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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."

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I'm not 100% sure that this fits [descriptive-set-theory], it does seems to be somewhere on the border though. –  Asaf Karagila May 11 '11 at 6:05
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Hausdorff will do (with compact it implies even normality); and T_1 will not (cofinite spaces, e.g.). –  Henno Brandsma May 11 '11 at 21:10
    
@Paul You have changed the question to be something that I wasn't asking. I don't believe countably compact is equivalent to compact and countable. –  J.K.T. May 19 '13 at 14:12
    
@J.K.T.: Sorry. It does. –  Paul May 20 '13 at 1:39

3 Answers 3

up vote 10 down vote accepted

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.

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I believe the name is BendixSon. –  Raeder May 10 '11 at 15:46
    
Thanks, I have corrected and added a reference article. –  JDH May 11 '11 at 20:18
    
The list of things that someone should have told me about in nine years of studies of higher mathematics continues to lengthen: this is a really pretty result! –  Pete L. Clark May 11 '11 at 20:36
    
The book by Semadeni, "Banach spaces of continuous functions" treats the countable compact Hausdorff spaces (the characterisation is needed for the treatment of the Banach space on it, as per the title), with complete proofs. This is where I first saw it. It's, fittingly, a book by a Pole. –  Henno Brandsma May 11 '11 at 21:09
    
Thanks very much, this is an excellent answer. –  J.K.T. May 12 '11 at 3:23

Call a topological space good if it's homeomorphic to a compact ordinal.

Lemma 1. Every countable compact Hausdorff space is first countable, zero-dimensional, and scattered.

Proof. These are well known facts.

Lemma 2. A closed subspace of a good space is good.

Proof. This follows from the fact that a closed subspace of an ordinal is homeomorphic to an ordinal.

Lemma 3. Let $S$ be a countable compact Hausdorff space. If each point of $S$ has a good clopen neighborhood, then $S$ is good.

Proof. By compactness, $S$ is covered by finitely many good clopen sets, which (by Lemma 2) can be made disjoint. Thus $S$ is the union of finitely many disjoint clopen sets $X_1,\dots,X_n$, where each $X_i$ is homeomorphic to a compact ordinal $\alpha_i$. It follows that $S$ is homeomorphic to the compact ordinal $\alpha=\alpha_1+\dots+\alpha_n$. (This is because $S$ is the topological sum of the $X_i$'s, and $\alpha$ is the topological sum of disjoint clopen subsets $A_i$, where $A_i$ is homeomorphic to the ordinal $\alpha_i$ and thus to $X_i$.)

Theorem. Every countable compact Hausdorff space is homeomorphic to an ordinal.

Proof. Let $S$ be a countable compact Hausdorff space. Assume for a contradiction that $S$ is not good. Let $X$ be the set of all points of $S$ which have no good clopen neighborhood. By Lemma 3, $X$ is nonempty. Since $S$ is scattered, $X$ has an isolated point $x$. (Of course $x$ is not isolated in $S$, since $\{x\}$ would then be a good clopen neighborhood of $x$.)

Let $U_1$ be a clopen neighborhood of $x$ such that $U_1\cap X=\{x\}$. Note that, by Lemmas 2 and 3, every clopen subset of $U_1$ which does not contain $x$ is good.

Let $U_1,U_2,\dots,U_n,\dots$ be a neighborhood base for $x$ such that each $U_n$ is clopen and properly contains $U_{n+1}$. Then the (nonempty) set $V_n=U_n\setminus U_{n+1}$ is a good clopen set, whence it is homeomorphic to a compact ordinal $\alpha_n$. It follows that the set $U_1\setminus X=V_1\cup V_2\cup\dots\cup V_n\cup\dots$, which is the topological sum of the $V_n$'s, is homeomorphic to the limit ordinal $\alpha=\alpha_1+\alpha_2+\dots+\alpha_n+\dots$. Hence $U_1$, the one-point compactification of $U_1\setminus\{x\}$, is homeomorphic to the ordinal $\alpha+1$, the one-point compactification of $\alpha$. Thus $U_1$ is a good clopen neighborhood of $x$, which was assumed to have no good clopen neighborhood. This contradiction proves the theorem.

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In fact, you can prove directly that any countable compact space $X$ is metrizable:

Let $A$ be a family of continuous functions from $X$ to $\mathbb{R}$ and let $e : \left\{ \begin{array}{ccc} X & \to & \mathbb{R}^A \\ x & \mapsto & (f(x)) \end{array} \right.$. If the family $A$ distinguishes points and closed sets in $X$, then $e$ embedds $X$ into $\mathbb{R}^A$; since $X$ is completely regular, such a family $A$ exists. Taking the linear subspace spanned by $e(X)$, you can suppose $A$ countable.

Therefore, $X$ is a subspace of $\mathbb{R}^{\omega}$; by compactness, $X$ is in fact a subspace of a product of intervals $\prod\limits_{\omega} I_k$. Since each $I_k$ is homeomorphic to $[0,1]$, $X$ is a subspace of $[0,1]^{\omega}$.

But $[0,1]^{\omega}$ is metrizable, $d : (x,y) \mapsto \sum\limits_{n \in \omega} \frac{1}{2^n}|x_n-y_n|$ defines a distance over it.


Moreover, since any metrizable space can be embedded into a normed space, you can use the proof given in [Stefan Mazurkiewicz and Wacław Sierpiński, Contribution à la topologie des ensembles dénombrables, Fundamenta Mathematicae 1, 17–27, 1920] to prove the classification you mention:

First notice that the Cantor-Bendixson rank of a countable compact space is a successor countable ordinal $\alpha+1$ and that $X_{\alpha}$ is finite. Then for any countable ordinal $\alpha$ and positive integer $n \geq 1$, let $P(\alpha,n)$ be the assertion: "Any countable compact space of Cantor-Bendixson rank $\alpha+1$ such that $\text{card}(X_{\alpha})=n$ is homeomorphic to $\omega^{\alpha} \cdot n+1$."

Clearly, $P(1,1)$ is true (here $X$ is just a converging sequence).

Step 1: If $P(\alpha,1)$ is true then $P(\alpha,n)$ is true.

Let $X_{\alpha}=\{p_1, \dots,p_n\}$. Then, viewing $X$ as a subspace of a normed space $Y$, there exist $n-1$ parallel hyperplanes $P_1$, ..., $P_{n-1}$ such that $Y \backslash \bigcup\limits_{i=1}^{n-1} P_i$ has $n$ connected components $D_1$, ..., $D_n$ with $p_k \in D_k$.

Because $P(\alpha,1)$ is true, each $X_k:= X \cap D_k$ is homeomorphic to $\omega^{\alpha}+1$. Therefore, $X$ is homeomorphic to $(\omega^{\alpha}+1) \cdot n= \omega^{\alpha} \cdot n+1$.

Step 2: If $P(\alpha,n)$ is true for any $\alpha<\alpha_0$ and $n \geq 1$, then $P(\alpha_0,1)$ is true.

Let $X_{\alpha_0}=\{p\}$ and let $(p_k)$ be a sequence in $X'$ converging to $p$ (without loss of generality, we suppose $\alpha_0 \geq 2$ so that $p \in X''$). Then, viewing $X$ as a subspace of a normed space $Y$, there exist a sequence of postive real numbers $(r_k)$ converging to zero such that the family of spheres $S(p,r_k)$ does not meet $X$ and $Y \backslash \bigcup\limits_{k \geq 1} S(p,r_k)$ has infinitely many connected components $D_1$, $D_2$, ... with $p_k \in D_k$.

By assumption, each $X_k:= X \cap D_k$ is homeomorphic to some $\omega^{\alpha_k} \cdot n_k+1$ with $\alpha_k <\alpha_0$ and $n_k \geq 1$. Therefore, $X$ is homeomorphic to $$\tau=[(\omega^{\alpha_1} \cdot n_1+1)+(\omega^{\alpha_2} \cdot n_2+1)+ \dots ]+1$$

But $\tau \leq \omega^{\alpha_0} +1$. If $\tau < \omega^{\alpha_0}+1$ then $\tau< \omega^{\alpha_0}$ because $\tau$ is compact, hence $X_{\alpha_0}= \emptyset$: a contradiction. Therefore, $\tau = \omega^{\alpha_0}+1$ and $P(\alpha_0,1)$ is true.

Step 3: We conclude by transfinite induction.

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