Let $A$ , $B$ be topological spaces such that there for some subset $D$ of $B$ there is a homeomorphism form $A$ to $D$ and for some subset $E$ of $A$ there is a homeomorphism form $B$ to $E$ ; then must there exist a homeomorphism from $A$ to $B$ ?
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$\begingroup$ If you simply google for cantor bernstein topological, you will find some related stuff on internet. You will see several hits which answer your question (basically given the same asnwers as already osted here). This post on MO might also be interesting in this context. $\endgroup$– Martin SleziakOct 22, 2014 at 21:14
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3 Answers
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No, consider $A=[0,1]$, $B=[0,1]\cup [2,3]$, $f:[0,1] \to [0,1]\cup [2,3] $ where $$ f(x) = x $$ and $g(x): [0,1]\cup [2,3]\to [0,1]$ with $$g(x) = \frac{x}{6}$$
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3$\begingroup$ And $A$ and $B$ aren't homeomorphic since one is connected while the other isn't. (I was just about to write this exact example, so I thought I would add just a little bit.) $\endgroup$ Oct 22, 2014 at 14:44
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Take $[0,1]$ and $(0,1)$ with their standard topologies in $\Bbb R$. Clearly they are not homeomorphic. But $[0,1]$ is homeomorphic to $[\frac14,\frac13]$ and $(0,1)$ is homeomorphic to itself.
answered Oct 22, 2014 at 14:46
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I asked a related property: are there non-homeomorphic $X$ and $Y$ such that there is a continuous bijection from $X$ to $Y$ and from $Y$ to $X$ as well?
And this mathoverflow answers shows that we can have such examples. They're a bit harder to come up with, I think.
answered Oct 22, 2014 at 14:55
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$\begingroup$ I'm not sure that property is stronger, just different. As far as I can tell, having a continuous bijection from X to Y does not imply an embedding of X into Y or vice versa. $\endgroup$ Oct 22, 2014 at 15:02
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1$\begingroup$ @NateEldredge Indeed, I will edit it. $\endgroup$ Oct 22, 2014 at 15:04
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$\begingroup$ Thank you for your perspective on this. I believe this was a useful remark, and added to the overall discussion. +1 $\endgroup$– RustynOct 22, 2014 at 16:46