Prove the image of separable space under continuous function is separable.

Question

Let $f:X\to Y$ be continuous. Show that if $X$ has a countable dense subset, then $f(X)$ satisfies the same condition.

Answer 1

HINT: This is a case in which the most obvious thing to try does work. $X$ is separable, so it has a countable dense subset $D$. Show that $f[D]$ is a countable dense subset of $f[X]$. This is extremely easy if you use the fact that $f$ is continuous if and only if $f^{-1}[U]$ is open in $X$ for each open $U\subseteq Y$.