# Proof of equivalence of definitions T3 spaces

I'm trying to prove that the following two definitions of $T_{3}$ spaces are equivalent, and hitting an impasse.

Definition 1: For any point and closed set not containing the point, there are disjoint neighborhoods of each.

Definition 2: Every neighborhood of a point contains the closure of a smaller neighborhood of the point.

What I have so far:

Assume definition 1, and let $x$ be ant point, $U$ any neighborhood of it, so therefore there is some open set $O$ such that $x \in O \subseteq U$. Then $x \not\in O^{C}$ and $O^{C}$ closed, so by being a $T_{3}$ space, we can infer that there exist neighborhoods $x\in U_{x}$ and $O^{C}\subseteq U_{O^{C}}$ such that $U_{x}\cap U_{O^{C}} = \emptyset$. Then $U_{x} \cap O^{C} = \emptyset$ and therefore $U_{x} \subseteq O \subseteq U$.

What remains for this half of the proof is to show that $\overline{U}_{x} \subseteq U$, but I'm just not seeing it. I've tried showing that each contact point of $U_{x} \in U$ but can't find a way--and I've tried using the fact that $\overline{U}_{x}$ is closed and therefore I could try to apply the first definition of a $T_{3}$ space again, but I don't see any relevant points that I can show are not in $\overline{U}_{x}$. I don't see how I could claim, at this time, that $\overline{U}_{x}\cap U_{O^{C}} = \emptyset$.

## 1 Answer

Suppose the space is regular. Then, take a point $x$ and a neighborhood of it, $U$. Then $U^C$ is closed. So by regularity, you can separate $x$ and $U^C$ by disjoint open sets, $A$ and $B$ respectively. Then, $A$ is a neighborhood of $x$, and $\overline{A}$ is disjoint from $U^C$, since for any point $b \in U^C$, $B$ is a neighborhood of $b$ that is disjoint from $A$.