Metrizability of Finite Topological Products

# Metrizability of Finite Topological Products

Suppose that $\{ X_1, X_2, ..., X_n \}$ is a finite collection of metrizable topological spaces. As we will see in the following theorem, the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ will also be metrizable.

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

**Proof:**Let $\{ X_1, X_2, ..., X_n \}$ be a finite collection of metrizable topological spaces. Then there exists metrics $d_i : X_i \times X_i \to [0, \infty)$ such that the open sets in the metric spaces $(X_i, d_i)$ give the topology on the topological spaces $(X_i, \tau_i)$ for all $i \in \{ 1, 2, ... n \}$.

- Consider the finite topological product $\displaystyle{\prod_{i=1}^{n} X_i}$. Define a metric $\displaystyle{d : \prod_{i=1}^{n} X_i \times \prod_{i=1}^{n} X_i \to [0, \infty)}$ for all $\displaystyle{\mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n) \in \prod_{i=1}^{n} X_i}$ by:

\begin{align} \quad d(\mathbf{x}, \mathbf{y}) = \sqrt{[d_1(x_1, y_1)]^2 + [d_2(x_2, y_2)]^2 + ... + [d_n(x_n, y_n)]^2} \end{align}

- It's not hard to check that $d$ is indeed a metric. We now need to show that the open sets in $\displaystyle{\left ( \prod_{i=1}^{n} X_i, \tau \right )}$ (where $\tau$ is the product topology) are the same as the sets in the metric space $\displaystyle{\left ( \prod_{i=1}^{n} X_i, d \right )}$.

- Let $\displaystyle{U = \prod_{i=1}^{n} U_i \subseteq \prod_{i=1}^{n} X_i}$ be an open set with respect to the metric space above. Let $\mathbf{x} \in U$. Since $U$ is open, there exists a ball centered at $\mathbf{x}$ with radius $r > 0$ such that:

\begin{align} \quad \mathbf{x} \in B(\mathbf{x}, r) = \left \{ \mathbf{y} \in \prod_{i=1}^{n} X_i : d(\mathbf{x}, \mathbf{y}) < r \right \} \subseteq U \end{align}

- So if $\mathbf{y} \in B(\mathbf{x}, r)$ then:

\begin{align} \quad d(\mathbf{x}, \mathbf{y}) = \sqrt{[d_1(x_1, y_1)]^2 + [d_2(x_2, y_2)]^2 + ... + [d_n(x_n, y_n)]^2} < r \end{align}

- This is satisfied if $d_1(x_i, y_i) < \frac{r}{n}$ for all $i \in \{ 1, 2, ..., n \}$. Take $r^* = \min \{ d_i(x_i, y_i) : i \in \{1, 2, ..., n\} \}$. Then $r^* \< \frac{r}{n}$. Since each $X_i$ is metrizable and since $U$ is open in $\displaystyle{\prod_{i=1}^{n} X_i}$ we have that each $U_i$ is open in $X_i$ and so for $\mathbf{x} \in U$ we have that $x_i \in U_i$, and that ball $B \left (x_i, r^* \right )$ is such that:

\begin{align} \quad x_i \in B \left ( x_i, r^* \right) \subseteq U_i \end{align}

- Let $\displaystyle{V = \prod_{i=1}^{n} B(x_i, r^*)}$. Then $V$ is an open neighbourhood of $\mathbf{x}$ in the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ and furthermore:

\begin{align} \quad \mathbf{x} \in V = \prod_{i=1}^{n} B(x, r^*) \subseteq \prod_{i=1}^{n} U_i = U \end{align}

- This shows that the open sets in the metric space $\displaystyle{\left ( \prod_{i=1}^{n} X_i, d \right )}$ are the same as the open sets in the topological product $\displaystyle{\left ( \prod_{i=1}^{n} X_i, \tau \right )}$, so $\displaystyle{\prod_{i=1}^{n} X_i}$ is metrizable.