I believe that you are looking for ideas from the Cantor Bendixson theorem.
The main idea of the proof is the Cantor-Bendixson derivative. Given a closed set $X$, the derived set $X'$ consists of all limit points of $X$. That is, one simply throws out the isolated points. Continuing in a transfinite sequence, one constructs $X_\alpha$ as follows:
- $X_0=X$, the original set.
- $X_{\alpha+1}=(X_\alpha)'$, the set of limit points of $X_\alpha$.
- $X_\lambda=\bigcap_{\alpha\lt\lambda}X_\alpha$, for limit ordinals $\lambda$.
Thus, $X_1$ consists of the limit points of $X$, and $X_2$ consists of the limits-of-limits, and so on. The set $X_\omega$ consists of points that are $n$-fold limits for any particular finite $n$, and $X_{\omega+1}$ consists of limits of those kind of points, and so on. The process continues transfinitely until a set is reached which has no isolated points; that is, until a perfect set is reached. The Cantor Bendixon rank of a set is the smallest ordinal $\alpha$ such that $X_\alpha$ is perfect.
The concept is quite interesting historically, since Cantor had undertaken this derivative before he developed his set theory and the ordinal concept. Arguably, it is this derivative concept that led Cantor to his transfinite ordinal concept.
It is easy to see that the ordinal $\omega^\alpha+1$ under the order topology has rank $\alpha+1$, and one can use this to prove a version of your desired theorem.
The crucial ingredients you need are the Cantor Bendixson rank of your space and the number of elements in the last nonempty derived set. From this, you can constuct the ordinal $(\omega^\alpha+1)\cdot n$ to which your space is homeomorphic.
Meanwhile, every countable ordinal is homeomorphic to a subspace of $\mathbb{Q}$, and is metrizable. The compact ordinals are precisely the successor ordinals (plus 0).
Update 5/11/2011. This brief article by Cedric Milliet contains a proof of the Mazurkiewicz-Sierpiński theorem (see
Stefan Mazurkiewicz and Wacław Sierpiński, Contribution à la topologie des
ensembles dénombrables, Fundamenta Mathematicae 1, 17–27, 1920), as follows:
Theorem 4. Every countable compact
Hausdorff space is homeomorphic to some well-ordered set with the order topology.
The article proves more generally that any two countable locally compact Hausdorff spaces $X$ and $Y$ of same Cantor-Bendixson rank and degree are homeomorphic. This is proved by transfinite induction on the rank, and the proof is given on page 4 of the linked article.
The argument is basically just fine, but the notation in that last line is a bit off: when you write $\bigcup_{n\in\mathbb{N}}$, you’re saying that you want to take the union of some sets indexed by natural numbers. What follows $\bigcup_{n\in\mathbb{N}}$ should be a typical one of these sets with index $n$. If you knew that every member of $\mathcal{B}$ was compact, for instance, you could write $X = \bigcup_{n\in\mathbb{N}}\operatorname{cl}B_n$. What you want here is simply $X = \bigcup \{\operatorname{cl}B:B \in \mathcal{B}\text{ and}\operatorname{cl}B\text{ is compact}\}$.
If fact, you never needed to index $\mathcal{B}$ in the first place. It’s perfectly fine to say this:
- Let $\mathcal{B}$ be a countable base. Since $X$ is locally compact, for each $x\in X$ there is an open nbhd $U_x$ of $x$ with compact closure, and since $\mathcal{B}$ is a base for $X$, there is some $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq U_x$. Clearly each $B_x$ has compact closure, and $X = \bigcup\limits_{x\in X}\operatorname{cl}B_x$. There are only countably many distinct members of $\mathcal{B}$, so there are only countably many different sets $B_x$, and we have therefore expressed $X$ as the union of countably many compact subsets, as desired.
Or you could replace that last sentence with something like this:
- $\{B_x:x\in X\} \subseteq \{B \in \mathcal{B}:\operatorname{cl}B\text{ is compact}\} \subseteq \mathcal{B}$, so $\{B_x:x\in X\}$ is countable, and $X = \bigcup\limits_{x\in X}\operatorname{cl}B_x$ therefore expresses $X$ as a countable union of compact subsets.
There are many other perfectly good ways to do it; I’m just trying to give you some idea of what’s possible, since you’re working on the proof-writing as much as you are on the mathematics itself.
Best Answer
In fact, thanks to Sierpinski-Mazurkiewicz theorem, one precisely knows all Hausdorff countable compact topological spaces; namely, any such space is homeomorphic to $\omega^{\alpha} \cdot n +1$ for some countable ordinal $\alpha$ and some integer $n \geq 1$. See for example my answer here. Now, it is not difficult to verify that $\omega^{\alpha} \cdot n +1$ is second-countable.