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If $D \subseteq X$ is any subspace of $X$ that is homeomorphic to a closed set $E \subseteq Y$ and $Cl_Y(Int_Y(E)) = E$, then does $Cl_X(Int_X(D)) =D?$

I asked a very similar question to this in the past except with the added condition that $D$ was a closed set in $X$.

I know that since $D$ is homeomorphic to $E$, their interiors are homeomorphic, that is $Int(D)$ is homeomorphic to $Int(E)$.

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    $\begingroup$ What about $\mathbb{R} \subseteq \mathbb{R}$ and $\mathbb{R} \times \lbrace 0 \rbrace \subseteq \mathbb{R}^2$? $\endgroup$ – Theo Bendit Feb 18 '18 at 11:52
  • $\begingroup$ @TheoBendit Ahh so (independently of the question I asked above) if $X$ is homeomorphic to $Y$, then $Int(X)$ need not be homeomorphic to $Int(Y)$ $\endgroup$ – Perturbative Feb 18 '18 at 12:36
  • $\begingroup$ Well, regardless if $X$ is homeomorphic to $Y$, then $\operatorname{Int} X = X$ and similarly for $Y$ (assuming $X$ and $Y$ still refer to the full topological spaces). So, if $X$ is homeomorphic to $Y$, then so are their interiors! $\endgroup$ – Theo Bendit Feb 18 '18 at 13:01
  • $\begingroup$ @TheoBendit What were you intending to show in your example in that case? Because in your example $\operatorname{Int}_{\mathbb{R}}(\mathbb{R}) = \mathbb{R}$ is homeomorphic to $\mathbb{R} \times \{0\}$ but $\operatorname{Int_{\mathbb{R}^2}}(\mathbb{R} \times \{0\}) = \emptyset$ so while $\mathbb{R}$ is homeomorphic to $\mathbb{R} \times \{0\}$ their interiors aren't homeomorphic $\endgroup$ – Perturbative Feb 18 '18 at 18:40
  • $\begingroup$ I was intending to show that, given two homeomorphic subspaces of (possibly not homeomorphic) topological spaces, the closure of their interiors need not be homeomorphic, which is what I thought the question was asking. $\endgroup$ – Theo Bendit Feb 18 '18 at 23:55
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Let $f$ be a homeomorphism from $X$ to $Y$. Indeed, as shown in this post, $f(\operatorname{Cl}(E))=\operatorname{Cl}(f(E))=\operatorname{Cl}(D).$ Therefore, as you noted that $f(\operatorname{Int}(E))=\operatorname{Int}(f(E))$, $$f(\operatorname{Cl}(\operatorname{Int}(E)))=\operatorname{Cl}(f(\operatorname{Int}(E)))=\operatorname{Cl}(\operatorname{Int}(f(E)))=\operatorname{Cl}(\operatorname{Int}(D))=f(E)=D.$$

If, however, $f$ is a homeomorphism from $E$ to $D$- and not necessarily from $X$ to $Y$-, things are different. Consider $[0,1)$ with the subspace topology of $\mathbb{R}$ and $(0,1]$ as a subset of $[0,1]$. Let, as a counter-example, $f:[0,1)\to[0,1]$ be an imbedding of $[0,1)$ to $(0,1]$. Indeed, $\mathrm{Cl}_{[0,1)}(\mathrm{Int}_{[0,1)}([0,1)))=[0,1)$, but $\mathrm{Cl}_{[0,1]}(\mathrm{Int}_{[0,1]}((0,1]))=\mathrm{Cl}_{[0,1]}((0,1])=[0,1]$.

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  • $\begingroup$ We need $f$ to be a homeomorphism from $D$ to $E$, not $X$ to $Y$. $\endgroup$ – Perturbative Feb 18 '18 at 18:54
  • $\begingroup$ Edited it with regards to this case! Hope I solved your question :) $\endgroup$ – TPace Feb 18 '18 at 19:21
  • $\begingroup$ I've added some more clarification with respect to the interiors (in case they weren't clear before). Could you indicate if possible with respect to which space you are taking the interiors in your question? $\endgroup$ – Perturbative Feb 18 '18 at 20:01
  • $\begingroup$ Yes, I can! They're both like you described in your question. $\endgroup$ – TPace Feb 18 '18 at 20:03
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Counter-example: $X=Y=[0,1]\cup [2,3)$ and $D=[0,1)$ and $E=[2,3).$

The function $f:D\to E,$ where $f(x)=x+2,$ is a homeomorphism.

Since $E$ is open and closed, $Cl(Int(E))=Cl(E)=E.$

But $Cl(Int(D))=Cl(D)=[0,1]\ne D.$

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Counter-example: $X=Y=[0,1]\cup [2,3)$ and $D=[0,1)$ and $E=[2,3).$

Since $E$ is open and closed, $Cl(Int(E))=Cl(E)=E.$

But $Cl(Int(D))=Cl(D)=[0,1]\ne D.$

An even simpler counter-example: Let $X=[0,1]$ and $Y=[0,1)$ with $D=E=[0,1).$.

In general, let $T_Y$ be the topology on $Y,$ with $Y\ne \emptyset.$ Let $X=Y\cup \{p\}$ with $p\not \in Y.$ Let the topology on $X$ be $T_Y\cup \{t\cup \{p\}: \emptyset \ne t\in T_Y\}.$ Let $D=E=Y.$

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