First Countability of Finite Topological Products

Suppose that we have a finite collection of topological spaces $\{ X_1, X_2, ..., X_n \}$ that are all first countable. As the following theorem will show, the finite topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ will also be first countable.

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

• Proof: Let $\{ X_1, X_2, …, X_n \}$ be a collection of first countable topological spaces. To show that $\displaystyle{\prod_{i=1}^{n} X_i}$ is a first countable topological space we need to show that there exists a countable local basis for each element $\displaystyle{\mathbf{x} = (x_1, x_2, …, x_n) \in \prod_{i=1}^{n} X_i}$.

• Let $\displaystyle{\mathbf{x} = (x_1, x_2, …, x_n) \in \prod_{i=1}^{n} X_i}$. Note that each $x_i$ has a countable local basis $\mathcal B_i$ for each $i \in \{ 1, 2, …, n \}$ and so construct a countable local basis for $\mathbf{x}$ as:

(1)

\begin{align} \quad \mathcal B_{\mathbf{x}} = \left \{ \prod_{i=1}^{n} B_i : B_i \in \mathcal B_i \right \} \end{align}

• Note that $\mathcal B_{\mathbf{x}}$ is indeed countable since $\mathcal B_{\mathbf{x}}$ is a finite product of countable sets.

• Now to show that this collection is a local basis of $\mathcal B_{\mathbf{x}}$, let $\displaystyle{U = U_1 \times U_2 \times … \times U_n = \prod_{i=1}^{n} U_i \subseteq \prod_{i=1}^{n} X_i}$ be any open neighbourhood of $\mathbf{x}$.

• Since $U_1$ is an open neighbourhood of the first coordinate $x_1$ we have that there exists a local basis element $B_1 \in \mathcal B_1$ such that $x_1 \in B_1 \subseteq U_1$. Since $U_2$ is an open neighbourhood of the second coordinate $x_2$ we have that there exists a local basis element $B_2 \in \mathcal B_2$ such that $x_2 \in B_2 \subseteq U_2$. Furthermore, since $U_i$ is an open neighbourhood of the $i^{\mathrm{th}}$ coordinate $x_i$ we have that there exists a local basis element $B_i \in \mathcal B_i$ such that $x_i \in B_i \subseteq U_i$ for each $i \in \{1, 2, …, n \}$. Note that $\displaystyle{\prod_{i=1}^{n} B_i = B_1 \times B_2 \times … \times B_n \in \mathcal B_{\mathbf{x}}}$ and is such that:

(2)

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

• So for every open neighbourhood $\displaystyle{U \subseteq \prod_{i=1}^{n} X_i}$ of $\mathbf{x}$ there exists an element $\displaystyle{\prod_{i=1}^{n} B_i \in \mathcal B_{\mathbf{x}}}$ such that $\displaystyle{\mathbf{x} \in \prod_{i=1}^{n} B_i \subseteq U}$, and so $\mathcal B_{\mathbf{x}}$ is a local basis for $\mathbf{x}$. This shows that $\displaystyle{\prod_{i=1}^{n} X_i}$ is first countable. $\blacksquare$