Separability of Finite Topological Products

Table of Contents

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.

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