Separability under Homeomorphisms on Topological Spaces

# Separability under Homeomorphisms on Topological Spaces

Recall from the Dense and Nowhere Dense Sets in a Topological Space page that if $(X, \tau)$ is a topological space then a subset $A \subseteq X$ is said to be dense in $X$ if for every open set $U$ in $X$ we have that:

(1)\begin{align} \quad A \cap U \neq \emptyset \end{align}

Also recall from the Separable Topological Spaces page that a topological space $(X, \tau)$ is said to be separable if it contains a countable and dense subset.

We will now see that if homeomorphisms preserve the separability property between topological spaces.

Theorem 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$ be a homeomorphism. If $X$ is a separable space then $Y$ is separable. |

**Proof:**Let $X$ be a separable topological space and let $f : X \to Y$ be a homeomorphism between $X$ and $Y$.

- Since $X$ is separable, there exists a subset $A \subseteq X$ that is both countable and dense, i.e., for all open sets $U$ in $X$ we have that:

\begin{align} \quad A \cap U \neq \emptyset \end{align}

- We claim that the set $f(A)$ is a countable and dense subset of $Y$.

- Since $f$ is surjective and $A$ is a countable subset of $X$ we see that $f(A)$ is a countable subset of $Y$.

- Now let $V$ be an open set in $Y$. Then since $f$ is continuous we have that $f^{-1}(V)$ is an open set in $X$ and since $A$ is dense in $X$, this implies that:

\begin{align} \quad A \cap f^{-1}(V) \neq \emptyset \end{align}

- Therefore:

\begin{align} \quad f(A \cap f^{-1}(V)) & \neq \emptyset \\ \quad f(A) \cap f(f^{-1}(V)) & \neq \emptyset \\ \quad f(A) \cap V & \neq \emptyset \end{align}

- But this holds for all open sets $V$ in $Y$, so $f(A)$ is dense in $Y$.

- Therefore $f(A)$ is a countable and dense subset of $Y$, so $Y$ is separable. $\blacksquare$