# $U$ be open , $a \in U$, does there exist a open set $A$ containing $a$, such that $\bar A \subset U$.

$$U$$ 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?

• 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 $U^\complement$). Nov 14, 2021 at 13:11
• 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. Nov 14, 2021 at 17:54

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