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I'd like to work with a topology that has the following property: for any open neighbourhood $U_z$ of a point $z$, there exists a closed set $K$ another open neighbourhood $V_z$ such that

$$ V_z \subset K \subset U_z.$$

Is there a name for such a topology?

Thanks for your help!

[edited to incorporate comment about $V$ non-empty]

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    $\begingroup$ Presumably, you want $V$ non-empty... $\endgroup$ – Thomas Andrews Jan 28 '16 at 3:06
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    $\begingroup$ Seems pretty close to the notion of a regular space. $\endgroup$ – hardmath Jan 28 '16 at 3:09
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    $\begingroup$ Actually it seems closer to a normal space. $\endgroup$ – Tim Raczkowski Jan 28 '16 at 3:13
  • $\begingroup$ @TimRaczkowski: Particularly now that point $z$ has been introduced. $\endgroup$ – hardmath Jan 28 '16 at 3:15
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Your condition is precisely equivalent to regularity, which can be defined as follows: $\langle X,\tau\rangle$ is regular if for each $x\in X$ and open nbhd $U$ of $x$ there is an open nbhd $V$ of $x$ such that $\operatorname{cl}V\subseteq U$.

You may instead know regularity by the following equivalent definition: if $x\in X$, and $H$ is a closed subset of $X$ not containing $x$, then there are disjoint open sets $U$ and $V$ such that $x\in U$ and $H\subseteq V$. It’s an easy exercise to prove that these definitions are equivalent.

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  • $\begingroup$ Thanks Brian, that's exactly what I wanted to hear. $\endgroup$ – Daniel Mansfield Jan 28 '16 at 3:20
  • $\begingroup$ @Daniel: You’re welcome. $\endgroup$ – Brian M. Scott Jan 28 '16 at 3:20
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Yes, it's called "a topology". In other words, every topological space $X$ has this property.

Just take $K= \overline{U_z}$ and $V_z= X$ .

EDIT: Oops, this is wrong, please disregard. I misread the question. I thought OP required $U_z\subset K\subset V_z$.

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  • $\begingroup$ Yes, I have misread the question. Sorry. $\endgroup$ – MPW Jan 28 '16 at 3:17

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