$U$ 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?

  • 3
    $\begingroup$ 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

1 Answer 1


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 $\overline A\subset X\setminus\{b\}$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.