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Let $(X,d)$ and $(Y,d')$ be metric spaces and $f:X \longrightarrow Y$ a continuous function. I need to show that, if $X$ is separable, so the image of $f$, with $d'$, is separable too.

I've though in showing it by using the fact that, if $X$ is separable, then there existis an open base of $(X,d)$ that is countable. Since I already know that $(X,d)$ is separable, there is an bijection $\phi: \mathbb{N} \longrightarrow X$ in the form $\phi(n)=B(a_n,\delta)$ and, as $f$ is continuous, it takes the $B(a_n, \delta)$ to $B(f(a_n), \varepsilon)$. With that, I've though to say that there is a bijection $\pi=f \circ \phi: \mathbb{N} \longrightarrow Y$ in the form $\pi(n)=B(f(a_n),\varepsilon)$ to show that that base of $f(X)$ is countable, but i'm unsure if this is enough, or even right to conclude.

3 Answers 3

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I feel that the proof using general topology is much simpler and more intuitive.

Let $X$ and $Y$ be topological spaces and $f: X \rightarrow Y$ a continuous function. We want to show that if $X$ is separable then $f(X)$ is separable.

By separability of $X$, there is a countable subset $S \subset X$ that is dense in $X$, that is, any open set $U \subset X$ has nonempty intersection with $S$. Then, $f(S)$ will be dense in $f(X)$: if $V \subset f(X)$ is open, its preimage $f^{-1}(V)$ is open by continuity of $f$, thus $f^{-1}(V)$ contains some element $s \in S$, but this means that $f(s) \in V$. So, any open subset of $f(X)$ contains an element from $f(S)$. Since $S$ is countable, $f(S)$ is countable, too.

We proved that $f(S)$ is countable and dense in $f(X)$. Thus, $f(X)$ is separable.

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Let $D \subset X$ be a countable dense set in $X$. Let $y \in f(X)$ and let $x \in X$ be such that $f(x)=y$. There is a sequence $(x_n)$ of elements of $D$ such that $x_n \to x$ as $n \to \infty$. Since $f$ is continuous $f(x_n) \to f(x)$. Thus $f(D)$ is dense in $f(X)$. Since $f(D)$ is countable, $f(X)$ is separable.

EDIT: Some typos.

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    I'll like to note that, while this proof works whenever $X$ is first countable, and in particular a metric space, it doesn't for $X$ a general topological space. – 2017-02-06
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One of several equivalent def'ns of continuity is that $f:X\to Y$ is continuous iff $Cl_Y(f(A))\supset f(Cl_X(A))$ for every $A\subset X.$

So if $A$ is a dense subset of $X$ and $f:X\to Y$ is continuous then $$Cl_{f(X)}(f(A))=f(X)\cap Cl_Y (f(A))\supset f(X)\cap f(Cl_X(A))=f(X)\cap f(X)=f(X).$$ That is, $f(A)$ is a dense subset of the space $f(X).$

Note that all of this applies to any spaces $X,Y$.