Regular implies preregular

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., regular space) must also satisfy the second topological space property (i.e., preregular space)
View all topological space property implications | View all topological space property non-implications
Get more facts about regular space|Get more facts about preregular space

Statement

Any regular space is a preregular space.

Definitions used

Term	Definition used
regular space	A topological space $X$ is termed regular if the following holds: given any point $x\in X$ and closed subset $A\subseteq X$ such that $x\notin A$, there exist disjoint open subsets $U,V$ of $X$ such that $x\in U,A\subseteq V$, and $U\cap V=\varnothing$.
preregular space	A topological space $X$ is termed preregular if the following holds: given any topologically distinguishable points $x,y\in X$, there exist disjoint open subsets $U\ni x,V\ni y$.

Proof

Given: A topological space $X$ that is regular. Topologically distinguishable points $x,y\in X$.

To prove: There exist disjoint open subsets $U\ni x,V\ni y$.

Proof: Because $x,y$ are topologically distinguishable, it is true that either $x\notin {\overline {\{y\}}}$ or $y\notin {\overline {\{x\}}}$ (both may also be true). it suffices to consider only the first case, because the result we want to prove is symmetric in $x$ and $y$.

In the first case, let $A={\overline {\{y\}}}$. Now use the definition of regularity on $x$ and $A$ to obtain the open subsets $U$ and $V$ satisfying the desired conditions.