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Let $(M,d_M)$ be a separable metric space. That is, there is a countable subset $S\subset M$ whose closure in $M$ is $M$ itself, $\text{cl}_M(S)=M$.

Now suppose there is a homeomorphism $f:M\to N$. Must $N$ be separable?


My thoughts are as follows. Since $S$ is countable, so is $F=f(S)$. Using this, I broke up the problem into four cases, the last two of which I have been unable to complete.

Case 1:

If $S$ is closed in $M$, then $\text{cl}_M(S)=S=M$ and $F=f(S)=f(M)=N$, meaning that $\text{cl}_{N}(F)=N$. Thus $N$ is separable.

Case 2:

If $S$ is clopen (closed and open) in $M$, it is closed in $M$, and $N$ is separable.

Case 3:

If $S$ is open in $M$, $F$ is open in $N$, but this doesn't seem to help.

Case 4:

If $S$ is neither open nor closed in $M$, I have no idea what to do.

There must be a better way to go about this. Could I have some help? Thanks.

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A homeomorphism is a bijection that takes open sets to open sets, closed sets to closed sets, so it follows that it also respects closure, therefore $$\mathrm{cl}_N(f(S)) \ =\ f(\mathrm{cl}_M(S)) \ =\ f(M) \ =\ N$$

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  • $\begingroup$ Wow, that was extremely easy. Thank you. $\endgroup$ – clathratus Oct 9 at 18:14
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    $\begingroup$ To show that $\overline{f(S)} = f(\overline S)$, note first that $S\subset\overline S$, so $f(S)\subset f(\overline S)$, which is closed, so $\overline{f(S)}\subset f(\overline S)$. For the converse, let $y\in f(\overline S)$, i.e., $y = f(x)$ with $x\in\overline S$. Then there is $(x_n)\subset S$ such that $x_n\to x$. By continuity, $f(x_n)\to f(x) = y$, so $y\in\overline{f(S)}$. $\endgroup$ – amsmath Oct 9 at 18:19

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