Second Countability under Homeomorphisms on Topological Spaces

# Second Countability under Homeomorphisms on Topological Spaces

Recall from the Homeomorphisms on Topological Spaces page that if $X$ and $Y$ are topological spaces then a bijective map $f : X \to Y$ is said to be a homeomorphism if it is continuous and open.

Furthermore, if such a homeomorphism exists then we say that $X$ and $Y$ are homeomorphic and write $X \simeq Y$.

Also, recall from the Second Countable Topological Spaces page that a topological space $(X, \tau)$ is said to be second countable if there exists a basis $\mathcal B$ of the topology on $X$ that is countable.

We will now look at a nice topological property which says that if $f$ is a homeomorphism from $X$ to $Y$ and if $X$ is a second countable topological space then $Y$ is a second countable topological space.

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

**Proof:**Let $X$ be second countable. Then there exists a countable basis $\mathcal B$ of the topology defined on $X$. Then for all open sets $U$ in $X$ there exists a subset $\mathcal B^* \subseteq \mathcal B$ such that:

\begin{align} \quad U = \bigcup_{B \in \mathcal B^*} B \end{align}

- So then:

\begin{align} \quad f(U) = f \left ( \bigcup_{B \in \mathcal B^*} B \right ) = \bigcup_{B \in \mathcal B^*} f(B) \end{align}

- Since $f$ is a homeomorphism we have that $f(U)$ is an open set of $Y$. We claim that the following set is a countable basis of the topology on $Y$:

\begin{align} \quad \tilde{\mathcal B} = \{ f(B) : B \in \mathcal B \} \end{align}

- Clearly $\tilde{\mathcal B}$ is a countable set since $\mathcal B$ is a countable set. We only need to show that $\tilde{\mathcal B}$ is a basis of the topology on $Y$. Now suppose that $\tilde{\mathcal B}$ is not a basis of the topology on $Y$. Then there exists an open set $V$ in $Y$ such that for all subsets $\mathcal B^* \subseteq \mathcal B$ we have that:

\begin{align} \quad V \neq \bigcup_{B \in \mathcal B^*} f(B) \end{align}

- Then for all subsets $\mathcal B^* \subseteq \mathcal B$ we have that:

\begin{align} \quad f^{-1}(V) \neq \bigcup_{B \in \mathcal B^*} B \end{align}

- But since $f$ is a homeomorphism and $V$ is an open set in $Y$ we must have that $f^{-1} (V)$ is an open set in $X$. But then the open set $f^{-1}(V)$ cannot be written as any subcollection of the sets in the basis $\mathcal B$ of the topology on $X$. This contradicts $\mathcal B$ being a basis of the topology on $X$ and so the assumption that $\tilde{\mathcal B}$ is not a basis of the topology on $Y$ is false.

- Hence $\tilde{\mathcal B}$ is a countable basis of the topology on $Y$ so $Y$ is second countable. $\blacksquare$