definition of separation axioms in topology

Question

I am learning the Separation Axioms and came across the definition of regular space. In the definition they say, "Suppose the one point sets are closed in $X$"

My question is: how can one point sets be closed in any case? Can't you always find an open neighborhood around them? Is there any way of looking at this intuitively?

Answer 1

Look at the case $\Bbb{R}$. Take a point $b$ on the real line, then clearly the two rays that make up the complement of $\{b\}$ form an open set.

The reason one-point sets closed is part of the hypothesis to regularity and normality (sep axioms; at least in one definition), is so that we can have the inclusion:

Normal spaces $\subset$ Regular spaces $\subset$ Hausdorff spaces

Let's look at the proof that every regular space is Hausdorff using the definition:

Def. $X$ is regular if one-point sets are closed and for all closed sets $C$ and points $p \notin C$ their exist two disjoint neighborhoods containing $C$ and $p$.

Proof. Since one point sets are closed, then let $C = q$, the second point, and we're done.

Thus, as you can see the one-point set closed property makes the proof one line.

Also, let's look at the proof that every normal space is regular.

Again, since one-point set are closed, in the definition of normal, which is:

Def. $X$ is a normal space if one-point sets are closed and for any two disjoint closed sets $C,D$ in $X$, there are two disjoint open sets containing them.

we can let $C = p$ since one point sets are closed, and we're done.

That's why one-point sets closed is used in the definitions of regular and normal.

Answer 2

A one-point set is closed if its complement is open. In $\Bbb R$, for example, $\{x\}$ is closed for any $x\in\Bbb R$, because $\Bbb R\setminus\{x\}$ is open: if $y\in\Bbb R\setminus\{x\}$, there is an open interval $(a,b)$ such that $y\in(a,b)\subseteq\Bbb R\setminus\{x\}$. In particular, if $y<x$, then $y\in(y-1,x)\subseteq\Bbb R\setminus\{x\}$, and if $y>x$, then $y\in(x,y+1)\subseteq\Bbb R\setminus\{x\}$.

Now suppose that $X=\{0,1\}$ with the open sets $\varnothing$, $\{1\}$, and $X$; this is the Sierpiński space, and in it the one-point set $\{1\}$ is not closed, because its complement, $\{0\}$, is not open.

If $X$ is any space, and $x\in X$ is any point of $X$, then of course $X$ is an open nbhd of $x$ and an open set such that $\{x\}\subseteq X$; this has nothing at all to do with whether or not $\{x\}$ is closed.