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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.

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  • $\begingroup$ 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? $\endgroup$ – bof Feb 14 at 23:56
  • $\begingroup$ 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 $C$. $\endgroup$ – user121882 Feb 15 at 0:10
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Your condition is equivalent to being regular.

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 $V^c = \overline{B} \subset A = F^c$.

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  • $\begingroup$ @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. $\endgroup$ – Nate Eldredge Feb 15 at 0:50

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