a space is $T_1$ if and only if distinct points are separated

Theorem 1.

Let $X$ be a topological space. Then $X$ is a $T_{\mathrm{1}}$-space if and only if sets $\mathrm{\{}x\mathrm{\}}$, $\mathrm{\{}y\mathrm{\}}$ are separated for all distinct $x\mathrm{,}y\mathrm{\in }X$.

Proof.

Suppose $X$ is a $T_1$-space. Then every singleton is closed and if $x,y\in X$ are distinct, then

$\mathrm{\{}x\}\cap \overline{\{y\}}$ $=$ $\mathrm{\{}x\}\cap \{y\}=\mathrm{\varnothing },$

$\overline{\{x\}}\cap \{y\}$ $=$ $\mathrm{\{}x\}\cap \{y\}=\mathrm{\varnothing },$

and $\{x\}$, $\{y\}$ are separated. On the other hand, suppose that $\{x\}\cap \overline{\{y\}}=\mathrm{\varnothing }$ for all $x\ne y$. It follows that $\overline{\{y\}}=\{y\}$, so $\{y\}$ is closed and $X$ is a $T_1$-space. ∎

Title	a space is $T_1$ if and only if distinct points are separated
Canonical name	ASpaceIsT1IfAndOnlyIfDistinctPointsAreSeparated
Date of creation	2013-03-22 15:16:49
Last modified on	2013-03-22 15:16:49
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	6
Author	matte (1858)
Entry type	Theorem
Classification	msc 54D10