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Consider the following statement (in $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$):

There exists $(C_\alpha, x_\alpha)_{\alpha \in \omega_1}$ s.t. $C_\alpha \subseteq \mathbb{N}^\mathbb{N}$ is closed and $\text{CB}^\alpha(C_\alpha) = \{x_\alpha\}$

where $\mathbb{N}^\mathbb{N}$ is the Baire space (the space of infinite sequences of natural numbers) with the usual topology and $\text{CB}^\alpha$ is the $\alpha$-th Cantor-Bendixon derivative.

My questions are:

  1. How much choice do we need at least to prove this statement? Or more generally what known weak axiom can be assumed (on top of $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$) to prove the statement?
  2. Is it consisent $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R}) + \neg$The above statement?
  3. If we assume the consistency of $\mathsf{ZF}+\mathsf{AD}$, is it consistent $\mathsf{ZF}+\mathsf{AD}\ +$ The above statement?

Thanks!

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  • $\begingroup$ Related $\endgroup$
    – Wojowu
    Nov 15, 2021 at 10:25
  • $\begingroup$ This is very similar to your last question, so I'm not going to write out a full proof, but this is also equivalent to "there is a function choosing an enumeration of each countable ordinal." If you have a countable closed set of rank $\alpha,$ you can canonically enumerate $\alpha$ by sending a basic open set $U$ to $\beta$ if there is a unique point of maximal rank in $U,$ and that rank is $\beta.$ $\endgroup$ Nov 15, 2021 at 22:52

1 Answer 1

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Note that these are all countable sets, so what you'd have here is a sequence of countable sets of reals indexed by $\omega_1$.

This is certainly inconsistent with $\sf AD$, since from such a sequence we can construct a subset of size $\aleph_1$. Therefore this is also consistently false with $\sf AC_\omega(\Bbb R)$, at least assuming an inaccessible, as that's the case in Solovay's model.

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    $\begingroup$ Seems like inaccessible is the exact consistency strength, since once $\omega_1$ is computed correctly in some inner model that satisfies choice, you can construct such a sequence there, and use absoluteness. $\endgroup$
    – Yair Hayut
    Nov 15, 2021 at 10:51
  • $\begingroup$ Good point, thanks! $\endgroup$
    – Asaf Karagila
    Nov 15, 2021 at 11:03

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