> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]], then it is **separable** if it has a [[Cardinality|countable]] [[Dense|dense]] subset. > [!theorem] > > Let $X$ be a separable [[Metric Space|metric space]] with no isolated points and $\seq{x_n} \subset X$ be dense, then $\bracs{x_n: n \ge N}$ is dense in $X$ for all $N \in \nat$. > > *Proof*. Let $y \in X$ and $\eps > 0$. Take > $ > \delta = \min\bracs{d(y, x_n): x_n \ne y, n \le N} > $ > Since $y$ is not isolated, there exists $y' \in B(y, \min(\eps, \delta))$ with $y' \ne y$. Let $\gamma = \min(d(y', y), \min(\eps, \delta) - d(y', y))$, then since $\seq{x_n}$ is dense, there exists $n \in \nat$ such that $d(y', x_n) < \gamma$. Since $d(y', x_n) < d(y', y)$, $x_n \ne y$. As $d(y, x_n) < \min(\eps, \delta) \le \delta$ as well, $n > N$. Therefore $d(y, x_n) < \eps$ with $n > N$. Thus $\bracs{x_n: n \ge N}$ is dense in $X$ as well.