Is separability a topological property?

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.

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 $$\mathrm{cl}_N(f(S)) \ =\ f(\mathrm{cl}_M(S)) \ =\ f(M) \ =\ N$$
• 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 $y\in\overline{f(S)}$. – amsmath Oct 9 at 18:19