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I've heard someone saying that in $L(\mathbb{R})$ the following form of $\text{CH}$ holds:

$L(\mathbb{R})\vDash \forall X\subseteq \mathbb{R}(X \text{ countable or } \mathbb{R}\le^* X)$, i.e. every uncountable subset of the reals surjects onto $\mathbb{R}$

Now I'm wondering whether this is a theorem of $\text{ZF(C)}$ or we need to assume some large cardinal to prove it (e.g. so to have $\text{AD}$).
I'm struggling to find some reference.
Any help?
Thanks!

Edit: It cannot be a theorem of $\text{ZF}$ as otherwise we could prove that $\text{ZF}+\text{DC}+\text{PSP}$ is equiconsistent with $\text{ZF}$, which is not. But I'm still not finding a reference for the result under $\text{AD}$ for example..

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    $\begingroup$ For the question in the edit: AD implies perfect set property, which means every subset of $\mathbb R$ is countable or has size continuum. $\endgroup$
    – Wojowu
    Jun 3, 2022 at 11:09
  • $\begingroup$ @Wojowu Right, right. Thanks! $\endgroup$
    – Lorenzo
    Jun 3, 2022 at 11:11
  • $\begingroup$ I'm confused about the edit. Why would that shot that ZF+DC+PSP is equiconsistent with ZF? PSP implies a stronger result than just $X$ can be mapped onto $\Bbb R$. $\endgroup$
    – Asaf Karagila
    Jun 3, 2022 at 11:35
  • $\begingroup$ @AsafKaragila I should have elaborated a bit more. It is because Truss constructed a model of $\mathtt{ZF}+\mathtt{DC}+ V=L(\mathbb{R})+$ "Every subset of reals of size continuum contains a perfect set", if the statement I wrote was a theorem of $\mathtt{ZF}$ then Truss' model would satisfy $\mathtt{PSP}$. $\endgroup$
    – Lorenzo
    Jun 3, 2022 at 12:56
  • $\begingroup$ Also if we add $\omega_2$ Cohen reals to $L$, we get a model of $V=L(\mathbb{R})$ in which $L\cap \mathbb{R}$ is uncountable and does not surject onto $\mathbb{R}$.. $\endgroup$
    – Lorenzo
    Jun 3, 2022 at 13:27

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