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.