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For 2 topological spaces $(X,T_X)$ and $(Y,T_Y)$, I write $X \simeq Y$ if $X$ and $Y$ are homeomorphic. If $A \subset X$, I always endow $A$ with the subspace topology.

I was led to consider the following assumptions.

(1) $(X,T_X)$ is a topological space such that for all $A$,$B$ $\subset X$ : $A$ closed and $A \simeq B$ $\Rightarrow$ $B$ closed

It is true if $X$ is discrete or compact Hausdorff, but it's not true if $X = \mathbb{R}^n$, $n \geq 1$ (because $\mathbb{R}^n \simeq (0,1)^n$), or if $X$ has a closed singleton and an other non-closed singleton (like $X=\{a,b\}$ with $T_X = \{ \varnothing, \{a\}, \{a,b\}$).

(2) $(X,T_X)$ is a topological space such that for all $A$,$B$ $\subset X$ : $A$ open and $A \simeq B$ $\Rightarrow$ $B$ open

Again, it is true if $X$ is discrete, or if $X = \mathbb{R}^n$ with $n \geq 0$ (invariance of domain), but it's not true if $X$ has an open singleton and an other non-open singleton (like $X=\{a,b\}$ with $T_X = \{ \varnothing, \{a\}, \{a,b\}$).

My question is :

What are the conditions on the space $X$ that guarantee that one of $(1)$ and $(2)$ is true, or at least some examples apart from the ones mentionned ?

A topic already exists (Homeomorphic closed subspaces.) that deals whith $(1)$ but they don't give the general answer there. However, they say in this topic that if $X$ is countably compact and first countable, $(1)$ holds : why ?

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    $\begingroup$ I'm confused. If $A \simeq B$, then there is a homeomoprhism between them $\phi$ which is both an open and a closed map. Continuous maps have open/closed preimages of open/closed sets. So suppose $A$ is open/closed. $\phi^{-1}$ is a homeomorphism because $\phi$ is. And the preimage of $\phi^{-1}$ of $A$ is $B$. This shows $B$ is open/closed, respectively. $\endgroup$ – Alfred Yerger Oct 9 '16 at 17:42
  • $\begingroup$ @AlfredYerger: The homeomorphism between them need not extend to a homeomorphism $X\to X$. $\endgroup$ – Eric Wofsey Oct 9 '16 at 17:57
  • $\begingroup$ @AlfredYerger : your reasoning shows that $B$ is open/closed in the subspace topology of $B$, which is of course true. Not in the topology of $X$. $\endgroup$ – Vandrin Oct 9 '16 at 18:00
  • $\begingroup$ Ah. I understand the question now. Thanks for clearing it up. I can't say that I can come up with any other examples than what has been listed in the linked question though. $\endgroup$ – Alfred Yerger Oct 9 '16 at 18:01
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I doubt that there is a nice answer to either question. However, $(1)$ holds if $X$ is Hausdorff, sequential, and countably compact; every first countable space is sequential, so this generalizes the result mentioned in the comment under the earlier question. (That comment did not specify Hausdorffness, but it is required.)

Suppose that $X$ is Hausdorff, sequential, and countably compact, $A$ is a closed subset of $X$, and $h:A\to B\subseteq X$ is a homeomorphism. If $B$ is not closed in $X$, then $B$ is not sequentially closed, so there is a sequence $\langle x_n:n\in\Bbb N\rangle$ converging to some $x\in X\setminus B$. Let $D=\{x_n:n\in\Bbb N\}$; since $X$ is Hausdorff, $\operatorname{cl}D=D\cup\{x\}$, and $D$ is therefore infinite, closed, and discrete in the subspace $B$. But then $h^{-1}[D]$ is infinite, closed, and discrete in the subspace $A$, which is impossible: $A$, being a closed subset of $X$, is countably compact and therefore contains no such subset.

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  • $\begingroup$ You wrote that $D$ is closed, this is a little bit strange to me. That would mean $x \in D \subset B$, but we have $x \in X \setminus B$. $\endgroup$ – Vandrin Oct 11 '16 at 12:17
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    $\begingroup$ @Vandrin: $D$ is closed in the subspace $B$. It is not closed in $X$. $\endgroup$ – Brian M. Scott Oct 11 '16 at 18:46
  • $\begingroup$ Right. Thank you for the proof. I am not accepting the answer in case someone has another view on (1) and (2). $\endgroup$ – Vandrin Oct 12 '16 at 12:39

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