characterization of $T2$ spaces

Proposition 1.

[1, 2]Suppose $X$ is a topological space. Then $X$ is a $T_{\mathrm{2}}$ space (http://planetmath.org/T2Space) if and only iffor all $x\mathrm{\in }X$, we have

$\mathrm{\{}x\}$$=$${\displaystyle \bigcap \{A\mid A\subseteq X\text{𝑐𝑙𝑜𝑠𝑒𝑑},\exists \mathit{\text{ open set}}U\mathit{\text{such that}}x\in U\subseteq A\}}.$(1)

Proof.

By manipulating the definition using de Morgan’s laws, the claimcan be rewritten as

$${\{x\}}^{\mathrm{\complement }}=\bigcup \{V\mid V\subseteq X\text{open},\exists \text{ open set}U\text{such that}x\in U\subseteq V^{\mathrm{\complement }}\}.$$

Suppose $y\in {\{x\}}^{\mathrm{\complement }}$. As $X$ is a $T_2$ space,there are open sets $U,V$ such that $x\in U,y\in V$, and $U\cap V=\mathrm{\varnothing }$.Thus, the inclusion from left to right holds.On the other hand, suppose $y\in V$ for some open $V$ such that$\{x\}\subseteq V^{\mathrm{\complement }}$. Then

$$y\in V\subseteq {\{x\}}^{\mathrm{\complement }}$$

and the claim follows.∎

Notes

If we adopt the notation that a neighborhood of $x$is any set containing an open set containing $x$, then the equation 1can be written as

$\mathrm{\{}x\}$$=$${\displaystyle \bigcap \{A\mid A\subseteq X\text{is a closed neighborhood of }x\}}.$