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

Say $X$ is a http://planetmath.org/node/1852$T_{1}$ 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,

 $U=\bigcup_{y\neq x}U_{y}.$

$\{x\}$ 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 $y\neq x$ 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 $x\neq y$, 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 ASpaceIsT1IfAndOnlyIfEverySingletonIsClosed 2013-03-22 14:20:15 2013-03-22 14:20:15 waj (4416) waj (4416) 7 waj (4416) Proof msc 54D10 ASpaceIsT1IfAndOnlyIfEverySubsetAIsTheIntersectionOfAllOpenSetsContainingA