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In Turing invariant sets and the perfect set property, Math. Log. Quart. 66 (2020), Hamel, Horowitz and Shelah, the authors work in ZF + DC. They claim that DC can be dispensed with, asserting:

if $V \models {\rm ZF}\: + $ “all Turing invariant sets have the perfect set property” and $X \in V$ is a set of reals, then ${\bf HOD}(\mathbb{R},X) \models {\rm ZF} + {\rm DC}\: +$ “all Turing invariant sets have the perfect set property”.

Question: Why would ${\bf HOD}(\mathbb{R},X) \models {\rm DC}$, here?

For context: the paper proves: If all Turing invariant sets have the perfect set property, then all sets of reals have the perfect set property. It's available on arXiv: https://arxiv.org/abs/1912.12558

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  • $\begingroup$ Note that this implies in particular that ZF + "all Turing invariant sets (of reals) have the perfect set property" proves DC$_\mathbb{R}$, since it holds in $\mathrm{HOD}(\mathbb{R},X)$ w.r.t. the arbitrary $X$ from $V$. It follows that ZF+AD proves DC$_\mathbb{R}$. That was an open problem in the 70s/80s; has it been solved? $\endgroup$
    – Farmer S
    Commented May 10, 2021 at 13:14
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    $\begingroup$ @Farmer S. I don't believe that problem has been solved. Hence my question. Kechris showed that, in 𝕃(ℝ), AD implies DC. $\endgroup$
    – Ramez L. Sami
    Commented May 10, 2021 at 13:33
  • $\begingroup$ Haim sometimes shows up on the site, hopefully he'll see this soon. $\endgroup$
    – Asaf Karagila ♦
    Commented May 10, 2021 at 14:36
  • $\begingroup$ Though the claim is not correct as pointed out by Asaf below, the major result in the paper can be proved within $ZF$. $\endgroup$
    – 喻 良
    Commented May 13, 2021 at 1:08
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    $\begingroup$ @喻良, I didn't realize you were the author of the recursion-theoretic version of the proof, and the extension to countable $\Sigma^1_1$ equivalence relations. Bravo! $\endgroup$
    – Ramez L. Sami
    Commented May 19, 2021 at 13:00

1 Answer 1

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Consider the "singular Solovay-style model" of John Truss from

Truss, John, Models of set theory containing many perfect sets, Ann. Math. Logic 7, 197-219 (1974). ZBL0302.02024.

In that model the following holds:

  1. $V=L(\Bbb R)$, and therefore for any set of reals, $X$, the equalities $\mathrm{HOD}(\Bbb R,X)=L(\Bbb R)=V$ hold.

  2. $\sf DC$ fails, since $\omega_1$ is singular.

  3. Every set of reals, Turing invariant or not, has the perfect set property.

If I understand their claim correctly, this is a counterexample.

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  • $\begingroup$ Thanks for the example. I did think, at first hand, that what is stated couldn't be literally true. It would imply "ZF + AD $\vdash \text{DC}_\mathbf{R}$". Which, as far I know, is a long standing open problem. This was also noticed above. My hesitation is due to the distinguished signatures, of the paper. $\endgroup$
    – Ramez L. Sami
    Commented May 10, 2021 at 17:53

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