Separable Topological Spaces

Recall from the Dense and Nowhere Dense Sets in a Topological Space page that if $(X, \tau)$ is a topological space then a set $A \subseteq X$ is said to be dense if the intersection of $A$ with every open set (excluding the empty set) is nonempty, that is, for all $U \in \tau \setminus \{ \emptyset \}$ we have that:

(1)

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

Furthermore we said that $A$ is said to be nowhere dense if the interior of the closure of $A$ is empty, that is:

(2)

\begin{align} \quad \mathrm{int} (\bar{A}) = \emptyset \end{align}

We will now classify a special type of topological space which contains a countable and dense subset.

Definition: Let $(X, \tau)$ be a topological space. Then $X$ is said to be Separable if $X$ contains a countable and dense subset.

Consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology of open intervals on $\mathbb{R}$. Recall that the set of rational numbers $\mathbb{Q}$ is dense in $\mathbb{R}$. This is because any open set of $\mathbb{R}$ is either the empty set or the union of an arbitrary collection of open intervals. However, every open interval contains a rational number and the intersection of $\mathbb{Q}$ with any open interval is nonempty. Therefore, the intersection of $\mathbb{Q}$ with any open set $U \in \tau$ is nonempty and so for all $U \in \tau \setminus \{ \emptyset \}$ we have that:

(3)

\begin{align} \quad \mathbb{Q} \cap U \neq \emptyset \end{align}

Furthermore the set of rational numbers is countable, so the subset $\mathbb{Q}$ of $\mathbb{R}$ is a countable and dense subset of $\mathbb{R}$ with respect to the topology $\tau$. Hence, $(\mathbb{R}, \tau)$ is a separable topological space.