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Let $X$ and $Y$ metric spaces and $f:X\rightarrow Y$ a homeomorphism. Prove that: $X$ is separable iff $Y$ is separable.

My thoughts are: f, as it is defined, is surjective so $f(X)=Y$..that is so far i get, i'm thinking in equivalences of X being separable, like Lindeloff propierty, but i don't how to use that...any suggestions?

Ariel Marcelo Pardo
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2 Answers2

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Hint: Let $D$ be a countable dense subset of $X$

  • note that $f(D)$ is countable
  • prove that $\overline{f(D)} = Y$
Ben Grossmann
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  • @iolig $\Bbb R$ is separable under the usual topology. For instance, $\Bbb Q$ is a countable dense set. – Ben Grossmann Apr 20 '17 at 16:46
  • Just a thought: if we know that $f$ is a homeomorphism, $f$ is continuous and therefore preserves separable sets ($D$ is separable so $f(D)$ is separable) and that since $f$ is a homeomorphism $f(\overline{D}) = \overline{f(D)}$ - then we know that if $M$ is separable so is $N$? – Taylor Rendon Nov 24 '20 at 20:13
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    @Taylor The fact that homeomorphisms preserve separable sets is equivalent to the question being asked – Ben Grossmann Nov 24 '20 at 20:33
  • Gotcha. I believe my reason for asking was because my textbook asks you to prove if $f$ is continuous, it preserves separable sets and then later in the chapter it asks this question. From what you just said, wouldn't that mean those two questions are essentially the same? – Taylor Rendon Nov 24 '20 at 20:54
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    @Taylor I suppose "essentially the same" is a matter of perspective. Certainly, since a textbook is geared towards getting comfortable with definitions, it can be argued that there is value in the exercise of noticing that a homeomorphism and its inverse are continuous (which means that the earlier conclusion can be applied to the situation) – Ben Grossmann Nov 24 '20 at 20:59
  • That makes sense! So would it be correct to say that since $f$ is a continuous bijection that (and using your $D$ that you mentioned) $f(X) = f(\overline{D}) = \overline{f(D)} = Y$ and since $D$ is countable, so is $f(D)$. So $Y$ is separable and the other direction of the proof would be similar? – Taylor Rendon Nov 24 '20 at 21:19
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    @Taylor Yes, that would be correct – Ben Grossmann Nov 24 '20 at 21:24
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Suppose $X$ is dense, and let $x_n$ a countable dense set. Lets prove that $y_n := f(x_n)$ is dense in $Y$.

If $y_n$ is not dense, there is an open set $V_Y$ in $Y$ that doesn't contain any $y_n$, so $f^{-1}(V_Y)$ is an open set (by the continuity of $f$), that doesn't contain any $x_n$.

Otherwise, $x_n \in f^{-1}(V_Y) \Rightarrow f(x_n) \in V_Y \Rightarrow y_n \in V_Y)$.

So, $Y$ is separable. The other implication is analogous, since $f^{-1}$ is also a continuous function.

dqc
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