TITLE: Dependent choices (DC) in ${\bf HOD}(\mathbb{R},X)$, where $X$ is a set of reals
QUESTION [5 upvotes]: 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

REPLY [5 votes]: 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:

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

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

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


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