Every countable compact Hausdorff space is metrizable
general-topology
Every countable compact Hausdorff space is metrizable?
How should I show it?
Best Answer
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.
Of course, if $X$ is metrizable then it is also locally metrizable.
For the other direction, you could really follow the hint. Supposed that $X$ is locally metrizable, there exists an open cover $\bigcup_{x\in X}U_x$ of $X$ where $U_x$ is a metrizable neighborhood of $x$. Because of compactness, it has a finite subcover: $\bigcup_{i<n} U_i=X$ where $U_i:=U_{x_i}$ for some $x_i$.
Because of metrizability, each $U_i$ has countable base $(V_{i,j})_j$ of open subsets, so that all finite intersections of these are still countable, and they give a base of $X$.
The space also has to satisfy the separation axiom $T_3$, but this holds as $X$ is Hausdorff and compact.
Since $X$ is second countable, all you need for $X^*$ to be second countable is a countable neighborhood base for the point at infinity, right? Since $X$ is locally compact and second countable, you can cover $X$ with countably many open sets whose closures are compact. Then every compact subset of $X$ is contained in a finite union of those open sets, and the complements of the closures of those unions will provide a countable neighborhood base for $\infty$.
Best Answer
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.