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Is it true that "If two metric spaces each of which is the image of the other under a bijective continuous function, then the two metric spaces are homeomorphic."?? Thank you so much!!

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  • $\begingroup$ [This example][1] from a question I asked at mathoverflow is metrisable. [1]: mathoverflow.net/a/30711/2060 $\endgroup$ – Henno Brandsma Dec 9 '14 at 18:07
  • $\begingroup$ @HennoBrandsma So, we have two comments posted as answers, and an answer (yours) posted as a comment. Please expand a little bit and post an answer. $\endgroup$ – user147263 Dec 9 '14 at 21:06
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I posted a very similar question on mathoverflow (link) and one of the replies there is a reasonably easy metrisable one:

Let $X$ and $Y$ be spaces with underlying set $\mathbb{R}$. On $X$ we put the topology that consists of the usual topology on $(0,\infty)$ and the discrete metric on all $x \le 0$. On $Y$ we put the topology that is the usual one on $(-1,0)$ and $(1,\infty)$ and is discrete everywhere else.

Now $f(x) = x+1$ is a continuous bijection from $X$ to $Y$ and also a continuous bijection from $Y$ to $X$.

But they are not homeomorphic as $X$ as one non-trivial component, and $Y$ has two.

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From the definition of two homeomorphic spaces, we would require that the bijection has a continuous inverse. If one of the spaces is compact, then we are guaranteed a continuous inverse.

So if one space is compact, then the bijection is indeed a homeomorphism. It then follows that, since compactness is a topological property (i.e., topologically invariant), the other space will be compact as well.

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  • $\begingroup$ What is the answer to the question posed, yes or no? You just quoted a standard fact that does not answer the question. $\endgroup$ – user147263 Dec 9 '14 at 15:41
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    $\begingroup$ You seem to be nit picking the answers here. Thanks for the downvote. If you can't see the answer in what I typed, then perhaps you should consider another field. $\endgroup$ – Sultan of Swing Dec 9 '14 at 21:04
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This is the case if one is compact. This is the case because metric spaces are Hausdorff. If you know one is compact then you only need the existence of one continues bijection. (the compactness of the other then follows from the homeomorphism). More detail can be found here: https://proofwiki.org/wiki/Continuous_Bijection_from_Compact_to_Hausdorff_is_Homeomorphism

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  • $\begingroup$ Beat me to it 😄 $\endgroup$ – Sultan of Swing Dec 9 '14 at 11:34
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    $\begingroup$ It was a sloppy choice of words, excuses me. I rectified it. I don't think it is a if and only if relation $\endgroup$ – D. Vente Dec 9 '14 at 11:39
  • $\begingroup$ So... what is the answer to the question posed, yes or no? You just quoted a standard fact that does not answer the question. $\endgroup$ – user147263 Dec 9 '14 at 15:41
  • $\begingroup$ @Behaviour If one space is compact, then yes. As you can see I quoted above. Otherwise I cannot say without knowing more. The existence of two continues bijections does not grantee the existence of a continues inverse. $\endgroup$ – D. Vente Dec 9 '14 at 15:56
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    $\begingroup$ I provided the best answer I could. If you don't find it satisfactory I suggest you provide an answer yourself. I don't think it's a yes or no answer. It might be true under some extra conditions one of which I provided $\endgroup$ – D. Vente Dec 9 '14 at 16:23

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