Second Countable Topological Spaces are Separable Topo. Spaces

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. $\blacksquare$
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