First Countability under Homeomorphisms on Topological Spaces

# First 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 First Countable Topological Spaces page that a topological space $(X, \tau)$ is said to be first countable if every $x \in X$ has a countable local basis.

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 first countable topological space then $Y$ is a first countable topological space.

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

**Proof:**Let $X$ be a first countable topological space. Then every $x \in X$ has a countable local basis. For each $x \in X$ denote a countable local basis of $x$ by $\mathcal B_x$.

- By definition, each countable local basis $\mathcal B_x$ for each $x \in X$ is a collection of open neighbourhoods of $x$ such that that for all open neighbourhoods $U$ in $X$ of $x$ we have that there exists a $B \in \mathcal B_x$ such that $x \in B \subseteq U$.

- Now, since each $B \in \mathcal B_x$ is an open set in $X$ and $x \in B$, then $f(x) \in f(B)$ for each $B \in \mathcal B_x$, and since $f$ is a homeomorphism we see that $f(B)$ is open in $Y$ for each $B \in \mathcal B_x$ containing $f(x)$ - i.e., an open neighbourhood of $f(x)$.

- Furthermore, since $f$ is a bijection we have that for each $y \in Y$ that $y = f(x)$ for some $x \in X$. So, consider the following set of open neighbourhoods of $y = f(x)$:

\begin{align} \quad \mathcal B_{y} = \{ f(B) : B \in \mathcal B_x \} \end{align}

- We claim that $\mathcal B_y$ is a countable local basis for each $y = f(x) \in Y$.

- Clearly $\mathcal B_y$ is countable since $\mathcal B_x$ is countable, so it remains to show that $\mathcal B_y$ is a local basis for $y = f(x)$.

- Suppose that $\mathcal B_y$ is not a local basis of $y = f(x)$. Then there exists an open neighbourhood $V$ in $Y$ of $y = f(x)$ such that for all $f(B) \in \mathcal B_y$ we have that $f(x) \in f(B) \not \subseteq V$. But then $x \in B \not \subseteq f^{-1}(V)$ for all $B \in \mathcal B_x$.

- However, since $V$ is an open neighbourhood of $y = f(x)$ we have that $y = f(x) \in V$ so $f^{-1}(y) = x \in f^{-1}(V)$, so $f^{-1}(V)$ is an open neighbourhood of $x$ such that there exists no element $B \in \mathcal B_x$ such that $x \in B \subseteq f^{-1}(V)$. But this contradicts the fact that $\mathcal B_x$ is a local basis of $x$. Therefore the assumption that $\mathcal B_y$ is not a local basis of $y = f(x)$ is false.

- Hence, for all $y \in Y$, the collection $\mathcal B_y$ of open neighbourhoods of $y$ is a countable local basis of $y$. Therefore, $Y$ is first countable. $\blacksquare$