> [!theorem] > > Let $(X, d)$ be a [[Metric Space|metric space]]. $X$ is **separable** if it has a countable [[Dense|dense]] subset. > [!theorem] > > Let $(X, d)$ be a separable metric space, then $X$ is [[Second Countable|second countable]]. > > *Proof*. Let $S$ be a countable dense subset of $X$, and $U \in \topo_X$ be any [[Open Set|open set]]. Let $x \in S \cap U$, then > $ > \exists \varepsilon > 0: B(x, \varepsilon) \subseteq U > $ > Take $s = \sup\bracs{\varepsilon > 0: B(x, \varepsilon) \subseteq U}$. Let $y \in B(x, s)$, then $d(x, y) < s$. Write $\delta = s - d(x, y)$, then $\exists \varepsilon: \bracs{\varepsilon > 0: B(x, \varepsilon) \subseteq U}$ such that $\varepsilon \in (s - d(x, y), s)$ and $x \in B(x, \varepsilon) \subseteq U$. Therefore $B(x, s) \subseteq U$. > > Let $s_x = \sup\bracs{\varepsilon > 0: B(x, \varepsilon) \subseteq U}$ and $\seq{\varepsilon_{x, i}}$ be an increasing sequence of [[Rational Numbers|rational numbers]] such that $\varepsilon_{x, i} \upto s$. Then > $ > \bigcup_{i \in \nat}B(x, \varepsilon_{x, i}) \subseteq B(x, s_x) > $ > and for any $y \in B(x, s_x)$, we can find some $\varepsilon_{x, i}$ such that $y \in B(x, \varepsilon_{x, i})$, like from earlier. Therefore > $ > \bigcup_{i \in \nat}B(x, \varepsilon_{x, i}) = B(x, s_x) > $ > > Let $B_x = B(x, s_x)$, then since $B_x \subseteq U$, we have > $ > \bigcup_{x \in S \cap U}B_x \subseteq U > $ > Now, let $y \in U$, then since $S$ is dense in $X$, $S \cap U$ is dense in $U$. We have $U$ being open implying that > $ > \exists \varepsilon > 0: B(y, \varepsilon) \subseteq U > $ > and $\exists x \in S \cap U: d(x, y) < \varepsilon / 2$. Let $z \in B(y, \varepsilon/2)$, then > $ > d(x, z) \le d(x, y) + d(y, z) < \varepsilon > $ > and $s_x \ge \varepsilon_2$ meaning that $x \in B(x, s_x)$ and > $ > \bigcup_{x \in S \cap U}B_x \supseteq U > $ > > This means that we can decompose any $U \in \topo_X$ into > $ > U = \bigcup_{x \in S \cap U}B_x = \bigcup_{x \in S \cap U}\bigcup_{i \in \nat}B(x, s_{x, i}) > $ > and we have found a base > $ > B = \bracs{B(x, r): x \in S, r \in \rational} > $ > that generates any open set. Since $|S| \le |\rational| = |\nat|$, $|B| \le |\nat \times \nat| = |\nat|$ is a countable base.