right)$ with $x$ [[Definition:Free Variable|free]] such that $\mathcal M \models \phi \left({b, c}\right)$ and $\left\{ {m \in \mathcal M: \mathcal M \models \phi \left({m, c}\right)}\right\}$ is [[Definition:Finite Set|finite]].
+Consequently, the subset $\{m\in \mathcal M : \mathcal M \models D \left({m}\right) \wedge \phi(m, c)\} = \{d\in D : \mathcal M \models \phi(d, c)\}$ must have some [[Definition:Finite Set|finite]] $n \in \N$ many elements.
+Let $\psi(y)$ be the formula
+:$\displaystyle D(y) \wedge \exists x_1 \cdots \exists x_n \left( \left( \bigwedge_{i=1,\dots,n} D(x_i) \wedge \phi(x_i, y) \right) \wedge \forall z \left( D(z) \wedge \phi(z, y) \rightarrow \left( \bigvee_{i=1,\dots,n} z = x_i \right) \right) \right)$
+which asserts that
+:$\{d\in D : \mathcal M \models \phi(d, y)\}$ has $n$ many elements.
+Note the dependence on the free variable $y$.
+We will argue that $\phi(b, x) \wedge \psi \left({x}\right))$ demonstrates the algebraicity of $c$ over $A\cup  \left\{ {b}\right\}$.
+Suppose (with the intent of deriving a contradiction) that $\{d\in D : \mathcal M \models \phi(b, d) \wedge \psi(d)\}$ is infinite.
+:Since $D$ is strongly minimal, $\{d\in D : \mathcal M \models \phi(b, d) \wedge \psi(d)\}$ is [[Definition:Cofinite Set|cofinite]] in $D$.
+:Thus $D - \{d\in D : \mathcal M \models \phi(b, d) \wedge \psi(d)\}$ has some [[Definition:Finite Set|finite]] $k \in \N$ many elements.
+:Define $\chi \left({x}\right)$ to be a formula which asserts that $D - \{d\in D : \mathcal M \models \phi(x, d) \wedge \psi(d)\}$ has $k$ many elements.  This can be done similarly to how $\psi(y)$ was defined above.
+:$\chi$ cannot define a [[Definition:Finite Set|finite]] subset of $D$, since $\chi$ involves only parameters from $A$ and $\mathcal M \models \chi(b)$, and so this would imply that $b$ is algebraic over $A$.
+:Thus, $\chi$ defines an infinite subset of $D$.
+:So, we may let $b_1, \dots, b_{n+1}$ be distinct elements of $D$ such that $\mathcal M \models \chi(b_i)$ for each $i = 1,\dots,n+1$.
+:For each $i=1,\dots,n+1$, define
+::$C_i = \{d\in D : \phi(b_i, d) \wedge \psi(d)\}$
+:Then each $C_i$ is cofinite in $D$, since $\mathcal M \models \chi(b_i)$ and hence $D - \{d\in D : \mathcal M \models \phi(b_i, d) \wedge \psi(d)\}$ has $k$ many elements.
+:It follows that $\displaystyle \bigcap_{i=1,\dots,n+1} C_i$ is nonempty, since $D$ is infinite and the intersection excludes at most $k \cdot (n+1)$ elements of $D$.
+:Let $\displaystyle \hat c \in \bigcap_{i \mathop = 1, \dots, n + 1} C_i$.
+:By definition of each $C_i$, this means that $\mathcal M \models \psi(\hat c)$ and $\mathcal M \models \phi(b_i, \hat c)$ for $i=1,\dots,n+1$.
+:But this is a contradiction, since the definition of $\psi$ gives us that $\mathcal M \models \phi(d, \hat c)$ for only $n$ many $d\in D$.
+Thus $\left\{ {d \in D : \mathcal M \models \phi(b, d) \wedge \psi(d) }\right\} = \left\{ {m \in \mathcal M : \mathcal M \models \phi(b, m)\wedge \psi \left({m}\right)}\right\}$ is [[Definition:Finite Set|finite]].
+Since $\mathcal M \models \phi \left({b, c}\right) \wedge \psi \left({c}\right)$, this means that $c$ is definable over $A \cup \left\{ {b}\right\}$.
+{{qed}}
+[[Category:Model Theory]]
+[[Category:Named Theorems]]
+9dt78vqot17m9hvup8ra16y3m6zb2kz
+\end{proof}<|endoftext|>
+\section{Uncountable Fort Space is not Perfectly Normal}
+Tags: Perfectly Normal Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is not a [[Definition:Perfectly Normal Space|perfectly normal space]].
+\end{theorem}
+
+\begin{proof}
+From [[Clopen Points in Fort Space]], $\set p$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+Consider a [[Definition:Countable Intersection|countable intersection]] of [[Definition:Open Set (Topology)|open sets]] of $T$ which contain $p$.
+By definition, all these are [[Definition:Cofinite Set|cofinite in $S$]] and so [[Definition:Uncountable Set|uncountable]].
+So this intersection must itself contain all but a [[Definition:Countable Set|countable number]] of points of $S$.
+So $\set p$ is not a [[Definition:G-Delta Set|$G_\delta$ set]].
+Hence $T$ is not a [[Definition:Perfectly Normal Space|perfectly normal space]] as not all its [[Definition:Closed Set (Topology)|closed sets]] is a [[Definition:G-Delta Set|$G_\delta$ set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Countable Fort Space is Perfectly Normal}
+Tags: Perfectly Normal Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on a [[Definition:Countably Infinite Set|countably infinite set]] $S$.
+Then $T$ is a [[Definition:Perfectly Normal Space|perfectly normal space]].
+\end{theorem}
+
+\begin{proof}
+We have from [[Closed Set of Countable Fort Space is G-Delta|Closed Set of Countable Fort Space is $G_\delta$]] that every [[Definition:Closed Set (Topology)|closed set]] in $T$ is a [[Definition:G-Delta Set|$G_\delta$ set]].
+From [[Fort Space is T5|Fort Space is $T_5$]] and [[T5 Space is T4 Space|$T_5$ Space is $T_4$ Space]], we have that a [[Definition:Fort Space|Fort space]] is a [[Definition:T4 Space|$T_4$ space]].
+From [[Fort Space is T1|Fort Space is $T_1$]] it follows by definition that $T$ is a [[Definition:Perfectly Normal Space|perfectly normal space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fort Space is Compact}
+Tags: Compact Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $T$ is a [[Definition:Compact Topological Space|compact space]].
+\end{theorem}
+
+\begin{proof}
+Let $\CC$ be an [[Definition:Open Cover|open cover]] of $T$.
+Then $\exists U \in \CC$ such that $p \in U$ and so $\relcomp S U$ is [[Definition:Finite Set|finite]].
+For each $x \in \relcomp S U$ there exists some $C_x \in \CC$ such that $x \in C$.
+So $U$, together with each of those $C_x \in \CC$, is a [[Definition:Finite Subcover|finite subcover]] of $\CC$.
+Hence the result by definition of [[Definition:Compact Topological Space|compact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fort Space is Sequentially Compact}
+Tags: Sequentially Compact Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $T$ is a [[Definition:Sequentially Compact Space|sequentially compact space]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}_{n \mathop \in \N}$ be an [[Definition:Infinite Sequence|infinite sequence]] in $T$.
+Suppose $\sequence {x_n}$ takes an [[Definition:Infinite Set|infinite number]] of [[Definition:Distinct|distinct]] values in $S$.
+Then there is an [[Definition:Infinite Sequence|infinite]] [[Definition:Subsequence|subsequence]] $\sequence {x_{n_r} }_{r \mathop \in \N}$ with [[Definition:Sequence of Distinct Terms|distinct terms]].
+Let $U$ be a [[Definition:Neighborhood of Point|neighborhood]] of $p$.
+Then $S \setminus U$ is a [[Definition:Finite Set|finite set]] by definition.
+Thus there exists $N \in \N$ such that $\forall r > N: x_{n_r} \in U$.
+Thus $\sequence {x_{n_r} }$ [[Definition:Convergent Sequence (Topology)|converges]] to $p$.
+Otherwise $\sequence {x_n}$ only takes a [[Definition:Finite Set|finite number]] of [[Definition:Distinct|distinct]] values.
+Then, since $\sequence {x_n}$ is [[Definition:Infinite Sequence|infinite]], there exists $x \in S$ such that:
+:$\forall N \in \N: \exists n > N: x = x_n$
+This implies that we can take a [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$ which is [[Definition:Constant Sequence|constant]], and which [[Definition:Convergent Sequence (Topology)|converges]] to that [[Definition:Constant Sequence|constant]].
+We can conclude then that, by definition, $T$ is a [[Definition:Sequentially Compact Space|sequentially compact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boubaker's Theorem/Proof of Uniqueness}
+Tags: Boubaker Polynomials
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Commutative Ring|commutative ring]].
+Let $\left({D, +, \circ}\right)$ be an [[Definition:Subdomain|integral subdomain]] of $R$ whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
+Let $X \in R$ be [[Definition:Transcendental over Integral Domain|transcendental]] over $D$.
+Let $D \left[{X}\right]$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] in $X$ over $D$.
+Consider the following properties:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \sum_{k \mathop = 1}^N {p_n \left({0}\right)}
+      | r = -2N
+}}
+{{eqn | n = 2
+      | l = \sum_{k \mathop = 1}^N {p_n \left({\alpha_k}\right)}
+      | r = 0
+}}
+{{eqn | n = 3
+      | l = \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_x \left({x}\right)} {\mathrm d x} }\right\vert_{x \mathop = 0}
+      | r = 0
+}}
+{{eqn | n = 4
+      | l = \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d {p_n}^2 \left({x}\right)} {\mathrm d x^2} }\right\vert_{x \mathop = 0}
+      | r = \frac 8 3 N \left({N^2 - 1}\right)
+}}
+{{end-eqn}}
+where, for a given [[Definition:Positive Integer|positive integer]] $n$, $p_n \in D \left[{X}\right]$ is a non-[[Definition:Null Polynomial over Ring|null polynomial]] such that $p_n$ has $N$ roots $\alpha_k$ in $F$.
+Then the [[Definition:Subsequence|subsequence]] $\left \langle {B_{4 n} \left({x}\right)}\right \rangle$ of the [[Definition:Boubaker Polynomials|Boubaker polynomials]] is the unique polynomial sequence of $D \left[{X}\right]$ which verifies simultaneously the four properties $(1) - (4)$.
+\end{theorem}
+
+\begin{proof}
+Let:
+: $\left({R, +, \circ}\right)$ be a [[Definition:Commutative Ring|commutative ring]]
+: $\left({D, +, \circ}\right)$ be an [[Definition:Subdomain|integral subdomain]] of $R$ whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$
+: $X \in R$ be [[Definition:Transcendental over Integral Domain|transcendental]] over $D$.
+It has been demonstrated that the [[Definition:Boubaker Polynomials|Boubaker Polynomials]] sub-sequence $B_{4 n} \left({x}\right)$, defined in $D \left[{X}\right]$ as:
+:$\displaystyle B_{4 n} \left({x}\right) = 4 \sum_{p \mathop = 0}^{2 n} \frac {n - p} {4 n - p} \binom {4 n - p} p \left({-1}\right)^p x^{2 \left({2n - p}\right)}$
+satisfies the properties:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \sum_{k \mathop = 1}^N {p_n \left({0}\right)}
+      | r = -2N
+}}
+{{eqn | n = 2
+      | l = \sum_{k \mathop = 1}^N {p_n \left({\alpha_k}\right)}
+      | r = 0
+}}
+{{eqn | n = 3
+      | l = \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_x \left({x}\right)} {\mathrm d x} }\right\vert_{x \mathop = 0}
+      | r = 0
+}}
+{{eqn | n = 4
+      | l = \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d {p_n}^2 \left({x}\right)} {\mathrm d x^2} }\right\vert_{x \mathop = 0}
+      | r = \frac 8 3 N \left({N^2 - 1}\right)
+}}
+{{end-eqn}}
+with $\left. {\alpha_k}\right\vert_{k \mathop = 1 \,.\,.\, N}$ roots of $B_{4 n}$.
+Suppose there exists another $4 n$-indexed polynomial $q_{4 n} \left({x}\right)$, with $N$ roots $\left.{\beta_k}\right\vert_{k \mathop = 1 \,.\,.\, N}$ in $F$ and which also satisfies simultaneously properties $(1)$ to $(4)$.
+Let:
+:$\displaystyle B_{4 n} \left({x}\right) = \sum_{p \mathop = 0}^{2 n} a_{4 n, p} x^{2 \left({2 n - p}\right)}$
+and:
+:$\displaystyle q_{4 n} \left({x}\right) = \sum_{p \mathop = 0}^{2 n} b_{4 n, p} x^{2 \left({2 n - p}\right)}$
+and:
+:$\displaystyle \mathrm d_{4 n, p} = a_{4 n, p} - b_{4 n, p}$ for $p = 0 \,.\,.\, 2 n$
+then, simultaneous expressions of conditions $(1)$ and $(3)$ give:
+:$ \quad \displaystyle \sum_{k \mathop = 1}^N \mathrm d_{4 n, 2 n} = 0$
+:$ \quad \displaystyle \sum_{k \mathop = 1}^N \mathrm d_{4 n, 2 n - 2} = 0$
+It has also been demonstrated that $ B_{4 n}$ has exactly $4 n - 2$ real roots inside the domain $\left[{-2 \,.\,.\, 2}\right]$.
+So application of conditions $(3)$ and $(4)$ give $4n-2$ linear equation with variables $\left.{ d_{4n,p}}\right|_{p \mathop = 0 \,.\,.\, 2n-3}$.
+Finally, since $B_{4 n}$ contains $2 n$ monomial terms (see definition), we obtain a Cramer system in variables $\left.{ d_{4 n, p}}\right\vert_{p \mathop = 0 \,.\,.\, 2 n}$, with evident solution:
+:$\left.{\mathrm d_{4 n, p}}\right\vert_{p \mathop = 0 \,.\,.\, 2 n} = 0 $
+and consequently:
+:$\left.{a_{4 n, p} }\right\vert_{p \mathop = 0 \,.\,.\, 2 n} = \left.{b_{4 n, p} }\right\vert_{p \mathop = 0 \,.\,.\, 2 n}$
+which means:
+:$q_{4 n} \left({x}\right) = B_{4 n} \left({x}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Fort Space is not First-Countable}
+Tags: First-Countable Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is not a [[Definition:First-Countable Space|first-countable space]].
+\end{theorem}
+
+\begin{proof}
+Let $\UU$ be a [[Definition:Countable Set|countable set]] of [[Definition:Open Neighborhood|open neighborhoods]] of $p$.
+Let $U \in \UU$.
+Then as $p \in U$, $p \notin \relcomp S U$ and so for $U$ to be [[Definition:Open Set (Topology)|open]] it must follow that $\relcomp S U$ is [[Definition:Finite Set|finite]].
+From [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection|De Morgan's Laws: Complement of Intersection]] we have:
+:$\displaystyle H := \bigcup_{U \mathop \in \UU} \relcomp S U = \relcomp S {\bigcap \UU}$
+From [[Countable Union of Countable Sets is Countable]] it follows that $H$ can be no more than [[Definition:Countable Set|countable]].
+So $\displaystyle \bigcap \UU = \relcomp S H$ must be [[Definition:Uncountable Set|uncountable]].
+So (trivially):
+:$H \ne \set p$
+So:
+:$\exists q \ne p: q \in \displaystyle \bigcap \UU$
+So $\relcomp S {\set q}$ is an [[Definition:Open Neighborhood|open neighborhood]] of $P$ which does not contain any of the elements of $\UU$.
+So $\UU$ is not a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] of $p$.
+Hence by definition $T$ is not a [[Definition:First-Countable Space|first-countable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Generating Function for Boubaker Polynomials}
+Tags: Boubaker Polynomials, Generating Functions
+
+\begin{theorem}
+The [[Definition:Boubaker Polynomials|Boubaker polynomials]], defined as:   
+:$B_n \left({x}\right) = \begin{cases}
+1 & : n = 0 \\
+x & : n = 1 \\
+x^2 + 2 & : n = 2 \\
+x B_{n-1} \left({x}\right) - B_{n-2} \left({x}\right) & : n > 2 \end{cases}$
+have as an [[Definition:Ordinary Generating Function|ordinary generating function]]:
+: $\displaystyle f_{B_n, \operatorname{ORD}} \left({x, t}\right) = \sum_{n \mathop = 0}^{\infty} B_n \left({x}\right) t^n = \frac {1 + 3 t^2} {1 + t \left({t - x}\right)}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = f \left({x, t}\right) 
+      | r = 1 + x t + \left({x^2 + 2}\right) t^2 + \sum_{n \mathop = 3}^\infty \left({x B_{n-1} \left({x}\right) t^n - B_{n-2} \left({x}\right) t^n}\right)
+      | c = Recursive definition of $B_n \left({x}\right)$ for $n > 2$
+}}
+{{eqn | r = 1 + x t + \left({x^2 + 2}\right) t^2 + x t \sum_{n \mathop = 3}^\infty B_{n-1} \left({x}\right) t^{n-1} - t^2 \sum_{n \mathop = 3}^\infty B_{n-2} \left({x}\right) t^{n-2}
+}}
+{{eqn | r = 1 + x t + \left({x^2 + 2}\right) t^2 + x t (f \left({x, t}\right) - 1 - x t) - t^2 (f \left({x, t}\right) - 1)
+}}
+{{end-eqn}}
+and then solve for $f \left({x, t}\right)$.
+{{qed}}
+[[Category:Boubaker Polynomials]]
+[[Category:Generating Functions]]
+pp1c9eihixqpx465ugtaci240rjz1s3
+\end{proof}<|endoftext|>
+\section{Exponential Generating Function for Boubaker Polynomials}
+Tags: Boubaker Polynomials, Generating Functions
+
+\begin{theorem}
+The [[Definition:Boubaker Polynomials|Boubaker polynomials]], defined as:           
+:$B_n \left({x}\right) = \begin{cases}
+1 & : n = 0 \\
+x & : n = 1 \\
+x^2+2 & : n = 2 \\
+x B_{n-1} \left({x}\right) - B_{n-2} \left({x}\right) & : n > 2 \end{cases}$
+have as an [[Definition:Exponential Generating Function|exponential generating function]]:
+: $\displaystyle f_{B_n, \operatorname{EXP}} \left({x, t}\right) = \sum_{n=0}^\infty B_n \left({x}\right) \frac{t^n}{n!} =
+4 x e^{t \frac x 2} \frac {\sin \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right)} {\sqrt {1 - \left({\frac x 2}\right)^2}} - 2 e^{t \frac x 2} \cos \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right) - 3$
+{{Proofread}}
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Boubaker Polynomials|Boubaker polynomials]], we have:    
+:$ \begin{cases}
+B_0 \left({x}\right) = 1 \\    
+B_1 \left({x}\right) = x \\    
+\end{cases}$
+We have also, for $T_n$, the [[Definition:Chebyshev Polynomials of the First Kind|Chebyshev polynomials of the first kind]] and $U_n$, the [[Definition:Chebyshev Polynomials of the Second Kind|Chebyshev polynomials of the second kind]]:        
+:$ \begin{cases}
+T_0 \left({x}\right) = 1 \\    
+T_1 \left({x}\right) = x \\    
+\end{cases}$
+and :
+:$ \begin{cases}
+U_{-1} \left({x}\right) = 0 \\    
+U_0 \left({x}\right) = 1 \\    
+U_1 \left({x}\right) = 2x \\    
+\end{cases}$
+Then, from the definition of the exponential generating functions of the [[Definition:Chebyshev Polynomials of the First Kind|Chebyshev polynomials of the first kind]]:
+: $\displaystyle f_{T_n, \operatorname{EXP}} \left({x, t}\right) = e^{t \frac x 2} \cos \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right) $
+and the [[Definition:Chebyshev Polynomials of the Second Kind|Chebyshev polynomials of the second kind]]: 
+: $\displaystyle f_{U_n, \operatorname{EXP}} \left({x, t}\right) = e^{t \frac x 2} \frac {\sin \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right)} {\sqrt {1 - \left({\frac x 2}\right)^2}} $
+   
+and from the relation:
+: $B_n \left({x}\right) = 4 x U_{n-1}\left({\frac x 2}\right) -2 T_n \left({\frac x 2}\right)$
+which is valid only for $n>1$, we have:
+: $\displaystyle B_0 \left({x}\right) \frac{t^0}{0!}+ B_1 \left({x}\right) \frac{t^1}{1!} + \sum_{n=2}^\infty B_n \left({x}\right) \frac{t^n}{n!} = 4x \sum_{n=2}^\infty U_{n-1} \left({x}\right) \frac{t^n}{n!} - 2 \sum_{n=2}^\infty T_n \left({x}\right) \frac{t^n}{n!}+B_0 \left({x}\right) \frac{t^0}{0!}+ B_1 \left({x}\right) \frac{t^1}{1!} -4x \sum_{n=0}^1 U_{n-1} \left({x}\right) \frac{t^n}{n!} - 2 \sum_{n=0}^1 T_n \left({x}\right) \frac{t^n}{n!} $     
+Finally, by simplifying the two sides and considering the [[Definition:Exponential Generating Function|exponential generating function]]s and the values of the first terms of $T_n$ and $U_n$, we obtain:      
+: $\displaystyle \sum_{n=0}^\infty B_n \left({x}\right) \frac{t^n}{n!} =   
+4xe^{t \frac x 2} \frac {\sin \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right)} {\sqrt {1 - \left({\frac x 2}\right)^2}} - 2 e^{t \frac x 2} \cos \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right) - 3$
+Hence the result.     
+{{qed}}
+{{MissingLinks}}
+ 
+[[Category:Boubaker Polynomials]]
+[[Category:Generating Functions]]
+m9fsgzrtn86mth7uyu6jjnvjf7o36dp
+\end{proof}<|endoftext|>
+\section{Clopen Points in Fort Space}
+Tags: Clopen Sets, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Let $q \in S: q \ne p$.
+Then $\set q$ is both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] in $T$.
+$\set p$ itself, on the other hand, is [[Definition:Closed Set (Topology)|closed]] but not [[Definition:Open Set (Topology)|open]].
+\end{theorem}
+
+\begin{proof}
+We have that $\set q$ is [[Definition:Finite Set|finite]] so $\relcomp S {\set q}$ is [[Definition:Cofinite Set|cofinite]].
+So $\relcomp S {\set q}$ is [[Definition:Open Set (Topology)|open]] and so $\set q$ is [[Definition:Closed Set (Topology)|closed]].
+Then we have that $p \notin \set q$ so $\set q$ is [[Definition:Open Set (Topology)|open]].
+However, $p \notin \relcomp S {\set p}$ and $\relcomp S {\set p}$ is [[Definition:Infinite Set|infinite]].
+So $\set p$ is not [[Definition:Open Set (Topology)|open]] in $S$.
+But $\set p$ is [[Definition:Finite Set|finite]], so $\relcomp S {\set p}$ is [[Definition:Cofinite Set|cofinite]].
+Hence, as for $\set q$, we have that $\set p$ is [[Definition:Closed Set (Topology)|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fort Space is Totally Separated}
+Tags: Totally Separated Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $T$ is a [[Definition:Totally Separated Space|totally separated space]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in S$ such that $a \ne b$.
+From [[Clopen Points in Fort Space]], all points in $S$ apart from $p$ are both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] in $T$.
+{{WLOG}}, suppose that $a \ne p$.
+Then as $\set a$ is [[Definition:Closed Set (Topology)|closed]], $\relcomp S {\set a}$ is [[Definition:Open Set (Topology)|open]] in $T$.
+So we have a [[Definition:Partition (Topology)|partition]] $\set a \mid \relcomp S {\set a}$ of $S$ such that $a \in \set a, b \in \relcomp S {\set a}$.
+So, by definition  $T$ is a [[Definition:Totally Separated Space|totally separated space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fort Space is not Extremally Disconnected}
+Tags: Extremally Disconnected Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $T$ is not an [[Definition:Extremally Disconnected Space|extremally disconnected space]].
+\end{theorem}
+
+\begin{proof}
+We note from [[Fort Space is Completely Normal]] that $T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+Let $H \subseteq S$ be an [[Definition:Infinite Set|infinite]] subset of $S$ such that $p \in \relcomp S H$ and $\relcomp S H$ is [[Definition:Infinite Set|infinite]].
+Then $H$ is open as $p \in \relcomp S H$.
+From [[Limit Points in Fort Space]], as $H$ is [[Definition:Infinite Set|infinite]], $p$ is the only [[Definition:Limit Point of Set|limit point]] of $H$.
+So $H^- = H \cup \set p$ where $H^-$ is the [[Definition:Closure (Topology)|closure]] of $H$.
+But as $\relcomp S {H^-}$ is [[Definition:Infinite Set|infinite]] and does not contain $p$, it follows that $H^-$ is not [[Definition:Open Set (Topology)|open]] in $T$.
+So by definition $T$ is not an [[Definition:Extremally Disconnected Space|extremally disconnected space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fort Space is Zero Dimensional}
+Tags: Zero Dimensional Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $T$ is a [[Definition:Zero Dimensional Space|zero dimensional space]].
+\end{theorem}
+
+\begin{proof}
+Let $q \in S$ such that $q \ne p$.
+Then from [[Clopen Points in Fort Space]], $\set q$ is [[Definition:Clopen Set|clopen]].
+So $\forall q \in S, q \ne p: \set {\set q}$ is a [[Definition:Local Basis|local basis]] for $q$.
+If we take the [[Definition:Open Neighborhood|open neighborhoods]] of $p$ we get a [[Definition:Local Basis|local basis]] $\UU_p$ with the following property:
+:Since $p \in U \in \UU_p$, its [[Definition:Set Complement|complement]] does not contain $p$ and so it is [[Definition:Open Set (Topology)|open]].
+This implies that $U$ is also [[Definition:Closed Set (Topology)|closed]].
+The [[Definition:Set Union|union]] of the [[Definition:Local Basis|local basis]] forms a [[Definition:Basis (Topology)|basis]] for the topology.
+This [[Definition:Basis (Topology)|basis]] is formed with [[Definition:Clopen Set|clopen sets]].
+So, by definition, $T$ is a [[Definition:Zero Dimensional Space|zero dimensional space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Formula and its Negation Cannot Both Cause Forking}
+Tags: Model Theory
+
+\begin{theorem}
+Let $T$ be a [[Definition:Theory (Logic)|complete $\mathcal{L}$-theory]].
+Let $\mathfrak{C}$ be a [[Definition:Monster Model|monster model]] for $T$.
+Let $A\subseteq B$ be [[Definition:Subset|subsets]] of the universe of $\mathfrak{C}$.
+Let $\pi(\bar x)$ be an [[Definition:Type|$n$-type]] over $B$.
+If $\pi$ does not [[Definition:Fork|fork]] over $A$, then for any formula $\phi(\bar x, \bar b)$, either $\pi \cup \{\phi\}$ or $\pi \cup \{\neg\phi\}$ does not fork over $A$.
+\end{theorem}
+
+\begin{proof}
+We prove the contrapositive.
+Suppose both $\pi \cup \{\phi\}$ and $\pi \cup \{\neg\phi\}$ fork over $A$.
+By definition of forking,
+:$\pi\cup\{\phi\}$ implies $\phi_1 (\bar x, \bar c_1) \vee \cdots \vee \phi_k (\bar x, \bar c_k)$ and
+:$\pi\cup\{\neg\phi\}$ implies $\psi_1 (\bar x, \bar d_1) \vee \cdots \vee \psi_k (\bar x, \bar d_h)$,
+where each $\phi_i$ and $\psi_j$ [[Definition:Divide (Model Theory)|divide]] over $A$.
+Then $\pi$ implies the disjunction
+:$\bigvee_i \phi(\bar x, \bar c_i) \vee \bigvee_j \psi(\bar x, \bar d_i)$
+with each component formula dividing over $A$.
+By definition, this means that $\pi$ forks over $A$.
+{{qed}}
+[[Category:Model Theory]]
+fuwj2wm9i715zb7ksuzsg8h7w6n2czx
+\end{proof}<|endoftext|>
+\section{Non-Forking Types have Non-Forking Completions}
+Tags: Model Theory
+
+\begin{theorem}
+Let $T$ be a [[Definition:Theory (Logic)|complete $\mathcal L$-theory]].
+Let $\mathfrak C$ be a [[Definition:Monster Model|monster model]] for $T$.
+Let $A\subseteq B$ be [[Definition:Subset|subsets]] of the universe of $\mathfrak C$.
+Let $\pi(\bar x)$ be an [[Definition:Type|$n$-type]] over $B$.
+If $\pi$ does not [[Definition:Fork|fork]] over $A$, then there is a [[Definition:Complete Type|complete $n$-type]] $p$ over $B$ such that $\pi \subseteq p$ and $p$ does not fork over $A$.
+\end{theorem}
+
+\begin{proof}
+Suppose $\pi$ does not fork over $A$.
+We will use [[Zorn's Lemma]] to find a candidate for the needed complete type.
+Consider the collection $\Pi$ of all non-forking sets $\pi'$ of $\mathcal L$-formulas with parameters from $B$ such that $\pi'$ contains $\pi$.
+Order $\Pi$ by subset inclusion.
+Since [[Forking is Local|a set forks iff a finite subset forks]], the union over any [[Definition:Chain (Set Theory)|chain]] is still a non-forking set, and hence is an upper bound for the chain.
+Thus, by [[Zorn's Lemma]], there is a maximal (with respect to subset inclusion) $p$ in $\Pi$.
+Suppose $p$ is not complete.
+:By definition, for some $\phi(\bar x, \bar b)$, $p$ contains neither $\phi(\bar x, \bar b)$ nor $\neg\phi(\bar x, \bar b)$.
+:Since $p$ is non-forking, by [[Formula and its Negation Cannot Both Cause Forking]], at least one of $p\cup\phi(\bar x, \bar b)$ or $p\cup\neg\phi(\bar x, \bar b)$ is non-forking as well.
+:Hence $p$ is not maximal in $\Pi$, contradicting the choice of $p$.
+Thus $p$ is complete.
+{{qed}}
+{{AoC|Zorn's Lemma}}
+[[Category:Model Theory]]
+ppskkx54i644iki7i51u3hpdka104hw
+\end{proof}<|endoftext|>
+\section{Forking is Local}
+Tags: Model Theory
+
+\begin{theorem}
+Let $T$ be a [[Definition:Theory (Logic)|complete $\mathcal{L}$-theory]].
+Let $\mathfrak{C}$ be a [[Definition:Monster Model|monster model]] for $T$.
+Let $A\subseteq B$ be [[Definition:Subset|subsets]] of the universe of $\mathfrak{C}$.
+Let $\pi(\bar x)$ be an [[Definition:Type|$n$-type]] over $B$.
+$\pi$ [[Definition:Fork|forks]] over $A$ if and only if a finite subset of $\pi$ forks over $A$.
+\end{theorem}
+
+\begin{proof}
+The proof is straightforward using the definition of forking and the fact that proofs in first-order logic are finite.
+Suppose $\pi$ forks over $A$.
+:By definition, $\pi$ implies a disjunction of formulas which each [[Definition:Divide (Model Theory)|divide]] over $A$.
+:Since proofs are finite, this means that there is a finite subset of $\pi$ which implies this disjunction, completing this direction of the proof.
+Suppose a finite subset of $\pi$ forks over $A$.
+:By definition, the finite subset implies a disjunction of formulas which each [[Definition:Divide (Model Theory)|divide]] over $A$.
+:But then $\pi$ clearly implies this disjunction as well, completing this direction of the proof.
+{{qed}}
+[[Category:Model Theory]]
+5xn75ly0rbxveqozd29d7zr8ixy1slr
+\end{proof}<|endoftext|>
+\section{Subset of Finite Set is Finite}
+Tags: Subsets, Finite Sets
+
+\begin{theorem}
+Let $X$ be a [[Definition:Finite Set|finite set]].
+If $Y$ is a [[Definition:Subset|subset]] of $X$, then $Y$ is also [[Definition:Finite Set|finite]].
+\end{theorem}
+
+\begin{proof}
+From the definition, $X$ is [[Definition:Finite Set|finite]] {{iff}} $\exists n \in \N$ such that there exists a [[Definition:Bijection|bijection]]:
+:$f: X \leftrightarrow \N_n$
+where $\N_n$ is the [[Definition:Initial Segment of Natural Numbers|set of all elements of $\N$ less than $n$]], that is:
+:$\N_n = \set {0, 1, 2, \ldots, n - 1}$
+The case in which $X$ is [[Definition:Empty Set|empty]] is trivial.
+We begin proving the following particular case:
+:''If $X$ is [[Definition:Finite Set|finite]] and $a \in X$, then $X \setminus \set a$ is also [[Definition:Finite Set|finite]].''
+From [[Bijection between Specific Elements]] there exists a [[Definition:Bijection|bijection]] $f: \N_n \to X$, which satisfies $\map f n = a$.
+Next we prove the general case by [[Principle of Mathematical Induction|induction]].
+=== Basis for the Induction ===
+If $n = 1$ then $X \setminus \set a = \O$ is finite.
+If $n > 1$, the [[Definition:Restriction of Mapping|restriction]] of $f$ to $\set {k \in \N: k \le n - 1}$ yields a [[Definition:Bijection|bijection]] into $X \setminus \set a$.
+Hence $X \setminus \set a$ is [[Definition:Finite Set|finite]] and has $n - 1$ [[Definition:Element|elements]].
+So, we have that if $n = 1$, then its [[Definition:Subset|subsets]] ($\O$ and $X$) are [[Definition:Finite Set|finite]].
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if our Theorem is valid for [[Definition:Set|sets]] with $n$ [[Definition:Element|elements]], it is also true for [[Definition:Set|sets]] with $n + 1$ [[Definition:Element|elements]].
+This is our [[Definition:Induction Hypothesis|induction hypothesis]].
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Let $X$ have $n + 1$ [[Definition:Element|elements]], and let $Y \subseteq X$.
+If $Y = X$, there is nothing to prove.
+Otherwise, $\exists a \in X \setminus Y$.
+This means that $Y$ is actually a [[Definition:Subset|subset]] of $X \setminus \set a$.
+Since $X \setminus \set a$ has $n$ [[Definition:Element|elements]], it follows that $Y$ is [[Definition:Finite Set|finite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Statements Equivalent to Non-Dividing Type}
+Tags: Model Theory
+
+\begin{theorem}
+Let $T$ be a [[Definition:Theory (Logic)|complete $\mathcal L$-theory]].
+Let $\mathfrak C$ be a [[Definition:Monster Model|monster model]] for $T$.
+Let $A$ be a [[Definition:Subset|subset]] of the universe of $\mathfrak C$.
+The following are equivalent:
+:$(1): \quad$ [[Definition:Type|$\operatorname{tp}(\bar c / A,\bar{b})$]] does not [[Definition:Divide (Model Theory)|divide]] over $A$.
+:$(2): \quad$ For every $\{\bar b_i : i \in I\}$ containing $\bar b$ which is [[Definition:Order Indiscernible|order indiscernible]] over $A$, there is some $\bar c'$ with $\operatorname{tp}(\bar c' / A,\bar b) = \operatorname{tp}(\bar c / A,\bar b)$ such that $\{\bar b_i : i \in I\}$ is order indiscernible over $A,\bar c'$.
+:$(3): \quad$ For every $\{\bar b_i : i \in I\}$ containing $\bar b$ which is [[Definition:Order Indiscernible|order indiscernible]]  over $A$, there is some $A,\bar b$-[[Definition:Automorphism (Model Theory)|automorphism]] $f$ such that $\{f(\bar b_i) : i \in I\}$ is order indiscernible over $A,\bar c$.
+\end{theorem}
+
+\begin{proof}
+$(1)\implies (2)$:
+Let $\{\bar b_i : i \in I\}$ containing $\bar b$ be [[Definition:Order Indiscernible|order indiscernible]] over $A$.
+Suppose $\displaystyle \bigcup_{i\in I} \operatorname{tp}(\bar c / A,\bar b_i)$ is not satisfiable.
+:By [[Compactness Theorem|compactness]], some finite subset $\{\phi_1(x, \bar b_{i_1}), \dots, \phi_k(x, \bar b_{i_k})\}$ is not satisfiable.
+:Let $\phi$ be $\phi_1 \wedge \cdots \wedge \phi_k$.
+:Note that $\phi(x, \bar b)$ is in $\operatorname{tp}(\bar c / A,\bar b)$.
+:$\{\phi(x, \bar b_i ) : i\in I\}$ is not satisfiable since it implies $\{\phi_1(x, \bar b_{i_1}), \dots, \phi_k(x, \bar b_{i_k}))$.
+:Again by [[Compactness Theorem|compactness]], some finite subset $\{\phi(x, \bar b_{j_1}), \dots, \phi(x, \bar b_{j_h})\}$ is inconsistent.
+:Since all of the $b_i$ are order-indiscernibles, this means that any [[Definition:Cardinality|cardinality]] $h$ subset of $\{\phi(x, \bar b_i ) : i\in I\}$ is not satisfiable.
+:This contradicts the assumption that $\operatorname{tp}(\bar c / A,\bar b)$ does not divide over $A$.
+Thus $\displaystyle \bigcup_{i\in I} \operatorname{tp}(\bar c / A,\bar b_i)$ is satisfiable, and since $\mathfrak C$ is a monster model, it is satisfiable in $\mathfrak C$.
+By [[Infinite Ramsey's Theorem]] and indiscernibility of $\{\bar b_i : i \in I\}$ over $A$, for each finite set $\Delta$ of formulas with parameters from $A$, there is some $\bar c_\Delta$ such that:
+:$\bar c_\Delta$ realizes $\displaystyle \bigcup_{i\in I} \operatorname{tp}(\bar c / A, \bar b_i)$, and
+:for any $i_1 < \cdots < i_k$ and $j_1 < \cdots < j_k$ in $I$, $\mathfrak C \models \phi(b_{i_1}, \dots, b_{i_k}, \bar c_\Delta) \leftrightarrow  \phi(b_{j_1}, \dots, b_{j_k}, \bar c_\Delta)$ for all $\phi \in \Delta$.
+Since $b$ is one of the $b_i$, we have that each $c_\Delta$ realizes $\operatorname{tp}(\bar c / A, \bar b)$.
+By [[Compactness Theorem|compactness]], since there is a $c_\Delta$ with $\mathfrak C \models \phi(b_{i_1}, \dots, b_{i_k}, \bar c_\Delta) \leftrightarrow  \phi(b_{j_1}, \dots, b_{j_k}, \bar c_\Delta)$ for all  $\phi \in \Delta$ for each finite $\Delta$, there is a $\bar c'$ with type $\operatorname{tp}(\bar c / A, \bar b)$ and which satisfies $\mathfrak C \models \phi(b_{i_1}, \dots, b_{i_k}, \bar c') \leftrightarrow \phi(b_{j_1}, \dots, b_{j_k}, \bar c')$ for all $\phi$.
+But this holding for every $\phi$ means that $\{b_i : i\in I\}$ is order-indiscernible over $A,\bar c'$.
+$(2)\implies (1)$:
+We prove this case by contradiction.
+Suppose $\operatorname{tp}(\bar c / A,\bar{b})$ [[Definition:Divide (Model Theory)|divides]] over $A$.
+By definition, this means that $\operatorname{tp}(\bar c / A,\bar{b})$ implies some $\phi(x, \bar{b})$ with parameters from $A$ which divides over $A$.
+By definition of dividing, this means that for some $k \in \N$ there is a [[Definition:Sequence|sequence]] $(\bar{b}_i)_{i\in \N}$  such that [[Definition:Type|$\operatorname{tp}(\bar{b}_i /  A) = \operatorname{tp}(\bar{b} / A)$]] for each $i\in\N$, and for any distinct $k$-many terms $\bar b_{i_1}, \dots, \bar b_{i_k}$ of  the sequence, the set $\{\phi(\bar{x}, \bar{b}_{i_1}), \dots,  \phi(\bar{x}, \bar{b}_{i_k})\}$ is not satisfiable in $\mathfrak C$.
+Since $\operatorname{tp}(\bar{b}_i /  A) = \operatorname{tp}(\bar{b} / A)$ for each $i$, we can assume (using an argument involving [[Infinite Ramsey's Theorem]] and partitions based on the formula satisfied by the $b_i$) that $\{\bar b_i : i\in I\}$ is order-indiscernible over $A$.
+Then, by $(2)$, there is $\bar c'$ with $\operatorname{tp}(\bar c' / A,\bar b) = \operatorname{tp}(\bar c / A,\bar b)$ such that $\{\bar b_i : i\in I\}$ is order-indiscernible over $A,\bar c'$.
+However, $\operatorname{tp}(\bar c' / A,\bar b) = \operatorname{tp}(\bar c / A,\bar b)$ gives us that $\mathfrak C \models \phi(\bar c', \bar{b})$.
+So, $\{\bar b_i : i\in I\}$ being order-indiscernible over $A,\bar c'$ gives us that $\mathfrak C \models \phi(\bar c', \bar{b_i})$ for all $i$.
+This contradicts the non-satisfiability of $\{\phi(\bar{x}, \bar{b}_{i_1}), \dots,  \phi(\bar{x}, \bar{b}_{i_k})\}$.
+Thus, $\operatorname{tp}(\bar c /  A,\bar{b})$ does not divide over $A$.
+$(2)\implies (3)$:
+Since $c$ and $c'$ have the same type over $A,b$, the function $\bar c' \mapsto \bar c$ is [[Definition:Partial Elementary Embedding|partial elementary]].
+Since $\mathfrak C$ is [[Definition:Homogeneous (Model Theory)|homogeneous]], there is an $A, b$-[[Definition:Automorphism (Model Theory)|automorphism]] $f$ which extends $\bar c' \mapsto \bar c$.
+Let $\{f(\bar b_i) : i \in I\}$ be the image of $\{\bar b_i : i \in I\}$.
+$\{f(\bar b_i) : i \in I\}$ is order-indiscernible over $A,\bar c$ since if $i_1 < \cdots < i_k$ and $j_1 < \cdots < j_k$, then:
+{{begin-eqn}}
+{{eqn | l = \mathfrak C \models \phi(f(\bar b_{i_1}), \dots, f(\bar b_{i_k}), \bar c)
+      | o = \leadstoandfrom
+      | r = \mathfrak C \models \phi(\bar b_{i_1}, \dots, \bar b_{i_k}, \bar c')
+      | c = as $f$ is an automorphism (fixing any parameters in $\phi$ from $A, b$)
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \mathfrak C \models \phi(\bar b_{j_1}, \dots, \bar b_{j_k}, \bar c')
+      | c = as the original sequence is order-indiscernible over $A, \bar c'$
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \mathfrak C \models \phi(f(\bar b_{j_1}), \dots, f(\bar b_{j_k}), \bar c)
+      | c = as $f$ is an [[Definition:Automorphism (Model Theory)|automorphism]]
+}} 
+{{end-eqn}}
+$(3)\implies (2)$:
+Let $\bar c' = f^{-1}(\bar c)$.
+$\{\bar b_i : i \in I\}$ is order-indiscernible over $A, \bar c'$ since if $i_1 < \cdots < i_k$ and $j_1 < \cdots < j_k$, then:
+{{begin-eqn}}
+{{eqn | l = \mathfrak C \models \phi(\bar b_{i_1}, \dots, \bar b_{i_k}, \bar c')
+      | o = \leadstoandfrom
+      | r = \mathfrak C \models \phi(f(\bar b_{i_1}), \dots, f(\bar b_{i_k}), \bar c)
+      | c = as $f$ is an [[Definition:Automorphism (Model Theory)|automorphism]] (fixing any parameters in $\phi$ from $A, b$)
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \mathfrak C \models \phi(f(\bar b_{j_1}), \dots, f(\bar b_{j_k}), \bar c)
+      | c = as the image sequence is order-indiscernible over $A, \bar c$.
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \mathfrak C \models \phi(\bar b_{j_1}, \dots, \bar b_{j_k}, \bar c')
+      | c = as $f$ is an [[Definition:Automorphism (Model Theory)|automorphism]]
+}} 
+{{end-eqn}}
+{{qed}}
+[[Category:Model Theory]]
+i7s6qjbnrrzd6csyj45oh1fvllrn4wh
+\end{proof}<|endoftext|>
+\section{Fortissimo Topology is Topology}
+Tags: Fortissimo Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
+Then $\tau_p$ is a [[Definition:Topology|topology]] on $T$.
+\end{theorem}
+
+\begin{proof}
+We have that $p \in \relcomp S \O = S$ so $\O \in \tau_p$.
+We have that $\relcomp S S = \O$ so $\O \in \tau_p$.
+Now consider $A, B \in \tau_p$, and let $H = A \cap B$.
+If $p \notin A$ or $p \notin B$ then $p \notin A \cap B$ and so $H \in \tau_p$.
+Now suppose $p \in A$ and $p \in B$.
+Then:
+{{begin-eqn}}
+{{eqn | l = H
+      | r = A \cap B
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp S H
+      | r = \relcomp S {A \cap B}
+      | c = 
+}}
+{{eqn | r = \relcomp S A \cup \relcomp S B
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection|De Morgan's Laws: Complement of Intersection]]
+}}
+{{end-eqn}}
+In order for $A$ and $B$ to be [[Definition:Open Set (Topology)|open sets]] it follows that $\relcomp S A$ and $\relcomp S B$ are both [[Definition:Countable Set|countable]].
+From [[Countable Union of Countable Sets is Countable]], their [[Definition:Set Union|union]] is also [[Definition:Countable Set|countable]] and so $\relcomp S H$ is [[Definition:Countable Set|countable]].
+So $H = A \cap B \in \tau_p$ as its [[Definition:Relative Complement|complement]] is [[Definition:Countable Set|countable]].
+Now let $\UU \subseteq \tau_p$.
+Then:
+:$\displaystyle \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
+by [[De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Union|De Morgan's Laws: Complement of Union]].
+We have either of two options:
+:$(1): \quad \forall U \in \UU: p \in \relcomp S U$
+in which case:
+:$\displaystyle p \in \bigcap_{U \mathop \in \UU} \relcomp S U$
+Or:
+:$(2): \quad \exists U \in \UU: \relcomp S U$ is [[Definition:Countable Set|countable]]
+in which case:
+:$\displaystyle \bigcap_{U \mathop \in \UU} \relcomp S U$ is [[Definition:Countable Set|countable]], from [[Intersection is Subset]].
+So in either case $\displaystyle \bigcup \UU \in \tau_p$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fortissimo Space is Completely Normal}
+Tags: Completely Normal Spaces, Fortissimo Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
+Then $T$ is a [[Definition:Completely Normal Space|completely normal space]].
+Consequently, $T$ satisfies all weaker [[Definition:Separation Axioms|separation axioms]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Fortissimo Space is T1|Fortissimo Space is $T_1$]]
+:[[Fortissimo Space is T5|Fortissimo Space is $T_5$]]
+and so by definition $T$ is [[Definition:Completely Normal Space|completely normal]].
+{{qed}}
+See [[Sequence of Implications of Separation Axioms]] for confirmation of the statement about weaker [[Definition:Separation Axioms|separation axioms]].
+\end{proof}<|endoftext|>
+\section{Fortissimo Space is T1}
+Tags: T1 Spaces, Fortissimo Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
+Then $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+From [[Fortissimo Space is Excluded Point Space with Countable Complement Space]], $T$ is an [[Definition:Expansion of Topology|expansion]] of a [[Definition:Countable Complement Space|countable complement space]].
+Then we have that a [[Countable Complement Space is T1|Countable Complement Space is $T_1$]].
+Then from [[Separation Properties Preserved by Expansion]] we have that as a [[Definition:Countable Complement Space|cuntable  complement space]] is a [[Definition:Fréchet Space (Topology)|$T_1$ space]], then so is $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fortissimo Space is Excluded Point Space with Countable Complement Space}
+Tags: Fortissimo Space, Excluded Point Topology, Countable Complement Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
+Then $\tau_p$ is the [[Definition:Minimal Topology|minimal topology]] that is [[Definition:Topology Generated by Synthetic Sub-Basis|generated]] by the [[Definition:Excluded Point Topology|excluded point topology]] and the [[Definition:Countable Complement Topology|countable complement topology]].
+\end{theorem}
+
+\begin{proof}
+Let $T_1 = \struct {S, \tau_1}$ be the [[Definition:Excluded Point Space|excluded point space]] on $S$ from $p$.
+Let $T_2 = \struct {S, \tau_2}$ be the [[Definition:Countable Complement Space|countable complement space]] on $S$.
+By definition:
+:$\tau_1 = \set {H \subseteq S: p \in \relcomp S H} \cup \set S$ 
+:$\tau_2 = \leftset {H \subseteq S: \relcomp S H}$ is [[Definition:Countable Set|countable]] $\rightset{} \cup \set \O$
+By definition of [[Definition:Fortissimo Space|Fortissimo space]], we have:
+:$U \in \tau_1 \implies U \in \tau_p$
+:$U \in \tau_2 \implies U \in \tau_p$
+So $\tau_1 \cup \tau_2 \subseteq \tau_p$.
+Similarly:
+:$U \in \tau_p \implies U \in \tau_1 \lor U \in \tau_2$
+and so $\tau_p \subseteq \tau_1 \cup \tau_2$.
+So $\tau_p = \tau_1 \cup \tau_2$ and the result follows from [[Union is Smallest Superset]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fortissimo Space is T5}
+Tags: T5 Spaces, Fortissimo Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $T$ is a [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $A, B \in \tau_p$ such that $A$ and $B$ are [[Definition:Separated Sets|separated]].
+If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both [[Definition:Open Set (Topology)|open]].
+Otherwise $p$ must be in exactly one of them, because if $p$ were in both they could not be separated.
+{{WLOG}}, suppose $p \in A$.
+Then $p \notin B$ so $B$ is [[Definition:Open Set (Topology)|open]].
+Now suppose $B$ were not [[Definition:Closed Set (Topology)|closed]].
+Then $B$ is [[Definition:Uncountable Set|uncountable]] by definition of [[Definition:Fortissimo Space|Fortissimo space]].
+But every [[Definition:Open Set (Topology)|open set]] not containing $p$ is [[Definition:Co-Countable Set|co-countable]] in $S$.
+So $p$ would be in $B^-$, where $B^-$ is the [[Definition:Closure (Topology)|closure]] of $B$.
+But then $p \in A$, [[Definition:By Hypothesis|by hypothesis]].
+So $A \cap B^- \ne \O$ and so $A$ and $B$ are not [[Definition:Separated Sets|separated]].
+So $B$ must be [[Definition:Closed Set (Topology)|closed]].
+Therefore $\relcomp S B$ is an [[Definition:Open Set (Topology)|open set]] such that $A \subseteq \relcomp S B$, and the [[Definition:T5 Space|$T_5$ separation axiom]] is satisfied.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fortissimo Space is Lindelöf}
+Tags: Lindelöf Spaces, Fortissimo Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
+Then $T$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+Let $\CC$ be an [[Definition:Open Cover|open cover]] of $T$.
+Then $\exists U \in \CC$ such that $p \in U$ and so $\relcomp S U$ is [[Definition:Countable Set|countable]].
+So $U$, together with an [[Definition:Open Neighborhood|open neighborhood]] of each of the elements of $\relcomp S U$, is a [[Definition:Countable Subcover|countable subcover]] of $\CC$.
+Hence the result by definition of [[Definition:Lindelöf Space|Lindelöf space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fortissimo Space is not Sigma-Compact}
+Tags: Fortissimo Space, Sigma-Compact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
+Then $T$ is not a [[Definition:Sigma-Compact Space|$\sigma$-compact space]].
+\end{theorem}
+
+\begin{proof}
+From [[Compact Sets in Fortissimo Space]] we have that the only [[Definition:Compact Topological Subspace|compact sets]] in $T$ are [[Definition:Finite Set|finite]].
+A [[Definition:Set Union|union]] of [[Definition:Countable Set|countably]] many [[Definition:Finite Set|finite]] sets can not be an [[Definition:Uncountable Set|uncountable]] set.
+Hence the result by definition of [[Definition:Sigma-Compact Space|$\sigma$-compact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fortissimo Space is not Separable}
+Tags: Separable Spaces, Fortissimo Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be the [[Definition:Fortissimo Space|fortissimo space]] on an [[Definition:Uncountable Set|uncountable set]] $S$.
+Then $T$ is not a [[Definition:Separable Space|separable space]].
+\end{theorem}
+
+\begin{proof}
+Let $U$ be a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$.
+By the definition of the [[Definition:Fortissimo Space|fortissimo space]], $U$ is [[Definition:Closed Set (Topology)|closed]].
+From [[Closed Set Equals its Closure]], $U^- = U \ne S$.
+Thus, by definition, $U$ is not [[Definition:Everywhere Dense|everywhere dense in $T$]].
+Thus, there exists no [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$ which is [[Definition:Everywhere Dense|everywhere dense in $T$]].
+So, by definition, $T$ is not a [[Definition:Separable Space|separable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fortissimo Space is not First-Countable}
+Tags: First-Countable Spaces, Fortissimo Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
+Then $T$ is not a [[Definition:First-Countable Space|first-countable space]].
+\end{theorem}
+
+\begin{proof}
+This proof follows the proof from [[Countable Complement Space is not First-Countable]].
+{{AimForCont}} that $p \in S$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]].
+That means:
+:there exists a [[Definition:Countable Set|countable]] [[Definition:Set of Sets|set of sets]] $\BB_p \subseteq \tau$
+such that:
+:$\forall B \in \BB_p: p \in B$
+and such that:
+:every [[Definition:Open Neighborhood|open neighborhood]] of $p$ contains some $B \in \BB_p$.
+So:
+{{begin-eqn}}
+{{eqn | l = \bigcap \BB_p
+      | r = \set p
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = S \setminus \set p
+      | r = S \setminus \bigcap \BB_p
+      | c = 
+}}
+{{eqn | r = \bigcup_{B \mathop \in \BB_p} \paren {S \setminus B}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{end-eqn}}
+By definition of [[Definition:Fortissimo Space|the fortissimo space]], each of $S \setminus B$ is [[Definition:Countable Set|countable]].
+From [[Countable Union of Countable Sets is Countable]] it follows that $\displaystyle \bigcup_{B \mathop \in \BB_p} \paren {S \setminus B}$ is also [[Definition:Countable Set|countable]].
+So $S \setminus \set p$ and therefore $S$ is also [[Definition:Countable Set|countable]].
+From this contradiction (as we have specified that $S$ is [[Definition:Uncountable Set|uncountable]]) it follows that our assumption that $p \in S$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]] must be false.
+Hence by definition $T$ can not be [[Definition:First-Countable Space|first-countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modified Fort Topology is Topology}
+Tags: Modified Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{a, b} }$  be a [[Definition:Modified Fort Space|modified Fort space]].
+Then $\tau_{a, b}$ is a [[Definition:Topology|topology]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Let $S = N \cup \set {a, b}$ where $N$ is [[Definition:Infinite Set|infinite]], $a \ne b$ and $a, b \notin N$.
+We have that $\O \subseteq N$ so $\O \in \tau_{a, b}$.
+We have that $a, b \in S, S \setminus S = \O$ and $\O$ is trivially [[Definition:Finite Set|finite]], so $S \in \tau_{a, b}$.
+Now consider $A, B \in \tau_{a, b}$, and let $H = A \cap B$.
+If $A \subseteq N$ and $B \subseteq N$ then $A \cap B \subseteq N$ from [[Intersection is Largest Subset]].
+So by definition $A \cap B \in \tau_{a, b}$
+Now suppose $A \cap \set {a, b} \ne \O$ and $B \cap \set {a, b} \ne \O$.
+Then:
+{{begin-eqn}}
+{{eqn | l = H
+      | r = A \cap B
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = N \setminus H
+      | r = N \setminus \paren {A \cap B}
+      | c = 
+}}
+{{eqn | r = \paren {N \setminus A} \cup \paren {N \setminus B}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{end-eqn}}
+In order for $A$ and $B$ to be [[Definition:Open Set (Topology)|open sets]] we have that $N \setminus A$ and $N \setminus B$ are both [[Definition:Finite Set|finite]].
+Hence their [[Definition:Set Union|union]] is also [[Definition:Finite Set|finite]] and so $N \setminus \paren {A \cap B}$ is [[Definition:Finite Set|finite]].
+So $H = A \cap B \in \tau_{a, b}$ as its [[Definition:Relative Complement|complement]] is [[Definition:Finite Set|finite]].
+Now let $\UU \subseteq \tau_{a, b}$.
+Then from [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union|De Morgan's Laws: Difference with Union]]:
+:$\displaystyle N \setminus \paren {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \paren {N \setminus U}$
+We have either of two options:
+:$(1): \quad \forall U \in \UU: U \subseteq N$
+in which case:
+:$\displaystyle \bigcup \UU \subseteq N$
+Or:
+:$(2): \quad \exists U \in \UU: N \setminus U$ is [[Definition:Finite Set|finite]]
+in which case:
+:$\displaystyle \bigcap_{U \mathop \in \UU} \paren {N \setminus U}$ is [[Definition:Finite Set|finite]], from [[Intersection is Subset]].
+So in either case:
+:$\displaystyle \bigcup \UU \in \tau_{a, b}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Naturals is Finite iff Bounded}
+Tags: Subsets, Natural Numbers
+
+\begin{theorem}
+Let $X$ be a [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|natural numbers]] $\N$.
+Then $X$ is [[Definition:Finite Set|finite]] {{iff}} it is [[Definition:Bounded Subset of Real Numbers|bounded]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|natural numbers]] is also a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers $\R$]].
+By definition, a [[Definition:Bounded Subset of Real Numbers|bounded subset of $\R$]] is [[Definition:Bounded Below Set|bounded below in $\R$]] and [[Definition:Bounded Above Set|bounded above in $\R$]].
+By the [[Well-Ordering Principle]], $X$ is [[Definition:Bounded Below Set|bounded below]]
+Thus $X$ is [[Definition:Bounded Subset of Real Numbers|bounded]] {{iff}} $X$ is [[Definition:Bounded Above Set|bounded above]].
+That is, {{iff}}:
+:$\exists p \in \N: \forall x \in X: x \le p$
+=== Necessary Condition ===
+Let $X$ be [[Definition:Finite Set|finite]].
+Then $X = \set {x_1, x_2, \ldots, x_n}$ for some $n \in \N$.
+Let $p = x_1 + x_2 + \ldots + x_n$.
+Thus $x \in X \implies x \le p$.
+Thus $X$ is [[Definition:Bounded Above Set|bounded above by $p$]].
+So by definition $X$ is [[Definition:Bounded Subset of Real Numbers|bounded]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $X \subseteq \N$ be [[Definition:Bounded Subset of Real Numbers|bounded]].
+Then for some $p \in \N$, it is contained in the [[Definition:Initial Segment of Natural Numbers|initial segment]] $\N_p = \set {0, 1, \ldots, p}$ of $\N$.
+It follows from [[Subset of Finite Set is Finite]] that $X$ is [[Definition:Finite Set|finite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Topology is Topology}
+Tags: Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $\tau$ is a [[Definition:Topology|topology]] on $T$.
+\end{theorem}
+
+\begin{proof}
+We have that $\tuple {0, 0} \notin \O$ so $\O \in \tau$.
+We have that $\forall m: \set {n: \tuple {m, n} \notin S} = \O$ which is [[Definition:Finite Set|finite]], so $S \in \tau$.
+Now consider $A, B \in \tau$, and let $H = A \cap B$.
+If $\tuple {0, 0} \notin A$ or $\tuple {0, 0} \notin B$ then $\tuple {0, 0} \notin A \cap B$ and so $H \in \tau_p$.
+Now suppose $\tuple {0, 0} \in A$ and $\tuple {0, 0} \in B$.
+Then:
+{{begin-eqn}}
+{{eqn | l = H
+      | r = A \cap B
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp S H
+      | r = \relcomp S {A \cap B}
+      | c = 
+}}
+{{eqn | r = \relcomp S A \cup \relcomp S B
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection|De Morgan's Laws: Complement of Intersection]]
+}}
+{{end-eqn}}
+In order for $A$ and $B$ to be [[Definition:Open Set (Topology)|open sets]] it follows that:
+:At most a [[Definition:Finite Set|finite number]] $m_A$ of sets $S_{m_A} = \set {n: \tuple {m_A, n} \notin A}$ are [[Definition:Infinite Set|infinite]]
+:At most a [[Definition:Finite Set|finite number]] $m_B$ of sets $S_{m_B} = \set {n: \tuple {m_B, n} \notin B}$ are [[Definition:Infinite Set|infinite]].
+Let $r \in \Z_{\ge 0}$ and let:
+:$S_{r_A} = \set {n: \tuple {r, n} \notin A}$
+:$S_{r_B} = \set {n: \tuple {r, n} \notin B}$
+We have that: 
+{{begin-eqn}}
+{{eqn | o =
+      | r = S_{r_{A \mathop \cap B} }
+      | c = 
+}}
+{{eqn | r = \set {n: \tuple {r, n} \notin A \cap B}
+      | c = 
+}}
+{{eqn | r = \set {n: \tuple {r, n} \notin A} \cup \set {n: \tuple {r, n} \notin B}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection|De Morgan's Laws: Difference with Intersection]]
+}}
+{{eqn | r = S_{r_A} \cup S_{r_B}
+      | c = 
+}}
+{{end-eqn}}
+If $S_{r_A}$ and $S_{r_B}$ are [[Definition:Finite Set|finite]] then so is $S_{r_A} \cup S_{r_B}$.
+If only a [[Definition:Finite Set|finite number]] of $S_{m_A}$ and $S_{m_B}$ are [[Definition:Infinite Set|infinite]] then it follows that only a [[Definition:Finite Set|finite number]] of $S_{m_{A \cap B}}$ are [[Definition:Infinite Set|infinite]].
+So $A \cap B \in \tau$.
+Now let $\UU \subseteq \tau$.
+Then:
+:$\displaystyle \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
+by [[De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Union|De Morgan's Laws: Complement of Union]].
+We have either of two options:
+:$(1): \quad \forall U \in \UU: \tuple {0, 0} \notin U$
+in which case:
+:$\displaystyle \tuple {0, 0} \notin \bigcap \UU$
+Or:
+:$(2): \quad$ At most a [[Definition:Finite Set|finite number]] $m$ of sets $S_m = \set {n: \tuple {m, n} \notin U}$ are [[Definition:Infinite Set|infinite]]
+in which case:
+:At most a [[Definition:Finite Set|finite number]] $m$ of sets $S_m = \set {n: \tuple {m, n} \notin \bigcap \UU}$ are [[Definition:Infinite Set|infinite]].
+So in either case $\displaystyle \bigcup \UU \in \tau$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is Expansion of Countable Fort Space}
+Tags: Fort Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]], where $S = \Z_{\ge 0} \times \Z_{\ge 0}$.
+Let $T_p = \struct {S, \tau_p}$ be the [[Definition:Fort Space|Fort space]] on $S$ where $p = \left({0, 0}\right)$.
+Then $\tau$ is an [[Definition:Expansion of Topology|expansion]] of $\tau_p$.
+Furthermore, $S$ is [[Definition:Countably Infinite Set|countably infinite]], so $T_p$ is a [[Definition:Countable Fort Space|countable Fort space]].
+\end{theorem}
+
+\begin{proof}
+Let $H \in \tau_p$ where $p = \tuple {0, 0}$.
+Then either:
+:$(1): \quad \tuple {0, 0} \in \relcomp S H$
+or:
+:$(2): \quad H$ is [[Definition:Cofinite Set|cofinite]] in $S$, that is $\relcomp S H$ is [[Definition:Finite Set|finite]].
+Case $(1)$ means that $\tuple {0, 0} \notin H$ and so $H \in \tau$ by definition of the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Suppose case $(2)$ applies.
+Then for all $m \in \Z_{\ge 0}$, the sets $S_m$ defined as $S_m = \set {n: \tuple {m, n} \notin H}$ are [[Definition:Finite Set|finite]].
+Thus in a [[Definition:Finite Set|finite number]] of these (that is, none of them) the set $S_m$ is [[Definition:Infinite Set|infinite]].
+So $H \in \tau$.
+So, if $H \in \tau_p$ it follows that $H \in \tau$ and so $\tau_p \subseteq \tau$.
+Hence the result by definition of [[Definition:Expansion of Topology|expansion]].
+Finally, from [[Arens-Fort Space is Countable]] we have that $S$ is [[Definition:Countably Infinite Set|countably infinite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is T1}
+Tags: T1 Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Arens-Fort Space is Completely Hausdorff]].
+Then from [[Sequence of Implications of Separation Axioms]], a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]] is a [[Definition:Fréchet Space (Topology)|$T_1$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is Completely Hausdorff}
+Tags: Completely Hausdorff Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Fort Space is Completely Normal]]
+:[[Arens-Fort Space is Expansion of Fort Space]].
+From [[Sequence of Implications of Separation Axioms]] we have that a [[Definition:Fort Space|Fort space]] is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ space]].
+Then in [[Separation Properties Preserved by Expansion]], it follows that as a [[Definition:Fort Space|Fort space]] is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ space]], then so is the [[Definition:Arens-Fort Space|Arens-Fort space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is T5}
+Tags: T5 Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is a [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $A, B \subseteq S$ such that $A$ and $B$ are [[Definition:Separated Sets|separated]].
+Let $p = \tuple {0, 0}$.
+If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both [[Definition:Open Set (Topology)|open]] and the problem is solved.
+Otherwise $p$ must be in exactly one of them, because if $p$ were in both they could not be [[Definition:Separated Sets|separated]].
+{{WLOG}}, suppose $p \in A$.
+Then $p \notin B$ so $B$ is [[Definition:Open Set (Topology)|open]].
+$B \cup \set p$ is closed by the definition of the [[Definition:Arens-Fort Space|Arens-Fort space]].
+From [[Set is Subset of its Topological Closure]] and [[Set Closure is Smallest Closed Set in Topological Space]]:
+:$B \subseteq B^- \subseteq B \cup \set p$
+where $B^-$ is the [[Definition:Closure (Topology)|closure]] of $B$.
+So $B$ is [[Definition:Closed Set (Topology)|closed]] or $p$ is in $B^-$.
+But then $p \in A$, [[Definition:By Hypothesis|by hypothesis]].
+So $A \cap B^- \ne \O$ and so $A$ and $B$ are not [[Definition:Separated Sets|separated]].
+So $B$ must be [[Definition:Closed Set (Topology)|closed]].
+Therefore $\relcomp S B$ is an [[Definition:Open Set (Topology)|open set]] such that $A \subseteq \relcomp S B$, and the [[Definition:T5 Space|$T_5$ separation axiom]] is satisfied.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is Completely Normal}
+Tags: Completely Normal Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is a [[Definition:Completely Normal Space|completely normal space]].
+Consequently, $T$ satisfies all weaker [[Definition:Separation Axioms|separation axioms]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Arens-Fort Space is T1|Arens-Fort Space is $T_1$]]
+:[[Arens-Fort Space is T5|Arens-Fort Space is $T_5$]]
+and so by definition $T$ is [[Definition:Completely Normal Space|completely normal]].
+{{qed}}
+See [[Sequence of Implications of Separation Axioms]] for confirmation of the statement about weaker [[Definition:Separation Axioms|separation axioms]].
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is not First-Countable}
+Tags: First-Countable Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is not a [[Definition:First-Countable Space|first-countable space]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $T$ is [[Definition:First-Countable Space|first-countable]].
+Then there exists a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local base]] $B_0 = \sequence {U_i}_{i \mathop = 1}^\infty$ for $\tuple {0, 0}$.
+Suppose there does not exist a point $\tuple {n_i, m_i} \in U_i$ such that $n_i > i$ and $m_i > i$.
+Then:
+:$\forall \tuple {n, m} \in U_i: n \le i$ or $m \le i$
+Now suppose that there does not exist some $\tuple {n, m} \in U_i$ with $n > i$.
+Let $S_k$ be defined as:
+:$S_k = \set {n: \tuple {k, n} \notin U_i}$
+The definition of the [[Definition:Arens-Fort Space|Arens-Fort space]] clearly stipulates that $S_k$ can only be a [[Definition:Infinite Set|infinite set]] for a [[Definition:Finite Set|finite number]] of $k \in \Z_{\ge 0}$.
+But:
+:$k > i \implies S_k = \Z_{\ge 0}$
+So we have shown that:
+:$\exists \tuple {n, m} \in U_i: n > i$
+So, let $n > i$, and let $m \le i$.
+Then $\set {m: m > i} \subseteq S_n$. 
+Thus $S_n = \set {m: \tuple {n, m} \notin U_i}$ is [[Definition:Infinite Set|infinite]] for all $n > i$.
+Again, this contradicts the definition of the [[Definition:Arens-Fort Space|Arens-Fort space]].
+So, from the definition of the [[Definition:Neighborhood of Point|neighborhood]] of $\tuple {0, 0}$, for each $U_i \in B_0$ there is a point $\tuple {n_i, m_i} \in U_i$ such that both $n_i > i$ and $m_i > i$.
+Now let the set $E$ be constructed as:
+:$E := X \setminus \sequence {\tuple {n_i, m_i} }_{i \mathop = 1}^\infty$
+We now prove that $E$ is a [[Definition:Neighborhood of Point|neighborhood]] of $\tuple {0, 0}$.
+By the definition of [[Definition:Arens-Fort Space|Arens-Fort space]], it is enough to show that:
+:$\forall i \in \N: S_i$ is [[Definition:Finite Set|finite]].
+We have that:
+{{begin-eqn}}
+{{eqn | l = S_i
+      | r = \set {m: \tuple {i, m} \notin E}
+      | c = 
+}}
+{{eqn | r = \set {m: \tuple {i, m} \in \sequence {\tuple {n_j, m_j} }_{j \mathop = 1}^\infty}
+      | c = 
+}}
+{{end-eqn}}
+But we defined $n_j > j$.
+So if $j > i \implies n_j > j > i$.
+Thus $S_i \subseteq \sequence {m_j}_{j \mathop = 1}^i$ is [[Definition:Finite Set|finite]]. 
+However, $\tuple {n_i, m_i} \notin E$.
+Therefore there exists no $U_i \subseteq E$.
+From this [[Definition:Contradiction|contradiction]] it follows that $T$ cannot be [[Definition:First-Countable Space|first-countable]].
+{{qed}}
+{{proofread}}
+\end{proof}<|endoftext|>
+\section{Countable Space is Separable}
+Tags: Separable Spaces, Countable Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] where $S$ is a [[Definition:Countable Set|countable set]].
+Then $T$ is a [[Definition:Separable Space|separable space]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Topological Space|topological space]] $T = \struct {S, \tau}$ is [[Definition:Separable Space|separable]] if there exists a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $S$ which is [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+The [[Definition:Closure (Topology)|closure]] of $S$ in $S$ is trivially $S$.
+So, by definition, $S$ is [[Definition:Everywhere Dense|everywhere dense]] in $S$.
+As $S$ is [[Definition:Countable Set|countable]] by definition, the result follows.
+{{qed}}
+[[Category:Separable Spaces]]
+[[Category:Countable Sets]]
+5sagq1qbpwtck3uym0x314bh0a672vu
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is Separable}
+Tags: Separable Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is a [[Definition:Separable Space|separable space]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Definition:Arens-Fort Space|Arens-Fort space]] is an [[Definition:Expansion of Topology|expansion]] of a [[Definition:Countable Fort Space|countable Fort space]].
+So $S$ is [[Definition:Countable Set|countable]].
+The result follows from [[Countable Space is Separable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is Sigma-Compact}
+Tags: Sigma-Compact Spaces, Arens-Fort Space, Arens-Fort Space is Sigma-Compact
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is a [[Definition:Sigma-Compact Space|$\sigma$-compact space]].
+\end{theorem}
+
+\begin{proof}
+From [[Arens-Fort Space is Countable]] we have that $S$ is [[Definition:Countably Infinite Set|countably infinite]].
+Let $\tau_d$ be the [[Definition:Discrete Space|discrete topology]] on $S$.
+Consider the [[Definition:Identity Mapping|identity function]] $I_S: \left({S, \tau_d}\right) \to \left({S, \tau}\right)$.
+From [[Mapping from Discrete Space is Continuous]], $I_S$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+From [[Identity Mapping is Surjection]], $I_S$ is a [[Definition:Surjection|surjection]].
+From [[Countable Discrete Space is Sigma-Compact]] and [[Sigma-Compactness is Preserved under Continuous Surjection]] we conclude that $T$ is a [[Definition:Sigma-Compact Space|$\sigma$-compact space]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The result follows from [[Arens-Fort Space is Countable]] and [[Countable Space is Sigma-Compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is Lindelöf}
+Tags: Lindelöf Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:The [[Arens-Fort Space is Sigma-Compact]].
+:A [[Sigma-Compact Space is Lindelöf|$\sigma$-compact space is Lindelöf]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is not Weakly Locally Compact}
+Tags: Weakly Locally Compact Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is not a [[Definition:Weakly Locally Compact Space|weakly locally compact space]].
+\end{theorem}
+
+\begin{proof}
+We have that [[Neighborhood of Origin in Arens-Fort Space is not Compact]].
+So $\tuple {0, 0}$ is a point in $S$ which is not contained in a [[Definition:Compact Topological Subspace|compact]] [[Definition:Neighborhood (Topology)|neighborhood]].
+Hence, by definition, $T$ is not [[Definition:Weakly Locally Compact Space|weakly locally compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is not Compact}
+Tags: Compact Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is not a [[Definition:Compact Topological Space|compact space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+: The [[Arens-Fort Space is not Weakly Locally Compact]].
+: A [[Compact Space is Weakly Locally Compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is not Countably Compact}
+Tags: Countably Compact Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is not a [[Definition:Countably Compact Space|countably compact space]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} the [[Definition:Arens-Fort Space|Arens-Fort space]] is [[Definition:Countably Compact Space|countably compact]].
+From [[Arens-Fort Space is Lindelöf]], it is also [[Definition:Lindelöf Space|Lindelöf]].
+From [[Countably Compact Lindelöf Space is Compact]], $T$ is [[Definition:Compact Topological Space|compact]].
+But this [[Definition:Contradiction|contradicts]]  [[Arens-Fort Space is not Compact]].
+Hence the result from [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lindelöf T3 Space is Paracompact}
+Tags: Lindelöf Spaces, T3 Spaces, Paracompact Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:T3 Space|$T_3$ space]] which is also [[Definition:Lindelöf Space|Lindelöf]].
+Then $T$ is [[Definition:Paracompact Space|paracompact]].
+\end{theorem}<|endoftext|>
+\section{Arens-Fort Space is Paracompact}
+Tags: Arens-Fort Space is Paracompact
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is a [[Definition:Paracompact Space|paracompact space]].
+\end{theorem}
+
+\begin{proof}
+Let $\CC$ be any [[Definition:Open Cover|open cover]] of $T$.
+Let $H \in \CC$ be any [[Definition:Open Set (Topology)|open set]] which contains $\tuple {0, 0}$.
+For all $s \in S$ such that $s \ne \tuple {0, 0}$, we have that $\set s$ is [[Definition:Open Set (Topology)|open]] in $T$ by definition of the [[Definition:Arens-Fort Space|Arens-Fort space]].
+So the [[Definition:Open Cover|open cover]] of $T$ which consists of $H$ together with all the [[Definition:Open Set (Topology)|open sets]] $\set s$, where $s \in S \setminus H$ is a [[Definition:Refinement of Cover|refinement]] of $T$ which is [[Definition:Locally Finite Cover|locally finite]].
+Hence the result, by definition of [[Definition:Paracompact Space|paracompact space]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that:
+:The [[Arens-Fort Space is Completely Normal]].
+:The [[Arens-Fort Space is Lindelöf]].
+From [[Sequence of Implications of Separation Axioms]], it follows that $T$ is a [[Definition:T3 Space|$T_3$ space]].
+The result follows from [[Lindelöf T3 Space is Paracompact|Lindelöf $T_3$ Space is Paracompact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First-Countability is not Continuous Invariant}
+Tags: First-Countable Spaces, Continuous Invariants
+
+\begin{theorem}
+Let $T_A = \struct {A, \tau_A}$ and $T_B = \struct {B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
+Let $\phi: T_A \to T_B$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+If $T_A$ is a [[Definition:First-Countable Space|first-countable space]], then it does not necessarily follow that $T_B$ is also [[Definition:First-Countable Space|first-countable]].
+\end{theorem}
+
+\begin{proof}
+Let $T_S = \struct {S, \tau_S}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Let $T_D = \struct {S, \tau_D}$ be the [[Definition:Discrete Space|discrete space]], also on $S$.
+Let $I_S: S \to S$ be the [[Definition:Identity Mapping|identity mapping]] on $S$.
+From [[Mapping from Discrete Space is Continuous]], we have that $I_S$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+Then we have:
+:[[Discrete Space is First-Countable]]
+:[[Arens-Fort Space is not First-Countable]]
+Thus we have demonstrated a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] from a [[Definition:First-Countable Space|first-countable space]] to a space which is not [[Definition:First-Countable Space|first-countable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second-Countability is not Continuous Invariant}
+Tags: Second-Countable Spaces, Continuous Invariants
+
+\begin{theorem}
+Let $T_A = \struct {A, \tau_A}$ and $T_B = \struct {B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
+Let $\phi: T_A \to T_B$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+If $T_A$ is a [[Definition:Second-Countable Space|second-countable space]], then it does not necessarily follow that $T_B$ is also [[Definition:Second-Countable Space|second-countable]].
+\end{theorem}
+
+\begin{proof}
+Let $T_S = \struct {S, \tau_S}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Let $T_D = \struct {S, \tau_D}$ be the [[Definition:Discrete Space|discrete space]], also on $S$.
+As $S$ is [[Definition:Countable Set|countable]], from [[Arens-Fort Space is Expansion of Countable Fort Space]], it follows that $T_D = \struct {S, \tau_D}$ is a [[Definition:Countable Discrete Space|countable discrete space]].
+Let $I_S: S \to S$ be the [[Definition:Identity Mapping|identity mapping]] on $S$.
+From [[Mapping from Discrete Space is Continuous]], we have that $I_S$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+Then we have that a [[Countable Discrete Space is Second-Countable]].
+We have that the [[Arens-Fort Space is not First-Countable]].
+It follows from [[Second-Countable Space is First-Countable]] that the [[Definition:Arens-Fort Space|Arens-Fort space]] is not [[Definition:Second-Countable Space|second-countable]] either.
+Thus we have demonstrated a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] from a [[Definition:Second-Countable Space|second-countable space]] to a space which is not [[Definition:Second-Countable Space|second-countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is not Connected}
+Tags: Connected Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is not a [[Definition:Connected Topological Space|connected space]].
+\end{theorem}
+
+\begin{proof}
+Consider $p \in S$ such that $p \ne \tuple {0, 0}$.
+From [[Clopen Points in Arens-Fort Space]], we have that $\set p$ is both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] in $T$.
+So by definition of [[Definition:Closed Set (Topology)|closed set]], $\relcomp S {\set p}$ is also both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] in $T$.
+So, by definition, $\set p \mid \relcomp S {\set p}$ is a [[Definition:Separation (Topology)|separation]] of $T$
+Hence the result, by definition of [[Definition:Connected Topological Space|connected space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is not Locally Connected}
+Tags: Locally Connected Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is not a [[Definition:Locally Connected Space|locally connected space]].
+\end{theorem}
+
+\begin{proof}
+Let $\UU_0$ be a [[Definition:Local Basis|local basis]] for $\tuple {0, 0}$.
+Let $U \in \UU_0$.
+By definition of [[Definition:Local Basis|local basis]], $U$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Clopen Points in Arens-Fort Space]], $\set {\tuple {0, 0} }$ is not [[Definition:Open Set (Topology)|open]] in $T$.
+So $U \ne \set {\tuple {0, 0} }$. 
+Therefore:
+:$\exists p \in U: p \ne \tuple {0, 0}$
+From [[Singleton of Element is Subset]]:
+:$\set p \subseteq U$
+From [[Clopen Points in Arens-Fort Space]] it follows that $\set p$ is [[Definition:Clopen Set|clopen]].
+As $U \in U_0$ it follows by definition of [[Definition:Local Basis|local basis]] that:
+:$\tuple {0, 0} \in U$
+and so:
+:$U \ne \set p$
+That is:
+:$\set p \subsetneq U$
+From [[Connected iff no Proper Clopen Sets]], the set $U$ is not [[Definition:Connected Topological Space|connected]].
+{{explain|Re-establish the definition of [[Definition:Connected Topological Space]] as the current flow does not make it obvious that a connected subset has no proper clopen sets, only that this condition applies to the full space.}}
+It is deduced that any [[Definition:Local Basis|local basis]] is formed with [[Definition:Disconnected Set (Topology)|disconnected sets]].
+Thus, by definition, $T$ is not [[Definition:Locally Connected Space|locally connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Clopen Points in Arens-Fort Space}
+Tags: Clopen Sets, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Let $q \in S: q \ne \tuple {0, 0}$.
+Then $\set q$ is both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] in $T$.
+$\set {\tuple {0, 0} }$ itself is [[Definition:Closed Set (Topology)|closed]], but not [[Definition:Open Set (Topology)|open]].
+\end{theorem}
+
+\begin{proof}
+We have that $\set q$ is [[Definition:Finite Set|finite]] so $\relcomp S {\set q}$ is [[Definition:Cofinite Set|cofinite]].
+So $\relcomp S {\set q}$ is [[Definition:Open Set (Topology)|open]] and so $\set q$ is [[Definition:Closed Set (Topology)|closed]].
+Then we have that $\tuple {0, 0} \notin \set q$ so $\set q$ is [[Definition:Open Set (Topology)|open]].
+However, $\tuple {0, 0} \notin \relcomp S {\set {\tuple {0, 0} } }$ and $\relcomp S {\set {\tuple {0, 0} } }$ is clearly not [[Definition:Open Set (Topology)|open]] in $T$.
+But as $\set {\tuple {0, 0} }$ is [[Definition:Finite Set|finite]] so $\relcomp S {\set {\tuple {0, 0} } }$ is [[Definition:Cofinite Set|cofinite]].
+Hence, as for $\set q$, we have that $\set {\tuple {0, 0} }$ is [[Definition:Closed Set (Topology)|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Neighborhood of Origin of Arens-Fort Space is Closed}
+Tags: Closed Sets, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Every [[Definition:Neighborhood of Point|neighborhood]] of $\tuple {0, 0}$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+\end{theorem}
+
+\begin{proof}
+Let $H \subseteq T$ such that:
+:$\exists U \in \tau: \tuple {0, 0} \in U \subseteq H \subseteq S$
+that is: such that $H$ is a [[Definition:Neighborhood of Point|neighborhood]] of $\tuple {0, 0}$ in $T$.
+As $\tuple {0, 0} \in H$ it follows that $\tuple {0, 0} \notin \relcomp S H$.
+So, by definition of the [[Definition:Arens-Fort Space|Arens-Fort space]], $\relcomp S H$ is [[Definition:Open Set (Topology)|open]] in $T$.
+So by definition, we have that $H$ is [[Definition:Closed Set (Topology)|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is Zero Dimensional}
+Tags: Zero Dimensional Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ denote the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is [[Definition:Zero Dimensional Space|zero dimensional]].
+\end{theorem}
+
+\begin{proof}
+Let $q \in S$ such that $q \ne \tuple {0, 0}$.
+Then from [[Clopen Points in Arens-Fort Space]], $\set q$ is [[Definition:Clopen Set|clopen]].
+So $\forall q \in S, q \ne \tuple {0, 0}: \set {\set q}$ is a [[Definition:Local Basis|local basis]] for $q$.
+If we take the [[Definition:Neighborhood (Topology)|neighborhoods]] of $\tuple {0, 0}$ that are [[Definition:Open Neighborhood|open]] we get that they are also [[Definition:Closed Set (Topology)|closed]] from [[Neighborhood of Origin of Arens-Fort Space is Closed]].
+The [[Definition:Set Union|union]] of the [[Definition:Local Basis|local basis]] forms a [[Definition:Basis (Topology)|basis]] for the topology.
+This [[Definition:Basis (Topology)|basis]] is formed with [[Definition:Clopen Set|clopen sets]].
+So, by definition, $T$ is a [[Definition:Zero Dimensional Space|zero dimensional space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is Totally Separated}
+Tags: Totally Separated Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is [[Definition:Totally Separated Space|totally separated]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:The [[Arens-Fort Space is Zero Dimensional]].
+:The [[Arens-Fort Space is T1|Arens-Fort Space is $T_1$]] and so by [[T1 Space is T0 Space|$T_1$ Space is $T_0$ Space]] is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+Then we have that a [[Zero Dimensional T0 Space is Totally Separated|Zero Dimensional $T_0$ Space is Totally Separated]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isolated Points in Arens-Fort Space}
+Tags: Isolated Points, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Let $q \in S: q \ne \tuple {0, 0}$.
+Then $q$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $T$.
+\end{theorem}
+
+\begin{proof}
+If $q \ne \tuple {0, 0}$ then from [[Clopen Points in Arens-Fort Space]] we have that $\set q$ is both [[Definition:Closed Set (Topology)|closed]] and [[Definition:Open Set (Topology)|open]] in $T$.
+In particular, $\set q$ is [[Definition:Open Set (Topology)|open]] in $T$.
+The result follows from [[Point in Topological Space is Open iff Isolated]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Point in Topological Space is Open iff Isolated}
+Tags: Open Sets, Isolated Points
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $x \in S$.
+Then $\left\{{x}\right\}$ is [[Definition:Open Set (Topology)|open]] in $T$ {{iff}} $x$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $T$.
+\end{theorem}
+
+\begin{proof}
+Let $\left\{{x}\right\}$ be [[Definition:Open Set (Topology)|open]] in $T$.
+Then we have that:
+:$\exists \left\{{x}\right\} \in \tau: x \in \left\{{x}\right\}\subseteq S$
+This is precisely the condition which ensures that $x$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $T$.
+Now suppose that $x$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $T$.
+Then by definition there exists a [[Definition:Open Set (Topology)|open set]] of $T$ containing no points other than $x$:
+:$\exists U \in \tau: U = \left\{{x}\right\}$
+That is, if $x$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $T$ then $\left\{{x}\right\}$ is [[Definition:Open Set (Topology)|open]] in $T$.
+{{qed}}
+[[Category:Open Sets]]
+[[Category:Isolated Points]]
+luapupy565wng4ac9qxl0b6ulox69fo
+\end{proof}<|endoftext|>
+\section{Arens-Fort Space is Scattered}
+Tags: Scattered Spaces, Arens-Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
+Then $T$ is a [[Definition:Scattered Space|scattered space]].
+\end{theorem}
+
+\begin{proof}
+Let $H \subseteq T$ such that $H \ne \set {\tuple {0, 0} }$.
+Then $\exists x \in H: x \ne \tuple {0, 0}$.
+From [[Clopen Points in Arens-Fort Space]], every point of $T$ apart from $\tuple {0, 0}$ is [[Definition:Open Point|open]] in $T$.
+So $\set x$ is an [[Definition:Open Set (Topology)|open set]] of $T$.
+So $H \cap \set x = \set x$ and so $x$ is [[Definition:Isolated Point of Subset|isolated in $H$]].
+Thus $H$ contains at least one point which is [[Definition:Isolated Point of Subset|isolated in $H$]].
+On the other hand, suppose $H = \set {\tuple {0, 0} }$.
+From [[Singleton Point is Isolated]], $\tuple {0, 0}$ is an [[Definition:Isolated Point of Subset|isolated point in $\set {\tuple {0, 0} } = H$]].
+So again $H$ contains at least one point which is [[Definition:Isolated Point of Subset|isolated in $H$]].
+So for all $H \subseteq S$ we have that $H$ contains at least one point which is [[Definition:Isolated Point of Subset|isolated in $H$]].
+Hence the result, by definition of [[Definition:Scattered Space|scattered space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fortissimo Space is not Metrizable}
+Tags: Metrizable Topologies, Fortissimo Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
+Then $\tau_p$ is not a [[Definition:Metrizable Topology|metrizable topology]].
+\end{theorem}
+
+\begin{proof}
+From [[Metric Space is First-Countable]] and [[Fortissimo Space is not First-Countable]], it is deduced that $T$ is not a [[Definition:Metrizable Topology|metrizable topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fortissimo Space is Paracompact}
+Tags: Paracompact Spaces, Fortissimo Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
+Then $T$ is a [[Definition:Paracompact Space|paracompact space]].
+\end{theorem}
+
+\begin{proof}
+Let $\CC$ be an [[Definition:Open Cover|open cover]] of $T$.
+Let $U_p \in \CC$ be an [[Definition:Open Set (Topology)|open set]] in that cover which contains $p$.
+All the points of $S \setminus U_p$ are [[Definition:Open Point|open points]] by definition of [[Definition:Fortissimo Space|Fortissimo space]].
+Then $\DD = \set {\set s: s \in S \setminus U_p} \cup \set {U_p}$ is an [[Definition:Open Refinement|open refinement]] of $\CC$.
+Take $k \in S$:
+:If $k = p$, then $U_p$ is a [[Definition:Neighborhood of Point|neighbourhood]] of $k$ which [[Definition:Set Intersection|intersects]] only with $U_p$ from all the [[Definition:Subset|subsets]] in $\DD$
+:If $k \ne p$, then $\set k$ is a [[Definition:Neighborhood of Point|neighbourhood]] of $k$ which [[Definition:Set Intersection|intersects]] only with $\set k$ from all the [[Definition:Subset|subsets]] in $\DD$
+Thus $\DD$ is by definition [[Definition:Locally Finite Cover|locally finite]].
+Hence, by definition, $T$ is a [[Definition:Paracompact Space|paracompact space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Clopen Points in Modified Fort Space}
+Tags: Modified Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{a, b} }$  be a [[Definition:Modified Fort Space|modified Fort space]].
+Then all points in $S \setminus \set {a, b}$ are both [[Definition:Open Point|open]] and [[Definition:Closed Point|closed]] in $T$.
+$a$ and $b$ themselves are not [[Definition:Open Point|open]] in $T$, but they are [[Definition:Closed Point|closed]] in $T$.
+\end{theorem}
+
+\begin{proof}
+Let $p \in S: p \notin \set {a, b}$.
+From the definition of [[Definition:Modified Fort Space|modified Fort space]], any subset of $S \setminus \set {a, b}$ is [[Definition:Open Set (Topology)|open]] in $T$.
+It follows directly that as $p \in S \setminus \set {a, b}$ we have that $\set p \subseteq S \setminus \set {a, b}$.
+Hence $p$ is [[Definition:Open Point|open]] in $T$.
+As for $a$ and $b$, we have that $S \setminus \set a$ and $S \setminus \set b$ are not [[Definition:Finite Set|finite]] and so $a$ and $b$ are not [[Definition:Open Point|open]] in $T$.
+{{qed|lemma}}
+For all points $p \in S$ (including $a$ and $b$), we have that $\set p$ is (trivially) [[Definition:Finite Set|finite]].
+It follows that while $S \setminus \set p$ contains either $a$ or $b$, it is [[Definition:Cofinite Set|cofinite]].
+Thus $S \setminus \set p$ is [[Definition:Open Set (Topology)|open]] in $T$.
+It follows by definition that $p$ is [[Definition:Closed Point|closed]] in $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modified Fort Space is Compact}
+Tags: Compact Spaces, Modified Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{a, b} }$ be a [[Definition:Modified Fort Space|modified Fort space]].
+Then $T$ is a [[Definition:Compact Topological Space|compact space]].
+\end{theorem}
+
+\begin{proof}
+Let $\CC$ be an [[Definition:Open Cover|open cover]] of $T$.
+Then $\exists U \in \CC: a \in U$.
+Then by definition of [[Definition:Modified Fort Space|modified Fort space]], $U$ is [[Definition:Cofinite Set|cofinite]].
+That is, $S \setminus U$ is [[Definition:Finite Set|finite]].
+Then $S \setminus U$ is covered by a [[Definition:Finite Subcover|finite subcover]] of $\CC$.
+Hence, by definition, $T$ is [[Definition:Compact Topological Space|compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modified Fort Space is T1}
+Tags: T1 Spaces, Modified Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{a, b} }$ be a [[Definition:Modified Fort Space|modified Fort space]].
+Then $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+Let $p, q \in S: p \ne q$.
+There are two options:
+:$(1): \quad$ At least one of $p$ and $q$ is not in $\set {a, b}$
+:$(2): \quad$ Both $p$ and $q$ are in $\set {a, b}$.
+$(1): \quad$ First let $p \notin \set {a, b}$ or $q \notin \set {a, b}$.
+{{WLOG}}, let $p \notin \set {a, b}$.
+From [[Clopen Points in Modified Fort Space]] both $\set p \in \tau_{a, b}$ and $S \setminus \set p \in \tau_{a, b}$.
+So:
+:$p \in \set p, q \notin \set p$
+:$p \notin S \setminus \set p, q \in S \setminus \set p$
+Similarly for $q \notin \set {a, b}$.
+$(2): \quad$ Next, let $p, q \in \set {a, b}$.
+Let $V_p \subseteq S$ such that $p \in V_p, q \notin V_p$ and $V_p$ is [[Definition:Finite Set|finite]].
+Then $S \setminus V_p$ is [[Definition:Open Set (Topology)|open]] in $T$ containing $q$ but not $p$.
+Similarly, let $V_q \subseteq S$ such that $q \in V_q, p \notin V_q$ and $V_q$ is [[Definition:Finite Set|finite]].
+Then $S \setminus V_q$ is [[Definition:Open Set (Topology)|open]] in $T$ containing $p$ but not $q$.
+Hence it can be seen that $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modified Fort Space is not T2}
+Tags: Hausdorff Spaces, Modified Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{a, b} }$ be a [[Definition:Modified Fort Space|modified Fort space]].
+Then $T$ is not a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+Consider $U, V \in \tau_{a, b}$ such that $a \in U, b \in V$.
+We have that both $U$ and $V$ are [[Definition:Cofinite Set|cofinite]].
+So $U$ and $V$ must be [[Definition:Infinite Set|infinite]].
+Suppose $U \cap V = \O$.
+Then from [[Intersection with Complement is Empty iff Subset]] it follows that $U \subseteq \relcomp S V$ and so $U$ is [[Definition:Finite Set|finite]].
+But this contradicts the fact that $U$ is [[Definition:Infinite Set|infinite]].
+So $U \cap V \ne \O$.
+Hence the result by definition of [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modified Fort Space is not T3, T4 or T5}
+Tags: T3 Spaces, T4 Spaces, T5 Spaces, Modified Fort Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_{a, b} }$ be a [[Definition:Modified Fort Space|modified Fort space]].
+Then $T$ is not a [[Definition:T3 Space|$T_3$ space]], [[Definition:T4 Space|$T_4$ space]] or [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+We have that a [[Modified Fort Space is T1|Modified Fort Space is $T_1$]].
+From [[T1 Space is T0 Space|$T_1$ Space is $T_0$ Space]], $T$ is a [[Definition:Kolmogorov Space|$T_0$ space]].
+We then have that a [[Modified Fort Space is not T2|Modified Fort Space is not $T_2$]].
+From [[Regular Space is T2 Space|Regular Space is $T_2$ Space]], $T$ is not a [[Definition:Regular Space|regular space]].
+By definition, a [[Definition:Regular Space|regular space]] is a [[Definition:Topological Space|space]] that is both a [[Definition:Kolmogorov Space|$T_0$ space]] and a [[Definition:T3 Space|$T_3$ space]].
+But $T$ is a [[Definition:Kolmogorov Space|$T_0$ space]] and not a [[Definition:Regular Space|regular space]].
+So it follows that $T$ can not be a [[Definition:T3 Space|$T_3$ space]].
+Next we have that a [[Normal Space is T3 Space|Normal Space is a $T_3$ Space]].
+But as $T$ is not a [[Definition:T3 Space|$T_3$ space]], $T$ can not be a [[Definition:Normal Space|normal space]].
+By definition, a [[Definition:Normal Space|normal space]] is a space that is both a [[Definition:Fréchet Space (Topology)|$T_1$ space]] and a [[Definition:T4 Space|$T_4$ space]].
+But $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ space]] and not a [[Definition:Normal Space|normal space]].
+So it follows that $T$ can not be a [[Definition:T4 Space|$T_4$ space]].
+Finally we have that a [[T5 Space is T4 Space|$T_5$ Space is a $T_4$ Space]].
+But as $T$ is not a [[Definition:T4 Space|$T_4$ space]], $T$ can not be a [[Definition:T5 Space|$T_5$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis for Euclidean Topology on Real Number Line}
+Tags: Real Number Line with Euclidean Topology, Topological Bases
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number|set of real numbers]].
+Let $\BB$ be the [[Definition:Set of Sets|set]] of [[Definition:Subset|subsets]] of $\R$ defined as:
+:$\BB = \set {\openint a b: a, b \in \R}$
+That is, $\BB$ is the set of all [[Definition:Open Real Interval|open real intervals]] of $\R$:
+:$\openint a b := \set {x \in \R: a < x < b}$
+Then $\BB$ forms a [[Definition:Basis (Topology)|basis]] for the [[Definition:Euclidean Topology on Real Number Line|Euclidean topology]] on $\R$.
+\end{theorem}
+
+\begin{proof}
+From [[Real Number Line is Metric Space]], one can define an [[Definition:Open Real Interval|open interval]] on the [[Definition:Real Number|set of real numbers]] in terms of an [[Definition:Epsilon-Neighborhood (Real Number Line)|$\epsilon$-neighborhood]].
+From [[Open Real Interval is Open Ball]], an [[Definition:Open Real Interval|open interval]] $\openint a b$ is the [[Definition:Open Ball of Metric Space|open $\epsilon$-ball]] $\map {B_\epsilon} \alpha$.
+Then from [[Metric Induces Topology]] we have that:
+:$\BB = \set {\openint a b: a, b \in \R}$
+forms a [[Definition:Topology|topology]] on $\R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Number Line satisfies all Separation Axioms}
+Tags: Real Number Line with Euclidean Topology, Separation Axioms
+
+\begin{theorem}
+Let $\struct {\R, \tau_d}$ be the [[Definition:Real Number Line with Euclidean Topology|the real number line with the usual (Euclidean) topology]].
+Then $\struct {\R, \tau_d}$ fulfils all [[Definition:Separation Axioms|separation axioms]]:
+:$\struct {\R, \tau_d}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]]
+:$\struct {\R, \tau_d}$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]]
+:$\struct {\R, \tau_d}$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]]
+:$\struct {\R, \tau_d}$ is a [[Definition:Semiregular Space|semiregular space]]
+:$\struct {\R, \tau_d}$ is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]] 
+:$\struct {\R, \tau_d}$ is a [[Definition:T3 Space|$T_3$ space]] 
+:$\struct {\R, \tau_d}$ is a [[Definition:Regular Space|regular space]]
+:$\struct {\R, \tau_d}$ is an [[Definition:Urysohn Space|Urysohn space]]
+:$\struct {\R, \tau_d}$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]]
+:$\struct {\R, \tau_d}$ is a [[Definition:Tychonoff Space|Tychonoff (completely regular) space]]
+:$\struct {\R, \tau_d}$ is a [[Definition:T4 Space|$T_4$ space]]
+:$\struct {\R, \tau_d}$ is a [[Definition:Normal Space|normal space]]
+:$\struct {\R, \tau_d}$ is a [[Definition:T5 Space|$T_5$ space]] 
+:$\struct {\R, \tau_d}$ is a [[Definition:Completely Normal Space|completely normal space]]
+:$\struct {\R, \tau_d}$ is a [[Definition:Perfectly T4 Space|perfectly $T_4$ space]]
+:$\struct {\R, \tau_d}$ is a [[Definition:Perfectly Normal Space|perfectly normal space]]
+\end{theorem}
+
+\begin{proof}
+* [[Real Number Line is Metric Space]]
+* [[Metric Space fulfils all Separation Axioms]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Elementary Amalgamation Theorem}
+Tags: Model Theory
+
+\begin{theorem}
+Let $\MM$ and $\NN$ be $\LL$-[[Definition:First-Order Structure|structures]].
+Let $B$ be a [[Definition:Subset|subset]] of the universe of $\MM$ such that there is a [[Definition:Partial Elementary Embedding|partial elementary embedding]] $f: B \to \NN$.
+There is an [[Definition:Elementary Extension|elementary extension]] $\AA$ of $\MM$ and an [[Definition:Elementary Embedding|elementary embedding]] $g: \NN \to \AA$ such that $\map g {\map f b} = b$ for all $b \in B$.
+$\array {
+& & \AA& & \\
+&\preceq & & \nwarrow \exists g & & \!\!\!\!\!\!\!\!\!\!\!\!\!: g f = \operatorname {id}_B \\
+\MM & & & &\!\!\!\!\!\!\!\! \NN \\
+& \supseteq & & \nearrow f \\
+& & B& & \\
+}$
+\end{theorem}
+
+\begin{proof}
+The proof is essentially a straightforward application of the [[Compactness Theorem]] on the collection of sentences with parameters true in each of $\MM$ and $\NN$, but it requires some tedious set-up in handling symbols of the language involved to make sure that we get the map $g$ with the properties we want in the end.
+First, we will define some new languages by adding constants for all of the elements in the structures involved.
+This well help us construct the claimed $\AA$.
+This first part of the proof is essentially just setting the stage for our argument, there is little to verify.
+Let $C_\MM = \set {c_m : m \in \MM}$ be [[Definition:Set|set]] of new constant symbols (not appearing in $\LL$) corresponding to each element of the universe of $\MM$.
+Similarly, let $C'_\NN = \set {c_n: n \in \NN}$ be new constant symbols (not appearing in $\LL$ or $C_\MM$) corresponding to elements of the universe of $\NN$.
+It will be beneficial to use the same constant symbols for both the elements of $B$ and their images in $f \sqbrk B \subseteq \NN$, so we will define:
+:$C_\NN = \paren {C'_\NN - \set {c_{\map f b}: b \in B} } \cup \set {c_b : b \in B}$
+Let $\LL_\MM$ be the [[Definition:Logical Language|language]] obtained from $\LL$ by adding the constant symbols in $C_\MM$
+Let $\LL_\NN$ be that obtained from $\LL$ by adding the constants in $C_\NN$.
+Note that $\LL_\MM$ and $\LL_\MM$ share the constant symbols $c_b$ for $b\in B$, and those constant symbols which already appear in $\LL$, but all of their other constant symbols are distinct.
+We will view $\MM$ as an $\LL_\MM$-structure by interpreting each constant $c_m$ as $m$.
+Similarly, we will view $\NN$ as an $\LL_\NN$-structure by interpreting each $c_n$ for $n\in \NN - f \sqbrk B$ as $n$, and each $c_b$ for $b \in B$ as $\map f b$.
+Let $\LL_{\MM \NN}$ be the [[Definition:Logical Language|language]] obtained from $\LL$ by adding the constant symbols in $C_\MM \cup C_\NN$
+Let $\map {\mathrm {eldiag} } \MM$ be the set of $\LL_\MM$-sentences which are satisfied by $\MM$.
+Similarly, let $\map {\mathrm {eldiag} } \NN$ be the set of $\LL_\NN$-sentences which are satisfied by $\NN$.
+(We are using this notation since these are essentially [[Definition:Elementary Diagram|elementary diagrams]], the only difference is that we have made some more specific decisions about the constants involved.)
+Let $T$ be the $\LL_{\MM \NN}$-[[Definition:Theory (Logic)|theory]] $\map {\mathrm {eldiag} } \MM \cup \map {\mathrm {eldiag} } \NN$.
+Now that we have all of these definitions, we will show that $T$ is [[Definition:Model (Logic)|satisfiable]].
+This will give us a model which we can use to obtain the one claimed by the theorem.  It will suffice to show that all finite subsets of $T$ are satisfiable, since we may then apply the [[Compactness Theorem]].
+Let $\Delta$ be a finite subset of $T$.
+:$\Delta$ is then a finite subset of $\map {\mathrm {eldiag} } \MM$, together with finitely many $\LL_\NN$-sentences from $\map {\mathrm {eldiag} } \MM$ whose conjunction we will write as $\phi(\bar c_b, \bar c_n)$, where $\phi(\bar x, \bar y)$ is an $\LL$-formula, $\bar b$ is a [[Definition:Ordered Tuple|ordered tuple]] from $B$, and $\bar n$ is a tuple of mutually distinct elements from $\NN$ not containing elements of $f(B)$.
+:Suppose $\Delta$ has no model.
+::In particular, every $\LL_{\MM \NN}$-model of $\map {\mathrm {eldiag} } \MM$ must fail to be model of $\Delta$.
+::Hence we have $\map {\mathrm {eldiag} } \MM \models \neg \map \phi {\bar c_b, \bar c_n}$.
+::But, since the constants in $\bar c_n$ are distinct from all of the constants corresponding to elements of $B$ and $\MM$ in general, it doesn't matter how a $\LL_{\MM \NN}$-model of $\map {\mathrm {eldiag} } \MM$ interprets $\bar c_n$:
+:::$\map {\mathrm {eldiag} } \MM \models \forall \bar y \neg \map \phi {\bar c_b, \bar y}$.
+::Since $\MM \models \map {\mathrm {eldiag} } \MM$, we obtain:
+:::$\MM \models \forall \bar y \neg \map \phi {\bar c_b, \bar y}$, and hence
+:::$\MM \models \forall \bar y \neg \map \phi {\bar b, \bar y}$ since $\MM$ interprets each $c_b$ as $b$.
+::Since $f$ is partial elementary on $B$, we get:
+:::$\NN \models \forall \bar y \neg \map \phi {\map f {\bar b}, \bar y}$.
+::But, this contradicts the choice of $\map \phi {\bar c_b, \bar c_n}$ as being a conjunction of sentences in $\map {\mathrm {eldiag} } \NN$, since $\NN$ interprets each $c_b$ as its corresponding element $\map f b$.
+:Thus, $\Delta$ has a model.
+Hence, by the [[Compactness Theorem]], $T$ is satisfiable.
+Now, let $\AA^*$ be a model of $T$, and let $\AA$ be the $\LL$-structure obtained by removing all of the interpretations of constants in $C_\MM \cup C_\NN$ from $\AA^*$.
+Since $\AA^* \models \map {\mathrm {eldiag} } \MM$, we can, after possibly creating an isomorphic copy of $\AA^*$ with new distinct elements, and replacing each element $a$ of the universe of $\AA^*$ such that $c_m^{\AA^*} = a$ with the element $m \in \MM$, view $\AA$ as an elementary extension of $\MM$ such that $c_m^\AA = m \in \AA$.
+Define $g : \NN \to \AA$ by $\map g n = c_n^{\AA^*}$, where $c_n$ is the constant symbol in $C_\NN$ which $\NN$ interprets as $n$.
+Since $\AA^* \models \map {\mathrm {eldiag} } \NN$, we have that $g$ is [[Definition:Elementary Embedding|elementary]].
+Let $b \in B$
+Recall that $\NN$ interprets $c_b$ as $\map f b$, and since $b \in B\subseteq \MM$, we have $c_b^{\AA^*} = b$.
+Thus:
+:$\map g {\map f b} = c_b^{\AA^*}$ = b
+So, $\AA$ and $g$ are as claimed by the theorem.
+{{qed}}
+{{MissingLinks}}
+[[Category:Model Theory]]
+7qtvqfgpx6lgsquq0ojtbsy7fxqvi1l
+\end{proof}<|endoftext|>
+\section{Real Number Line is Non-Meager}
+Tags: Real Number Line with Euclidean Topology, Non-Meager Spaces
+
+\begin{theorem}
+Let $\struct {\R, \tau_d}$ be the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]].
+Then $\struct {\R, \tau_d}$ is [[Definition:Non-Meager Space|non-meager]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Real Number Line is Complete Metric Space]].
+From the [[Baire Category Theorem]], a [[Definition:Complete Metric Space|complete metric space]] is also a [[Definition:Baire Space (Topology)|Baire space]].
+The result follows from [[Baire Space is Non-Meager]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Baire Space iff Open Sets are Non-Meager}
+Tags: Baire Spaces, Non-Meager Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then $T$ is a [[Definition:Baire Space (Topology)|Baire space]] {{iff}} every [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open]] set of $T$ is [[Definition:Non-Meager Space|non-meager]] in $T$.
+\end{theorem}
+
+\begin{proof}
+We prove the [[Definition:Contrapositive Statement|contrapositive]]:
+:$T$ is not a [[Definition:Baire Space (Topology)|Baire space]] {{iff}} there exists a [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open]] set of $T$ which is [[Definition:Meager Space|meager]] in $T$.
+We have by [[Definition:Baire Space (Topology)/Definition 3|definition of Baire space]]:
+:$T$ is a Baire space {{iff}} the [[Definition:Interior (Topology)|interior]] of the [[Definition:Set Union|union]] of any [[Definition:Countable|countable set]] of [[Definition:Closed Set (Topology)|closed sets]] of $T$ which are [[Definition:Nowhere Dense|nowhere dense]] is [[Definition:Empty Set|empty]].
+Therefore:
+:$T$ is not a Baire space {{iff}} the [[Definition:Interior (Topology)|interior]] of the [[Definition:Set Union|union]] of some [[Definition:Countable|countable set]] of [[Definition:Closed Set (Topology)|closed sets]] of $T$ which are [[Definition:Nowhere Dense|nowhere dense]] is [[Definition:Non-Empty Set|non-empty]].
+We have by [[Definition:Meager Space|definition of meager]]:
+:$A$ is meager in $T$ {{iff}} it is a [[Definition:Countable Union|countable union]] of [[Definition:Subset|subsets]] of $S$ which are [[Definition:Nowhere Dense|nowhere dense]] in $T$.
+Here $I$ denotes a [[Definition:Countable|countable]] indexing set.
+=== Sufficient Condition ===
+Suppose $T$ is not a [[Definition:Baire Space (Topology)|Baire space]].
+Then there is a [[Definition:Countable|countable set]] of [[Definition:Closed Set (Topology)|closed sets]] of $T$ which are [[Definition:Nowhere Dense|nowhere dense]] with the [[Definition:Interior (Topology)|interior]] of its [[Definition:Set Union|union]] [[Definition:Non-Empty Set|non-empty]].
+Let $\FF = \set {F_i: i \in I}$ be such a set.
+Then $\paren {\bigcup \FF}^\circ$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open set]].
+Consider the set $\FF' = \set {F_i \cap \paren {\bigcup \FF}^\circ: i \in I}$.
+From [[Intersection is Subset]]:
+:$F_i \cap \paren {\bigcup \FF}^\circ \subseteq F_i$ for any $i \in I$
+Since $F_i$ are [[Definition:Nowhere Dense|nowhere dense]], so is $F_i \cap \paren {\bigcup \FF}^\circ$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \bigcup \FF'
+      | r = \bigcup_{i \mathop \in I} \paren {F_i \cap \paren {\bigcup \FF}^\circ}
+}}
+{{eqn | r = \paren {\bigcup \FF}^\circ \cap \bigcup_{i \mathop \in I} F_i
+      | c = [[Intersection Distributes over Union/Family of Sets|Intersection Distributes over Union]]
+}}
+{{eqn | r = \paren {\bigcup \FF}^\circ \cap \bigcup \FF
+}}
+{{eqn | r = \paren {\bigcup \FF}^\circ
+      | c = [[Set Interior is Largest Open Set]]; [[Intersection with Subset is Subset]]
+}}
+{{end-eqn}}
+Therefore $\bigcup \FF'$ is [[Definition:Meager Space|meager]] in $T$.
+Hence we have shown the existence of a [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open]] set of $T$ which is [[Definition:Meager Space|meager]] in $T$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Suppose there exists a [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open]] set of $T$ which is [[Definition:Meager Space|meager]] in $T$.
+Let $\displaystyle A = \bigcup_{i \in I} A_i$ be such a set.
+Then $A$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open set]], and $A_i$ are [[Definition:Nowhere Dense|nowhere dense]] for every $i \in I$.
+Thus for every $i$, $\paren {A_i^-}^\circ = \O$.
+By [[Closure of Topological Closure equals Closure]]:
+:$\paren {\paren {A_i^-}^-}^\circ = \paren {A_i^-}^\circ = \O$.
+By [[Set is Closed iff Equals Topological Closure]], $A_i^-$ are [[Definition:Closed Set (Topology)|closed sets]] of $T$ which are [[Definition:Nowhere Dense|nowhere dense]].
+We also have:
+{{begin-eqn}}
+{{eqn | l = A
+      | r = A^\circ
+      | c = [[Interior of Open Set]]
+}}
+{{eqn | r = \paren {\bigcup_{i \in I} A_i}^\circ
+}}
+{{eqn | o = \subseteq
+      | r = \paren {\bigcup_{i \in I} A_i^-}^\circ
+      | c = [[Set is Subset of its Topological Closure]]; [[Set Union Preserves Subsets/Families of Sets|Set Union Preserves Subsets]]; [[Interior of Subset]]
+}}
+{{end-eqn}}
+Hence there is a [[Definition:Countable|countable set]] of [[Definition:Closed Set (Topology)|closed sets]] of $T$ which are [[Definition:Nowhere Dense|nowhere dense]] with the [[Definition:Interior (Topology)|interior]] of its [[Definition:Set Union|union]] [[Definition:Non-Empty Set|non-empty]].
+Therefore $T$ is not a [[Definition:Baire Space (Topology)|Baire space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Extension Realizing All Types}
+Tags: Model Theory
+
+\begin{theorem}
+Let $\MM$ be an $\LL$-[[Definition:First-Order Structure|structure]].
+Let $M$ be its [[Definition:Universe (Model Theory)|universe]].
+There is an [[Definition:Elementary Extension|elementary extension]] $\NN$ of $\MM$ such that every [[Definition:Type|type]] over $M$ (relative to $\MM$) is [[Definition:Realization|realized]] in $\NN$.
+\end{theorem}
+
+\begin{proof}
+Let $S \left({\MM}\right)$ denote the [[Definition:Set|set]] containing all [[Definition:Complete Type|complete types]] over $M$ of every number of [[Definition:Free Variable|free variables]].
+Let $\kappa = \left\vert{S \left({\MM}\right)}\right\vert$.
+Use a [[Definition:Bijection|bijection]] between $\kappa$ and $S \left({\MM}\right)$ to write the [[Definition:Element|elements]] of $S \left({\MM}\right)$ as $p_\alpha$ for $\alpha < \kappa$.
+For each $\alpha < \kappa$, let $N_\alpha$ be an [[Definition:Elementary Extension|elementary extension]] of $\MM$ which has an [[Definition:Ordered Tuple|ordered tuple]] $\bar a_\alpha$ [[Definition:Realization|realizing]] $p_\alpha$.
+Such [[Definition:Elementary Extension|extensions]] and [[Definition:Ordered Tuple|tuples]] exist by [[Type is Realized in some Elementary Extension]].
+We will construct the [[Definition:Elementary Extension|extension]] claimed by the theorem as the [[Definition:Union|union]] over a [[Definition:Chain of Elementary Extensions|chain of elementary extensions]] defined using [[Transfinite Induction]].
+At each step, we use the [[Elementary Amalgamation Theorem]] to add on $N_\alpha$.
+Base case $\alpha = 0$:
+Let $\BB_0 = \MM$.
+Note that $\BB_0$ is an [[Definition:Elementary Extension|elementary extension]] of $\MM$ by choice of $\NN_0$.
+[[Definition:Limit Ordinal|Limit ordinals]] $\alpha \le \kappa$:
+Let $\displaystyle \BB_\alpha = \bigcup_{\beta \mathop < \alpha} \BB_\beta$
+$\BB_\alpha$ is an [[Definition:Elementary Extension|elementary extension]] of $\MM$ by [[Union of Elementary Chain is Elementary Extension]].
+[[Definition:Successor Ordinal|Successor ordinals]] $\alpha + 1$ for $\alpha < \kappa$:
+We have that $\BB_\alpha$ and $\NN_\alpha$ are both [[Definition:Elementary Extension|elementary extensions]] of $\MM$.
+By the [[Elementary Amalgamation Theorem]], there is:
+:an [[Definition:Elementary Extension|elementary extension]] $\BB_{\alpha + 1}$ of $\BB_\alpha$
+and:
+:an [[Definition:Elementary Embedding|elementary embedding]] $g_\alpha: \NN_\alpha \to \BB_{\alpha + 1}$ which is the [[Definition:Identity Mapping|identity]] on $M$ viewed as a [[Definition:Subset|subset]] of $\NN_\alpha$.
+Note that $\map {g_\alpha} {\bar a_\alpha}$ [[Definition:Realization|realizes]] $p_\alpha$ in $\BB_{\alpha + 1}$ since $g_\alpha$ is [[Definition:Elementary Embedding|elementary]].
+Thus $\BB_{\alpha + 1}$ is an [[Definition:Elementary Extension|elementary extension]] of $\MM$ which contains $\map {g_\alpha} {\bar a_\alpha}$ [[Definition:Realization|realizing]] $p_\alpha$.
+Now, let $\displaystyle \BB = \bigcup_{\alpha < \kappa} \BB_\alpha$.
+$\BB$ is an [[Definition:Elementary Extension|elementary extension]] of each $\BB_\alpha$ by [[Union of Elementary Chain is an Elementary Extension]].
+In particular, $\BB$ is an [[Definition:Elementary Extension|elementary extension]] of $\BB_0 = \MM$
+We have that:
+:each $p_\alpha$ is [[Definition:Realization|realized]] in $\BB_{\alpha + 1}$ by the corresponding $\map {g_\alpha} {\bar a_\alpha}$
+and:
+:$\BB$ is an [[Definition:Elementary Extension|elementary extension]] of each $\BB_{\alpha + 1}$
+Thus we have that $\map {g_\alpha} {\bar a_\alpha}$ [[Definition:Realization|realizes]] $p_\alpha$ in $\BB$.
+{{qed}}
+[[Category:Model Theory]]
+nydyqgv6evfgf48xvhhom5vczk5retb
+\end{proof}<|endoftext|>
+\section{Baire Space is Non-Meager}
+Tags: Baire Spaces, Non-Meager Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Baire Space (Topology)|Baire space]] (in the context of [[Definition:Topology (Mathematical Branch)|topology]]).
+Then $T$ is [[Definition:Non-Meager Space|non-meager]] in $T$.
+\end{theorem}
+
+\begin{proof}
+From [[Baire Space iff Open Sets are Non-Meager]], all [[Definition:Open Set (Topology)|open sets]] of $T$ are [[Definition:Non-Meager Space|non-meager]] in $T$.
+But $T$ itself is an [[Definition:Open Set (Topology)|open set]] of $T$ by definition of [[Definition:Topological Space|topological space]].
+Hence the result.
+{{qed}}
+[[Category:Baire Spaces]]
+[[Category:Non-Meager Spaces]]
+kzh9g5nfshlvd9lkp0529aeaoy5y7a1
+\end{proof}<|endoftext|>
+\section{Real Number Line is Separable}
+Tags: Real Number Line with Euclidean Topology, Separable Spaces, Real Number Line is Separable
+
+\begin{theorem}
+Let $\struct {\R, \tau_d}$ be the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]].
+Then $\struct {\R, \tau_d}$ is [[Definition:Separable Space|separable]].
+\end{theorem}
+
+\begin{proof}
+The [[Rational Numbers form Metric Space|rational numbers $\Q$ form a metric space]].
+We have that the [[Rationals are Everywhere Dense in Topological Space of Reals]].
+We also have that the [[Rational Numbers are Countably Infinite]].
+The result follows from the definition of [[Definition:Separable Space|separable space]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Follows from:
+:[[Real Number Line is Second-Countable]]
+:[[Second-Countable Space is Separable]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Separation Axioms on Double Pointed Topology/T3 Axiom}
+Tags: Separation Axioms on Double Pointed Topology, T3 Spaces
+
+\begin{theorem}
+Let $T_1 = \struct {S, \tau_S}$ be a [[Definition:Topological Space|topological space]].
+Let $D = \struct {A, \set {\O, A} }$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on an arbitrary [[Definition:Doubleton|doubleton]] $A = \set {a, b}$.
+Let $T = \struct {T_1 \times D, \tau}$ be the [[Definition:Double Pointed Topology|double pointed topological space]] on $T_1$.
+Then $T \times D$ is a [[Definition:T3 Space|$T_3$ space]] {{iff}} $T$ is also a [[Definition:T3 Space|$T_3$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $S' = S \times \set {a, b}$.
+Let $F' \subseteq S'$ such that $F'$ is [[Definition:Closed Set (Topology)|closed]] in $T \times D$.
+Then $F' = F \times \set {a, b}$ or $F' = F \times \O$ by definition of the [[Definition:Double Pointed Topology|double pointed topology]].
+If $F' = F \times \O$ then $F' = \O$ from [[Cartesian Product is Empty iff Factor is Empty]], and the result is trivial.
+So suppose $F' = F \times \set {a, b}$.
+From [[Open and Closed Sets in Multiple Pointed Topology]] it follows that $F$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+Let $y' = \tuple {y, q} \in \relcomp {S'} {F'}$.
+Then $y \notin F$.
+Suppose that $T$ is a [[Definition:T3 Space|$T_3$ space]].
+Then by definition:
+:For any [[Definition:Closed Set (Topology)|closed set]] $F$ of $T$ and any point $y \in S$ such that $y \notin F$ there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.
+Then $y' \in V \times \set {a, b}$ and $F' \subseteq U \times \set {a, b}$ and:
+:$U \times \set {a, b} \cap V \times \set {a, b} = \O$
+demonstrating that $T \times D$ is a [[Definition:T3 Space|$T_3$ space]].
+Now suppose that $T \times D$ is a [[Definition:T3 Space|$T_3$ space]].
+Then $\exists U', V' \in S': y' \in V'$ and $F' \subseteq U'$ such that $U' \cap V' = \O$.
+As $D$ is the [[Definition:Indiscrete Topology|indiscrete topology]] it follows that:
+:$U' = U \times \set {a, b}$
+:$V' = V \times \set {a, b}$
+for some $U, V \subseteq T$.
+From [[Open and Closed Sets in Multiple Pointed Topology]] it follows that $U$ and $V$ are [[Definition:Open Set (Topology)|open]] in $T$.
+As $U' \cap V' = \O$ it follows that $U \cap V = \O$.
+It follows that $F$ and $y$ fulfil the conditions that make $T$ a [[Definition:T3 Space|$T_3$ space]].
+Hence the result.
+\end{proof}<|endoftext|>
+\section{Separation Axioms on Double Pointed Topology/T4 Axiom}
+Tags: Separation Axioms on Double Pointed Topology, T4 Spaces
+
+\begin{theorem}
+Let $T_1 = \struct {S, \tau_S}$ be a [[Definition:Topological Space|topological space]].
+Let $D = \struct {A, \set {\O, A} }$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on an arbitrary [[Definition:Doubleton|doubleton]] $A = \set {a, b}$.
+Let $T = \struct {T_1 \times D, \tau}$ be the [[Definition:Double Pointed Topology|double pointed topological space]] on $T_1$.
+Then $T \times D$ is a [[Definition:T4 Space|$T_4$ space]] {{iff}} $T$ is also a [[Definition:T4 Space|$T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $S' = S \times \set {a, b}$.
+Let $H' \subseteq S'$ such that $H$ is [[Definition:Closed Set (Topology)|closed]] in $T \times D$.
+Then $H' = H \times \set {a, b}$ or $H' = H \times \O$ by definition of the [[Definition:Double Pointed Topology|double pointed topology]].
+If $H' = H \times \O$ then $H' = \O$ from [[Cartesian Product is Empty iff Factor is Empty]], and the result is trivial.
+So suppose $H' = H \times \set {a, b}$.
+From [[Open and Closed Sets in Multiple Pointed Topology]] it follows that $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+Suppose that $T$ is a [[Definition:T4 Space|$T_4$ space]].
+Then by definition:
+:For any two [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Set (Topology)|closed sets]] $A, B \subseteq S$ there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ containing $A$ and $B$ respectively.
+Then $A \times \set {a, b} \subseteq U \times \set {a, b}$ and $B \times \set {a, b} \subseteq V \times \set {a, b}$ and:
+:$U \times \set {a, b} \cap V \times \set {a, b} = \O$
+demonstrating that $T \times D$ is a [[Definition:T4 Space|$T_4$ space]].
+Now suppose that $T \times D$ is a [[Definition:T4 Space|$T_4$ space]].
+Then $\exists U', V' \in S': A' \subseteq U'$ and $B' \subseteq V'$ such that $U' \cap V' = \O$.
+As $D$ is the [[Definition:Indiscrete Topology|indiscrete topology]] it follows that:
+:$U' = U \times \set {a, b}$
+:$V' = V \times \set {a, b}$
+for some $U, V \subseteq T$.
+From [[Open and Closed Sets in Multiple Pointed Topology]] it follows that $U$ and $V$ are [[Definition:Open Set (Topology)|open]] in $T$.
+As $U' \cap V' = \O$ it follows that $U \cap V = \O$.
+It follows that $A$ and $B$ fulfil the conditions that make $T$ a [[Definition:T4 Space|$T_4$ space]].
+Hence the result.
+\end{proof}<|endoftext|>
+\section{Separation Axioms on Double Pointed Topology/T5 Axiom}
+Tags: Separation Axioms on Double Pointed Topology, T5 Spaces
+
+\begin{theorem}
+Let $T_1 = \struct {S, \tau_S}$ be a [[Definition:Topological Space|topological space]].
+Let $D = \struct {A, \set {\O, A} }$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on an arbitrary [[Definition:Doubleton|doubleton]] $A = \set {a, b}$.
+Let $T = \struct {T_1 \times D, \tau}$ be the [[Definition:Double Pointed Topology|double pointed topological space]] on $T_1$.
+Then $T \times D$ is a [[Definition:T5 Space|$T_5$ space]] {{iff}} $T$ is also a [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose that $T \times D$ is a [[Definition:T5 Space|$T_5$ space]].
+Let $A, B \subseteq T$ be two [[Definition:Separated Sets|separated sets]].
+First, we will show that $A \times D$ and $B \times D$ are also [[Definition:Separated Sets|separated]].
+To this end, observe that:
+{{begin-eqn}}
+{{eqn | l = \paren {A \times D}^- \cap \paren {B \times D}
+      | r = \paren {A^- \times D} \cap \paren {B \times D}
+      | c = [[Closure of Subset of Double Pointed Topological Space]]
+}}
+{{eqn | r = \paren {A^- \cap B} \times D
+      | c = [[Cartesian Product Distributes over Intersection]]
+}}
+{{eqn | r = \O \times D
+      | c = $A$ and $B$ are [[Definition:Separated Sets|separated]]
+}}
+{{eqn | r = \varnothing
+      | c = [[Cartesian Product is Empty iff Factor is Empty]]
+}}
+{{end-eqn}}
+[[Definition:Mutatis Mutandis|Mutatis mutandis]], also:
+:$\paren {B \times D}^- \cap \paren {A \times D} = \O$
+Thus $A \times D$ and $B \times D$ are [[Definition:Separated Sets|separated]].
+Since $T \times D$ is assumed to be a [[Definition:T5 Space|$T_5$ space]], find [[Definition:Open Set (Topology)|open sets]] $U, V$ such that:
+:$U \cap V = \O$
+:$A^- \times D \subseteq U$
+:$B^- \times D \subseteq V$
+By [[Open Sets of Double Pointed Topology]], there are [[Definition:Open Set (Topology)|open sets]] $U', V'$ of $\tau$ such that:
+:$U = U' \times D$
+:$V = V' \times D$
+We have, furthermore:
+{{begin-eqn}}
+{{eqn | l = \varnothing
+      | r = U \cap V
+}}
+{{eqn | r = \paren {U' \times D} \cap \paren {V' \times D}
+}}
+{{eqn | r = \paren {U' \cap V'} \times D
+      | c = [[Cartesian Product Distributes over Intersection]]
+}}
+{{end-eqn}}
+From [[Cartesian Product is Empty iff Factor is Empty]], $U' \cap V' = \O$ since $D$ is [[Definition:Non-Empty Set|non-empty]].
+By [[Cartesian Product of Subsets]], we also infer:
+:$A^- \subseteq U'$
+:$B^- \subseteq V'$
+It follows that $T$ is a [[Definition:T5 Space|$T_5$ space]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $T$ is a [[Definition:T5 Space|$T_5$ space]].
+Let $X, Y \subseteq T \times D$ be [[Definition:Separated Sets|separated sets]].
+By [[Closure of Subset of Double Pointed Topological Space]], we know that:
+:$X^- = {\pr_1 \sqbrk X}^- \times D$
+:$Y^- = {\pr_1 \sqbrk Y}^- \times D$
+where $\pr_1$ is the [[Definition:First Projection|first projection]].
+Suppose now that for some $s \in S$, one has:
+:$s \in {\pr_1 \sqbrk X}^- \cap \pr_1 \sqbrk Y$
+Then it would be that for some $d \in D$:
+:$\tuple {s, d} \in Y$
+However, by the expression for $X^-$ above, also $\tuple {s, d} \in X^-$.
+This is a contradiction, for $X$ and $Y$ were assumed [[Definition:Separated Sets|separated]].
+Thus:
+:${\pr_1 \sqbrk X}^- \cap \pr_1 \sqbrk Y = \O$
+and analogously, we derive:
+:$\pr_1 \sqbrk X \cap {\pr_1 \sqbrk Y}^- = \O$
+that is, $\pr_1 \sqbrk X$ and $\pr_1 \sqbrk Y$ are [[Definition:Separated Sets|separated]].
+Since $T$ is a [[Definition:T5 Space|$T_5$ space]], we find [[Definition:Open Set (Topology)|open sets]] $U, V$ of $\tau$ such that:
+:$U \cap V = \varnothing$
+:${\pr_1 \sqbrk X}^- \subseteq U$
+:${\pr_1 \sqbrk Y}^- \subseteq V$
+By [[Cartesian Product of Subsets]], we have:
+:$X^- = {\pr_1 \sqbrk X}^- \times D \subseteq U \times D$
+:$Y^- = {\pr_1 \sqbrk Y}^- \times D \subseteq V \times D$
+By [[Open Sets of Double Pointed Topology]], $U \times D$ and $V \times D$ are also [[Definition:Open Set (Topology)|open sets]] in $S \times D$.
+Finally, we compute:
+{{begin-eqn}}
+{{eqn | l = \paren {U \times D} \cap \paren {V \times D}
+      | r = \paren {U \cap V} \times D
+      | c = [[Cartesian Product Distributes over Intersection]]
+}}
+{{eqn | r = \O \times D
+      | c = by choice of $U$ and $V$
+}}
+{{eqn | r = \O
+      | c = [[Cartesian Product is Empty iff Factor is Empty]]
+}}
+{{end-eqn}}
+That is, $T \times D$ is a [[Definition:T5 Space|$T_5$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Separation Axioms on Double Pointed Topology}
+Tags: Double Pointed Topologies, Separation Axioms, Separation Axioms on Double Pointed Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau_S}$ be a [[Definition:Topological Space|topological space]].
+Let $D = \struct {\set {a, b}, \tau_D}$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on two points.
+Let $T \times D$ be the [[Definition:Double Pointed Topology|double pointed topology]] on $T$.
+Then:
+:$T \times D$ is not a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]], a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]], a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]] or a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+:$T \times D$ is a [[Definition:T3 Space|$T_3$ space]], a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]], a [[Definition:T4 Space|$T_4$ space]] or a [[Definition:T5 Space|$T_5$ space]] {{iff}} $T$ is.
+\end{theorem}
+
+\begin{proof}
+=== $T_0$, $T_1$, $T_2$ and $T_{2 \frac 1 2}$ Axioms ===
+From:
+:[[Double Pointed Topology is not T0|Double Pointed Topology is not $T_0$]]
+:[[T1 Space is T0 Space|$T_1$ Space is $T_0$ Space]]
+:[[T2 Space is T1 Space|$T_2$ Space is $T_1$ Space]]
+:[[Completely Hausdorff Space is Hausdorff Space|Completely Hausdorff Space is $T_2$ Space]]
+we have that $T \times D$ is not [[Definition:Kolmogorov Space|$T_0$]], [[Definition:Fréchet Space (Topology)|$T_1$]] or [[Definition:Hausdorff Space|$T_2$]] or [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$]].
+{{qed|lemma}}
+From:
+:[[Separation Axioms on Double Pointed Topology/T3 Axiom|Separation Axioms on Double Pointed Topology: $T_3$ Axiom]]
+$T \times D$ is a [[Definition:T3 Space|$T_3$ space]] {{iff}} $T$ is.
+{{qed|lemma}}
+From:
+:[[Separation Axioms on Double Pointed Topology/T3.5 Axiom|Separation Axioms on Double Pointed Topology: $T_{3 \frac 1 2}$ Axiom]]
+$T \times D$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]] {{iff}} $T$ is.
+{{qed|lemma}}
+From:
+:[[Separation Axioms on Double Pointed Topology/T4 Axiom|Separation Axioms on Double Pointed Topology: $T_4$ Axiom]]
+$T \times D$ is a [[Definition:T4 Space|$T_4$ space]] {{iff}} $T$ is.
+{{qed|lemma}}
+From:
+:[[Separation Axioms on Double Pointed Topology/T5 Axiom|Separation Axioms on Double Pointed Topology: $T_5$ Axiom]]
+$T \times D$ is a [[Definition:T5 Space|$T_5$ space]] {{iff}} $T$ is.
+{{qed}}
+{{expand|investigate whether we can directly use [[Separation Properties Preserved under Topological Product]]}}
+\end{proof}<|endoftext|>
+\section{Separation Axioms on Double Pointed Topology/T3.5 Axiom}
+Tags: Separation Axioms on Double Pointed Topology, T3 1: 2 Spaces
+
+\begin{theorem}
+Let $T_1 = \struct {S, \tau_S}$ be a [[Definition:Topological Space|topological space]].
+Let $D = \struct {A, \set {\O, A} }$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on an arbitrary [[Definition:Doubleton|doubleton]] $A = \set {a, b}$.
+Let $T = \struct {T_1 \times D, \tau}$ be the [[Definition:Double Pointed Topology|double pointed topological space]] on $T_1$.
+Then $T \times D$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]] {{iff}} $T$ is also a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+\end{theorem}<|endoftext|>
+\section{Square of Vandermonde Matrix}
+Tags: Vandermonde Matrices
+
+\begin{theorem}
+The [[Definition:Square (Algebra)|square]] of the [[Definition:Vandermonde Matrix|Vandermonde matrix of order $n$]]:
+: $\mathbf V = \begin{bmatrix}
+  x_1 & x_2 & \cdots & x_n \\
+  x_1^2 & x_2^2 & \cdots & x_n^2 \\
+\vdots & \vdots & \ddots & \vdots \\
+  x_1^n & x_2^n & \cdots & x_n^n
+\end{bmatrix}$
+is symmetrical in $x_1, \ldots, x_n$.
+{{questionable|The case $n {{=}} 2$ left me clueless to what could possibly be intended here; only $\mathbf {V V}^T$ is trivially seen symmetric in the $x_n$, but this can hardly be called a square}}
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+[[Category:Vandermonde Matrices]]
+91falks8quapb23z4vth44nc0obh7vf
+\end{proof}<|endoftext|>
+\section{Open and Closed Sets in Multiple Pointed Topology}
+Tags: Multiple Pointed Topology, Open Sets, Closed Sets, Clopen Sets
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $A$ be a [[Definition:Finite Set|finite set]] whose [[Definition:Cardinality|cardinality]] is greater than $1$.
+Let $D = \left({A, \left\{{\varnothing, A}\right\}}\right)$ be the [[Definition:Indiscrete Space|indiscrete space]] on $A$.
+Let $T \times D$ be a [[Definition:Multiple Pointed Topology|multiple pointed topological space]] generated from $T$ and $D$.
+Let $H \subseteq T$.
+Then:
+: $H \times A$ is [[Definition:Open Set (Topology)|open in $T \times D$]] {{iff}} $H$ is [[Definition:Open Set (Topology)|open in $T$]]
+: $H \times A$ is [[Definition:Closed Set (Topology)|closed in $T \times D$]] {{iff}} $H$ is [[Definition:Closed Set (Topology)|closed in $T$]]
+: $H \times \varnothing$ is [[Definition:Clopen Set|both open and closed in $T \times D$]]
+Let $\varnothing \subset H_A \subset A$, that is, let $H_A$ be a [[Definition:Proper Subset|proper subset]] of $A$.
+Then $H \times H_A$ is neither [[Definition:Open Set (Topology)|open]] nor [[Definition:Closed Set (Topology)|closed]] in $T \times D$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Multiple Pointed Topology|multiple pointed topology]], $T \times D$ is the [[Definition:Product Space (Topology)|product space]] of $T$ with $D$:
+:$T \times D = \left({S \times A, \tau}\right)$
+where $\tau$ is the [[Definition:Tychonoff Topology|Tychonoff topology]] on $S \times A$.
+Consider the set $H \times A$.
+We have that:
+:$H \times A = \operatorname{pr}_1^{-1} \left({H}\right)$
+where $\operatorname{pr}_1: T \times D \to T$ is the [[Definition:First Projection|first projection]] from $T \times D$ to $T$.
+By definition of the [[Definition:Tychonoff Topology|Tychonoff topology]], $\operatorname{pr}_1: T \times D \to T$ is a [[Definition:Continuous Mapping (Topology)|continuous mapping]].
+=== Open Sets ===
+By definition of [[Definition:Tychonoff Topology|Tychonoff topology]], if $H$ is [[Definition:Open Set (Topology)|open in $T$]] then $H \times A$ is [[Definition:Open Set (Topology)|open in $T \times D$]].
+As the [[Tychonoff Topology is Coarsest Topology such that Projections are Continuous]], it follows that if $H$ is not [[Definition:Open Set (Topology)|open in $T$]] then $H \times A$ is not [[Definition:Open Set (Topology)|open in $T \times D$]].
+{{qed|lemma}}
+=== Closed Sets ===
+Let $H \subseteq S$ be [[Definition:Closed Set (Topology)|closed in $T$]].
+We have that $\operatorname{pr}_1: T \times D \to T$ is a [[Definition:Continuous Mapping (Topology)|continuous mapping]].
+By [[Continuity Defined from Closed Sets]], it follows that $H \times A$ is [[Definition:Closed Set (Topology)|closed in $T \times D$]].
+As the [[Tychonoff Topology is Coarsest Topology such that Projections are Continuous]], it follows that if $H$ is not [[Definition:Closed Set (Topology)|closed in $T$]] then $H \times A$ is not [[Definition:Closed Set (Topology)|closed in $T \times D$]].
+{{qed|lemma}}
+=== Cartesian Product with Empty Set ===
+From [[Cartesian Product is Empty iff Factor is Empty]], $H \times \varnothing = \varnothing$.
+From [[Open and Closed Sets in Topological Space]], $\varnothing$ is [[Definition:Clopen Set|both open and closed in $T \times D$]].
+{{qed|lemma}}
+=== Neither Open Nor Closed ===
+Let $\varnothing \subset H_A \subset A$.
+That is, $H_A \ne \varnothing$ and $H_A \ne A$.
+So $H_A$ is neither [[Definition:Open Set (Topology)|open]] nor [[Definition:Closed Set (Topology)|closed]] in the [[Definition:Indiscrete Topology|indiscrete topology]] on $A$
+We have that:
+:$S \times H_A = \operatorname{pr}_2^{-1} \left({H_A}\right)$
+where $\operatorname{pr}_2: T \times D \to D$ is the [[Definition:Second Projection|second projection]] from $T \times D$ to $D$.
+By definition of the [[Definition:Tychonoff Topology|Tychonoff topology]], $\operatorname{pr}_2: T \times D \to D$ is a [[Definition:Continuous Mapping (Topology)|continuous mapping]].
+We have that the [[Tychonoff Topology is Coarsest Topology such that Projections are Continuous]].
+So if $H_A$ is neither [[Definition:Open Set (Topology)|open]] nor [[Definition:Closed Set (Topology)|closed]] in $D$ then $S \times H_A$ is neither [[Definition:Open Set (Topology)|open]] nor [[Definition:Closed Set (Topology)|closed]] in $T \times D$.
+Hence, by definition of the [[Definition:Tychonoff Topology|Tychonoff topology]], nor is $H \times H_A$.
+{{qed|lemma}}
+{{qed}}
+[[Category:Multiple Pointed Topology]]
+[[Category:Open Sets]]
+[[Category:Closed Sets]]
+[[Category:Clopen Sets]]
+93opzeu96qlq6kz77rctynhnws9au81
+\end{proof}<|endoftext|>
+\section{Basis for Box Topology}
+Tags: Box Topology, Topological Bases
+
+\begin{theorem}
+Let $\left \langle {\left({S_i, \tau_i}\right)} \right \rangle_{i \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Topological Space|topological spaces]].
+Let $S$ be the [[Definition:Cartesian Product of Family|cartesian product]] of $\left \langle {S_i} \right \rangle_{i \mathop \in I}$.
+That is:
+:$\displaystyle S := \prod_{i \mathop \in I} S_i$
+Define:
+:$\displaystyle \mathcal B := \left\{{\prod_{i \mathop \in I} U_i: \forall i \in I: U_i \in \tau_i}\right\}$
+Then $\mathcal B$ is a [[Definition:Synthetic Basis|synthetic basis]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Let us check the two conditions for $\mathcal B$ to be a [[Definition:Synthetic Basis|synthetic basis]] in turn.
+=== $(B1)$: Covering ===
+From [[Definition:Open Set Axioms|open set axiom]] $(O3)$, we have $S_i \in \tau_i$ for all $i \in I$.
+Thus $S = \displaystyle \prod_{i \mathop \in I} S_i \in \mathcal B$.
+Hence by [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]], we have:
+:$S \subseteq \displaystyle \bigcup \mathcal B$
+so $\mathcal B$ is a [[Definition:Cover of Set|cover]] for $S$.
+{{qed|lemma}}
+=== $(B2)$: Intersections are Unions ===
+Let $A = \displaystyle \prod_{i \mathop \in I} U_i$ and $B = \displaystyle \prod_{i \mathop \in I} V_i$ be in $\mathcal B$.
+Then by [[Cartesian Product of Intersections]], we have:
+:$A \cap B = \displaystyle \prod_{i \mathop \in I} \left({U_i \cap V_i}\right)$
+By [[Definition:Open Set Axioms|open set axiom]] $(O2)$, $U_i \cap V_i \in \tau_i$ for all $i \in I$.
+Hence $A \cap B \in \mathcal B$, and in particular, by [[Union of Singleton]]:
+:$A \cap B = \displaystyle \bigcup \left\{{A \cap B}\right\}$
+{{qed|lemma}}
+Having verified both conditions, we conclude $\mathcal B$ is a [[Definition:Synthetic Basis|synthetic basis]] on $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Space with Open Point is Non-Meager}
+Tags: Non-Meager Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $x \in S$ be an [[Definition:Open Point|open point]].
+Then $T$ is a [[Definition:Non-Meager Space|non-meager space]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$ be an [[Definition:Open Point|open point]] of $T$.
+That is:
+:$\set x \in \tau$
+Recall that:
+:a [[Definition:Topological Space|topological space]] is [[Definition:Non-Meager Space|non-meager]] if it is not [[Definition:Meager Space|meager]]
+and:
+:a [[Definition:Topological Space|topological space]] is [[Definition:Meager Space|meager]] {{iff}} it is a [[Definition:Countable Union|countable union]] of [[Definition:Subset|subsets]] of $S$ which are [[Definition:Nowhere Dense|nowhere dense]] in $S$.
+{{AimForCont}} that $T$ is [[Definition:Meager Space|meager]].
+Let:
+:$\displaystyle T = \bigcup \SS$
+where $\SS$ is a [[Definition:Countable Set|countable set]] of [[Definition:Subset|subsets]] of $S$ which are [[Definition:Nowhere Dense|nowhere dense]] in $S$.
+Then:
+:$\exists H \in \S S: x \in H$
+and so:
+:$\set x \subseteq H$
+We have that $H$ is [[Definition:Nowhere Dense/Definition 2|nowhere dense]] in $T$.
+By definition, its [[Definition:Closure (Topology)|closure]] $H^-$ contains no [[Definition:Open Set (Topology)|open set]] of $T$ which is [[Definition:Non-Empty Set|non-empty]].
+But from [[Set is Subset of its Topological Closure]] we have that:
+:$H \subseteq H^-$
+So by [[Subset Relation is Transitive]]:
+:$\set x \subseteq H^-$
+So $H$ is not [[Definition:Nowhere Dense|nowhere dense]].
+Therefore $T$ cannot be a [[Definition:Countable Union|countable union]] of [[Definition:Subset|subsets]] of $S$ which are [[Definition:Nowhere Dense|nowhere dense]] in $S$.
+That is, $T$ is not [[Definition:Meager Space|meager]].
+Hence the result by definition of [[Definition:Non-Meager Space|non-meager]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Trivial Topological Space is Non-Meager}
+Tags: Non-Meager Spaces, Trivial Topological Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Trivial Topological Space|trivial topological space]].
+Then $T$ is [[Definition:Non-Meager Space|non-meager]].
+\end{theorem}
+
+\begin{proof}
+As $T$ is a [[Definition:Trivial Topological Space|trivial topological space]], by definition $S$ is a [[Definition:Singleton|singleton]]: $S = \set s$, say.
+Then $\set s$ is an [[Definition:Open Set (Topology)|open set]].
+That is, $s$ is an [[Definition:Open Point|open point]].
+The result follows from [[Space with Open Point is Non-Meager]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Mapping on Finite Union of Closed Sets}
+Tags: Continuous Mappings, Closed Sets
+
+\begin{theorem}
+Let $T = \struct {X, \tau}$ and $S = \struct {Y,\sigma}$ be [[Definition:Topological Space|topological spaces]].
+For all $i \in \set {1, 2, \ldots, n}$, let $C_i$ be [[Definition:Closed Set (Topology)|closed]] in $T$.
+Let $f: X \to Y$ be a [[Definition:Mapping|mapping]] such that the [[Definition:Restriction of Mapping|restriction]] $f \restriction_{C_i}$ is [[Definition:Continuous Mapping (Topology)|continuous]] for all $i$.
+Then $f$ is [[Definition:Continuous Mapping (Topology)|continuous]] on $C = \displaystyle \bigcup_{i \mathop = 1}^n C_i$, that is, $f \restriction_C$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+If $\family {C_i}$ is [[Definition:Infinite Set|infinite]], the result does not necessarily hold.
+\end{theorem}
+
+\begin{proof}
+Let $V \subset S$ be a [[Definition:Closed Set (Topology)|closed set]].
+By [[Continuity Defined from Closed Sets]], we have that $U_i = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is also [[Definition:Closed Set (Topology)|closed]].
+From the definition of a [[Definition:Restriction of Mapping|restriction]], we have that $U_i = C_i \cap f^{-1} \sqbrk V$.
+Therefore, we can compute:
+{{begin-eqn}}
+{{eqn | l = \paren {f \restriction_{C_i} }^{-1} \sqbrk V
+      | r = C \cap f^{-1} \sqbrk V
+      | c = {{Defof|Restriction of Mapping}}
+}}
+{{eqn | r = \paren {\bigcup_{i \mathop = 1}^n C_i} \cap f^{-1} \sqbrk V
+      | c = Definition of $C$
+}}
+{{eqn | r = \bigcup_{i \mathop = 1}^n \paren {C_i \cap f^{-1} \sqbrk V}
+      | c = [[Intersection Distributes over Union]]
+}}
+{{eqn | r = \bigcup_{i \mathop = 1}^n U_i
+      | c = Definition of $U_i$
+}}
+{{end-eqn}}
+That is, $U = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is the [[Definition:Set Intersection|intersection]] of finitely many [[Definition:Closed Set (Topology)|closed sets]].
+Therefore, $U$ is itself [[Definition:Closed Set (Topology)|closed]] by definition of a [[Definition:Topology|topology]].
+It follows by [[Continuity Defined from Closed Sets]] that $f \restriction_C$ is also continuous.
+{{qed}}
+{{finish|Counterexample for infinitely many closed sets.}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of T2 Space}
+Tags: Hausdorff Spaces, Separation Axioms
+
+\begin{theorem}
+{{TFAE|def = T2 Space|view = $T_2$ (Hausdorff) space}}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+=== Definition 1 implies Definition 2 ===
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] for which:
+:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ 
+Let us take any arbitrary $x, y \in S: x \ne y$.
+Let $\mathcal C_x$ be the set of all [[Definition:Closed Neighborhood|closed neighborhoods]] of $x$:
+:$\mathcal C_x = \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$
+where $\relcomp S H$ is the [[Definition:Relative Complement|complement of $H$ in $S$]].
+We need to demonstrate that the only element in the [[Definition:Set Intersection|intersection]] of $\mathcal C_x$ is $x$:
+:$\bigcap \mathcal C_x = \set x$
+and to do that we show that if $y \ne x$ then $y \notin \bigcap \mathcal C_x$.
+Let $C = \bigcap C_x$.
+Clearly $x \in C$ and so $\set x \subseteq C_x$.
+We have that $\exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ by hypothesis.
+As $x \in U$ it follows that $x \notin V$ and so $x \in \relcomp S V$.
+Thus $x \in U \subseteq \relcomp S V$.
+That is, $\relcomp S V$ is a [[Definition:Closed Neighborhood|closed neighborhood]] of $x$ and so $\relcomp S V \in \mathcal C_x$.
+As $y \in V$ it follows that $y \notin \relcomp S V$.
+So $\relcomp S V$ is a [[Definition:Closed Neighborhood|closed neighborhood]] of $x$ which does not contain $y$.
+So $y \notin \bigcap C_x$.
+As $y$ is arbitrary:
+:$\forall y \in S, y \ne x: \exists H: \relcomp S H \in \tau: y \notin H$
+and so $C_x \subseteq \set x$.
+That is:
+:$\displaystyle \forall x \in S: \set x = \bigcap \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$
+or, each point is the [[Definition:Set Intersection|intersection]] of all its [[Definition:Closed Neighborhood|closed neighborhoods]].
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] for which each point is the [[Definition:Set Intersection|intersection]] of all its [[Definition:Closed Neighborhood|closed neighborhoods]].
+Let $x, y \in S: x \ne y$.
+Let $\mathcal C_x$ be the set of all [[Definition:Closed Neighborhood|closed neighborhoods]] of $x$:
+:$\mathcal C_x = \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$
+where $\relcomp S H$ is the [[Definition:Relative Complement|complement of $H$ in $S$]].
+This arises from the definition of a [[Definition:Closed Set (Topology)|closed set]] as the [[Definition:Relative Complement|complement in $S$]] of an [[Definition:Open Set (Topology)|open set]].
+We have that:
+:$\displaystyle \set x = \bigcap \set {H: \relcomp S H \in \tau, \exists U \in \tau: x \in U \subseteq H}$
+Then as $y \notin \set x$ it is not the case that $\forall H \in C_x: y \in H$.
+So for some $H \in C_x$ it must be the case that $y \in \relcomp S H = V$.
+But $V = \relcomp S H \in \tau$, that is, $V$ is [[Definition:Open Set (Topology)|open]] in $T$.
+Also, as $U \subseteq H$, it must follow that $U \cap V = \O$.
+So:
+:$\exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
+As $x$ and $y$ are arbitrary, it follows that:
+:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ 
+{{qed|lemma}}
+=== Definition 1 implies Definition 3 ===
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] for which:
+:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ 
+Let $x, y \in S: x \ne y$ be arbitrary.
+From [[Set is Open iff Neighborhood of all its Points]], $U$ and $V$ are [[Definition:Neighborhood of Point|neighborhoods]] of $x$ and $y$.
+Thus as $x$ and $y$ are arbitrary:
+ 
+:$\forall x, y \in S, x \ne y: \exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x \cap N_y = \O$
+{{qed|lemma}}
+=== Definition 3 implies Definition 1 ===
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] for which:
+:$\forall x, y \in S, x \ne y: \exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x \cap N_y = \O$
+Let $x, y \in S: x \ne y$ be arbitrary.
+We have that:
+: $\exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x \cap N_y = \O$
+{{AimForCont}} $\exists z \in S: z \in U \cap V$.
+Then $z \in U, z \in V$
+{{begin-eqn}}
+{{eqn | l = z
+      | o = \in
+      | r = U
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | lo= \land
+      | l = z
+      | o = \in
+      | r = V
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | ll= \leadsto
+      | l = z
+      | o = \in
+      | r = N_x
+      | c = {{Defof|Subset}}
+}}
+{{eqn | lo= \land
+      | l = z
+      | o = \in
+      | r = N_y
+      | c = {{Defof|Subset}}
+}}
+{{eqn | ll= \leadsto
+      | l = z
+      | o = \in
+      | r = N_x \cap N_y
+      | c = {{Defof|Set Intersection}}
+}}
+{{end-eqn}}
+From this contradiction it follows that $U \cap V = \O$.
+As $x$ and $y$ are arbitrary, it follows that:
+:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of T3 Space}
+Tags: T3 Spaces
+
+\begin{theorem}
+{{TFAE|def = T3 Space|view = $T_3$ space}}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+=== Definition by Open Sets implies Definition by Closed Neighborhoods ===
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] for which:
+:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$ 
+Let $U \in \tau$, and let $x \in U$.
+Then:
+:$\relcomp S U$ is [[Definition:Closed Set (Topology)|closed]]
+and:
+:$x \notin \relcomp S U$
+We have by hypothesis:
+:$\exists A, B \in \tau: \relcomp S U \subseteq A, x \in B: A \cap B = \O$ 
+It follows that:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = B
+      | c = by hypothesis
+}}
+{{eqn | o = \subseteq
+      | r = \relcomp S A
+      | c = [[Empty Intersection iff Subset of Complement]]: from $A \cap B$
+}}
+{{eqn | o = \subseteq
+      | r = \relcomp S {\relcomp S U}
+      | c = [[Set Complement inverts Subsets]]: from $\relcomp S U \subseteq A$
+}}
+{{eqn | r = U
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{end-eqn}}
+That is:
+:$x \in \relcomp S A \subseteq U$
+So we have demonstrated that there exists a [[Definition:Closed Neighborhood|closed neighborhood]] $\relcomp S A$ of $x$ contained in $U$.
+As $U, x$ are arbitrary:
+:$\forall U \in \tau: \forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq U$
+{{qed|lemma}}
+=== Definition by Closed Neighborhoods implies Definition by Intersection of Closed Neighborhoods ===
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] for which:
+:$\forall U \in \tau: \forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq U$
+where $N_x$ denotes a [[Definition:Neighborhood of Point in Topological Space|neighborhood]] of $x$.
+Let $H \subseteq S$ such that $\relcomp S H \in \tau$. 
+Let $C_H$ be the [[Definition:Set|set]] of all [[Definition:Closed Neighborhood|closed neighborhoods]] of $H$:
+:$C_H = \set {N_H: \relcomp S {N_H} \in \tau, \exists V \in \tau: H \subseteq V \subseteq N_H}$
+By construction:
+:$H \subseteq \bigcap C_H$
+Let $x \notin H$.
+Then:
+:$x \in \relcomp S H \in \tau$
+We have by hypothesis:
+:$\exists N_x: \relcomp S {N_x} \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq \relcomp S H$
+It follows that:
+{{begin-eqn}}
+{{eqn | l = H
+      | r = \relcomp S {\relcomp S H}
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{eqn | o = \subseteq
+      | r = \relcomp S {N_x}
+      | c = [[Set Complement inverts Subsets]]: from $N_x \subseteq \relcomp S H$
+}}
+{{eqn | o = \subseteq
+      | r = \relcomp S V
+      | c = [[Set Complement inverts Subsets]]: from $V \subseteq N_x$
+}}
+{{end-eqn}}
+Therefore by construction of $C_H$:
+:$\relcomp S V \in C_H$
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = V
+      | c = by hypothesis
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \notin
+      | r = \relcomp S V
+      | c = {{Defof|Relative Complement}}
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \notin
+      | r = \bigcap C_H
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \in
+      | r = \relcomp S {\bigcap C_H}
+      | c = {{Defof|Relative Complement}}
+}}
+{{end-eqn}}
+That is:
+{{begin-eqn}}
+{{eqn | l = x \in \relcomp S H
+      | o = \implies
+      | r = x \in \relcomp S {\bigcap C_H}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp S H
+      | o = \subseteq
+      | r = \relcomp S {\bigcap C_H}
+      | c = {{Defof|Subset}}
+}}
+{{eqn | ll= \leadsto
+      | l = \bigcap C_H
+      | o = \subseteq
+      | r = H
+      | c = [[Set Complement inverts Subsets]]
+}}
+{{end-eqn}}
+We have shown that:
+:$\bigcap C_H \subseteq H$
+and:
+:$H \subseteq \bigcap C_H$
+So by definition of [[Definition:Set Equality|set equality]]:
+:$\bigcap C_H = H$
+As $H$ is arbitrary:
+:$\forall H \subseteq S: \relcomp S H \in \tau: H = \bigcap \set {N_H: \relcomp S {N_H} \in \tau, \exists V \in \tau: H \subseteq V \subseteq N_H}$
+{{qed|lemma}}
+=== Definition by Intersection of Closed Neighborhoods implies Definition by Open Sets ===
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] for which:
+:$\forall H \subseteq S: \relcomp S H \in \tau: H = \bigcap \set {N_H: \relcomp S {N_H} \in \tau, \exists V \in \tau: H \subseteq V \subseteq N_H}$
+Let $F \subseteq S$ and $\relcomp S F \in \tau$.
+We have by hypothesis:
+:$F = \bigcap \set {N_F: \relcomp S {N_F} \in \tau, \exists V \in \tau: F \subseteq V \subseteq N_F}$
+Pick arbitrary $x \notin F$. 
+Then:
+:$\exists N \subseteq S: \relcomp S N \in \tau, \exists V \in \tau: F \subseteq V \subseteq N$
+Because $x \notin F \subseteq N$:
+:$x \in \relcomp S N$
+Because $V \subseteq N$, it follows from [[Empty Intersection iff Subset of Complement]] that:
+:$V \cap \relcomp S N = \O$
+Therefore we have:
+{{begin-eqn}}
+{{eqn | l = \relcomp S N, V
+      | o = \in
+      | r = \tau
+}}
+{{eqn | l = x
+      | o = \in
+      | r = \relcomp S N
+}}
+{{eqn | l = F
+      | o = \subseteq
+      | r = V
+}}
+{{eqn | l = V \cap \relcomp S N
+      | r = \O
+}}
+{{end-eqn}}
+As $F$ and $x$ are arbitrary:
+:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$ 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Big Implies Saturated}
+Tags: Model Theory
+
+\begin{theorem}
+Let $\mathcal M$ be an $\mathcal L$-[[Definition:First-Order Structure|structure]].
+Let $\kappa$ be a [[Definition:Cardinal|cardinal]].
+If $\mathcal M$ is $\kappa$-[[Definition:Big Model|big]], then it is $\kappa$-[[Definition:Saturated Model|saturated]].
+\end{theorem}
+
+\begin{proof}
+The idea of the proof is:
+: to go to some elementary equivalent structure where a type is realized
+and:
+: to interpret a new relation symbol as the singleton containing some realization of the type.
+This lets us write sentences about the extension saying that:
+:there must exist an element satisfying the relation
+and:
+:every element satisfying the relation satisfies all of the formulas in the type.
+Then using bigness, these sentences carry over into $\mathcal M$.
+Let $p$ be a [[Definition:Complete Type|complete $n$-type over $A$]], where $A$ is a subset of the universe of $\mathcal M$ with [[Definition:Cardinality|cardinality]] strictly less than $\kappa$.
+We are to show that $p$ is [[Definition:Realization|realized]] in $\mathcal M$.
+Let $\mathcal L_A$ be the [[Definition:Logical Language|language]] obtained from $\mathcal L$ by adding [[Definition:Constant Symbol|constant symbols]] for each element of $A$.
+Viewing $\mathcal M$ as an $\mathcal L_A$-structure in the natural way, let $\operatorname {Th}_A (\mathcal M)$ be the collection of $\mathcal L_A$-[[Definition:Sentence|sentences]] satisfied by $\mathcal M$.
+By definition of [[Definition:Type|type]], $p \cup \operatorname {Th}_A \left({\mathcal M}\right)$ is [[Definition:Model (Logic)|satisfiable]] by some $\mathcal L_A$-structure $\mathcal N$.
+Since $\mathcal N$ satisfies $p$, some $n$-[[Definition:Ordered Tuple|ordered tuple]] $\bar b$ in it realizes $p$.
+Now, let $\mathcal L_A^*$ be the language obtained from $\mathcal L_A$ by adding a new relation symbol $R$.
+We can extend $\mathcal N$ to be an $\mathcal L_A^*$-structure by interpreting the symbol $R$ as the [[Definition:Singleton|singleton]] $R^\mathcal N = \left\{{\bar b}\right\}$.
+Since $\mathcal N \models \operatorname {Th}_A \left({\mathcal M}\right)$, we have that $\mathcal M$ and $\mathcal N$ are [[Definition:Elementary Equivalence|elementary equivalent]] as $\mathcal L_A$-structures.
+We have that $\mathcal M$ is $\kappa$-[[Definition:Big Model|big]] by assumption.
+Hence there is some relation $R^\mathcal M$ on the universe of $\mathcal M$ such that:
+:$\left({\mathcal M, R^\mathcal M}\right)$ is elementary equivalent to $\left({\mathcal N, R^\mathcal N}\right)$ as $\mathcal L_A^*$-structures.
+But $\left({\mathcal N, R^\mathcal N}\right)$ satisfies the $\mathcal L_A^*$-sentences:
+:$\exists \bar x R \left({\bar x}\right)$
+and
+:$\forall \bar x \left({R \left({\bar x}\right) \to \phi \left({\bar x}\right)}\right)$
+for each $\phi \left({\bar x}\right)$ in $p$.
+Hence $\left({\mathcal M, R^\mathcal M}\right)$ also satisfies these sentences, and so there must be some $\bar d \in R^\mathcal M$ realizing $p$.
+So $\mathcal M$ realizes $p$.
+{{qed}}
+[[Category:Model Theory]]
+78f6cxjvoukbaly9ux17vxwtj90s35c
+\end{proof}<|endoftext|>
+\section{Limit Points in T1 Space}
+Tags: Limit Points, T1 Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which satisfies the [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) axiom]].
+Let $H \subset S$ be any [[Definition:Subset|subset]] of $S$.
+Let $x \in H$.
+Then $x$ is a [[Definition:Limit Point of Set|limit point]] of $H$ {{iff}} every [[Definition:Neighborhood of Point|neighborhood]] of $x$ contains [[Definition:Infinite Set|infinitely many]] points of $H$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose every [[Definition:Neighborhood of Point|neighborhood]] of $x$ contains [[Definition:Infinite Set|infinitely many]] points of $H$.
+Let $N$ be an [[Definition:Open Neighborhood|open neighborhood]] of $x$.
+Then $H \cap N$ is [[Definition:Infinite Set|infinite]], and so is $H \cap \paren {N \setminus \set x}$.
+Since $N$ is arbitrary, by {{Defof|Limit Point of Set}}, $x$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+We prove the [[Definition:Contrapositive Statement|contrapositive]]:
+Suppose there exists some [[Definition:Neighborhood of Point|neighborhood]] $V$ of $x$ containing only [[Definition:Finite Set|finitely many]] points of $H$.
+Then there exists some [[Definition:Open Set (Topology)|open set]] $N \subseteq V$ such that $x \in N$.
+By [[Set Intersection Preserves Subsets]], $N \cap H \subseteq V \cap H$.
+By [[Subset of Finite Set is Finite]], $N \cap H$ is [[Definition:Finite Set|finite]].
+For each $y \in H \cap N$ and $y \ne x$, since $T$ is [[Definition:Fréchet Space (Topology)|$T_1$]]:
+:$\exists U_y \in \tau: x \in U_y, y \notin U_y$
+Consider the set $\displaystyle U = N \cap \bigcap_{y \mathop \in H \cap \paren {N \setminus \set x} } U_y$.
+This is a [[Definition:Finite Set|finite]] [[Definition:Set Intersection|intersection]] of [[Definition:Open Set (Topology)|open sets]].
+By [[General Intersection Property of Topological Space]], $U \in \tau$.
+By {{Defof|Set Intersection}}, since $x \in N$ and $x \in U_y$ for all $y \in H \cap \paren {N \setminus \set x}$, we have $x \in U$.
+Therefore $U$ is an [[Definition:Open Neighborhood|open neighborhood]] of $x$.
+Let $z \in H \cap \paren {U \setminus \set x}$.
+Then:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = z \in H \land z \in U \setminus \set x
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | o = \leadsto
+      | r = z \in H \land z \notin \set x \land z \in U
+      | c = {{Defof|Set Complement}}
+}}
+{{eqn | o = \leadsto
+      | r = z \in H \land z \notin \set x \land z \in N \land \paren {\forall y} \paren {y \in H \cap \paren {N \setminus \set x} \implies z \in U_y}
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | o = \leadsto
+      | r = z \in H \land z \in N \setminus \set x \land \paren {\forall y} \paren {y \in H \cap \paren {N \setminus \set x} \implies z \in U_y}
+      | c = {{Defof|Set Complement}}
+}}
+{{eqn | o = \leadsto
+      | r = z \in H \cap \paren {N \setminus \set x} \land \paren {\forall y} \paren {y \in H \cap \paren {N \setminus \set x} \implies z \in U_y}
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | o = \leadsto
+      | r = z \in H \cap \paren {N \setminus \set x} \land z \in U_z
+      | c = [[Universal Instantiation]]
+}}
+{{end-eqn}}
+But from definition of $U_z$: 
+:$z \in H \cap \paren {N \setminus \set x} \implies z \notin U_z$
+Which is a [[Definition:Contradiction|contradiction]].
+This shows that $H \cap \paren {U \setminus \set x} = \O$.
+By {{Defof|Limit Point of Set}}, $x$ is not a [[Definition:Limit Point of Set|limit point]] of $H$.
+Thus the [[Definition:Contrapositive Statement|contrapositive]] is proved.
+{{qed}}
+[[Category:Limit Points]]
+[[Category:T1 Spaces]]
+blp2niyph0ixy0brxggaf1uwx1ct2v8
+\end{proof}<|endoftext|>
+\section{Singleton Point is Isolated}
+Tags: Isolated Points, Singletons
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $x \in S$.
+Then $x$ is an [[Definition:Isolated Point (Topology)|isolated point]] of the [[Definition:Singleton|singleton set]] $\left\{{x}\right\}$, but not necessarily an [[Definition:Isolated Point (Topology)|isolated point of $T$]].
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$ be an [[Definition:Open Set (Topology)|open set]] of $T$ such that $x \in T$.
+The fact that such a $U$ exists follows from the fact that [[Definition:Open Set Axioms|$S$ is open in $T$]] and $x \in S$.
+Then as $\left\{{x}\right\} \subseteq S$ it follows that $\left\{{x}\right\} \cap U = \left\{{x}\right\}$ from [[Intersection with Subset is Subset]].
+So by definition, $x$ is an [[Definition:Isolated Point (Topology)|isolated point of $\left\{{x}\right\}$]].
+From [[Topological Space is Discrete iff All Points are Isolated]], unless $T$ is the [[Definition:Discrete Space|discrete space]] on $S$, not all elements of $T$ are [[Definition:Isolated Point (Topology)|isolated points of $T$]].
+{{qed}}
+[[Category:Isolated Points]]
+[[Category:Singletons]]
+1xustjkquotedgz9fhc6j1rit9rvbxb
+\end{proof}<|endoftext|>
+\section{Singleton Set is not Dense-in-itself}
+Tags: Denseness, Singletons
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $x \in S$.
+Then the [[Definition:Singleton|singleton set]] $\left\{{x}\right\}$ is not [[Definition:Dense-in-itself|dense-in-itself]].
+\end{theorem}
+
+\begin{proof}
+From [[Singleton Point is Isolated]], $x$ is [[Definition:Isolated Point (Topology)|isolated]] in $\left\{{x}\right\}$.
+So by definition $\left\{{x}\right\}$ is not [[Definition:Dense-in-itself|dense-in-itself]].
+{{qed}}
+[[Category:Denseness]]
+[[Category:Singletons]]
+akms0y3s9xd4yoyqbiuid33kcvymq78
+\end{proof}<|endoftext|>
+\section{Tietze Extension Theorem}
+Tags: Normal Spaces, Continuous Mappings
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Normal Space|normal]].
+Let $A \subseteq S$ be a [[Definition:Closed Set (Topology)|closed set]] in $T$.
+Let $f: A \to \R$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] from $A \subseteq S$ to the [[Definition:Real Number Line|real number line]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
+Then there exists a [[Definition:Continuous Extension|continuous extension]] $g: S \to \R$, i.e. such that:
+:$\forall s \in A: \map f s = \map g s$
+\end{theorem}
+
+\begin{proof}
+=== [[Tietze Extension Theorem/Lemma|Lemma]] ===
+{{:Tietze Extension Theorem/Lemma}}{{qed|lemma}}
+First suppose that for any [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] on a [[Definition:Closed Set (Topology)|closed set]] there is a [[Definition:Continuous Extension|continuous extension]].
+Let $C$ and $D$ be [[Definition:Disjoint Sets|disjoint sets]] which are [[Definition:Closed Set (Topology)|closed in $S$]].
+Define $f: C \cup D \to \R$ by:
+:$\map f x = \begin {cases}
+0 & : x \in C \\
+1 & : x \in D \end {cases}$
+Now $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] and we can extend it to a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] $g: X \to \R$.
+By [[Urysohn's Lemma]], $S$ is [[Definition:Normal Space|normal]] because $g$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] such that $\map g x = 0$ for $x \in C$ and $\map g x = 1$ for $x \in D$.
+Conversely, let $S$ be [[Definition:Normal Space|normal]] and $A$ be [[Definition:Closed Set (Topology)|closed in $S$]].
+By the [[Tietze Extension Theorem/Lemma|lemma]], there exists a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] $g_0: S \to \R$ such that:
+:$\forall x \in S: \size {\map {g_0} x} \le \dfrac 1 3$
+:$\forall x \in A: \size {\map f x − \map {g_0} x} \le \dfrac 2 3$
+Since $\paren {f − g_0}: A \to \R$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]], the [[Tietze Extension Theorem/Lemma|lemma]] tells us there is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] $g_1: X \to \R$ such that:
+:$\forall x \in S: \size {\map {g_1} x} \le \dfrac 1 3 \paren {\dfrac 2 3}$
+:$\forall x \in A: \size {\map f x − \map {g_0} x - \map {g_1} x} \le \dfrac 2 3 \paren {\dfrac 2 3}$
+By repeated application of the [[Tietze Extension Theorem/Lemma|lemma]] we can construct a [[Definition:Sequence|sequence]] of [[Definition:Everywhere Continuous Mapping (Topology)|continuous mappings]] $g_0, g_1, g_2, \ldots$ such that:
+:$\forall x \in S: \size {\map {g_n} x} \le \dfrac 1 3 \paren {\dfrac 2 3}^n$
+:$\forall x \in A: \size {\map f x − \map {g_0} x − \map {g_1} x - \map {g_2} x - \cdots - \map {g_n} x} \le \dfrac 2 3 \paren {\dfrac 2 3}^n$
+Define:
+:$\displaystyle \map g x = \sum_{n \mathop = 0}^\infty \map {g_n} x$
+We have that:
+:$\size {\map {g_n} x} \le \dfrac 1 3 \paren {\dfrac 2 3}^n$
+and:
+:$\displaystyle \sum_{n \mathop = 0}^\infty \dfrac 1 3 \paren {\dfrac 2 3}^n$
+[[Sum of Infinite Geometric Sequence|converges as a geometric series]].
+Thus $\displaystyle \sum_{n \mathop = 0}^\infty \map {g_n} x$ [[Definition:Absolutely Convergent Series|converges absolutely]] and [[Definition:Uniform Convergence|uniformly]].
+So $g$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] defined everywhere.
+Also, $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac 1 3 \paren {\dfrac 2 3}^n = 1$ implies that $\size {\map g x} \le 1$.
+Now for $x \in A$, we have that:
+:$\displaystyle \size {\map f x − \sum_{n \mathop = 0}^k \map {g_n} x} \le \paren {\dfrac 2 3}^{k + 1}$
+As $k \to \infty$, the {{RHS}} $\to 0$ and so the sum goes to $\map g x$.
+Thus:
+:$\size {\map f x − \map g x} = 0$
+Therefore $g$ extends $f$. 
+{{qed}}
+{{Namedfor|Heinrich Franz Friedrich Tietze|cat = Tietze}}
+\end{proof}<|endoftext|>
+\section{Hermitian Operators have Real Eigenvalues}
+Tags: Linear Algebra
+
+\begin{theorem}
+Hermitian [[Definition:Operation|operations]] have [[Definition:Real Number|real]] eigenvalues.
+\end{theorem}
+
+\begin{proof}
+Let $\hat H$ be a Hermitian operator on an [[Definition:Inner Product Space|inner product space]] $V$ over the [[Definition:Field (Abstract Algebra)|field]] of [[Definition:Complex Number|complex numbers]], $\C$. That is, $\hat H = \hat H^\dagger$.
+Then for an eigenvector $\left\vert{x}\right\rangle \in V, \left\vert{x}\right\rangle \ne \left\vert{0}\right\rangle$ and eigenvalue $\lambda \in \C$:
+:$\left.{\hat H}\middle\vert{x}\right\rangle = \left.{\lambda}\middle\vert{x}\right\rangle$ 
+We know for a general operator $\hat A$ on $V$, the following holds:
+:$\forall \left\vert{x}\right\rangle, \left\vert{y}\right\rangle \in V: \left\langle{x}\middle\vert{\hat A}\middle\vert{y}\right\rangle = \left\langle{y}\middle\vert{\hat A^\dagger}\middle\vert{x }\right\rangle^*$
+where $^*$ denotes the [[Definition:Complex Conjugate|complex conjugate]].
+Noting $\hat H = \hat H^\dagger$ gives:
+:$\left\langle{x}\middle\vert{\hat H}\middle\vert{y}\right\rangle = \left\langle{y}\middle\vert{\hat H}\middle\vert{x}\right\rangle^*$
+Now we compute:
+{{begin-eqn}}
+{{eqn | l = \left\langle{x}\middle\vert{\hat H}\middle\vert{x}\right\rangle
+      | r = \left\langle{x}\middle\vert{ \left({\hat H}\middle\vert{x}\right\rangle}\right)
+      | c = 
+}}
+{{eqn | r = \left\langle{x}\middle\vert{\lambda}\middle\vert{x}\right\rangle
+      | c = 
+}}
+{{eqn | r = \lambda \left\langle{x}\middle\vert{x}\right\rangle
+      | c = 
+}}
+{{end-eqn}}
+Using our previous result:
+{{begin-eqn}}
+{{eqn | l = \left\langle{x}\middle\vert{\hat H}\middle\vert{x}\right\rangle
+      | r = \left\langle{x}\middle\vert{\hat H}\middle\vert{x}\right\rangle^*
+      | c = 
+}}
+{{eqn | r = \left({\lambda \left\langle{x}\middle\vert{x}\right\rangle}\right)^*
+      | c = 
+}}
+{{end-eqn}}
+Equating the two previous equations gives:
+:$\lambda \left\langle{x}\middle\vert{x}\right\rangle = \left({\lambda \left\langle{x}\middle\vert{x}\right\rangle}\right)^*$
+Recalling the conjugate symmetry property of the [[Definition:Inner Product|inner product]], we can see that:
+:$\left\langle{x}\middle\vert{x}\right\rangle = \left\langle{x}\middle\vert{x}\right\rangle^*$ 
+which is true {{iff}} $\left\langle{x}\middle\vert{x}\right\rangle \in \R$.
+So
+:$\lambda \left\langle{x}\middle\vert{x}\right\rangle = \lambda^* \left\langle{x}\middle\vert{x}\right\rangle$
+and so:
+:$\lambda = \lambda^*$
+Therefore $\lambda \in \R$.
+{{qed}}
+[[Category:Linear Algebra]]
+fvvr0rwh5xnun62ea9bf3lt0ll3idue
+\end{proof}<|endoftext|>
+\section{Hermitian Operators have Orthogonal Eigenvectors}
+Tags: Linear Algebra
+
+\begin{theorem}
+The eigenvectors of a Hermitian [[Definition:Operation|operation]] are [[Definition:Orthogonal (Linear Algebra)|orthogonal]].
+\end{theorem}
+
+\begin{proof}
+Let $\hat H$ be a Hermitian operator on an [[Definition:Inner Product Space|inner product space]] $V$ over the [[Definition:Complex Number|complex numbers]] $\C$, with a simple spectrum:
+:$\hat H \left\vert{x_i}\right\rangle = \lambda_i \left\vert{x_i}\right\rangle$ 
+:$\lambda_i \ne \lambda_j$  
+:$\forall i, j \in \N: i \ne j$
+Now we compute the following:
+{{begin-eqn}}
+{{eqn | l = \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle
+      | r = \left\langle{x_j}\middle \vert{\left({\hat H}\middle \vert{x_i}\right\rangle}\right)
+      | c = 
+}}
+{{eqn | r = \left\langle{x_j}\middle \vert{\lambda_i}\middle \vert{x_i}\right\rangle
+      | c = 
+}}
+{{eqn | r = \lambda_i \left\langle{x_j}\middle \vert{x_i}\right\rangle
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^*
+      | r = \left({\left\langle{x_i}\middle \vert{\left({\hat H}\middle \vert{x_j}\right\rangle}\right)}\right)^*
+      | c = 
+}}
+{{eqn | r = \left\langle{x_i}\middle \vert{\lambda_j}\middle \vert{x_j}\right\rangle^*
+      | c = 
+}}
+{{eqn | r = \left({\lambda_j \left\langle{x_i}\middle \vert{x_j}\right\rangle}\right)^*
+      | c = 
+}}
+{{end-eqn}}
+{{explain|Justification for the above is needed.}}
+From the property $\lambda_j = \lambda_j^*$ and the [[Definition:Complex Conjugate|conjugate]] symmetry of the [[Definition:Inner Product|inner product]]:
+:$\left\langle{x_i}\middle \vert{x_j}\right\rangle = \left\langle{x_j}\middle \vert{x_i}\right\rangle^*$
+this becomes:
+:$\left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^* = \lambda_j \left\langle{x_j}\middle \vert{x_i}\right\rangle$
+It can be shown that the following relation holds since $\hat H = \hat H^\dagger$:
+:$\left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^*$
+{{handwaving|"can be shown" -- link to a page which demonstrates this}}
+This now gives us the equations:
+:$(1): \quad \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \lambda_i \left\langle{x_j}\middle \vert{x_i}\right\rangle$
+:$(2): \quad \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \lambda_j \left\langle{x_j}\middle \vert{x_i}\right\rangle$
+Subtracting $(2)$ from $(1)$ gives:
+:$\left({\lambda_i - \lambda_j}\right) \left\langle{x_j}\middle \vert{x_i}\right\rangle = 0$
+Note that $\left({\lambda_i - \lambda_j}\right) \ne 0$ since we were given $\lambda_i \ne \lambda_j$.
+Therefore:
+:$\left\langle{x_j}\middle \vert{x_i}\right\rangle = 0$
+Two vectors have inner product $0$ if and only if they are orthogonal.
+Therefore the eigenvectors of $\hat H$ are orthogonal.
+{{qed}}
+[[Category:Linear Algebra]]
+tvwvf85maehs2tcdqnbwhyifkqvys85
+\end{proof}<|endoftext|>
+\section{Hausdorff Space iff Diagonal Set on Product is Closed}
+Tags: Hausdorff Spaces
+
+\begin{theorem}
+Let $T = \struct{S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\Delta_S$ be the [[Definition:Diagonal Relation|diagonal set]] on $S$:
+:$\Delta_S = \set{\tuple{x, x} \in S \times S: x \in S}$
+where $S \times S$ is the [[Definition:Cartesian Product|Cartesian product]] of $S$ with itself.
+Let $T^2 = \struct{S \times S, \TT}$ be the [[Definition:Product Space (Topology)|product space]] with [[Definition:Tychonoff Topology|Tychonoff topology]] $\TT$.
+Then $T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]] iff $\Delta_S$ is [[Definition:Closed Set (Topology)|closed]] in $T^2$.
+\end{theorem}
+
+\begin{proof}
+Suppose $T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+Let $\Delta: S \to S \times S$ be the [[Definition:Diagonal Mapping|diagonal mapping]] on $S$.
+We have that $\map \Delta S$ consists of all the elements of $S \times S$ whose coordinates are equal.
+:$\map \Delta S = \set{\tuple{x, x} \in S \times S: x \in S} = \set{\tuple{x, y} \in S \times S: x = y}$
+By definition, $\Delta_S$ is the same set as $\map \Delta S$.
+Now consider $H := \relcomp {S \times S} {\map \Delta S}$, its [[Definition:Relative Complement|complement in $S \times S$]].
+We have that $H$ consists of all the elements of $S \times S$ whose coordinates are different:
+:$H = \set{\tuple{x, y} \in S \times S: x \ne y}$
+Let $\tuple{x, y} \in H$.
+Because $T$ is a [[Definition:Hausdorff Space|$T_2$ space]]:
+:$\exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
+Because $U \cap V = \O$ it follows that $U \times V \cap \map \Delta S = \O$.
+That is:
+:$U \times V \subseteq H$
+Now from [[Natural Basis of Tychonoff Topology of Finite Product]], we have that $U \times V$ is an element of the [[Definition:Natural Basis|natural basis]] of $T^2 = \struct{S \times S, \TT}$.
+Hence $H$ is [[Definition:Open Set (Topology)|open]] in $T^2$.
+Therefore, by definition, $\Delta_S$ is [[Definition:Closed Set (Topology)|closed]] in $T^2 = \struct{S \times S, \TT}$.
+{{qed|lemma}}
+Now suppose $\Delta_S$ is [[Definition:Closed Set (Topology)|closed]] in $T^2 = \struct{S \times S, \TT}$.
+Then $H = \set{\tuple{x, y} \in S \times S: x \ne y}$ is [[Definition:Open Set (Topology)|open]] in $T^2$.
+Let $\tuple{x, y} \in H$.
+We have that $x \ne y$ by the nature of $H$.
+Then there exists an element $U \times V$ in the [[Definition:Natural Basis|natural basis]] of $T^2$, and therefore [[Definition:Open Set (Topology)|open]] in $T^2$.
+As $U \times V \subseteq H$ it follows that $U \cap V = \O$.
+From [[Natural Basis of Tychonoff Topology of Finite Product]], as $U \times V$ is [[Definition:Open Set (Topology)|open]] in $T^2$, then $U$ and $V$ are both [[Definition:Open Set (Topology)|open]] in $T$.
+So, for any $\tuple{x, y}$ we have $U, V \subseteq S$ such that $x \in U, y \in V, U \cap V = \O$.
+So $T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T0 Property is Hereditary}
+Tags: T0 Spaces, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a [[Definition:Topological Subspace|subspace]] of $T$.
+Then $T_H$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T$ be a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+That is:
+:$\forall x, y \in S$ such that $x \ne y$, either:
+::$\exists U \in \tau: x \in U, y \notin U$
+:or:
+::$\exists U \in \tau: y \in U, x \notin U$
+We have that the set $\tau_H$ is defined as:
+:$\tau_H := \set {U \cap H: U \in \tau}$
+Let $x, y \in H$ such that $x \ne y$.
+Then as $x, y \in S$ we have that:
+:$\exists U \in \tau: x \in U, y \notin U$
+or:
+:$\exists U \in \tau: y \in U, x \notin U$
+Then either:
+:$U \cap H \in \tau_H: x \in U \cap H, y \notin U \cap H$
+or:
+:$U \cap H \in \tau_H: y \in U \cap H, x \notin U \cap H$
+and so the [[Definition:Kolmogorov Space|$T_0$ axiom]] is satisfied.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T1 Property is Hereditary}
+Tags: T1 Spaces, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+Let $T_H = \struct {H, \tau_H}$, where $\varnothing \subset H \subseteq S$, be a [[Definition:Topological Subspace|subspace]] of $T$.
+Then $T_H$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T$ be a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+That is:
+:$\forall x, y \in S$ such that $x \ne y$, both:
+::$\exists U \in \tau: x \in U, y \notin U$
+:and:
+::$\exists U \in \tau: y \in U, x \notin U$
+We have that the set $\tau_H$ is defined as:
+:$\tau_H := \set {U \cap H: U \in \tau}$
+Let $x, y \in H$ such that $x \ne y$.
+Then as $x, y \in S$ we have that:
+:$\exists U \in \tau: x \in U, y \notin U$
+and:
+:$\exists U \in \tau: y \in U, x \notin U$
+Then both:
+:$U \cap H \in \tau_H: x \in U \cap H, y \notin U \cap H$
+and:
+:$U \cap H \in \tau_H: y \in U \cap H, x \notin U \cap H$
+and so the [[Definition:Fréchet Space (Topology)|$T_1$ axiom]] is satisfied.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T2 Property is Hereditary}
+Tags: Hausdorff Spaces, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a [[Definition:Topological Subspace|subspace]] of $T$.
+Then $T_H$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+That is, the property of being a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]] is [[Definition:Hereditary Property (Topology)|hereditary]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+Then:
+:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ 
+That is, for any two [[Definition:Distinct Elements|distinct elements]] $x, y \in S$ there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ containing $x$ and $y$ respectively.
+We have that the set $\tau_H$ is defined as:
+:$\tau_H := \set {U \cap H: U \in \tau}$
+Let $x, y \in H$ such that $x \ne y$.
+Then as $x, y \in S$ we have that:
+:$\exists U, V \in \tau: x \in U, y \in V, U \cap V = \O$
+As $x, y \in H$ we have that:
+:$x \in U \cap H, y \in V \cap H: \paren {U \cap H} \cap \paren {V \cap H} = \O$
+and so the [[Definition:Hausdorff Space|$T_2$ axiom]] is satisfied in $H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Completely Hausdorff Property is Hereditary}
+Tags: Completely Hausdorff Spaces, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a [[Definition:Topological Subspace|subspace]] of $T$.
+Then $T_H$ is a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) space]].
+Then:
+:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$ 
+That is, for any two [[Definition:Distinct Elements|distinct elements]] $x, y \in S$ there exist [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ containing $x$ and $y$ respectively whose [[Definition:Closure (Topology)|closures]] are [[Definition:Disjoint Sets|disjoint]].
+We have that the set $\tau_H$ is defined as:
+:$\tau_H := \set {U \cap H: U \in \tau}$
+Let $x, y \in H$ such that $x \ne y$.
+Then as $x, y \in S$ we have that:
+:$\exists U, V \in \tau: x \in U, y \in V, U^- \cap V^- = \O$
+As $x, y \in H$ we have that:
+:$x \in U \cap H, y \in V \cap H: \paren {U^- \cap H} \cap \paren {V^- \cap H} = \O$
+From [[Closure in Subspace/Corollary 1|Corollary to Closure in Subspace]] we have that:
+:$\paren {U \cap H}^- \subseteq U^- \cap H$
+:$\paren {V \cap H}^- \subseteq V^- \cap H$
+where $\paren{U \cap H}^-$ and $\paren{V \cap H}^-$ are the [[Definition:Closure (Topology)|closures]] in $H$ of the respective [[Definition:Set|sets]].
+So:
+:$x \in U \cap H, y \in V \cap H: \paren {U \cap H}^- \cap \paren {V \cap H}^- = \O$
+and so the [[Definition:Completely Hausdorff Space|$T_{2 \frac 1 2}$ (completely Hausdorff) axiom]] is satisfied in $H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure in Subspace}
+Tags: Set Closures, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a [[Definition:Topological Subspace|subspace]] of $T$.
+Let $U \subseteq H$ and let $\map {\cl_T} U$ be the [[Definition:Closure (Topology)|closure]] of $U$ in $T$.
+Then $\map {\cl_T} U \cap H$ is the [[Definition:Closure (Topology)|closure]] of $U$ in $T_H$.
+That is:
+:$\map {\cl_T} U \cap H = \map {\cl_H} U$
+where $\map {\cl_H} U$ denotes the [[Definition:Closure (Topology)|closure]] of $U$ in $T_H$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathbb K$ be defined as:
+:$\mathbb K := \set{K \subseteq S: U \subseteq K, K \text { closed in } T}$
+Then from [[Set Closure as Intersection of Closed Sets]]:
+:$\displaystyle \map {\cl_T} U = \bigcap \mathbb K$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\cl_T} U \cap H
+      | r = \paren{\bigcap \mathbb K} \cap H
+      | c =
+}}
+{{eqn | r = \bigcap_{K \mathop \in \mathbb K} \paren{K \cap H}
+      | c = [[Intersection is Idempotent]]
+}}
+{{end-eqn}}
+From [[Closed Set in Topological Subspace]] it follows that $K \cap H$ is [[Definition:Closed Set (Topology)|closed in $T_H$]] for all $K \in \mathbb K$.
+Let $K'$ be a [[Definition:Closed Set (Topology)|closed set in $T_H$]] containing $U$.
+From [[Closed Set in Topological Subspace]], there exists a [[Definition:Closed Set (Topology)|closed set]] $K$ in $T$ such that $K' = K \cap H$.
+Since $U \subseteq K' \subseteq K$ then $K \in \mathbb K$.
+Thus $\set{K \cap H : K \in \mathbb K}$ is the set of all [[Definition:Closed Set (Topology)|closed sets]] of $T_H$ which contain $U$.
+From [[Set Closure as Intersection of Closed Sets]] $\displaystyle \bigcap_{K \mathop \in \mathbb K} \paren {K \cap H}$ is the [[Definition:Closure (Topology)|closure]] of $U$ in $T_H$.
+The result follows.
+{{qed}}
+[[Category:Set Closures]]
+[[Category:Topological Subspaces]]
+kq23ugifg8ryrgroq24jy8q5vi0cg49
+\end{proof}<|endoftext|>
+\section{T3 Property is Hereditary}
+Tags: T3 Spaces, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is a [[Definition:T3 Space|$T_3$ space]].
+Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a [[Definition:Topological Subspace|subspace]] of $T$.
+Then $T_H$ is a [[Definition:T3 Space|$T_3$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:T3 Space|$T_3$ space]].
+Then:
+:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$ 
+That is, for any [[Definition:Closed Set (Topology)|closed set]] $F \subseteq S$ and any point $y \in S$ such that $y \notin F$ there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.
+We have that the set $\tau_H$ is defined as:
+:$\tau_H := \set {U \cap H: U \in \tau}$
+Let $F \subseteq H$ such that $F$ is [[Definition:Closed Set (Topology)|closed]] in $H$.
+Let $y \in H$ such that $y \notin F$.
+From [[Closed Set in Topological Subspace]] $F$ is also [[Definition:Closed Set (Topology)|closed]] in $T$.
+Because $T$ is a [[Definition:T3 Space|$T_3$ space]], we have that:
+:$\exists U, V \in \tau: F \subseteq U, y \in V, U \cap V = \O$
+As $F \subseteq H$ and $y \in H, y \notin F$ we have that:
+:$F \subseteq U \cap H, y \in V \cap H: \paren {U \cap H} \cap \paren {V \cap H} = \O$
+and so the [[Definition:T3 Space|$T_3$ axiom]] is satisfied in $H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T3 1/2 Property is Hereditary}
+Tags: T3 1: 2 Spaces, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a [[Definition:Topological Subspace|subspace]] of $T$.
+Then $T_H$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+Then:
+:For any [[Definition:Closed Set (Topology)|closed set]] $F \subseteq S$ and any point $y \in S$ such that $y \notin F$, there exists an [[Definition:Urysohn Function|Urysohn function]] for $F$ and $\set y$.
+We have that the set $\tau_H$ is defined as:
+:$\tau_H := \set {U \cap H: U \in \tau}$
+Let $F \subseteq H$ such that $F$ is [[Definition:Closed Set (Topology)|closed]] in $H$.
+Let $y \in H$ such that $y \notin F$.
+From [[Closed Set in Topological Subspace]] $F$ is also [[Definition:Closed Set (Topology)|closed]] in $T$.
+Because $T$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]], we have that there exists an [[Definition:Urysohn Function|Urysohn function]] for $F$ and $\set y$:
+That is, there exists a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]] $f: S \to \closedint 0 1$, where $\closedint 0 1$ is the [[Definition:Closed Unit Interval|closed unit interval]], such that:
+:$f {\restriction_F} = 0, f {\restriction_{\set y} } = 1$
+where $f {\restriction_F}$ denotes the [[Definition:Restriction of Mapping|restriction]] of $f$ to $F$.
+That is:
+:$\forall a \in F: \map f a = 0$
+:$\forall b \in \set y: \map f b = 1$
+From [[Continuity of Composite with Inclusion]], as $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] on $S$, then $f {\restriction_H}$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] on $H$.
+Thus $f {\restriction_H}$ is an [[Definition:Urysohn Function|Urysohn function]] for $F$ and $\set y$ in $H$.
+So the [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ axiom]] is satisfied in $H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T5 Property is Hereditary}
+Tags: T5 Spaces, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]] which is a [[Definition:T5 Space|$T_5$ space]].
+Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a [[Definition:Topological Subspace|subspace]] of $T$.
+Then $T_H$ is a [[Definition:T5 Space|$T_5$ space]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:T5 Space|$T_5$ space]].
+Then:
+:$\forall A, B \subseteq S, \map {\cl_S} A \cap B = A \cap \map {\cl_S} B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ 
+where $\map {\cl_S} A$ denotes the [[Definition:Closure (Topology)|closure]] of $A$ in $S$.
+That is:
+:For any two [[Definition:Separated Sets|separated sets]] $A, B \subseteq S$ there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] $U, V \in \tau$ containing $A$ and $B$ respectively.
+We have that the set $\tau_H$ is defined as:
+:$\tau_H := \set {U \cap H: U \in \tau}$
+Let $A, B \subseteq H$ such that $\map {\cl_H} A \cap B = A \cap \map {\cl_H} B = \O$.
+That is, $A$ and $B$ are [[Definition:Separated Sets|separated]] in $H$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\cl_S} A \cap B
+      | r = \map {\cl_S} A \cap \paren {H \cap B}
+      | c = as $B \subseteq H$
+}}
+{{eqn | r = \paren {\map {\cl_S} A \cap H} \cap B
+      | c = [[Intersection is Associative]]
+}}
+{{eqn | r = \map {\cl_H} A \cap B
+      | c = [[Closure in Subspace]]
+}}
+{{eqn | r = \O
+      | c = Assumption
+}}
+{{end-eqn}}
+Similarly:
+:$A \cap \map {\cl_S} B = \O$
+So $A$ and $B$ are separated in $S$.
+Because $T$ is a [[Definition:T5 Space|$T_5$ space]], we have that:
+:$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
+It follows that:
+:$\exists U \cap H, V \cap H \in \tau_H : A \subseteq U \cap H, B \subseteq V \cap H, \paren {U \cap H} \cap \paren {V \cap H} = \O$
+and so the [[Definition:T5 Space|$T_5$ axiom]] is satisfied in $H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T5 Space iff Every Subspace is T4}
+Tags: T5 Spaces, T4 Spaces, Topological Subspaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then $T$ is a [[Definition:T5 Space|$T_5$ space]] {{iff}} every [[Definition:Topological Subspace|subspace]] of $T$ is a [[Definition:T4 Space|$T_4$ space]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose $\struct {S, \tau}$ is $T_5$. 
+Take an arbitrary [[Definition:Topological Subspace|subspace]] $\struct {Y, \tau_Y}$ of $S$. 
+Let $A, B \subset Y$ be [[Definition:Closed Set (Topology)|closed sets]] with $A \cap B = \O$. 
+Let $R^-$ denote the [[Definition:Closure (Topology)|closure]] of $R \subset Y$. 
+From [[Closure in Subspace|Closure in Subspace]] we have that:
+:$\paren {A \cap Y}^- = A^- \cap Y$
+But as $A \subseteq Y$ we have from [[Intersection with Subset is Subset]] that $A \cap Y = A$, and so:
+:$A^- = A^- \cap Y$
+Since $A, B \subset S$, we look at $A^- \cap B$.
+Since $A, B \subset Y$, we have: 
+:$A^- \cap B = A^- \cap B \cap Y \subset A^- \cap Y = A^-$.
+But from [[Closed Set Equals its Closure]] we have $A^- = A$, since $A$ is [[Definition:Closed Set (Topology)|closed]] in $Y$.
+Thus we obtain: 
+:$A^- \cap B \subset A$ 
+and similarly: 
+:$A \cap B^- \subset B$ 
+Since $A \cap B = \O$, we obtain:
+:$A^- \cap B = \O$
+and similarly:
+:$A \cap B^- = \O$
+Thus we now have:
+:$A, B \subset S$
+and:
+:$A^- \cap B = B^- \cap A = \O$
+Thus, since $S$ is $T_5$, there exist $U_0, U_1$ disjoint [[Definition:Open Set (Topology)|open sets]] in $S$, such that $A \subset U_0$ and $B \subset U_1$.
+Then $V_i = U_i \cap Y$ is an [[Definition:Open Set (Topology)|open set]] in $Y$ for $i = 0, 1$.
+Also:
+:$V_0 \cap V_1 = \O$
+and trivially:
+:$A \subset V_0$
+and:
+:$B \subset V_1$
+Thus $Y$ is $T_4$.
+Since $Y \subset S$ was arbitrary, we have found that every subspace of $\struct {S, \tau}$ is $T_4$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose every [[Definition:Topological Subspace|subspace]] of $\struct {S, \tau}$ is $T_4$.
+Let $A, B \subset S$ with $A^- \cap B = B^- \cap A = \O$.
+Then because every [[Definition:Topological Subspace|subspace]] of $\struct {S, \tau}$ is $T_4$, the [[Definition:Topological Subspace|subspace]]:
+:$\struct {S \setminus \paren {A^- \cap B^-}, \tau_{S \setminus \paren {A^- \cap B^-} } }$
+is $T_4$.
+Because we have $A^- \cap B = B^- \cap A = \O$, from [[Set is Subset of its Topological Closure]] we see that:
+:$A \cap B = \O$
+Also:
+:$A \subset S \setminus \paren {A^- \cap B^-}$
+and:
+:$B \subset S \setminus \paren {A^- \cap B^-}$
+Then, because:
+:$\struct {S \setminus \paren {A^- \cap B^-}, \tau_{S \setminus \paren {A^- \cap B^-} } }$
+is $T_4$, there exist $U_0, U_1 \subset S \setminus \paren {A^- \cap B^-}$ [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open sets]] in $S \setminus \paren {A^- \cap B^-}$ such that $A \subset U_0$ and $B \subset U_1$. 
+Because:
+:$U_0 = V_0 \cap \paren {S \setminus \paren {A^- \cap B^-} }$ for some [[Definition:Open Set (Topology)|open set]] $V_0 \subset S$ in $S$
+and similarly:
+:$U_1 = V_1 \cap \paren {S \setminus \paren {A^- \cap B^-} }$ for some [[Definition:Open Set (Topology)|open set]] $V_1 \subset S$ in $S$
+we find that:
+:$A \subset V_0$
+and:
+:$B \subset V_1$
+and:
+:$V_0 \cap V_1 = \O$
+Thus $\struct {S, \tau}$ is a $T_5$-space.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product Space is T0 iff Factor Spaces are T0}
+Tags: T0 Spaces, Product Spaces
+
+\begin{theorem}
+Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
+Let $T = T_A \times T_B$ be the [[Definition:Product Space (Topology)|product space]] formed from $T_A$ and $T_B$.
+Then $T$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]] {{iff}} $T_A$ and $T_B$ are themselves both [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) spaces]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition === 
+Let $T$ be a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+From:
+:[[Subspace of Product Space Homeomorphic to Factor Space]]
+:[[Separation Properties Preserved in Subspace]]
+:[[Separation Axioms Preserved under Homeomorphism]]
+it follows that $T_A$ and $T_B$ are both $T_0$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Now suppose that $T$ is not $T_0$.
+Then $\exists p, q \in S_A \times S_B$ such that:
+:$p \ne q$
+and:
+:$\forall U \in \tau: \set {p, q} \subseteq U$ or $\set {p, q} \cap U = \O$
+If $p \ne q$, then they are different in at least one coordinate.
+{{WLOG}}, let $p_A \ne q_A$.
+Let $V \in \tau_A$ such that $p_A \in V$.
+Then from [[Projection from Product Topology is Continuous]]:
+:$\pr_A^{-1} \sqbrk V \in \tau$.
+Since $p \in \pr_A^{-1} \sqbrk V$, we have that:
+:$q \in \pr_A^{-1} \sqbrk V$
+From [[Projection is Surjection]]:
+:$\map {\pr_i} q = q_A \in V$
+So:
+:$\forall V \in \tau_A: \set {p_A, q_A} \subseteq V$
+or:
+:$\set {p_A, q_A} \cap V = \O$
+Thus, by definition, $T_A$ is not $T_0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product Space is T1 iff Factor Spaces are T1}
+Tags: T1 Spaces, Product Spaces
+
+\begin{theorem}
+Let $\mathbb S = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] for $\alpha$ in some [[Definition:Indexing Set|indexing set]] $I$.
+Let $\displaystyle T = \struct {S, \tau} = \displaystyle \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the [[Definition:Product Space of Topological Spaces|product space]] of $\mathbb S$.
+Then $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]] {{iff}} each of $\struct {S_\alpha, \tau_\alpha}$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose that for some $\beta$, $\struct {S_\beta, \tau_\beta}$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ space]].
+Then $\exists a, b \in S_\beta$ such that $\forall U_\beta \in \tau_\beta$, $a \in U_\beta \implies b \in U_\beta$.
+Consider the elements $y, z \in S$ defined as:
+:$y = \family {x_\alpha}: x_\alpha = \begin{cases}
+s_\alpha & : \alpha \ne \beta \\
+a & : \alpha = \beta \end{cases}$
+:$z = \family {x_\alpha}: x_\alpha = \begin{cases}
+s_\alpha & : \alpha \ne \beta \\
+b & : \alpha = \beta \end{cases}$
+That is, $y$ and $z$ match on all coordinates except that for $\beta$.
+Let $H \subseteq S: y \in H$ be [[Definition:Open Set (Topology)|open]].
+Then $z \in H$ as $\forall U_\beta \in \map {\pr_\beta} H: a \in U_\beta \implies b \in U_\beta$
+So $T$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose $T$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+Then $\exists a, b \in S, a \ne b$ such that for all $U \in \tau$, $a \in U \implies b \in U$.
+Then $a$ and $b$ are different in at least one coordinate.
+Suppose, $a_\alpha = p, b_\alpha = q$ for some coordinate $\alpha$.
+Then for each $U_\alpha \in \tau_\alpha$:
+:$\displaystyle a \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta \implies b \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta$
+But:
+:$\displaystyle a \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta$ {{iff}} $p \in U_\alpha$
+:$\displaystyle b \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta$ {{iff}} $q \in U_\alpha$
+from which we infer, for each $U_\alpha \in \tau_\alpha$:
+:$p \in U_\alpha \implies q \in U_\alpha$
+It follows that $\struct {S_\alpha, \tau_\alpha}$ is not a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gödel's Incompleteness Theorems/First}
+Tags: Gödel's Incompleteness Theorems
+
+\begin{theorem}
+Let $T$ be the [[Definition:Set|set]] of [[Definition:Theorem of Logic|theorems]] of some [[Definition:Recursive Set|recursive set]] of [[Definition:Sentence|sentences]] in the [[Definition:Language of Arithmetic|language of arithmetic]] such that $T$ contains [[Definition:Minimal Arithmetic|minimal arithmetic]].
+$T$ cannot be both [[Definition:Consistent (Logic)|consistent]] and [[Definition:Complete Theory|complete]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that such a $T$ is [[Definition:Consistent (Logic)|consistent]] and [[Definition:Complete Theory|complete]].
+By the [[Undecidability Theorem]], since $T$ is [[Definition:Consistent (Logic)|consistent]] and contains $Q$, it is not [[Definition:Recursive Set|recursive]].
+But, by [[Complete Recursively Axiomatized Theories are Recursive]], since $T$ is [[Definition:Complete Theory|complete]] and is the set of theorems of a [[Definition:Recursive Set|recursive set]], it is [[Definition:Recursive Set|recursive]].
+The result follows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Undecidability Theorem}
+Tags: Mathematical Logic
+
+\begin{theorem}
+Let $T$ be the set of [[Definition:Theorem of Logic|theorems]] of some [[Definition:Consistent (Logic)|consistent theory]] in the [[Definition:Language of Arithmetic|language of arithmetic]] which contains [[Definition:Minimal Arithmetic|minimal arithmetic]] $Q$.
+$T$ is not [[Definition:Recursive Set|recursive]].
+\end{theorem}
+
+\begin{proof}
+Let $\Theta$ be the set of [[Definition:Gödel Number|Gödel numbers]] of the theorems in $T$.
+We have that $T$ is a [[Definition:Consistent (Logic)|consistent]] extension of $Q$.
+By [[Set of Gödel Numbers of Arithmetic Theorems Not Definable in Arithmetic]], $\Theta$ is not a [[Definition:Definable#Definable Set|definable set]] in $T$.
+Since [[Recursive Sets are Definable in Arithmetic]], this means that $\Theta$ is not [[Definition:Recursive Set|recursive]].
+{{AimForCont}} $T$ were [[Definition:Recursive Set|recursive]].
+Then by [[Gödel Numbering is Recursive]], we could recursively go from $\Theta$ to $T$.
+Then using the recursivity of $T$, we could then recursively go back to $\Theta$.
+Thus $\Theta$ would be [[Definition:Recursive Set|recursive]].
+This would be a [[Definition:Contradiction|contradiction]].
+Thus, by [[Proof by Contradiction]], $T$ is not [[Definition:Recursive Set|recursive]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Gödel Numbers of Arithmetic Theorems Not Definable in Arithmetic}
+Tags: Mathematical Logic
+
+\begin{theorem}
+Let $T$ be the set of [[Definition:Theorem of Logic|theorems]] of some [[Definition:Consistent (Logic)|consistent theory]] in the [[Definition:Language of Arithmetic|language of arithmetic]] which contains [[Definition:Minimal Arithmetic|minimal arithmetic]].
+The set of [[Definition:Gödel Number|Gödel numbers]] of the theorems of $T$ is not [[Definition:Definable#Definable Set|definable]] in $T$.
+\end{theorem}
+
+\begin{proof}
+{{Questionable|Not only the proof is faulty, this theorem is wrong. If we have a recursively enumerable set A of axioms, then the set of theorems (and hence, the Gödel numbers of those) proven by A is recursively enumerable. Minimal arithmetic and PA are both recursively enumerable, and hence their theorems have a recursively enumerable set of Gödel numbers. Any recursively enumerable set is arithmetical with degree Sigma^0_1 in the arithmetical hierarchy. Moreover, there is already a well known provability predicate Pr(x) in the language of Peano arithmetic. Indeed, if we use the diagonal lemma on ~Pr(x) to get a sentence with T proves G iff ~Pr(G), then we will have an unprovable sentence.|Burak}}
+The proof is by contradiction.
+Let $\Theta$ be the set of [[Definition:Gödel Number|Gödel numbers]] of the theorems of $T$.
+Suppose it is defined in $T$ by the formula $\map \theta y$.
+Since $T$ contains $Q$, we may apply the [[Diagonal Lemma]] to $\neg \map \theta y$.
+This gives us a sentence $G$ such that
+:$T \vdash G \leftrightarrow \neg \map \theta {\hat G}$
+where $\hat G$ is the Gödel number of $G$ (more accurately, it is the term in the language of arithmetic obtained by applying the function symbol $s$ to $0$ this many times).
+{{Questionable|The below argument is not correct. $T$ does not necessarily prove that $\neg \map \theta {\hat G}$ just because it does not prove $G$. PA has a provability predicate $\map \Pr x$ which satisfies the property that if PA proves G, then PA proves $\map \Pr {\hat G}$). The step used here assumes a somewhat similar property in the opposite direction that our provability predicate satisfies if T does not prove G, then T proves $\neg \map \theta {\hat G}$, which we did not have as an assumption on $\map \Theta x$. |Burak}}
+Suppose $G$ is not a theorem of $T$.
+:Then the Gödel number of $G$ is not in $\Theta$.
+:Since $\theta$ defines $\Theta$ in $T$, this means that:
+::$T \vdash \neg \map \theta {\hat G}$
+:But, by choice of $G$ (specifically, the bi-implication above), this gives us:
+::$T\vdash G$
+:which contradicts $G$ not being a theorem of $T$
+Thus, $G$ is a theorem of $T$.
+:But, then the Gödel number of $G$ is in $\Theta$, and
+:since $\theta$ defines $\Theta$ in $T$, this means that
+::$T \vdash \map \theta {\hat G}$
+:But, then this gives us
+::$T \vdash \neg G$
+:which contradicts $G$ being a theorem of $T$, ''since $T$ is consistent''.
+Since assuming $\Theta$ was definable in $T$ necessarily leads to a contradiction, we conclude that it is impossible.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Diagonal Lemma}
+Tags: Mathematical Logic
+
+\begin{theorem}
+Let $T$ be the set of [[Definition:Theorem of Logic|theorems]] of some [[Definition:Theory (Logic)|theory]] in the [[Definition:Language of Arithmetic|language of arithmetic]] which contains [[Definition:Minimal Arithmetic|minimal arithmetic]].
+For any [[Definition:Logical Formula|formula]] $\map B y$ in the language of arithmetic, there is a sentence $G$ such that
+:$T \vdash G \leftrightarrow \map B {\hat G}$
+where $\hat G$ is the Gödel number of $G$ (more accurately, it is the term in the language of arithmetic obtained by applying the function symbol $s$ to $0$ this many times).
+{{Disambiguate|Definition:Logical Formula}}
+\end{theorem}
+
+\begin{proof}
+There is a primitive recursive function $\mathrm {diag}$ which is defined by:
+:$\map {\mathrm {diag} } n = \widehat {\map A {\hat A} }$
+where:
+:$\map A x$ is the formula such that $\hat A = n$.
+:the $\hat{ }$ sign denotes the Gödel number of the contained formula (and we are not being formal about distinguishing between integers and symbols in the language).
+Informally, $\mathrm{diag}$ takes a Gödel number, decodes it to a formula, plugs in the Gödel number for that formula in place of a free variable, and encodes this new formula back to a new Gödel number.
+Since $T$ contains $Q$, by [[Recursive Sets are Definable in Arithmetic]] applied to the graph of $\mathrm {diag}$, we have that there is some formula $\map {\mathrm {Diag} } {x, y}$ which [[Definition:Definable#Definable Set|defines]] the graph of $\mathrm {diag}$ in $T$.
+That is:
+:$\map {\mathrm {diag} } n = m$ if and only if $T\vdash \map {\mathrm {Diag} } {n, m}$
+Let $\map A x$ be the formula:
+:$\exists y \paren {\map {\mathrm {Diag} } {x, y} \land \map B y}$
+Let $G$ be $\map A {\hat A}$.
+We then have $T \vdash \map {\mathrm {Diag} } {\hat A, \hat G}$, by checking the definitions.
+Let $T' = T \cup \set G$.
+Then:
+:$T' \vdash \map A {\hat A}$
+Hence:
+:$T' \vdash \exists y \paren {\map {\mathrm {Diag} } {\hat A, y} \land \map B y}$
+But since $\hat G$ is the only number such that $T \vdash \map {\mathrm {Diag} } {\hat A, \hat G}$, this gives us:
+:$T' \vdash \map B {\hat G}$
+Thus:
+:$T \vdash G \rightarrow \map B {\hat G}$
+Let $T' = T \cup \set {\map B {\hat G} }$.
+Again, we have:
+:$T \vdash \map {\mathrm {Diag} } {\hat A, \hat G}$
+so this gives us:
+:$T' \vdash \map {\mathrm {Diag} } {\hat A, \hat G} \land \map B {\hat G}$
+and hence:
+:$T' \vdash \exists y \paren {\map {\mathrm {Diag} } {\hat A, y} \land \map B y}$
+But this is the same thing as:
+:$T' \vdash G$
+Thus:
+:$T \vdash \map B {\hat G} \rightarrow G$
+Thus $G$ is as claimed.
+{{qed}}
+{{MissingLinks|first line of proof}}
+\end{proof}<|endoftext|>
+\section{Tarski's Undefinability Theorem}
+Tags: Mathematical Logic
+
+\begin{theorem}
+Let $\ZZ$ be the standard structure $\struct {\Z, +, \cdot, s, <, 0}$ for the [[Definition:Language of Arithmetic|language of arithmetic]].
+Let $\operatorname {Th}_\ZZ$ be the [[Definition:Sentence|sentences]] which are [[Definition:Satisfiable|true]] in $\ZZ$.
+Let $\Theta$ be the set of [[Definition:Gödel Number|Gödel numbers]] of those sentences in $\operatorname {Th}_\ZZ$.
+$\Theta$ is not [[Definition:Definable#Definable Set|definable]] in $\operatorname {Th}_\ZZ$.
+\end{theorem}
+
+\begin{proof}
+$\operatorname {Th}_\ZZ$ is easily seen to be a [[Definition:Consistent (Logic)|consistent]] extension of [[Definition:Minimal Arithmetic|minimal arithmetic]].  (In fact, the axioms in minimal arithmetic were selected based on the behavior of standard arithmetic.)
+Thus, the theorem is a special case of [[Set of Gödel Numbers of Arithmetic Theorems Not Definable in Arithmetic]] (and can be seen to follow immediately).
+{{qed}}
+{{Namedfor|Alfred Tarski|cat = Tarski}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Limit Point}
+Tags: Limit Points
+
+\begin{theorem}
+{{TFAE|def = Limit Point of Set|view = limit point}}
+That is, let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$.
+Let $H^{\complement}$ denote the [[Definition:Relative Complement|relative complement]] of $H$ in $S$.
+Then the following conditions are [[Definition:Logical Equivalence|equivalent]] for any [[Definition:Element|point]] $x \in S$:
+:$(1): \quad$ Every [[Definition:Open Neighborhood|open neighborhood]] $U$ of $x$ satisfies $H \cap \paren {U \setminus \set x} \ne \O$.
+:$(2): \quad x$ [[Definition:Element|belongs to]] the [[Definition:Closure (Topology)|closure]] of $H$ but is not an [[Definition:Isolated Point (Topology)|isolated point]] of $H$.
+:$(3): \quad x$ is an [[Definition:Adherent Point|adherent point]] of $H$ but is not an [[Definition:Isolated Point (Topology)|isolated point]] of $H$.
+:$(4): \quad H^{\complement} \cup \set x$ is ''not'' a [[Definition:Neighborhood of Point|neighborhood]] of $x$.
+\end{theorem}
+
+\begin{proof}
+=== $({1}) \iff ({2})$ ===
+The [[Definition:Closure (Topology)|closure]] of $H$ is defined as the [[Definition:Set Union|union]] of the set of all [[Definition:Isolated Point (Topology)|isolated points]] of $H$ and the set of all [[Definition:Limit Point of Set|limit points]] of $H$.
+The rest then follows directly from [[Equivalence of Definitions of Isolated Point]].
+{{qed|lemma}}
+=== $({2}) \iff ({3})$ ===
+Follows directly from [[Equivalence of Definitions of Adherent Point]].
+{{qed|lemma}}
+=== $({1}) \iff ({4})$ ===
+The following [[Definition:Logical Equivalence|equivalence]] holds:
+{{begin-eqn}}
+{{eqn | o =
+      | c = There exists an [[Definition:Open Neighborhood|open neighborhood]] $U$ of $x$ such that $H \cap \paren {U \setminus \set x} = \O$
+}}
+{{eqn | o = \leadstoandfrom
+      | c = There exists an [[Definition:Open Neighborhood|open neighborhood]] $U$ of $x$ such that $U \subseteq H^{\complement} \cup \set x$
+      | cc = [[Modus Ponendo Tollens]]
+}}
+{{eqn | o = \leadstoandfrom
+      | c = $H^{\complement} \cup \set x$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$
+      | cc = {{Defof|Neighborhood of Point}}
+}}
+{{end-eqn}}
+The result follows from the [[Rule of Transposition]].
+{{qed}}
+{{refactor|implement $({1}) \iff ({4})$ in house style}}
+=== $({1}) \iff ({4})$, Proof Variant ===
+The following [[Definition:Logical Equivalence|equivalence]] holds:
+There exists an [[Definition:Open Neighborhood|open neighborhood]] $U$ of $x$ such that $H \cap \paren {U \setminus \set x} = \O$
+{{begin-eqn}}
+{{eqn | l = \O
+      | o = =
+      | r = H \cap \paren {U \setminus \set x}
+}}
+{{eqn | ll = \leadstoandfrom
+      | l  = \O
+      | o  = =
+      | r  = \paren {U \cap H} \setminus \set x
+      | cc = [[Intersection with Set Difference is Set Difference with Intersection]]
+}}
+{{eqn | ll = \leadstoandfrom
+      | l  = \O
+      | o  = =
+      | r  = \paren {H \cap U} \setminus \set x
+      | cc = [[Intersection is Commutative]]
+}}
+{{eqn | ll = \leadstoandfrom
+      | l  = \O
+      | o  = =
+      | r  = U \cap \paren {H \setminus \set x}
+      | cc = [[Intersection with Set Difference is Set Difference with Intersection]]
+}}
+{{eqn | ll = \leadstoandfrom
+      | l  = \O
+      | o  = =
+      | r  = U \cap \map \complement { \map \complement {H \setminus \set x} }
+      | cc = [[Complement of Complement]]
+}}
+{{eqn | ll = \leadstoandfrom
+      | l  = U
+      | o  = \subseteq
+      | r  = \map \complement {H \setminus \set x}
+      | cc = [[Intersection with Complement is Empty iff Subset]]
+}}
+{{eqn | ll = \leadstoandfrom
+      | l  = U
+      | o  = \subseteq
+      | r  = \map \complement {H \cap \map \complement {\set x} }
+      | cc = [[Set Difference as Intersection with Complement]]
+}}
+{{eqn | ll = \leadstoandfrom
+      | l  = U
+      | o  = \subseteq
+      | r  = \map \complement H \cup \map \complement {\map \complement {\set x} }
+      | cc = [[De Morgan's Laws (Set Theory)/Set Complement]]
+}}
+{{eqn | ll = \leadstoandfrom
+      | l  = U
+      | o  = \subseteq
+      | r  = \map \complement H \cup \set x
+      | cc = [[Complement of Complement]]
+}}
+{{end-eqn}}
+By {{Defof|Neighborhood of Point}}, $\map \complement H \cup \set x$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$
+The result follows from the [[Rule of Transposition]].
+{{qed|lemma}}
+\end{proof}<|endoftext|>
+\section{Denseness Preserved in Coarser Topology}
+Tags: Denseness
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $T' = \struct {S, \tau'}$ be another [[Definition:Topological Space|topological space]] on $S$ such that $\tau'$ is [[Definition:Coarser Topology|coarser]] than $\tau$.
+Let $H \subseteq S$ be [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+Then $H$ is also [[Definition:Everywhere Dense|everywhere dense]] in $T'$.
+\end{theorem}
+
+\begin{proof}
+Let $H \subseteq S$ be [[Definition:Everywhere Dense|everywhere dense]] in $T$. 
+As $\tau'$ is [[Definition:Coarser Topology|coarser]] than $\tau$, by definition $\tau' \subseteq \tau$.
+Let $x \in S$.
+As $H \subseteq S$ is [[Definition:Everywhere Dense|everywhere dense]] in $T$, either $x \in H$ or $x$ is a [[Definition:Limit Point of Set|limit point of $H$]] in $T$.
+Suppose $x \notin H$.
+Then $x$ is a [[Definition:Limit Point of Set|limit point of $H$]] in $T$.
+Then by definition of [[Definition:Limit Point of Set|limit point of $H$]]:
+:$\forall U \in \tau, x \in U: \exists y \in H: y \in U, y \ne x$ 
+As $\tau' \subseteq \tau$ we have that $U \in \tau' \implies U \in \tau$.
+So:
+:$\forall U \in \tau', x \in U: \exists y \in H: y \in U, y \ne x$ 
+That is, $x$ is a [[Definition:Limit Point of Set|limit point of $H$]] in $T\ '$.
+As $x$ is arbitrary, this follows for all $x \in S, x \notin H$.
+So, for all $x \in S$, either $x \in H$ or $x$ is a [[Definition:Limit Point of Set|limit point of $H$]] in $T'$.
+Thus, by definition, $H \subseteq S$ is [[Definition:Everywhere Dense|everywhere dense]] in $T'$.
+{{qed}}
+[[Category:Denseness]]
+js5kepz4ph9o4ylhmlhwnvpvqnnonr1
+\end{proof}<|endoftext|>
+\section{Rationals are Dense in Compact Complement Topology}
+Tags: Rational Number Space, Compact Complement Topology, Denseness
+
+\begin{theorem}
+Let $T = \struct {\R, \tau^*}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
+Let $\Q$ be the [[Definition:Rational Number|set of rational numbers]].
+Then $\Q$ is [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+\end{theorem}
+
+\begin{proof}
+We have that the [[Compact Complement Topology is Coarser than Euclidean Topology]].
+The result follows from [[Denseness Preserved in Coarser Topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Complement Topology is Separable}
+Tags: Separable Spaces, Compact Complement Topology, Compact Complement Topology is Separable
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Compact Complement Topology|compact complement topology]] on $\R$.
+Then $T$ is a [[Definition:Separable Space|separable space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+: [[Compact Complement Topology is Second-Countable]]
+: [[Second-Countable Space is Separable]]
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that:
+: [[Rationals are Dense in Compact Complement Topology]]
+: [[Rational Numbers are Countably Infinite]]
+Hence the result by definition of [[Definition:Separable Space|separable space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is Locally Connected}
+Tags: Locally Connected Spaces, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $T$ is a [[Definition:Locally Connected Space|locally connected space]].
+\end{theorem}
+
+\begin{proof}
+:[[Either-Or Topology is Locally Path-Connected]]
+:[[Locally Path-Connected Space is Locally Connected]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is not Locally Arc-Connected}
+Tags: Locally Arc-Connected Spaces, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $T$ is not a [[Definition:Locally Arc-Connected Space|locally arc-connected space]].
+\end{theorem}
+
+\begin{proof}
+Let $\BB$ be a [[Definition:Basis (Topology)|basis]] for $\tau$.
+Suppose that $0 \in B$, where $B \in \BB$.
+Then by definition of the [[Definition:Either-Or Topology|either-or topology]], $\openint {-1} 1 \subseteq B$.
+In particular, $\dfrac 1 2 \in B$.
+Let $f: \closedint 0 1 \to B$ be any [[Definition:Injection|injection]] such that:
+:$\map f 0 = 0$
+:$\map f 1 = \dfrac 1 2$
+Since $0 \notin \set {\dfrac 1 2}$, it follows that $\set {\dfrac 1 2}$ is [[Definition:Open Set (Topology)|open]] in $\tau$.
+As $f$ is an [[Definition:Injection|injection]], it follows that:
+:$f^{-1} \sqbrk {\set {\dfrac 1 2} } = \set 1$
+By [[Closed Real Interval is not Open Set]], $\set 1 = \closedint 1 1$ is not [[Definition:Open Set (Topology)|open]] in $\closedint 0 1$.
+Thus, $f$ cannot be [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Hence, no [[Definition:Arc (Topology)|arc]] from $0$ to $\dfrac 1 2$ can exist.
+It follows that $B$ is not [[Definition:Arc-Connected Set|arc-connected]], and hence $T$ is not [[Definition:Locally Arc-Connected Space|locally arc-connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Points in Uncountable Fort Space}
+Tags: Limit Points, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be an [[Definition:Uncountable Fort Space|uncountable Fort space]].
+Let $U \subseteq S$ be a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Subset|subset]] of $S$.
+Then $p$ is the only [[Definition:Limit Point of Set|limit point]] of $U$.
+\end{theorem}
+
+\begin{proof}
+Suppose $y \in S, y \ne p$.
+We have by definition of [[Definition:Fort Space|Fort space]] that $\set y$ is [[Definition:Open Set (Topology)|open]] in $T$.
+So there is no $z \in \set y: z \ne y, z \in U$. 
+Hence $y$ can not be a [[Definition:Limit Point of Set|limit point of $U$]].
+Suppose $p \in V \in \tau_p$ for some $V \subseteq S$.
+From {{Defof|Fort Space}}, $\relcomp S V$ is [[Definition:Finite Set|finite]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \relcomp S {V \setminus \set p}
+      | r = \paren {\relcomp S V} \cup \paren {V \cap \set p}
+      | c = [[Set Difference with Set Difference is Union of Set Difference with Intersection]]
+}}
+{{eqn | o = \subseteq 
+      | r = \paren {\relcomp S V} \cup \set p
+      | c = [[Set Union Preserves Subsets]]: $V \cap \set p \subset \set p$: [[Intersection is Subset]]
+}}
+{{end-eqn}}
+Thus $\relcomp S {V \setminus \set p}$ is also [[Definition:Finite Set|finite]].
+Since $U$ is [[Definition:Countably Infinite Set|countably infinite]]:
+:$U \not \subseteq \relcomp S {V \setminus \set p}$.
+From [[Empty Intersection iff Subset of Complement]]:
+:$U \cap \paren {V \setminus \set p} \ne \O$.
+Since $V$ is arbitrary, every [[Definition:Open Neighborhood|open neighborhood]] of $p$ has non-empty intersection with $U$.
+Thus $p$ is the only [[Definition:Limit Point of Set|limit point]] of $U$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fort Space is Scattered}
+Tags: Fort Space is Scattered, Scattered Spaces, Fort Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Then $T$ is a [[Definition:Scattered Space|scattered space]].
+\end{theorem}
+
+\begin{proof}
+Let $H \subseteq T$ such that $H \ne \varnothing$ and $H \ne \set p$.
+Then $\exists x \in H: x \ne p$.
+From [[Clopen Points in Fort Space]], every point of $T$ apart from $p$ is [[Definition:Open Point|open]] in $T$.
+So $\set x$ is an [[Definition:Open Set (Topology)|open set]] of $T$.
+So $H \cap \set x = \set x$ and so $x$ is [[Definition:Isolated Point of Subset|isolated in $H$]].
+Thus $H$ contains at least one point which is [[Definition:Isolated Point of Subset|isolated in $H$]].
+On the other hand, suppose $H = \set p$.
+From [[Singleton Point is Isolated]], $p$ is an [[Definition:Isolated Point of Subset|isolated point in $\set p = H$]].
+So again $H$ contains at least one point which is [[Definition:Isolated Point of Subset|isolated in $H$]].
+So for all $H \subseteq S$ we have that $H$ contains at least one point which is [[Definition:Isolated Point of Subset|isolated in $H$]].
+Hence the result, by definition of [[Definition:Scattered Space|scattered space]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Suppose that $H \subseteq T$ has no [[Definition:Isolated Point of Subset|isolated points of $H$]].
+So, by definition, $H$ is [[Definition:Dense-in-itself|dense in itself]].
+We have that:
+:a [[Fort Space is T1|Fort Space is $T_1$]]
+:a [[Dense-in-itself Subset of T1 Space is Infinite|Dense-in-itself Subset of $T_1$ Space is Infinite]].
+So $H$ is [[Definition:Infinite Set|infinite]], and so contains more than one point.
+So $\exists q \in H: q \ne p$.
+But, from [[Clopen Points in Fort Space]], $\set q$ is [[Definition:Open Set (Topology)|open]] in $T$.
+So $H \cap \set q = \set q$ and so by definition $q$ is [[Definition:Isolated Point of Subset|isolated in $H$]].
+From this contradiction it follows that $H$ is not [[Definition:Dense-in-itself|dense in itself]] and contains at least one [[Definition:Isolated Point of Subset|isolated point]].
+Hence the result, by definition of [[Definition:Scattered Space|scattered space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dense-in-itself Subset of T1 Space is Infinite}
+Tags: Denseness, T1 Spaces, Infinite Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet)]].
+Let $H \subseteq T$ be [[Definition:Dense-in-itself|dense-in-itself]].
+Then $H$ is [[Definition:Infinite Set|infinite]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Contradiction]]:
+{{AimForCont}} $H$ is [[Definition:Finite Set|finite]].
+From [[Finite T1 Space is Discrete|Finite $T_1$ Space is Discrete]], $H$ has the [[Definition:Discrete Topology|discrete topology]].
+From [[Discrete Space is not Dense-In-Itself]] it then follows that $H$ can not be [[Definition:Dense-in-itself|dense-in-itself]].
+So for $H$ to be [[Definition:Dense-in-itself|dense-in-itself]], it must be [[Definition:Infinite Set|infinite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equal Chords in Circle}
+Tags: Circles
+
+\begin{theorem}
+In a [[Definition:Circle|circle]], equal [[Definition:Chord of Circle|chords]] are equally distant from the [[Definition:Center of Circle|center]], and [[Definition:Chord of Circle|chords]] that are equally distant from the [[Definition:Center of Circle|center]] are equal in [[Definition:Length (Linear Measure)|length]].
+{{:Euclid:Proposition/III/14}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Euclid-III-14.png|300px]]
+Let $ABDC$ be a [[Definition:Circle|circle]], and let $AB$ and $CD$ be equal [[Definition:Chord of Circle|chords]] on it.
+From [[Finding Center of Circle]], let $E$ be the [[Definition:Center of Circle|center]] of $ABDC$.
+Construct $EF$, $EG$ [[Definition:Perpendicular|perpendicular]] to $AB$ and $CD$, and join $AE$ and $EC$.
+From [[Conditions for Diameter to be Perpendicular Bisector]], $EF$ [[Definition:Bisect|bisects]] $AD$.
+So $AF = FB$ and so $AB = 2 AF$.
+Similarly $CD = 2CG$.
+As $AB = CD$ it follows that $AE = CG$.
+Since $AE = EC$ the [[Definition:Square (Geometry)|square]] on $AE$ equals the [[Definition:Square (Geometry)|square]] on $EC$.
+We have that $\angle AFE$ is a [[Definition:Right Angle|right angle]].
+So from [[Pythagoras's Theorem]] the [[Definition:Square (Geometry)|square]] on $AE$ equals the sum of the [[Definition:Square (Geometry)|squares]] on $AF$ and $FE$.
+Similarly, the [[Definition:Square (Geometry)|square]] on $EC$ equals the sum of the [[Definition:Square (Geometry)|squares]] on $EG$ and $GC$.
+Thus the [[Definition:Square (Geometry)|square]] on $EF$ equals the [[Definition:Square (Geometry)|square]] on $EG$.
+So $EF = EG$.
+So from [[Definition:Euclid's Definitions - Book III/4 - Equally Distant in Circle|Definition III: 4]], [[Definition:Chord of Circle|chords]] are at equal distance from the center when the perpendiculars drawn to them from the center are equal.
+So $AB$ and $CD$ are at equal distance from the [[Definition:Center of Circle|center]].
+{{qed|lemma}}
+Now suppose that $AB$ and $CD$ are at equal distance from the [[Definition:Center of Circle|center]].
+That is, let $EF = EG$.
+Using the same construction as above, it is proved similarly that $AB = 2AF$ and $CD = 2CG$.
+Since $AE = CE$, the [[Definition:Square (Geometry)|square]] on $AE$ equals the [[Definition:Square (Geometry)|square]] on $CE$.
+From [[Pythagoras's Theorem]]:
+: the [[Definition:Square (Geometry)|square]] on $AE$ equals the sum of the [[Definition:Square (Geometry)|squares]] on $AF$ and $FE$
+: the [[Definition:Square (Geometry)|square]] on $EC$ equals the sum of the [[Definition:Square (Geometry)|squares]] on $EG$ and $GC$.
+So the sum of the [[Definition:Square (Geometry)|squares]] on $AF$ and $FE$ equals the sum of the [[Definition:Square (Geometry)|squares]] on $EG$ and $GC$.
+But the [[Definition:Square (Geometry)|square]] on $FE$ equals the [[Definition:Square (Geometry)|square]] on $EG$.
+So the [[Definition:Square (Geometry)|square]] on $AF$ equals the [[Definition:Square (Geometry)|square]] on $CG$, and so $AF = CG$.
+As $AB = 2AF$ and $CD = 2CG$ it follows that $AB = CD$.
+Hence the result.
+{{qed}}
+{{Euclid Note|14|III}}
+\end{proof}<|endoftext|>
+\section{Relative Lengths of Chords of Circles}
+Tags: Circles
+
+\begin{theorem}
+Of [[Definition:Chord of Circle|chords in a circle]], the [[Definition:Diameter of Circle|diameter]] is the greatest, and of the rest the nearer to the [[Definition:Center of Circle|center]] is always greater than the more remote.
+{{:Euclid:Proposition/III/15}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Euclid-III-15.png|300px]]
+Let $ABCD$ be a [[Definition:Circle|clrcle]], let $AD$ be its [[Definition:Diameter of Circle|diameter]] and $E$ the [[Definition:Center of Circle|center]].
+Let $BC$ and $FG$ be [[Definition:Chord of Circle|chords]] of $ABCD$, where $BC$ is nearer to the [[Definition:Center of Circle|center]] than $FG$.
+Let $EH$ and $EK$ be drawn [[Definition:Perpendicular|perpendicular]] to $BC$ and $FG$ respectively.
+Because $BC$ is nearer to the [[Definition:Center of Circle|center]] than $FG$, it follows from [[Definition:Euclid's Definitions - Book III/5 - Greater Distance in Circle|Book III: Definition 5]] that $EK$ is longer than $EH$.
+Position $L$ on $EK$ so that $EL = EH$, and draw $MN$ through $L$ [[Definition:Perpendicular|perpendicular]] to $FG$.
+Join $EM$, $EN$, $EF$ and $EG$.
+Since $EH = EL$ it follows from [[Equal Chords in Circle]] that $BC = MN$.
+Also, since $AE = EM$ and $ED = EN$, $AD = ME + EN$.
+But $ME + EN > MN$ from [[Sum of Two Sides of Triangle Greater than Third Side]], and $MN = BC$.
+So $AD > BC$.
+Thus we have that the [[Definition:Diameter of Circle|diameter]] is greater than any other [[Definition:Chord of Circle|chord of a circle]].
+{{qed|lemma}}
+Since $ME = FE$ and $EN = EG$ and $\angle MEN > \angle FEG$, from the [[Hinge Theorem]] it follows that $MN > FG$.
+But $MN = BC$.
+So a [[Definition:Chord of Circle|chord]] nearer the [[Definition:Center of Circle|center]] is greater than one more remote.
+{{qed}}
+{{Euclid Note|15|III}}
+\end{proof}<|endoftext|>
+\section{Construction of Tangent from Point to Circle}
+Tags: Circles, Construction of Tangent from Point to Circle
+
+\begin{theorem}
+From a given [[Definition:Point|point]] outside a given [[Definition:Circle|circle]], it is possible to draw a [[Definition:Tangent Line|tangent]] to that [[Definition:Circle|circle]].
+{{:Euclid:Proposition/III/17}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Euclid-III-17.png|400px]]
+Let $A$ be the given [[Definition:Point|point]] and let $BCD$ be the given [[Definition:Circle|circle]].
+It is required that a [[Definition:Straight Line|straight line]] be drawn from $A$ to $BCD$.
+Let [[Finding Center of Circle|the center $E$ of $BCD$ be found]].
+Join $AE$ and draw the [[Definition:Circle|circle]] $AFG$ with center $E$ and [[Definition:Radius of Circle|radius]] $AE$.
+From $D$ [[Construction of Perpendicular Line|let $DF$ be drawn perpendicular to $EA$]].
+Join $EF$ and let $B$ be the point at which $EF$ joins the [[Definition:Circle|circle]] $BCD$.
+Join $AB$.
+Then $AB$ is the required [[Definition:Tangent Line|tangent]] to $BCD$.
+=== Proof of Construction ===
+Since $E$ is the [[Definition:Center of Circle|center]] of the [[Definition:Circle|circles]] $BCD$ and $AFG$, $EA = EF$ and $ED = EB$.
+Therefore the two [[Definition:Side of Polygon|sides]] $AE, EB$ equal the two sides $FE, ED$, and they contain a common [[Definition:Angle|angle]] at $E$.
+So by [[Triangle Side-Angle-Side Equality]], $\triangle DEF = \triangle BEA$.
+Therefore $\angle EDF = \angle EBA$.
+But $\angle EDF$ is a [[Definition:Right Angle|right angle]], so $\angle EBA$ is also a [[Definition:Right Angle|right angle]].
+But $EB$ is a [[Definition:Radius of Circle|radius]] of [[Definition:Circle|circle]] $BCD$.
+It follows from the [[Line at Right Angles to Diameter of Circle/Porism|porism to Line at Right Angles to Diameter of Circle]] that $AB$ is [[Definition:Tangent Line|tangent]] to $BCD$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+:[[File:Euclid-III-17a.png|400px]]
+Let $BCD$ with [[Definition:Center of Circle|center]] $A$ be the [[Definition:Circle|circle]], and let $E$ be the exterior point from which a [[Definition:Tangent Line|tangent]] is to be drawn.
+[[Bisection of Straight Line|Bisect]] $AE$ at $F$.
+Then draw a [[Definition:Circle|circle]] $AEG$ whose [[Definition:Center of Circle|center]] is $F$ and whose [[Definition:Radius of Circle|radius]] is $AF$.
+The point $G$ is where $AEG$ intersects $BCD$.
+The [[Definition:Straight Line|line]] $EG$ is the required [[Definition:Tangent Line|tangent]].
+=== Proof of Construction ===
+By the method of construction, $AE$ is the [[Definition:Diameter of Circle|diameter]] of $AEG$.
+By [[Thales' Theorem]] $\angle AGE$ is a [[Definition:Right Angle|right angle]].
+But $AG$ is a [[Definition:Radius of Circle|radius]] of $BCD$.
+The result follows from [[Radius at Right Angle to Tangent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Construction of Circle from Segment}
+Tags: Circles, Construction of Circle from Segment
+
+\begin{theorem}
+{{:Euclid:Proposition/III/25}}
+\end{theorem}
+
+\begin{proof}
+Let $ABC$ be the given [[Definition:Segment of Circle|segment of a circle]] whose [[Definition:Base of Segment|base]] is $AC$.
+Bisect $AC$ at $D$, draw $DB$ [[Definition:Perpendicular|perpendicular]] to $AC$, and join $AB$.
+First suppose that $ABC$ is such that $\angle ABD > \angle BAD$:
+:[[File:Euclid-III-25a.png|250px]]
+On $BA$, construct $\angle BAE$ equal to $\angle ABD$.
+Join $BD$ through to $E$ and join $EC$.
+Then $E$ is the [[Definition:Center of Circle|center]] of the required [[Definition:Circle|circle]].
+Second, suppose that $ABC$ is such that $\angle ABD = \angle BAD$:
+:[[File:Euclid-III-25b.png|250px]]
+Then $D$ is the [[Definition:Center of Circle|center]] of the required [[Definition:Circle|circle]].
+Finally, suppose that $ABC$ is such that $\angle ABD < \angle BAD$:
+:[[File:Euclid-III-25c.png|250px]]
+On $BA$, construct $\angle BAE$ equal to $\angle ABD$.
+The point $E$, which falls on $BD$, is the [[Definition:Center of Circle|center]] of the required [[Definition:Circle|circle]].
+=== Proof of Construction ===
+First suppose that $ABC$ is such that $\angle ABD > \angle BAD$:
+:[[File:Euclid-III-25a.png|250px]]
+Since $\angle ABE = \angle BAE$, from [[Triangle with Two Equal Angles is Isosceles]] we have that $EB = EA$.
+Since $AD = DC$ and $DE$ is common, and $\angle ADE$ is a [[Definition:Right Angle|right angle]], by [[Triangle Side-Angle-Side Equality]] we have that $\triangle ADE = \triangle CDE$.
+Hence $AE = CE$ both of which are equal to $BE$ from above.
+So from [[Condition for Point to be Center of Circle]] $E$ is the [[Definition:Center of Circle|center]] of the required [[Definition:Circle|circle]].
+Second, suppose that $ABC$ is such that $\angle ABD = \angle BAD$:
+:[[File:Euclid-III-25b.png|250px]]
+From [[Triangle with Two Equal Angles is Isosceles]] we have that $AD = DB$ and so also equal to $DC$.
+So from [[Condition for Point to be Center of Circle]] $D$ is the [[Definition:Center of Circle|center]] of the required [[Definition:Circle|circle]].
+Incidentally, note that in this case [[Definition:Segment of Circle|segment]] $ABC$ is actually a [[Definition:Semicircle|semicircle]].
+Finally, suppose that $ABC$ is such that $\angle ABD < \angle BAD$:
+:[[File:Euclid-III-25c.png|250px]]
+The same proof applies:
+Since $\angle ABE = \angle BAE$, from [[Triangle with Two Equal Angles is Isosceles]] we have that $EB = EA$.
+Since $AD = DC$ and $DE$ is common, and $\angle ADE$ is a [[Definition:Right Angle|right angle]], by [[Triangle Side-Angle-Side Equality]] we have that $\triangle ADE = \triangle CDE$.
+Hence $AE = CE$ both of which are equal to $BE$ from above.
+So from [[Condition for Point to be Center of Circle]] $E$ is the [[Definition:Center of Circle|center]] of the required [[Definition:Circle|circle]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+:[[File:Euclid-III-25d.png|250px]]
+Choose any point $C$ on the [[Definition:Circumference of Circle|circumference]].
+[[Definition:Bisect|Bisect]] $AC$ at $D$ and $BC$ at $E$ and construct a [[Definition:Perpendicular|perpendicular]] $DF$ and $EF$ from each through the point of bisection.
+The point of intersection $F$ is the [[Definition:Center of Circle|center]] of the required [[Definition:Circle|circle]].
+$AFC$ and $CFB$ are [[Definition:Isosceles Triangle|isosceles triangles]] and so $AF, CF$ and $BF$ are all equal.
+The result follows from [[Condition for Point to be Center of Circle]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equal Angles in Equal Circles}
+Tags: Circles
+
+\begin{theorem}
+In equal [[Definition:Circle|circles]], equal [[Definition:Angle|angles]] stand on equal [[Definition:Arc of Circle|arcs]], whether at the [[Definition:Center of Circle|center]] or at the [[Definition:Circumference of Circle|circumference]] of those circles.
+{{:Euclid:Proposition/III/26}}
+\end{theorem}
+
+\begin{proof}
+Let $ABC$ and $DEF$ be equal [[Definition:Circle|circles]].
+Let $\angle BGC = \angle EHF$ and $\angle BAC = \angle EDF$.
+:[[File:Euclid-III-26.png|500px]]
+Let $BC$ and $EF$ be joined.
+Since the [[Definition:Circle|circles]] $ABC$ and $DEF$ are equal, their [[Definition:Radius of Circle|radii]] are equal.
+So $BG = EH$ and $CG = FH$.
+We also have [[Definition:By Hypothesis|by hypothesis]] that $\angle BGC = \angle EHF$.
+So from [[Triangle Side-Angle-Side Equality]] it follows that $BC = EF$.
+Since $\angle BAC = \angle EDF$ we have from {{EuclidDefLink|III|11|Similar Segments}} that [[Definition:Segment of Circle|segment]] $BAC$ is [[Definition:Similar Segments|similar]] to [[Definition:Segment of Circle|segment]] $EDF$.
+Moreover, these [[Definition:Segment of Circle|segments]] have equal [[Definition:Base of Segment|bases]].
+So from [[Similar Segments on Equal Bases are Equal]], [[Definition:Segment of Circle|segment]] $BAC$ is equal to [[Definition:Segment of Circle|segment]] $EDF$. 
+But as $ABC$ and $DEF$ are equal [[Definition:Circle|circles]], it follows that [[Definition:Arc of Circle|arc]] $BKC$ equals [[Definition:Arc of Circle|arc]] $ELF$.
+{{qed}}
+{{Euclid Note|26|III|{{EuclidNoteConverse|prop=27|title=Angles on Equal Arcs are Equal}}}}
+\end{proof}<|endoftext|>
+\section{Angles on Equal Arcs are Equal}
+Tags: Circles
+
+\begin{theorem}
+In equal [[Definition:Circle|circles]], [[Definition:Angle|angles]] standing on equal [[Definition:Arc of Circle|arcs]] are equal to one another, whether at the [[Definition:Center of Circle|center]] or at the [[Definition:Circumference of Circle|circumference]] of those [[Definition:Circle|circles]].
+{{:Euclid:Proposition/III/27}}
+\end{theorem}
+
+\begin{proof}
+Let $ABC$ and $DEF$ be equal [[Definition:Circle|circles]].
+Let $\angle BGC$ and $\angle EHF$ stand on the equal [[Definition:Arc of Circle|arcs]] $BC$ and $EF$.
+:[[File:Euclid-III-27.png|500px]]
+Suppose $\angle BGC \ne \angle EHF$.
+Then one of them is bigger.
+Suppose $\angle BGC > \angle EHF$.
+On the [[Definition:Straight Line|straight line]] $BG$, [[Construction of Equal Angle|construct $\angle BGK$ equal to $\angle EHF$]].
+From [[Equal Angles in Equal Circles]], these [[Definition:Angle|angles]] stand on equal [[Definition:Arc of Circle|arcs]].
+So [[Definition:Arc of Circle|arcs]] $BK = EF$.
+But $EF = BC$ and so $BK = BC$, which is impossible.
+So $\angle BGC$ and $\angle EHF$ are not unequal, therefore $\angle BGC = \angle EHF$.
+From the [[Inscribed Angle Theorem]], $2 \angle BAC = \angle BGC$ and $2 \angle EDF = \angle EHF$.
+Therefore $2 \angle BAC = \angle EHF$.
+{{qed}}
+{{Euclid Note|27|III|{{EuclidNoteConverse|prop=26|title=Equal Angles in Equal Circles}}}}
+\end{proof}<|endoftext|>
+\section{Bisection of Arc}
+Tags: Circles
+
+\begin{theorem}
+It is possible to [[Definition:Bisection|bisect]] an [[Definition:Arc of Circle|arc of a circle]].
+{{:Euclid:Proposition/III/30}}
+\end{theorem}
+
+\begin{proof}
+Join $AD$ and $BD$.
+We have that $AC = CB$ and $CD$ is common.
+We also have that $\angle ACD = \angle BCD$ as they are both [[Definition:Right Angle|right angles]].
+So from [[Triangle Side-Angle-Side Equality]] $\triangle ACD = \triangle BCD$ and so $AD = BD$.
+But from [[Straight Lines Cut Off Equal Arcs in Equal Circles]], the [[Definition:Arc of Circle|arc]] $AD$ equals the [[Definition:Arc of Circle|arc]] $BD$.
+That is, the [[Definition:Arc of Circle|arc]] $AB$ has been [[Definition:Bisection|bisected]] at $D$.
+{{qed}}
+{{Euclid Note|30|III}}
+\end{proof}<|endoftext|>
+\section{Relative Sizes of Angles in Segments}
+Tags: Circles
+
+\begin{theorem}
+In a [[Definition:Circle|circle]]:
+: the [[Definition:Angle in Segment|angle]] in a [[Definition:Semicircle|semicircle]] is [[Definition:Right Angle|right]]
+: the [[Definition:Angle in Segment|angle]] in a [[Definition:Segment of Circle|segment]] greater than a [[Definition:Semicircle|semicircle]] is [[Definition:Acute Angle|acute]]
+: the [[Definition:Angle in Segment|angle]] in a [[Definition:Segment of Circle|segment]] less than a [[Definition:Semicircle|semicircle]] is [[Definition:Obtuse Angle|obtuse]].
+Further:
+: the [[Definition:Angle of Segment|angle]] of a [[Definition:Segment of Circle|segment]] greater than a [[Definition:Semicircle|semicircle]] is [[Definition:Obtuse Angle|obtuse]]
+: the [[Definition:Angle of Segment|angle]] of a [[Definition:Segment of Circle|segment]] less than a [[Definition:Semicircle|semicircle]] is [[Definition:Acute Angle|acute]].
+{{:Euclid:Proposition/III/31}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Euclid-III-31.png|300px]]
+Let $ABCD$ be a [[Definition:Circle|circle]] whose [[Definition:Diameter of Circle|diameter]] is $BC$ and whose [[Definition:Center of Circle|center]] is $E$.
+Join $AB$, $AC$, $AD$, $DC$ and $AE$.
+Let $BA$ be [[Definition:Production|produced]] to $F$.
+Since $BE = EA$, from [[Isosceles Triangle has Two Equal Angles]] it follows that $\angle ABE = \angle BAE$.
+Since $CE = EA$, from [[Isosceles Triangle has Two Equal Angles]] it follows that $\angle ACE = \angle CAE$.
+So from $\angle BAC = \angle ABE + \angle ACE = \angle ABC + \angle ACB$.
+But from [[Sum of Angles of Triangle Equals Two Right Angles]] $\angle FAC = \angle ABC + \angle ACB$.
+So $\angle BAC = \angle FAC$, and so from [[Definition:Euclid's Definitions - Book I/10 - Right Angle|Book I Definition 10]] each one is a [[Definition:Right Angle|right angle]].
+So the [[Definition:Angle in Segment|angle]] in the [[Definition:Semicircle|semicircle]] $BAC$ is a [[Definition:Right Angle|right angle]].
+{{qed|lemma}}
+From [[Two Angles of Triangle Less than Two Right Angles]], in $\triangle ABC$, $\angle ABC + \angle BAC$ is less than two [[Definition:Right Angle|right angles]].
+As $\angle BAC$ is a [[Definition:Right Angle|right angle]], it follows that $\angle ABC$ is less than a [[Definition:Right Angle|right angle]].
+It is also the [[Definition:Angle in Segment|angle]] in a [[Definition:Segment of Circle|segment]] $ABC$ greater than a [[Definition:Semicircle|semicircle]].
+Therefore the [[Definition:Angle in Segment|angle]] in a [[Definition:Segment of Circle|segment]] greater than a [[Definition:Semicircle|semicircle]] is [[Definition:Acute Angle|acute]].
+{{qed|lemma}}
+We have that $ABCD$ is a [[Definition:Cyclic Quadrilateral|cyclic quadrilateral]].
+From [[Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles]], $\angle ABC + \angle ADC$ equals two [[Definition:Right Angle|right angles]].
+As $\angle ABC$ is less than a [[Definition:Right Angle|right angle]], it follows that $\angle ADC$ is greater than a [[Definition:Right Angle|right angle]].
+But $\angle ADC$ is the [[Definition:Angle in Segment|angle]] in a [[Definition:Segment of Circle|segment]] $ADC$ less than a [[Definition:Semicircle|semicircle]].
+Therefore the [[Definition:Angle in Segment|angle]] in a [[Definition:Segment of Circle|segment]] less than a [[Definition:Semicircle|semicircle]] is [[Definition:Obtuse Angle|obtuse]].
+{{qed|lemma}}
+Since the [[Definition:Angle in Segment|angle]] contained by the [[Definition:Straight Line|straight lines]] $BA$ and $AC$ is a [[Definition:Right Angle|right angle]], the [[Definition:Angle in Segment|angle]] contained by the [[Definition:Arc of Circle|arc]] $ABC$ and the [[Definition:Straight Line|straight line]] $AC$ is greater than a [[Definition:Right Angle|right angle]].
+So the [[Definition:Angle of Segment|angle]] of a [[Definition:Segment of Circle|segment]] greater than a [[Definition:Semicircle|semicircle]] is [[Definition:Obtuse Angle|obtuse]].
+{{qed|lemma}}
+Since the [[Definition:Angle in Segment|angle]] contained by the [[Definition:Straight Line|straight lines]] $AC$ and $AF$ is a [[Definition:Right Angle|right angle]], the [[Definition:Angle in Segment|angle]] contained by the [[Definition:Arc of Circle|arc]] $ADC$ and the [[Definition:Straight Line|straight line]] $AC$ is less than a [[Definition:Right Angle|right angle]].
+So the [[Definition:Angle of Segment|angle]] of a [[Definition:Segment of Circle|segment]] less than a [[Definition:Semicircle|semicircle]] is [[Definition:Acute Angle|acute]].
+{{qed}}
+{{Euclid Note|31|III}}
+\end{proof}<|endoftext|>
+\section{Angles made by Chord with Tangent}
+Tags: Circles
+
+\begin{theorem}
+Let $EF$ be a [[Definition:Tangent Line|tangent]] to a [[Definition:Circle|circle]] $ABCD$, touching it at $B$.
+Let $BD$ be a [[Definition:Chord of Circle|chord]] of $ABCD$.
+Then:
+: the [[Definition:Angle in Segment|angle]] in [[Definition:Segment of Circle|segment]] $BCD$ equals $\angle DBE$
+and: 
+: the [[Definition:Angle in Segment|angle]] in [[Definition:Segment of Circle|segment]] $BAD$ equals $\angle DBF$.
+{{:Euclid:Proposition/III/32}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Euclid-III-32.png|250px]]
+Draw $BA$ [[Definition:Perpendicular|perpendicular]] to $EF$ through $B$.
+Let $C$ be selected on the [[Definition:Circle|circle]] on the [[Definition:Arc of Circle|arc]] $BD$.
+Join $AD, DC, CB$.
+From [[Right Angle to Tangent of Circle goes through Center]], the [[Definition:Center of Circle|center]] of the circle lies on $AB$.
+By definition, then, $AB$ is a [[Definition:Diameter of Circle|diameter]] of the circle.
+From [[Relative Sizes of Angles in Segments]], it follows that $\angle ADB$ is a [[Definition:Right Angle|right angle]].
+Therefore from [[Sum of Angles of Triangle Equals Two Right Angles]] $\angle BAD + \angle ABD$ equals a [[Definition:Right Angle|right angle]].
+But $\angle ABF$ is also a [[Definition:Right Angle|right angle]].
+So $\angle ABF = \angle BAD + \angle ABD$
+Subtracting $\angle ABD$ from each, it follows that $\angle DBF = \angle BAD$.
+{{qed|lemma}}
+Next we have that $ABCD$ is a [[Definition:Cyclic Quadrilateral|cyclic quadrilateral]].
+From [[Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles]] $\angle BAD + \angle BCD$ equals two [[Definition:Right Angle|right angles]].
+But from [[Two Angles on Straight Line make Two Right Angles]], so does $\angle DBE + \angle DBF$.
+But as $\angle BAD = \angle DBF$ it follows that $\angle BCD = \angle DBE$.
+{{qed}}
+{{Euclid Note|32|III}}
+\end{proof}
\ No newline at end of file