I was wandering whether this weak form of $\text{CH}$ holds in $L(\mathbb{R})$ provably in $\text{ZF}+\text{DC}$



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> $(\text{ZF}+\text{DC}) \ L(\mathbb{R})\vDash \forall X\subseteq\mathbb{R} ( X \text{ countable} \lor \mathcal{P}(X)\not\le^\* \mathbb{R})$
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where $\mathcal{P}$ is the powerset and $x\le^\* y$ means that $y$ surjects onto $x$.


Is the above statement true? Or can we prove that there is a model of $\text{ZF}+\text{DC}+ V=L(\mathbb{R})$ such that the above form of $\text{CH}$ does not hold, i.e. such that there is an uncountable subset of $\mathbb{R}$ with $\mathbb{R}$ surjecting onto its powerset?


Ideas?  

Thanks!



Weak form of $\text{CH}$ in $L(\mathbb{R})$
No, this can fail. Force $\mathrm{MA}\_{\omega\_1}$ over $L$ by ccc forcing and let $M$ be the $L(\mathbb R)$ of the extension $L[G]$. Note that $M$ has the same cardinals as $L$. We have $X=\mathbb R\cap L\in M$ is of size $\omega\_1$, however $M$ has a surjection $f:\mathbb R\rightarrow \mathcal P(\omega\_1)\cap M$: Let $\langle a\_\beta\mid\beta<\omega\_1\rangle$ be an almost disjoint sequence in $[\omega]^\omega$. Such a sequence exists in $L$ and thus in $M$. Define $f$ via $\beta\in f(x)$ iff $x\cap a\_\beta$ is finite. By $\mathrm{MA}\_{\omega\_1}$ in $L[G]$, $f$ is surjective.


On the other hand, if $\mathrm{AD}$ holds in $L(\mathbb R)$ then this weak form of $\mathrm{CH}$ holds in $L(\mathbb R)$, simply because all sets of reals in $L(\mathbb R)$ have the perfect set property then, so one can only produce counterexamples by starting with models that do not have (inner models with) too many large cardinals.


