Let $X,Y$ be topological spaces and $f \colon X \rightarrow Y $ continuous surjection. If $X$ is separable, is $Y$ separable?

Question

I'm not sure if this is valid statement. I tried to show it directly but I got stuck. Then I tried to think of an counterexample but I couldn't. Any hint helps! P.S. separable

Consider the image under $f$ of a countable dense subset of $X$. Is that dense in $Y$? — Angina Seng, Jun 01 '19 at 10:02

score 5 · Accepted Answer · answered Jun 01 '19 at 10:07

Let $D\subseteq X$ be countable and dense.

Then $f(D)$ is automatically countable and can be proved to be dense in $Y$.

If $U$ is a non-empty open set in $Y$ then $f^{-1}(U)$ is also open in $X$ (because $f$ is continuous) and secondly it is not empty (because $f$ is surjective).

So for some $x\in X$ we have $x\in f^{-1}(U)\cap D$ and consequently $f(x)\in U\cap f(D)$.

score 1 · Answer 2 · answered Jun 01 '19 at 11:02

1

Continuity of $f$ is equivalent to $\forall A \subseteq X: f[\overline{A}]\subseteq \overline{f[A]}$. So if $D$ is dense in $X$ and $f$ is onto:

$$Y=f[X]=f[\overline{D}]\subseteq \overline{f[D]} \subseteq Y$$

so $f[D]$ is dense in $Y$, and countable whenever $D$ is.

answered Jun 01 '19 at 11:02

Henno Brandsma

242,131

How is $\overline{f(D)} \subseteq Y$? – Ryukendo Dey Jun 09 '23 at 17:24

Let $X,Y$ be topological spaces and $f \colon X \rightarrow Y $ continuous surjection. If $X$ is separable, is $Y$ separable?

2 Answers2