> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. $X$ is **second countable** if $X$ has a [[Cardinality|countable]] [[Base|base]]. > [!theorem] > > Let $(X, \topo)$ be a second countable space. Then it is also [[Separable Topological Space|separable]]. > > *Proof*. Let $\ce$ be a countable base for $\topo$, pick $x_U \in U$ for every $U \in \ce$, and let $E = \bracs{x_U: U \in \ce}$ be a collection of all such points, which would be countable since $\ce$ is countable. > > Let $x \in X$ and $V$ be a [[Neighbourhood|neighbourhood]] of $x$, then it contains an open set $U \in \topo$ where $x \in U \subset V$. Since $U \in \topo$ is a union of $\ce$ sets, $\exists W \in \ce$ such that $x, x_W \in W \subset U \subset V$ and $V \cap E \ne \emptyset$. So $x$ is an [[Accumulation Point|accumulation point]] of $E$. > > Since $\ol{E} = E \cup \text{acc}(E)$, $x \in \ol{E}$ and $X = \ol{E}$. Therefore $E$ is a countable dense subset, and $(X, \topo)$ is separable.