Hausdorffness is closure-local

This article gives the statement, and possibly proof, of a topological space property (i.e., Hausdorff space) satisfying a topological space metaproperty (i.e., closure-local property of topological spaces)
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
Get more facts about Hausdorff space |Get facts that use property satisfaction of Hausdorff space | Get facts that use property satisfaction of Hausdorff space|Get more facts about closure-local property of topological spaces

Statement

Suppose $X$ is a topological space such that for any point $x\in X$, there exists an open subset $U\ni x$ such that the closure $\overline{U}$ is a Hausdorff space. Then, $X$ is also a Hausdorff space.

Related facts

• Hausdorffness is not local
• Hausdorffness is hereditary

Facts used

1. Openness is transitive

Proof

Given: A topological space $X$ such that for any point $x\in X$, there exists an open subset $U\ni x$ such that the closure $\overline{U}$ is a Hausdorff space. We are given two distinct points $x,y\in X$.

To prove: We can find disjoint open subsets $V,W$ of $X$ such that $V\ni x,W\ni y$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	There exists an open subset $U\ni x$ such that the closure $\overline{U}$ is a Hausdorff space.		given data directly used		direct
2	If $y\notin \overline{U}$, then we can take $V=U$ and $W=X\setminus \overline{U}$.			Step (1)
3	If $y\in \overline{U}$, there exist open subsets $A,B$ of $\overline{U}$ such that $x\in A,y\in B$ and $A\cap B$ is empty.
4	Continuing with the assumption and notation from Step (3), then there exists $W$ open in $X$ such that $B=W\cap \overline{U}$. In particular, $y\in W$.	definition of subspace topology		Step (3)
5	Continuing with the assumption and notation from Step (3), the set $V=A\cap U$ is open in $X$ and contains $x$.	Fact (1)		Steps (1), (3)	By definition of subspace topology going from $\overline{U}$ to $U$, $V$ is open in $U$. Thus, by Fact (1), $V$ is open in $X$. Also, $x\in V$ because $x\in A$ (Step (3)) and $x\in U$ (Step (1)).
6	Continuing with the assumption and notation from Steps (3)-(5), the sets $V$ and $W$ are disjoint.			Steps (3), (4) ,(5)	$V$ is contained in $U$ by Step (5) and hence in $\overline{U}$. So, $V\cap W=V\cap (W\cap \overline{U})=V\cap B\subseteq A\cap B$. $A$ and $B$ are disjoint by Step (3), hence the intersection is trivial.
7	$V,W$ are the desired disjoint open subsets.			Steps (4), (5), (6)	Step-combination direct