Second Countable Topological Spaces are Separable Topo. Spaces

Recall from the Second Countable Topological Spaces page that the topological space $(X, \tau)$ is said to be second countable if there exists a countable basis $\mathcal B$ of $\tau$.

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

We will now look at a rather nice theorem which says that every second countable topological space is a separable topological space.

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

• Proof: Let $(X, \tau)$ be a second countable topological space. Then there exists a countable basis $\mathcal B = \{ B_1, B_2, ..., B_n, ... \}$ of $\tau$. Since $\mathcal B$ of $\tau$ is a basis of $\tau$ we have that every open set $U \in \tau$ can be expressed as the union of sets in some subcollection $\mathcal B^* \subseteq \mathcal B$. In particular:

(1)

\begin{align} \quad U = \bigcup_{B \in \mathcal B^*} B \end{align}

• We must now construct a countable dense subset of $X$. Assume that $\mathcal B$ does not contain the empty set. If it does contain the empty set then we can discard it. Then for each $B_n \in \mathcal B$ take $x \in B_n$ and define the set $A$ as:

(2)

\begin{align} \quad A = \{ x_n : x_n \: \mathrm{is \: any \: element \: in \:} B_n, n = 1, 2, ... \} \end{align}

• Then $A$ is a countable subset of $X$ since we take one element from each set in the countable basis.

• Furthermore, for all $U \in \tau \setminus \{ \emptyset \}$ we have that $A \cap U \neq \emptyset$ because $A$ contains one element from each of the basis sets and $U$ is the union of some subcollection of the basis sets. Therefore $A$ is a dense subset of $X$.

• Hence $A$ is a countable dense subset of $X$, so $(X, \tau)$ is a separable topological space. $\blacksquare$

Note: The above theorem tells us that every second countable topological space is separable. The converse is NOT true in general! There are separable topological spaces that are not second countable.

However, if $X$ is a metric space then separability and second countability are equivalent properties of topological spaces.