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>



Dependent choices (DC) in ${\bf HOD}(\mathbb{R},X)$, where $X$ is a set of reals
Consider the "singular Solovay-style model" of John Truss from



> 
> *Truss, John*, [**Models of set theory containing many perfect sets**](http://dx.doi.org/10.1016/0003-4843(74)90015-1), Ann. Math. Logic 7, 197-219 (1974). [ZBL0302.02024](https://zbmath.org/?q=an:0302.02024).
> 
> 
> 


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.


