Countable Topological Spaces are Separable Topological Spaces

Recall from the Separable Topological Spaces page that a topological space $(X, \tau)$ is said to be separable if it contains a countable and dense subset.

We will now look at an extremely simple theorem which says that if $X$ is a countable set, then $(X, \tau)$ is a separable topological space regardless of the topology $\tau$.

Theorem 1: Let $(X, \tau)$ be a topological space. If $X$ is a countable set then $(X, \tau)$ is a separable topological space.

• Proof: Let $X$ be a countable set and let $\tau$ be any topology on $X$. By definition, the topology $\tau$ must contain the whole set $X$.

• Let $A = X$. Then $A$ is a countable set. Furthermore, for all $U \in \tau \setminus \{ \emptyset \}$ we have that $U \subseteq A$. Therefore:

(1)

\begin{align} \quad A \cap U \neq \emptyset \end{align}

• So $A$ is also a dense set. Therefore, $A$ is a countable set subset (namely the whole set) of $X$, so $(X, \tau)$ is a separable topological space. $\blacksquare$