I heard at some point (without seeing a proof) that every countable, compact space $X$ is homeomorphic to a countable successor ordinal with the usual order topology. Is this true? Perhaps someone can offer a sketch of the proof or suggest a topology/ordinal text which treats countable spaces in depth. I also wonder if these spaces are metrizable or if they can be embedded into $\mathbb{R}$.
By the way, I'm assuming $X$ is Tychonoff but perhaps this can be reduced to a weaker separation axiom.
Edit: Just to clarify and avoid erroneous future editing,"countable compact space" means "compact space, whose underlying set is countable." This is different from "countably compact space."
Best Answer
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:
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.