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For context, the Cantor-Bendixson theorem states that a closed subset $A$ of a Polish space can be written as the union of a perfect subset and a countable set $A=P\cup C$.

Now, I know two proofs of this theorem: one of them uses condensation points, and the other uses Cantor-Bendixson rank; but in the both cases the proof has a step that says "$C$ is a countable union of countable sets, and therefore countable...", which famously relies on AC$_\omega(^\omega\omega)$. But is this really necessary? If this is the case, one reasonable attempt is to construct a Polish space from a countable family of countable sets, but I don't see immediately how that can be carried out.

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    $\begingroup$ I don't know enough axiomatic set theory to be of much help, but in case you're interested, there is a 3rd proof: (from a lengthy late-1990s manuscript of mine that might one day see the light of day) "Three techniques have arisen to prove the Cantor-Bendixson theorem: transfinite iteration of the derived set operator (Cantor and Bendixson), the use of condensation points (Lindelöf), and the use a maximal dense-in-itself set (Hausdorff). Each of these methods has subsequently generated, through abstraction, its own circle of ideas and areas of applicability." $\endgroup$
    – Dave L. Renfro
    Commented Jun 10 at 18:21

2 Answers 2

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This is a ZF theorem, and is in fact equivalent to the reverse mathematical system $\Pi^1_1-\mathrm{CA}_0$ over the base theory $\mathrm{RCA}_0$ (this is in Simpson's reverse math textbook). It's worth reading Wikipedia's article "Reverse mathematics" — it has a great summary of such facts.

The Cantor-Bendixson rank argument is essentially already a ZF proof. Define a surjective partial map $f: \omega \cong \mathbb{Q}^2 \rightharpoonup C$ by sending $(q_1, q_2)\mapsto p$ if $p$ is the unique highest rank point in $C \cap (q_1, q_2).$

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  • $\begingroup$ I see, thanks! The concrete construction from rank argument makes it a lot easier to keep track of the axioms used $\endgroup$
    – Akira Satou
    Commented Jun 11 at 15:01
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The answer by Glazer is incomplete in that open sets are coded/represented (in second-order reverse math and attendant lands) as countable unions of basic open balls. Such a representation is rather hard to obtain.

The full answer is as follows:

  1. One can convert "third-order sets that are open" to the typical second-order representation (countable unions of basic open balls) in ZF. Actually, Kleene's quantifier $\exists^3$ is enough following [1].

  2. Using the representation from 1), the second-order results provide the required decomposition, not using AC.

Similar results hold for lower semi-continuous functions, as the characteristic functions of open sets are among those (see [2]).

References

[1] Dag Normann and Sam Sanders, Open sets in computability theory and Reverse Mathematics, JLC, https://arxiv.org/abs/1910.02489.

[2] Dag Normann and Sam Sanders, The Biggest Five of Reverse Mathematics, JML.

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  • $\begingroup$ Thanks for the reference! $\endgroup$
    – Akira Satou
    Commented Jun 11 at 15:02

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