Second Countability of Finite Topological Products

# Second Countability of Finite Topological Products

Recall from the First Countability of Finite Topological Products page that if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of first countable topological spaces then the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ is also first countable.

Suppose that $\{ X_1, X_2, ..., X_n \}$ is a finite collection of second countable topological spaces. In the following theorem we will similarly show that the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ inherits the second countable property.

Theorem 1: Let $\{ X_1, X_2, …, X_n \}$ be a finite collection of topological spaces. If $X_i$ is second countable for each $i \in I$ then the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ is also second countable. |

**Proof:**Let $\{ X_1, X_2, …, X_n \}$ be a finite collection of second countable topological spaces. We will show that $\displaystyle{\prod_{i=1}^{n} X_i}$ is also a second countable topological space by showing that there exists a countable basis for the product topology.

- Since each $X_i$ is second countable we have that the topology $\tau_i$ for each $i \in I$ has a countable basis, call it $\mathcal B_i$. Let $\mathcal B$ be the Cartesian product of all of these countable bases:

\begin{align} \quad \mathcal B = \prod_{i=1}^{n} \mathcal B_i \end{align}

- We claim that $\mathcal B$ is a countable basis of the product topology. Clearly it is countable since it is a finite product of countable sets.

- To show that this set is a basis of the product topology, let $\displaystyle{U \subset \prod_{i=1}^{n} X_i}$ be an open set in the product topology. Then $U = U_1 \times U_2 \times … \times U_n$ where $U_i$ is open in $X_i$ for each $i \in I$. So there exists a basis element $B_i \in \mathcal B_i$ such that $B_i \subseteq U_i$. Furthermore:

\begin{align} \quad \prod_{i=1}^{n} B_i \subseteq \prod_{i=1}^{n} U_i = U \end{align}

- So for every open set $U$ in the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ there exists an open set $\displaystyle{\prod_{i=1}^{n} B_i} \in \mathcal B$ such that $\displaystyle{\prod_{i=1}^{n} B_i} \subseteq U$. This shows that $\displaystyle{\prod_{i=1}^{n} X_i}$ is a second countable topological space. $\blacksquare$