Proposition

Suppose $X$ is a second countable space. Then $X$ contains a countable dense subset.

Solution

Let $\mathcal{B}$ be a countable basis for $X$. For each nonempty $B \in \mathcal{B}$, we will pick a point $x_B \in B$ arbitrarily. We claim that $A = \{ x_B \mid B \in \mathcal{B}, B \ne \emptyset \}$ is a countable dense subset.

• $A$ is countable because the map $x_B \mapsto B$ is an injection from $A$ into $\mathcal{B}$.
• We will show that $A$ is dense. Let $x \in X$. If $x \in \overline{A}$, we are done. Suppose otherwise. Since $\overline{A}$ is the intersection of all closed subsets of $X$ containing $A$, this implies the existence of a closed set $C \subset X$ such that $x \notin C$ and $A \subset C$. Since $C$ is closed, $C^c$ is open. Moreover, $C^c$ is an open neighborhood of $x$. Since $C^c$ is an open set and $x \in C^c$, there must exist a basis element $B \in \mathcal{B}$ such that $x \in B \subset C^c$. Then $x_B \in C^c$. However, this is impossible because $x_B \in A \subset C$. Therefore, $x$ must be in $\overline{A}$, so $A$ is indeed dense.

Therefore, we showed the existence of a countable dense subset of $X$.