Separability of Finite Topological Products

Separability of Finite Topological Products

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

We will now look at another property that is inherited from a finite collection of topological spaces - separability!

 Theorem 1: Let $\{ X_1, X_2, …, X_n \}$ be a finite collection of topological spaces. If $X_i$ is separable for each $i \in I$ then the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ is also separable.
• Proof: Let $\{ X_1, X_2, ..., X_n \}$ be a finite collection of separable topological spaces. Then each $X_i$ contains a countable and dense subset $A_i$.
• We form a countable and dense subset of $\displaystyle{\prod_{i=1}^{n} X_i}$ as the product of all of these countable dense subsets:
(1)
\begin{align} \quad A = \prod_{i=1}^{n} A_i \end{align}
• Clearly $A$ is countable since it is the product of a finite collection of countable sets. Furthermore, we claim that $A$ is dense in $\displaystyle{\prod_{i=1}^{n} A_i}$. To show this, let $U = U_1 \times U_2 \times ... \times U_n \in \prod_{i=1}^{n} X_i$ be open. Then $U_i$ is open in $X_i$ for each $i \in I$. Since $A_i$ is dense in $U_i$ we have that:
(2)
\begin{align} \quad A_i \cap U_i \neq \emptyset \end{align}
• So there exists a point $a_i \in A_i \cap U_i$ for each $i \in I$. So the point $\mathbf{a} = (a_1, a_2, ..., a_n) \in A \cap U$ which shows that $A \cap U \neq \emptyset$ for each open set $U$ in the topological product.
• Therefore $\displaystyle{\prod_{i=1}^{n} X_i}$ is separable. $\blacksquare$