Let $X$ be an arbitrary topological space, and $U,V\subseteq X$ two subspaces of $X$ such that $U\cong V$ ($U$ and $V$ are homeomorphic) with respect the subspace topology of $X$. I know examples where $U$ is closed in $X$ and $V$ is not. Is there some conditions to guarantee the next statement:

If $U$ is closed in $X$, then $V$ es closed in $X$.

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    $\begingroup$ When $X$ is compact Hausdorff $\endgroup$ – Henno Brandsma Nov 20 '14 at 20:34
  • $\begingroup$ Curious to know a counterexample in the general case. $\endgroup$ – user4894 Nov 20 '14 at 20:37
  • $\begingroup$ @user4894 $\Bbb R \cong (0,1)$ $\endgroup$ – user98602 Nov 20 '14 at 20:42
  • $\begingroup$ Following the path of @HennoBrandsma, also if $X$ is countably compact and first countable. The problem is that I have an example where exactly $X$ is not compact and not first countable. So maybe there is no hope :(. $\endgroup$ – Asupollo Nov 20 '14 at 20:51
  • $\begingroup$ @MikeMiller Silly of me to ask before thinking. Thanks. $\endgroup$ – user4894 Nov 20 '14 at 21:05

If $X$ is compact and Hausdorff, then a subspace is closed if and only if it is compact. Since compactness is a topological invariant, $X$ being compact Hausdorff ensures that if two subspaces are homeomorphic and one is closed, then so is the other.

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