(Note: I asked this question at MSE days ago and received no answer, so I'm now reposting it here.)

I want to prove the following statement:

Let $K_1$ and $K_2$ be two countable, compact sets of $\mathbb R$ such that $K_1'$ and $K_2'$ are homeomorphic, then $K_1$ and $K_2$ are homeomorphic.

More generally, I want to prove that if $K_1$ and $K_2$ be two countable, compact sets of $\mathbb R$ have the same Cantor–Bendixson rank, $\alpha+1$, and $|K_1^{(\alpha)}|=|K_2^{(\alpha)}|$, then $K_1$ and $K_2$ are homeomorphic.

I know that a proof is given by Mazurkiewicz and Sierpinski in "Contribution à la topologie des ensembles dénombrables" (1920) (http://matwbn.icm.edu.pl/ksiazki/fm/fm1/fm114.pdf). And by Millient in "A remark in Cantor derivative" (http://arxiv.org/pdf/1104.0287v1.pdf). But I want a more direct proof in $\mathbb{R}$; can anybody help me? Thanks.

  • 5
    $\begingroup$ As a consequence of Mazurkiewicz-Sierpinski theorem, any countable compact space is homeomorphic to some $\omega^{\alpha} \cdot n +1$, where $\alpha$ is a countable ordinal and $n \geq 1$; in particular, every such space is homeomorphic to a subspace of $\mathbb{R}$. Therefore, it cannot be simpler to prove Mazurkiewicz-Sierpinsky theorem restricted to subspaces of $\mathbb{R}$. $\endgroup$
    – Seirios
    Jul 1 '14 at 7:11
  • 1
    $\begingroup$ @Seirios : I'm not sure this conclusion is inevitable (of course, this is an ill-defined philosophical question). The fact that MS for $\mathbb R$ implies MS, but by MS itself, does not exclude the possibility that MS for $\mathbb R$ might be "easier". Or here's a (very silly) analogy: The fact that every finite dimensional vector space $V$ has a basis implies that $V\cong F^n$. However, it's still "easier" to find a basis for $F^n$ than for $V$. $\endgroup$ Jul 1 '14 at 17:22
  • $\begingroup$ @ChristianRemling: Right, so in fact it depends on the difficulty to prove that a compact countable space is homeomorphic to a subspace of $\mathbb{R}$; fortunately, Joseph Van Name gave a simple argument in his answer. Therefore, the two problems are roughly the same. $\endgroup$
    – Seirios
    Jul 2 '14 at 7:15
  • 1
    $\begingroup$ By the way, I find the original proof of Mazurkiewicz and Sierpinsky already simple. Their article is written in French, but I wrote a part of their argument here: chiasme.wordpress.com/2013/06/19/… $\endgroup$
    – Seirios
    Jul 2 '14 at 7:17
  • $\begingroup$ Thanks. I think that using the ordinal topology is a bit complicated. I found another proof that seems simpler (math.stackexchange.com/questions/853500/…), but there are some parts that I don't understand, can you help me please? $\endgroup$
    – MateAndres
    Jul 2 '14 at 15:35

Let me give a proof that makes use of the ordering inherited from $\mathbb{R}$.

Suppose that $A\subseteq\mathbb{R}$ is a countable compact space. Then $A$ has no subset $B$ order isomorphic to the rational numbers (otherwise $A$ would be uncountable). I now claim that the order topology on $A$ is isomorphic to the subspace topology on $A$. To avoid confusion, let $\mathcal{O}$ be the order topology on $A$ and let $\mathcal{S}$ be the subspace topology on $A$. It is well known that for every ordered set, the subspace topology is finer than the order topology, i.e. $\mathcal{O}\subseteq\mathcal{S}$. However, since $(A,\mathcal{S})$ is compact, we have $\mathcal{O}=\mathcal{S}$, so the order topology and the subspace topology coincide on $A$. In particular, since $A$ is complete as an ordered set, $A$ is a complete lattice by this answer of mine.

Let $\mathcal{H}_{0}$ be the set of all finite and countable successor ordinals. Let $\mathcal{H}_{\alpha}$ be the collection of all sums $\sum_{i\in B}B_{i}$ where $B_{i}\in\bigcup_{\beta<\alpha}\mathcal{H}_{\beta}$ for each $\beta<\alpha$ and where $B=\omega+1$,$B=(\omega+1)^{*}$ or $B$ is finite.

I claim that $\bigcup_{\alpha<\omega_{1}}\mathcal{H}_{\alpha}$ is precisely the collection of all countable complete linear orders (this is a slight modification of a result by Hausdorff proven here). Suppose that $X$ is a countable compact linear ordering that is not in $\bigcup_{\alpha<\omega_{1}}\mathcal{H}_{\alpha}$. Then it is easy to show that there exists some $x\in X$ with $[0,x_{1/2}],[x_{1/2},1]\not\in\bigcup_{\alpha<\omega_{1}}\mathcal{H}_{\alpha}$. In other words, we can cut $X$ into two pieces not in $\bigcup_{\alpha<\omega_{1}}\mathcal{H}_{\alpha}$. Now cut each piece $[0,x_{1/2}],[x_{1/2},1]$ into pieces $[0,x_{1/4}],[x_{1/4},x_{1/2}],[x_{1/2},x_{3/4}],[x_{3/4},1]$. Continue this process until we have obtained an $x_{r/2^{n}}$ for all $n\in\omega$ and $0<r<\frac{1}{2^{n}}$ and we have obtained an isomorphic copy of the rational numbers. This is a contradiction since $X$ cannot contain a copy of the rational numbers.

I claim that for each ordinal $\alpha$, each $X\in\mathcal{H}_{\alpha}$ is homeomorphic to some countable ordinal, and this claim shall be proven by transfinite induction. Suppose that $\alpha$ is an ordinal and assume that whenever $\beta<\alpha$ each $X\in\mathcal{H}_{\beta}$ is homeomorphic to some ordinal. Now assume that $X\in\mathcal{H}_{\alpha}$. If $X=\sum_{i\in B}B_{i}$ where $B$ is finite and for each $i\in B$ we have $B_{i}\in\mathcal{H}_{\beta_{i}}$ for some $\beta_{i}<\alpha$, then clearly $X$ is homeomorphic to some ordinal. Therefore without loss of generality, we can assume that $X=\sum_{\alpha\in\omega+1}B_{\alpha}$. Then let $W_{n}$ be a well ordered set and let $h_{n}:B_{n}\rightarrow W_{n}$ be a homeomorphism for all $n\in\omega$.

Let $0_{\omega}$ be the least element in $B_{\alpha}$.

If $0_{\omega}$ is not a limit point of $B_{\alpha}$, then since $B_{\alpha}$ is homeomorphic to a well ordered set, one can easily show that there exists an isomorphism $h_{\omega}:0_{\omega}\rightarrow W_{\omega}$ such that $0_{\omega}$ is the least element of $W_{\omega}$. Then define a mapping $h:\sum_{\alpha\in\omega+1}B_{\alpha}\rightarrow \sum_{\alpha\in\omega+1}W_{\alpha}$ by letting $h(x,\alpha)=(h_{\alpha}(x),\alpha)$. Then $H$ can easily be seen to be a homeomorphism.

Now assume that $0_{\omega}$ is a limit point of $B_{\alpha}$. Then let $(x_{n})_{n\in\omega}$ be a descending sequence in $B_{\omega}$ that converges to $0_{\omega}$ and such that $x_{0}$ is the greatest element of $B_{\omega}$ and where if $n\neq 0$, then $x_{n}$ has an immediate successor $y_{n-1}$.

For all $n\in\omega$, let $Y_{n}$ be a well ordered set and let $j_{n}:[y_{n},x_{n}]\rightarrow Y_{n}$ be an homeomorphism.

Let $Z_{2n}=W_{n}$ and let $Z_{2n+1}=Y_{n}$ for each $n\in\mathbb{N}$, and let $Z_{\omega}=\{0^{\omega}\}$ for some element $0^{\omega}$. Let

$k:\sum_{\alpha\in\omega+1}B_{\alpha}\rightarrow\sum_{\alpha\in\omega+1}Z_{\alpha}$ be the mapping where $k(x,n)=(h_{n}(x),2n)$ for $n\in\omega$, $k(x,\omega)=(j_{n}(x),2n+1)$ for $x\in[y_{n},x_{n}]$ and $k(0_{\omega},\omega)=(0^{\omega},\omega)$. Then $k$ is a homeomorphism. Therefore each countable compact subspace $A\subseteq\mathbb{R}$ is homeomorphic to some successor ordinal.

Suppose that $\omega^{\alpha_{n}}a_{n}+...+\omega^{\alpha_{0}}a_{0}$ is an ordinal in Cantor's normal form. Then by putting the initial segment $\omega^{\alpha_{n}}a_{n}+1$ last instead of first, we conclude that $\omega^{\alpha_{n}}a_{n}+...+\omega^{\alpha_{0}}a_{0}$ is homeomorphic to $\omega^{\alpha_{n}}a_{n}+1$. In particular, if two countable compact sets $X,Y$ have the same Cantor-Bendixson rank $\alpha+1$ and $|X^{(\alpha)}|=|Y^{(\alpha)}|$, then $X\simeq Y$.

Also take note that it is easy to prove that every compact countable metric space is isomorphic to a subspace of $\mathbb{R}$. If $X$ is a countable compact space, then $C(X)$ is a Banach space. However, whenever $x,y\in X$ are distinct elements, then $A_{x,y}\subseteq\{f\in C(X)|f(x)=f(y)\}$ is a closed nowhere dense subspace of $C(X)$. Therefore by the Baire category theorem, $\bigcup_{x\neq y}A_{x,y}$ contains no open subset of $C(X)$ and if $f\in C(X)\setminus\bigcup_{x\neq y}A_{x,y}$, then $f$ is injective. Since $X$ is compact, the set $X$ and its image $f[X]$ are homeomorphic.

  • $\begingroup$ I find your proof that a compact countable metric space is homeomorphic to a subspace of $\mathbb{R}$ very nice. A possibility to prove that a compact countable metric space is metrizable is to introduce $$e : \left\{ \begin{array}{ccc} X & \to & \mathbb{R}^A \\ x & \mapsto & (f(x)) \end{array} \right.$$ where $A$ is the set of continuous functions $X \to \mathbb{R}$. It is an embedding, and because $X$ is countable and compact, we may deduce an embedding into $[0,1]^{\omega}$, which is metrizable. $\endgroup$
    – Seirios
    Jul 2 '14 at 7:24
  • 1
    $\begingroup$ By the way, I find the original argument of Mazurkiewicz and Sierpinski more elementary. $\endgroup$
    – Seirios
    Jul 2 '14 at 7:25
  • $\begingroup$ @Seirios In your post, why do we need the sequence $(p_k)$ in $X^{(1)}$ converging to $p$? $\endgroup$
    – MateAndres
    Jul 3 '14 at 13:01
  • $\begingroup$ The sequence is just useful to exhibit the "onion structure"; then, the induction hypothesis may be applied to any connected component. Perhaps a more appropriate place to deal with such questions would be my blog itself; you can write your questions as comments. $\endgroup$
    – Seirios
    Jul 3 '14 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.