a space is T1 if and only if every singleton is closed

Say $X$ is a http://planetmath.org/node/1852${\mathrm{T}}_{\mathrm{1}}$ topological space. Let’s show that $\{x\}$ is closed for every $x\in X$:

The $T_1$ axiom (http://planetmath.org/T1Space) gives us, for every $y$ distinct from $x$, an open $U_y$ that contains $y$ but not $x$. Since we’re in a topological space, we can take the union of all these open sets to get a new open set,

$$U=\bigcup _{y\ne x}{U}_y.$$

$\{x\}$ is the complement of $U$, closed because $U$ is open: None of the $U_y$ contain $x$, so $U$ doesn’t contain $x$. But any $y\ne x$ is in $U$, since $y\in U_y\subset U$. That takes care of that.

Now let’s say we have a topological space $X$ in which $\{x\}$ is closed for every $x\in X$. We’d like to show that $T_1$ holds:

Given $x\ne y$, we want to find an open set that contains $x$ but not $y$. $\{y\}$ is closed by hypothesis, so its complement is open, and our search is over.

Title	a space is T1 if and only if every singleton is closed
Canonical name	ASpaceIsT1IfAndOnlyIfEverySingletonIsClosed
Date of creation	2013-03-22 14:20:15
Last modified on	2013-03-22 14:20:15
Owner	waj (4416)
Last modified by	waj (4416)
Numerical id	7
Author	waj (4416)
Entry type	Proof
Classification	msc 54D10
Related topic	ASpaceIsT1IfAndOnlyIfEverySubsetAIsTheIntersectionOfAllOpenSetsContainingA