> [!theoremb] Theorem > > Let $X$ be a [[Second Countable|second countable]] [[Regular Space|regular space]], then > 1. $X$ is [[Normal Space|normal]]. > 2. There exists a countable family $\cf \subset C(X, [0, 1])$ that separates points and closed sets. > 3. $X$ is metrisable. > > *Proof*. Let $\cb$ be a countable base for $X$. > > $(1)$: Let $A, B \subset X$ be disjoint [[Closed Set|closed]] sets. For each $x \in A$, there exists an open [[Neighbourhood|neighbourhood]] $U_x \in \cn^o(x)$ such that $\overline{U_x} \cap B = \emptyset$, and an open set $U_x' \in \cn^o(x) \cap \cb$ with $U_x' \subset U_x$. Since $\cb$ is countable, $\bracs{U_x': x \in A}$ is countable. Enumerate $\seq{U_j} = \bracs{U_x': x \in A}$, then $\bigcup_{n \in \nat}U_n \supset A$. Similarly, there exists $\seq{V_j} \subset \cb$ such that $\bigcup_{n \in \nat}V_n \supset B$ and $\overline{V_n} \cap A = \emptyset$. > > For each $n \in \nat$, define > $ > U_n' = U_n \setminus \bigcup_{j = 1}^n \overline{V_j} \quad V_n' = V_n \setminus \bigcup_{j = 1}^n \overline{U_j} > $ > then $U_n'$ and $V_n'$ are both open. Moreover, for any $m, n \in \nat$ with $m \le n$, $\overline{U_m} \subset (V_n')^c$, and $U_m' \cap V_n' = \emptyset$. Now, let > $ > U' = \bigcup_{n \in \nat}U_n' \quad V' = \bigcup_{n \in \nat}V_n' > $ > then $U' \in \cn^o(A)$, $V' \in \cn^o(B)$ with $U' \cap V' = \emptyset$. > > $(2)$: Let > $ > \cl = \bracsn{(E, F) \in \cb^2: \overline{E} \subset F} > $ > By [[Urysohn's Lemma]], for each $(E, F) \in \cl$, there exists $f_{(E, F)} \in C(X, [0, 1])$ such that $f|_{E} = 1$ and $f|_{F^c} = 0$. > > Let $x \in X$ and $K \subset X$ be closed with $x \not\in K$. Since $X$ is normal, there exists $F \in \cn^o(x) \cap \cb$ such that $F \subset K^c$, and $E \in \cn^o(x) \cap \cb$ such that $\overline{E} \subset F$. In which case, $f_{(E, F)}$ separates $x$ from $K$. > > Therefore $\cf = \bracs{f_{(E, F)}: (E, F) \in \cl}$ is a desired countable family. > > $(3)$: Since $X$ is $T^1$ and $\cf$ separates points and closed sets, the [[Embedding in Cube|embedding into a cube]] $e: X \to [0, 1]^\cf$ is a [[Homeomorphism|homeomorphism]] onto its image. As $\cf$ is countable, $[0, 1]^\cf$ is a metric space, thus $X$ is metrisable.