TITLE: What happens if singleton set is not closed
QUESTION [8 upvotes]: I'm reading Munkres and now is learning the separation axioms.
When he starts to discuss regularity and normality, he says "Suppose one-point sets are closed in $X$". Our Prof. also didn't explain much in the class.
So I'm quite curious what happens if singleton set is not closed. Will it be open in some cases or neither open nor closed?
Also I can't come up with an example of such space.
For $ x\in X$, $\overline{\lbrace x\rbrace}=\lbrace x\rbrace$ does not seem to depend on the topology or the set.
Any explanation will be appreciated.

REPLY [6 votes]: A nice counterexample is the Sierpiński space $S=\{0,1\}$ with the topology $\big\{\varnothing,\{1\},\{0,1\}\big\}$: in this space the set $\{0\}$ is closed, but the set $\{1\}$ is not: the only open nbhd of $0$ is $\{0,1\}$, which contains $1$, so $0\in\operatorname{cl}\{1\}$.
A space $X$ with the property that $\{x\}$ is closed for every $x\in X$ is said to be a $T_1$-space, though the $T_1$ property is usually not defined this way.

Definition. A space $X$ is $T_1$ if whenever $x$ and $y$ are distinct points of $X$, there are open sets $U$ and $V$ such that $x\in U$ and $y\notin U$, and $y\in V$ and $x\notin V$.

A good (and quite easy) exercise is to prove that this is equivalent to the statement that $\{x\}$ is closed for each $x\in X$.

REPLY [3 votes]: One example is the indiscrete space, where only the whole space and the empty set are open (and hence closed) - so, if there is more than one point, singletons would not be closed. This space has the property that all pairs of points $x,y$ are topologically indistinguishable - meaning if $S$ is an open set then $x\in S$ if and only if $y\in S$ - that is, the topology gives no information to tell them apart, which makes it somewhat trivial - however, one can find spaces where the points are topologically distinguishable, but where singletons are not closed (and plenty of the other answers do this, so I won't provide additional examples).
I doubt you'd have many familiar examples of such topologies if you haven't been introduced to more "pathological" topologies that like to be counterexamples to every statement you want. One should note that the condition that singletons are closed is (equivalent to) a separation axiom making a space a $T_1$ space, which is a fairly weak separation axiom.

REPLY [3 votes]: In many fields of mathematics, most of the spaces in consideration are such that singletons are indeed closed subsets. 
However, there are also theories for which the previous statement is false. For example, in algebraic geometry one of the basic concepts is a scheme, and the topology of a general scheme is such that singletons are not necessarily closed.

REPLY [3 votes]: First a bit of terminology: a T1 space is a space where points are closed. Now, the simplest example of a space that is not T1 is just a set with more than one element with the trivial topology (the only open sets are the whole set and the empty set). In this case, the closure of a singleton space is the whole space. You can find more elaborate examples in the wiki page:
http://en.wikipedia.org/wiki/T1_space
Though not precisely related to your question, note that the interesting distinction is between T1 spaces and Hausdorff spaces, not T1 vs not T1. T1 but not Hausdorff show up often enough, algebraic varieties with the Zarisky topology being prime examples.

REPLY [2 votes]: According to the usual definitions of the concept of separation in topology. All points are closed if and only if your topological space is T1 separated or a T1 space.
It might happen that you have a T0 space which is not T1. The following example comes from the Steen-Seebach's counter examples in topology (by the way, if you are interested by pathological spaces, you want to read this book) :
Take $X:=[-1,1]$. Its open sets are : $[-1,b)$ for $b>0$ and $(a,1]$ for $a<0$ and $(a,b)$. This is a topology.
You can see that the point $0$ won't be closed so is not T1. Nevertheless it is easily seen to be T0.