Let $X$ be a regular Hausdorff topological space without isolated point. We say that $X$ is ultraregular if the following condition holds: for every subset $A \subset X$, such that $A$ and $\complement A$ are both without isolated point, then $A$ (and $\complement A$) are open sets.

Let $A$ be an everywhere dense set (i.e. $\overline A=X$). I would like to prove that the interior $\mathring A$ of $A$ is everywhere dense as well, i.e. $\overline{\mathring A}=X$.

It is not difficult to see that $\mathring A$ is not empty, and that $\mathring A$ is not closed. But I'm not able to prove the stronger fact that $\mathring A$ is everywhere dense.

It could help to know the following fact: if $A$ is a set without isolated point, then $\overline A$ is an open set.

Thank you.

  • $\begingroup$ Just wondering, are there any such spaces? Could you give examples? $\endgroup$ Sep 18, 2014 at 21:36
  • $\begingroup$ Let $X$ be a regular space without isolated point. Denote by $T_0$ the topology of $X$. It is easy to see that the set of topologies $T$ which are finer than $T_0$ and regular without isolated point is inductive. So we can find a maximal topology, which is finer than $T_0$ and regular without isolated point. We readily see that such topologies are exactly the ultraregular topologies. $\endgroup$
    – timofei
    Sep 19, 2014 at 17:14
  • $\begingroup$ Ok. So there only non-constructive examples? Or also explicit ones? $\endgroup$ Sep 19, 2014 at 20:21
  • $\begingroup$ I don't know of any explicit ultraregular space. $\endgroup$
    – timofei
    Sep 20, 2014 at 16:46

1 Answer 1


Let $D$ be a dense subset of $X$, and let $U=\operatorname{int}D$. Let $V=X\setminus\operatorname{cl}U$; then $\operatorname{cl}U$ and $V$ are clopen in $X$, and we wish to show that $V=\varnothing$. Let $E=D\cap V$; note that $\operatorname{int}E=\varnothing$. Clearly $E$ is dense in $V$, which is ultraregular, so $E$ has no isolated points. Let $A=V\setminus E$, and suppose that $p$ is an isolated point of $A$. Then there is a $W$ open in $X$ such that $p\in W\subseteq E\cup\{p\}$. Clearly $W\cap E\ne\emptyset$, so $W\setminus\{p\}$ is a non-empty open subset of $E$, which is impossible. Thus, $A$ has no isolated points, and $E$ is open in $V$ and hence in $X$, which is possible only if $E=\varnothing$ and hence $V=\operatorname{cl}E=\varnothing$.

Probably this is just repeating your argument that $U$ is non-empty, but relativized to $V$.

  • 11
    $\begingroup$ Glad to see you, you have been missed! $\endgroup$
    – Daniel R
    Sep 25, 2014 at 10:09
  • 6
    $\begingroup$ Nice seeing you back Brian!! +1 $\endgroup$
    – ILoveMath
    Sep 27, 2014 at 3:20
  • 2
    $\begingroup$ Thank you for your answer. I was about to answer my own question with something quite similar to this. Indeed the basic (and easy) fact is that any open subspace in an ultraregular space is ultraregular. $\endgroup$
    – timofei
    Sep 29, 2014 at 19:18
  • $\begingroup$ @timofei: It’s always more satisfying to answer one’s own question: I’m glad that you could. $\endgroup$ Sep 30, 2014 at 20:40

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