# What separation axiom is necessary for existence of neighborhood which closure is a subset of another given neighborhood

I'm looking for the weakest separation axiom, which gives the following property:

Let $$A$$ be neighborhood of the point $$x$$. Then there exists another neighborhood $$B$$ of $$x$$, such $$\overline{B}\subset A$$.

I thought that $$T_{3}$$ would suffice, but I was able to get only closed set $$C$$ such $$x\in C\subset A$$

I could take $$B=Int(C)$$, but then I do not see that $$B$$ will be neighborhood of $$x$$, i.e. $$x\in B$$.

If no separation axiom per se, then in what space this property would hold? The more abstract, the better.

• To get a closed set $C$ such that $x\in C\subset A$ the $T_1$ axiom is enough. What formulation of $T_3$ are you using? – bof Feb 14 at 23:56
• en.wikipedia.org/wiki/Regular_space Mind that I'm not looking for closed $C$ but for neighborhood $B$ of $x$ for which closure would be $C$. – user121882 Feb 15 at 0:10

Suppose $$X$$ is regular. Let $$A$$ be a neighborhood of $$x$$ and suppose without loss of generality that $$A$$ is open. Then regularity lets us choose disjoint open sets $$U \ni x$$ and $$V \supset A^c$$. That is, $$U \subset V^c$$ where $$V$$ is closed, so $$\overline{U} \subset V^c \subset A$$. Thus $$B = U$$ is the desired neighborhood of $$x$$.
Conversely, suppose $$X$$ has your property. Let $$x \in X$$ and let $$F$$ be a closed set not containing $$x$$. Then $$A = F^c$$ is a neighborhood of $$x$$. Suppose $$B$$ is a neighborhood of $$x$$ with $$\overline{B} \subset A$$. Set $$U = B^\circ$$, the interior of $$B$$, and $$V = (\overline{B})^c$$. Then $$U,V$$ are open and disjoint, $$x \in U$$, and $$F \subset V$$ since $$V^c = \overline{B} \subset A = F^c$$.
• @bof: Yes, I should have said "regular" instead of $T_3$, and I've fixed it now. They are trivially equivalent and I've given the trivial proof. – Nate Eldredge Feb 15 at 0:50