# 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)
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## 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.

## 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