Answer to Question #217683 in Differential Geometry | Topology for Prathibha Rose

Question

Prove that seperability is a topological property

Answer 1

Suppose that $X$ is a separable space and $Y$ a topological space with $f:X\to Y$ a homeomorphism between them. Let us prove that $Y$ is separable.

As $X$ is separable, there exists a countable dense subset $D$. We claim that $D':=f(D)$ is a countable dense subset of $Y$ First of all, as $f$ is a homeomorphism, it is, in particular, a bijection, so $D'$ is countable. Secondly, we remark that for a continuous function $h:A\to B$ we have $\overline{h^{-1}(C)}\subseteq h^{-1}(\overline{C})$ for any $C\subseteq B$. Indeed, as $h$ is continuous, $h^{-1}(\overline{C})$ is a closed set containing $h^{-1}(C)$, it contains, in particular, its closure. Finally, we apply this as following : $\overline{f^{-1}(D')} \subseteq f^{-1}(\overline{ D'})$, but $\overline{f^{-1}(D')}=\overline{D}=X$ (as $f$ is bijective, $f^{-1}(D')=D$) , so $X\subseteq f^{-1}(\overline{D'})\subseteq X$ and therefore $f^{-1}(\overline{D'})=X$. Applying $f$ to both parts, remembering that it is bijective, we conclude that $\overline{D'}=Y$. Therefore, $D'$ is a countable dense subset of $Y$, so it is separable.