Every countable compact Hausdorff space is metrizable?

Question

Every countable compact Hausdorff space is metrizable? How should I show it?

Answer 1

Let $X=\{x_n: n \in \Bbb N\}$ be an enumeration of $X$. For each pair $(n,m)$ with $n \neq m$ pick disjoint open sets $U_{n,m}, V_{n,m} \in \mathcal{T}$ such that $x_n \in U_{n,m}, x_m \in V_{n,m}$. This can by done as $(X,\mathcal{T}$ is Hausdorff.

Then denote by $\mathcal{T}'$ the topology on $X$ that is generated by the subbasis $\{U_{n,m}, V_{n,m}: n,m \in \Bbb N, n \neq m\}$ and note that $\mathcal{T}' \subseteq \mathcal{T}$, so that $1_X: (X, \mathcal{T}) \to (X,\mathcal{T}')$ is continuous. And so we have a bijection (clearly) from a compact space $(X,\mathcal{T})$ to a Hausdorff space $(X,\mathcal{T}')$, so this map is a homeomorphism (a closed and continuous bijection) and so $1_X$ si open as well, and this immediately implies $\mathcal{T}=\mathcal{T}'$. But the latter has a countable subbase, and hence a countable base, and hence $(X,\mathcal{T})$ is second countable, and it's also regular (even normal, being compact and Hausdorff), so Urysohn's metrisation theorem tells us that $X$ is metrisable.

Answer 2

For variety, here is another proof of the same result, not as slick as Henno's proof though.

Compact Hausdorff spaces are regular. And countable spaces that are first countable are also second countable. So by the Urysohn metrization theorem it is enough to show that $X$ is first countable.

Fix a point $a\in X$. Let $(x_n)_{n\in\mathbb{N}}$ be an enumeration of all the points different from $a$. We can pick an open nbhd $V_1$ of $a$ not containing $x_1$. Then for each $n>1$, by regularity plus $T_2$ we can choose a nbhd $V_n$ of $a$ not containing $x_n$ and such that $V_n\subseteq\overline{V_n}\subseteq V_{n-1}$. This gives a decreasing sequence of nbhds of $a$ such that $\bigcap_nV_n=\bigcap_n\overline{V_n}=\{a\}$. Now given an arbitrary open nbhd $U$ of $a$, the intersection $\bigcap_n\overline{V_n}$ is contained in $U$, so $\bigcap_n(\overline{V_n}\setminus U)=\emptyset$. Each of the $\overline{V_n}\setminus U$ is closed, and they are nested, so by the finite intersection property of compact spaces, one of the $\overline{V_n}\setminus U$ is empty, and hence $V_n\subseteq\overline{V_n}\subseteq U$.