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$U$ be open , $a \in U$, does there exist a open set $A$ containing $a$, such that $\bar A \subset U$, where $\bar A$ denotes the closure of $A$.

I was doing a exercise in Several variable Analysis that uses this fact, which is obvious in $\Bbb R^n$ (with the usual metric). I was thinking if the above statement has a generalisation in arbitrary topological spaces?

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    $\begingroup$ This property essentially defines the concept of a regular topological space. (For comparison with the Wikipedia definition, note that $\bar A^\complement$ is an open neighborhood of $U^\complement$). $\endgroup$ Nov 14, 2021 at 13:11
  • $\begingroup$ I suggest you do a quick read of the definition of a topological space and of the "separation properties" $T_0, T_1,T_2,....$ etc. that any space may or may not have. $\endgroup$ Nov 14, 2021 at 17:54

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This is not true in general. Take an infinite set $X$ endowed with the co-finite topology: $S\subset X$ is open if and only if $S=\emptyset$ or $S^\complement$ is finite. Take $a,b\in X$, with $a\ne b$. Then $a\in X\setminus\{b\}$, which is an open set. But there is no open set $A$ such that $a\in A$ and that $\overline A\subset X\setminus\{b\}$.

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  • $\begingroup$ Another counter-example is Sierpinski space $S=\{a,b\}$ where $\{a\}=U$ is open but $\{b\}$ is not open. $\endgroup$ Nov 14, 2021 at 17:58

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