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This is perhaps a more general question, but is there any name for a topological space which supports the following structure? Given any open set $A$ in the topology and some point $x$, there exists a closed set $B$ and another open set $A'$ which satisfy

$$ x \in A' \subset B \subset A $$ ?

That such a structure is supported is clear in, e.g., metric spaces (with the usual topology generated by the distance function) by letting $S$ be the open ball which contains $x$ and picking $B$ to be the closed ball centered at $x$ with half of the distance of $x$ to the boundary of $A$. Then $A'$ can simply be, say, the interior of $B$.

Perhaps there may not be a name for this particular case, but is there a name for an equivalent (or slightly stronger) property of a space?

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    $\begingroup$ That is the exact definition of a regular topological space. $\endgroup$ – bof Apr 19 '18 at 5:40
  • $\begingroup$ Ah, I see; the construction I was thinking about actually requires compactness (! which I didn't state in the question, so it is besides the point), but yes, thank you! $\endgroup$ – Guillermo Angeris Apr 19 '18 at 5:49
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An Example:

Locally compact Hausdorff space $X$: For every $x\in X$ and any open set $G$ that containing $x$, there is a compact set $K$ and an open set $H$ such that $x\in H\subseteq K\subseteq G$.

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  • $\begingroup$ Brilliant, this is exactly what I needed. $\endgroup$ – Guillermo Angeris Apr 19 '18 at 5:41
  • $\begingroup$ "Locally compact Hausdorff" implies "regular" but is far from being equivalent. $\endgroup$ – bof Apr 19 '18 at 5:41
  • $\begingroup$ So I have given a stronger topological space which has the given property. $\endgroup$ – user284331 Apr 19 '18 at 5:43
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    $\begingroup$ @GuillermoAngeris, regular topology is the one you need, not locally compact Hausdorff space. $\endgroup$ – user284331 Apr 19 '18 at 5:43

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