The aim of this note is to classify the countable (Hausdorff) compact topological spaces; namely, we prove Sierpinski-Mazurkiewicz theorem following  [Stefan Mazurkiewicz and Wacław Sierpiński, Contribution à la topologie des ensembles dénombrables, Fundamenta Mathematicae 1, 17–27, 1920].

First, let us define Cantor-Bendixson rank of some topological space X by transfinite induction:

  • X^{(0)}=X,
  • X^{(\alpha+1)} is the set of limit points of X^{(\alpha)} for any limit ordinal \alpha,
  • X^{(\lambda)}= \bigcap\limits_{ \beta < \lambda} X^{(\beta)} for any limit ordinal \lambda.

The Cantor-Bendixson rank of X, denoted by \text{rk}(X), is the smallest ordinal \alpha satisfying X^{(\alpha+1)}=X^{(\alpha)}.

Lemma: Let X be a countable (Hausdorff) compact topological space. Then \text{rk}(X) is a countable successor ordinal, say \text{rk}(X)= \alpha+1, and X^{(\alpha)} is finite nonempty.

Proof. For any ordinal \kappa, \coprod\limits_{\beta<\kappa} X^{(\beta)} \backslash X^{(\beta+1)} \subset X. Because X is countable, there exists a countable ordinal \gamma such that X^{(\gamma)} \backslash X^{(\gamma+1)}= \emptyset ie. X^{(\gamma+1)}=X^{(\gamma)}.

Let \kappa be the Cantor-Bendixson rank of X. Since any perfect set in a compact space has cardinality at least 2^{\aleph_0} (proved below), X^{(\kappa)}=\emptyset. In particular, if \kappa is a limit ordinal, \bigcap\limits_{\beta<\kappa} X^{(\kappa)} is a decreasing sequence of nonempty closed sets converging to \emptyset, which is impossible since X is compact.

Therefore, \kappa is a successor ordinal, say \kappa=\alpha+1, and X^{(\alpha)} is nonempty and finite (otherwise, X^{(\alpha+1)} would be nonempty by compactness). \square

Definition: Let X be a countable (Hausdorff) compact topological space such that \text{rk}(X)=\alpha+1 and X^{(\alpha)} has cardinality n \geq 1. We say that (\alpha,n) is the characteristic system of X.

As preliminaries, we show that for all countable ordinal \alpha and integer n \geq 1, there exists a countable compact topological space of characteristic system (\alpha,n), namely \omega^{\alpha} \cdot n+1.

Lemma: Let \alpha be a countable ordinal and n \geq 1 be a positive integer. Then \omega^{\alpha} \cdot n+1 is a countable compact topological space of characteristic system (\alpha,n).

Proof. First, we show that (\alpha,1) is the characteristic system of \omega^{\alpha}+1 by transfinite induction over \alpha. For \alpha=0, it is clear.

Then, writing [0,\omega^{\alpha+1}]= \{0\} \coprod \left( \coprod\limits_{k \geq 0} [\omega^{\alpha} \cdot k+1,\omega^{\alpha} \cdot (k+1) ] \right) \coprod \{\omega^{\alpha+1}\} and noticing that the closed sets [\omega^{\alpha} \cdot k+1,\omega^{\alpha} \cdot (k+1) ] are isomorphic to \omega^{\alpha}+1 and that they can be separated by open sets, we deduce that [0,\omega^{\alpha}]^{(\alpha+1)}= \{\omega^{\alpha+1}\}.

Let \lambda be a limit ordinal. In the same way, writing [0,\omega^{\lambda}]= \{0\} \coprod\left( \coprod\limits_{\beta<\lambda} [\omega^{\beta}+1,\omega^{\beta+1}] \right) \coprod \{\omega^{\lambda}\} and noticing that that the closed sets [\omega^{\beta}+1,\omega^{\beta+1}] are isomorphic to \omega^{\beta+1}+1 and that they can be separated by open sets, we deduce that [0,\omega^{\alpha}]^{(\lambda)}=\{\omega^{\lambda}\} since \sup \text{rk} (\omega^{\beta+1}+1)=\lambda.

We conclude the proof with again the same argument: writing [0,\omega^{\alpha} \cdot n]= \{0\} \coprod \left( \coprod\limits_{k=0}^{n-1} [\omega^{\alpha} \cdot k+1,\omega^{\alpha} \cdot (k+1) ] \right) as a disjoint union of closed sets isomorphic to \omega^{\alpha}+1 and separated by open sets, we deduce that [0,\omega^{\alpha}]^{(\alpha)}= \{\omega^{\alpha}, \omega^{\alpha} \cdot 2,\dots, \omega^{\alpha} \cdot n\}. \square

Now we can state and prove our main theorem:

Theorem: (Sierpinski-Mazurkiewicz) Let X be a countable compact topological space of characteristic system (\alpha,n). Then X is homeomorphic to \omega^{\alpha} \cdot n+1.

Proof: First, we prove that any countable compact space X is metrizable:

Let A be the set 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..Then e is continuous and, because A distinguishes points and closed sets in X (since X is completely regular), e is injective and open, so e embedds X into \mathbb{R}^A. Taking the linear subspace spanned by e(X), we can suppose that X is a subspace of \mathbb{R}^{\omega}.

By compactness, X is in fact a subspace of a product of closed intervals \prod\limits_{\omega} I_k, and since each I_k is homeomorphic to [0,1], we can view X as a subspace of [0,1]^{\omega}.

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

Lemma: Any metric space is isometric to a subspace of a (complete) normed space.

Proof. Let (X,d) be a metric space and F=C_b(X) be the set of continuous bounded functions X \to \mathbb{R} with sup norm. Fix x_0 \in X. Then \phi : x \mapsto d(x, \cdot)-d(x_0,\cdot) is an isometry from X into F. \square

Therefore, any countable compact space can be viewed as a subspace of a normed vector space.

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.”

Step 1: P(1,1) is true.

Let X be a countable compact topological space of characteristic system (1,1). Let X^{(1)} = \{p\} and let f : X \backslash X^{(1)} \to \omega be a bijection. The extension \tilde{f} : X \to \omega+1 defined by f(p)=\omega is a homeomorphism.

Step 2: If P(\alpha,1) is true then P(\alpha,n) is true.

Let X be a countable compact topological space of characteristic system (\alpha,n) and 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 X_1+ \dots + X_n \simeq (\omega^{\alpha}+1) \cdot n= \omega^{\alpha} \cdot n+1.

Step 3: If P(\alpha,n) is true for any \alpha<\alpha_0 and n \geq 1, then P(\alpha_0,1) is true.

Let X be a countable compact topological space of characteristic system (\alpha_0,n) and X^{(\alpha_0)}=\{p\}. Without loss of generality, we can suppose \alpha_0 \geq 2 so that p \in X^{(2)}, ie. there exists a sequence (p_k) in X^{(1)} converging to p. Then, viewing X as a subspace of a normed space Y, there exists 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.

First, notice that \tau \leq \omega^{\alpha_0}+1. If \tau< \omega^{\alpha_0}+1 then \tau< \omega^{\alpha_0} because \tau is compact; consequently, there exist \beta<\alpha_0 and m \geq 1 such that \tau < \omega^{\beta} \cdot m+1, hence \text{rk}(\tau) \leq \text{rk}(\omega^{\beta} \cdot m+1)=\beta+1 \leq \alpha_0, a contradiction with \text{rk}(X)= \alpha_0+1. Therefore, \tau=\omega^{\alpha_0}+1 and P(\alpha_0,1) is true.

Step 4: We conclude the proof by transfinite induction. \square

Corollary: Two countable compact topological spaces are homeomorphic if, and only if, they have the same characteristic system.

Corollary: Any countable compact topological space is homeomorphic to a well-orded set.

Corollary: Any countable compact topological space is homeomorphic to a subspace of \mathbb{R}.

Corollary: There exist \aleph_1 countable compact topological spaces up to homeomorphism.

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