Exercise 30.4

Proof. For given $n\in {\mathbb{Z}}_{\text{+}}$, we have an open cover of $X$ by $B(x,1∕n)$ for each $x\in X$; since $X$ is compact, let $A_n$ be the finite subcover. ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n{A}_n$ is countable since it is the countable union of finite sets; we claim it is a basis for $X$. Let $U\subseteq X$ open and $x\in U$. Since $X$ is metrizable, there exists $B(x,\delta )\subseteq U$ for some $\delta >0$. Let $N$ such that $2∕N<\delta$. Since $A_N$ covers $X$, there exists $B(y,1∕N)\ni x$. $B(y,1∕N)\subseteq B(x,\delta )$, for if we choose $z\in B(y,1∕N)$, $d(x,z)\le d(x,y)+d(y,z)\le 1∕N+1∕N<\delta$. Thus, $x\in B(y,1∕N)\subseteq U$, and so ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_n{A}_n$ is a countable basis by Lemma $13.2$. □