TITLE: If $S$ is an infinite $\sigma$ algebra on $X$ then $S$ is not countable
QUESTION [94 upvotes]: I am going over a tutorial in my real analysis course. There is
an proof in which I don't understand some parts of it.
The proof relates to the following proposition:
($S$ - infinite $\sigma$-algebra on $X$) $\implies $ $S$ is uncountable.
Proof:

Assume: $S=\{A_{i}\}_{i=1}^{+\infty}$. $\forall x\in X:
B_{x}:=\cap_{x\in A_{i}}A_{i}$. [Note: $B_{x}\in S$ $\impliedby$ ($B_{x}$ - countable intersection].
Lemma: $B_{x}\cap B_{y}\neq\emptyset \implies B_{x}=B_{y}$.
Proof(of lemma):
$z\in B_{x}\cap B_{y} \implies B_{z}\subseteq B_{x}\cap B_{y}$.
1.$x\not\in B_{z} \implies x\in B_{x}\setminus B_{z} \wedge B_{x}\setminus B_{z} \subset S \wedge  B_{x}\setminus B_{z} \subset B_{x}$
(contradiction:$\space$ definition of $B_{x}$) $\implies$ $B_{z}=B_{x}$
2.$y\not\in B_{z} \implies y\in B_{y} \setminus B_{z} \space \wedge \space B_{y}\setminus B_{z} \subset S \space\wedge\space B_{y} \setminus B_{z}\subset B_{y} $(contradiction: definition of $B_{y}$) $\implies$ $B_{z}=B_{y}$ $\implies B_{x}=B_{y} \space \square$
Consider: $\{B_{x}\}_{x\in X}$. If: there are finite sets of the
form $B_{x}$ then: $S$ is a union of a finite number of disjoint sets
$\implies$ $S$ is finite $\implies$ there is an infinite number of sets of
the form $B_{x}$. $\implies$ $|\bigcup\limits_{i\in A \subseteq\mathbb{N}}B_{x_{i}}| \geq \aleph_{0}$.(contradiction) $\square$

There are couple of things I don't understand in this proof:

Why the fact that we found a set ($B_{x}\setminus B_{z}$) in $S$
containing $x$ and is strictly contained in $B_{x}$ a contradiction
?

Why if there are only a finite number of different sets of the
form $B_{x}$ then $S$ is a union of a finite number of disjoint
sets and is finite ?

REPLY [4 votes]: The proof above is terse. I've filled out the missing bits, any error corrections are greatly appreciated. 
What we want to prove: A sigma algebra $\mathcal{F}$  of set $X$ is either finite or uncountable.
Proof:
First suppose $X$ is finite. Then, the largest sigma algebra of $X$, $\mathcal{P}(X)$, is finite. Now, suppose $X$ is infinite. We assume that $\mathcal{F}$ is countably infinite, and derive a contradiction. 
As $\mathcal{F}$ is countably infinite, denote $\mathcal{F} = \{A_i : i \in \mathbb{N} \}$. For $\forall x \in X$, define $B_x := \cap_{x \in A_i} A_i$. As $\mathcal{F}$ is countable, note that $B_x$ is a countable intersection of elements of $\mathcal{F}$, so is an element of $\mathcal{F}$. Therefore, for $\forall x, y,z \in X$, $B_x \cap B_y \in \mathcal{F}$, and $B_x \backslash B_z \in \mathcal{F}$.
I claim that $B_x \cap B_y \neq \emptyset \implies B_x = B_y$. As $B_x \cap B_y \in \mathcal{F}$, $\exists i_0$ such that $B_x \cap B_y = A_{i_0}$. Let $z \in B_x \cap B_y$. Then, 
$$B_z = \cap_{z \in A_i} A_i =  \left( \cap_{\substack{z \in A_i \\ i\neq i_0}} A_i\right) \cap A_{i_0} \subset A_{i_0} = B_x \cap B_y.$$
Suppose that $ x \notin B_z$. Then, $x \in B_x \backslash B_z \in \mathcal{F}$. This is a contradiction because as $z \in B_x, B_z$, we know $B_x \backslash B_z$ is strictly smaller than $B_x$. However, $B_x$ is defined to be the intersection of all sets in $\mathcal{F}$ containing $x$, so is the `smallest' set in $\mathcal{F}$ containing $x$ which is also a subset of all other sets in $\mathcal{F}$ containing $x$. Therefore, we conclude $x \in B_z$. As $B_z$ also contains $x$, we obtain $B_x \subset B_z$.
Thus, we have
$$
B_x \subset B_z \subset B_x \cap B_y \subset B_x.
$$
We can conclude $B_x = B_z = B_x \cap B_y$. By symmetry, $B_y = B_z = B_x \cap B_y$. Thus, $B_x = B_y$, and our claim is proven. 
Let $B = \{ B_x\}_{x \in X}$. If $|B| = n < \infty$, then we have that $\mathcal{F}$ is finite. This is because for any $A \in \mathcal{F}$, $A$ can be expressed $A = \cup_{a \in A} B_a$ (as $B_a \subset A,$ $\forall a \in A$, and $B_x$ exists and is non-empty for $\forall x \in X$ as $x \in X \in \mathcal{F}$). So, there can exist at most $2^n$ elements of $\mathcal{F}$, a contradiction to the assumption that  $\mathcal{F}$ is countably infinite. Now assume that $B$ is uncountably infinite. As we established $B_x \in \mathcal{F}, \forall x \in X$ (so $B \subset \mathcal{F}$), we have that $\mathcal{F}$ is uncountably infinite, a contradiction to the assumption that  $\mathcal{F}$ is countably infinite.
Finally, assume that $B$ is countably infinite. Using the claim, we can let $B = \{ C_i \}_{i \in \mathbb{N}} \cup C_0$ where $C_0 = \emptyset$, and $C_i$ are disjoint non-empty sets such that $C_i = C_j \implies i = j$, $\forall i, j \in \mathbb{Z}_{\ge 1}$. Consider the set $C = \{\cup_{i \in I} C_i: I \subset \mathbb{N}\}$. Note that $I \neq J \implies \cup_{i \in I} C_i \neq \cup_{i \in J} C_i$. As $I$ runs over all elements of the power set of $\mathbb{N}$, $C$ has uncountably infinitely many elements. However, as each element of $C$ is composed of a countable union of elements of $\mathcal{F}$, $C \subset \mathcal{F}$. This is a contradiction to the assumption that $\mathcal{F}$ is countably infinite.