a compact metric space is second countableCompactMetricSpacesAreSecondCountable2007-04-28 18:45:392008-06-20 02:30:50Theoremsecond countablecompactmetric spacemetrizableopen ballbasisopen covercountable%%packages
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\begin{proposition*}
Every compact metric space is second countable.
\end{proposition*}
\begin{proof}
Let $(X,d)$ be a compact metric space, and for each $n\in\mathbb{Z}^+$ define $\mathcal{A}_n=\{B(x,1/n):x\in X\}$, where $B(x,1/n)$ denotes the open ball centered about $x$ of \PMlinkid{radius}{1296} $1/n$. Each such collection is an open cover of the compact space $X$, so for each $n\in\mathbb{Z}^+$ there exists a finite collection $\mathcal{B}_n\subseteq\mathcal{A}_n$ that \PMlinkescapetext{covers} $X$. Put $\mathcal{B}=\bigcup_{n=1}^\infty\mathcal{B}_n$. Being a countable union of finite sets, it follows that $\mathcal{B}$ is countable; we assert that it forms a basis for the metric topology on $X$. The first property of a basis is satisfied trivially, as each set $\mathcal{B}_n$ is an open cover of $X$. For the second property, let $x,x_1,x_2\in X$, $n_1,n_2\in\mathbb{Z}^+$, and suppose $x\in B(x_1,1/n_1)\cap B(x_2,1/n_2)$. Because the sets $B(x_1,1/n_1)$ and $B(x_2,1/n_2)$ are open in the metric topology on $X$, their intersection is also open, so there exists $\epsilon>0$ such that $B(x,\epsilon)\subseteq B(x_1,1/n_1)\cap B(x_2,1/n_2)$. Select $N\in\mathbb{Z}^+$ such that $1/N<\epsilon$. There must exist $x_3\in X$ such that $x\in B(x_3,1/2N)$ (since $\mathcal{B}_{2N}$ is an open cover of $X$). To see that $B(x_3,1/2N)\subseteq B(x_1,1/n_1)\cap B(x_2,1/n_2)$, let $y\in B(x_3,1/2N)$. Then we have
\begin{equation}
d(x,y)\leq d(x,x_3)+d(x_3,y)<\dfrac{1}{2N}+\dfrac{1}{2N}=\dfrac{1}{N}<\epsilon\text{,}
\end{equation}
so that $y\in B(x,\epsilon)$, from which it follows that $y\in B(x_1,1/n_1)\cap B(x_2,1/n_2)$, hence that $B(x_3,1/2N)\subseteq B(x_1,1/n_1)\cap B(x_2,1/n_2)$. Thus the countable collection $\mathcal{B}$ forms a basis for a topology on $X$; the verification that the topology \PMlinkescapetext{induced} by $\mathcal{B}$ is in fact the metric topology follows by an \PMlinkescapetext{argument similar} to that used to verify the second property of a basis, and completes the proof that $X$ is second countable.
\end{proof}\
It is worth nothing that, because a countable union of countable sets is countable, it would have been sufficient to assume that $(X,d)$ was a Lindel\"{o}f space.