Let $X$ be a topological space and $U \subseteq X$ open. Then $U \subseteq \operatorname{int}(\operatorname{cl}(U))$.

I am looking for known assumptions on $X$ and $U$ such that one of the following properties holds:

  1. $U$ and $\operatorname{int}(\operatorname{cl}(U))$ are homeomorphic
  2. if $U$ and $\operatorname{int}(\operatorname{cl}(U))$ are homeomorphic then $U = \operatorname{int}(\operatorname{cl}(U))$, i.e. $U$ is a regular open set.

Examples and counterexamples:

  1. If $X = \mathbb{R}$ then $U = (1,2) \cup (2,3)$ is not homeomorphic to $\operatorname{int}(\operatorname{cl}(U)) = (1,3)$.
  2. If $X = \mathbb{R}^n$ and $U$ convex then $U$ is regularly open.

For my purposes, $X$ can be assumed to be locally compact and metrizable.

  • 1
    $\begingroup$ Another example: $\mathbb R^n\setminus(\{0\}^{n-1}\times[0,\infty))$ is homeomorphic to $\mathbb R^n$. (Added parentheses to remove ambiguity.) $\endgroup$
    – Dejan Govc
    May 19, 2016 at 14:17
  • $\begingroup$ Property 2 holds for $X = \mathbb{R}$ (but not for $\mathbb{R}^n$, $n \geq 2$ as shown by Dejan Govc). For a proof, represent $U$ and $\textrm{int}(\textrm{cl}(U))$ as a (countable) disjoint union of open intervals. The homeomorphism between $U$ and $\textrm{int}(\textrm{cl}(U))$ induces a bijection between the set of these intervals and homeomorphisms between each such pair of intervals. If $U$ is the disjoint union of intervals $I_k$ then $\textrm{cl}(U)$ is the disjoint union of $U$ and $\partial U$ which in turn is the union of $\partial I_k$ and a countable set $C$ of accumulation points. $\endgroup$
    – yada
    May 21, 2016 at 9:11
  • $\begingroup$ Now if $x \in \textrm{int}(\textrm{cl}(U))$ argue that if $x \in \partial I_k$ then there is no other interval $I_{k'}$ such that $x \in \partial I_{k'}$. But then $x$ is in the boundary of an open interval of $\textrm{int}(\textrm{cl}(U))$, contradiction. Also argue that $x \in C$ leads to a contradiction. Thus $x \in U$. $\endgroup$
    – yada
    May 21, 2016 at 9:16
  • 1
    $\begingroup$ I'm not sure property 2 holds for $X=\mathbb R$. Consider $U=\bigcup_{k\in\mathbb Z}\left((4k,4k+1)\cup(4k+1,4k+2)\right)$. Then $U$ and $\operatorname{int}(\operatorname{cl}(U))=\bigcup_{k\in\mathbb Z}(4k,4k+2)$ are homeomorphic but not equal. (The issue here is the two intervals $(4k,4k+1)$ and $(4k+1,4k+2)$ do indeed have a boundary point in common.) $\endgroup$
    – Dejan Govc
    May 22, 2016 at 10:09
  • 1
    $\begingroup$ I'm surprised this hasn't been asked before. $\endgroup$ Jun 3, 2016 at 2:39


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