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I am curious what is the relation between the inner model $L(\mathbb R)$ and the axiom of dependent choice $\mathsf{DC}$. Do we have $$L(\mathbb R)\models \mathsf{DC},$$ when $V\models \mathsf{DC}$?

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    $\begingroup$ The wiki page you link to claims that the answer to your question is 'Yes', though does not provide a reference. $\endgroup$
    – Hayden
    May 3, 2020 at 12:51
  • $\begingroup$ I can prove this for the Chang model (i.e., replace $\omega^\omega$ with $\rm Ord^\omega$). $\endgroup$
    – Asaf Karagila
    May 3, 2020 at 14:01
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    $\begingroup$ Yes, this is classical. The point is that it is enough to have $\mathsf{DC}_{\mathbb R}$ in $L(\mathbb R)$ to conclude $\mathsf{DC}$ in there. And the former holds because it holds in $V$. I believe Moschovakis's book on Descriptive set theory includes a proof. $\endgroup$ May 3, 2020 at 17:41

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The inductive construction of $L(\mathbb R)$ shows that every element, say $x\in L_\alpha(\mathbb R)$, has the form $\{y\in L_\beta(\mathbb R):L_\beta(\mathbb R)\models\phi(y,\vec p)\}$ for some $\beta<\alpha$ and some parameters $p\in L_\beta(\mathbb R)$. Each parameter in $\vec p$ can itself be so expressed, with even smaller $\beta$. Continue thus until you get $x$ expressed in terms of finitely many ordinals, finitely many reals, and finitely many formulas. By coding, you get $x=F(\gamma,r)$ for an ordinal $\gamma$, a real $r$, and a definable class-function $F$ with a sufficiently simple definition to be absolute for $L(\mathbb R)$.

Now suppose you're given, in $L(\mathbb R)$, a binary relation $Q$ on some set $A$, such that $(\forall x\in A)(\exists y\in A)\,Q(x,y)$. You want an $\omega$-sequence $s$ such that, for all $n$, $Q(s(n),s(n+1))$. Define a new, smaller relation on $A$ by letting $Q'(x,y)$ mean that $Q(x,y)$ and $y=F(\gamma,r)$ for some $\gamma$ and $r$ such that no $\gamma'<\gamma$ and $r'$ have $Q(x,F(\gamma',r'))$. In other words, among all options $y$ for a fixed $x$, keep only those that can be represented as $F(\gamma,r)$ with minimum possible $\gamma$. Note that this $\gamma$ is determined by $x$; I'll write it as $\gamma(x)$.

This reduces the problem to dependently choosing reals $r$ to go with these minimal $\gamma$'s. In detail, starting with any specified $x_0\in A$, you want to choose an $\omega$-sequence $t$ of reals so that the sequence $s$ defined by $s(0)=x_0$ and $s(n+1)=F(\gamma(s(n)),t(n))$ satisfies $Q(s(n),s(n+1))$ (and in fact $Q'(s(n),s(n+1))$) for all $n$.

The task of producing such a $t$ is an instance of dependent choice (of reals), so it can be carried out in the real world $V$, by hypothesis. But $t$ is an $\omega$-sequence of reals, so it can be coded (in an absolute way) by a single real. Therefore $t$ is available in $L(\mathbb R)$, and, using it, we immediately get the desired $s$.

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  • $\begingroup$ +1. To the OP: as a technical coda, note that a more complicated argument shows that $\mathsf{ZF}+\mathsf{AD}$ in the ambient model is also enough (more snappily: $\mathsf{ZF}+\mathsf{AD}\vdash \mathsf{DC}^{L(\mathbb{R})}$). This was proved by Kechris. The general question of whether $\mathsf{ZF}+\mathsf{AD}\vdash\mathsf{DC}$ is still wildly open; note that Kechris' result can be rephrased equivalently as $\mathsf{ZF}+\mathsf{AD}+V=L(\mathbb{R})\vdash\mathsf{DC}$, which is how he phrases it. $\endgroup$ May 4, 2020 at 0:38
  • $\begingroup$ Andreas, I feel like there is a broader theorem here involving SVC somehow (perhaps with a condition on the seed as well). No? $\endgroup$
    – Asaf Karagila
    May 5, 2020 at 10:29

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