# Dependent choice and $L(\mathbb R)$.

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 $$V\models \mathsf{DC}$$?

• The wiki page you link to claims that the answer to your question is 'Yes', though does not provide a reference. May 3, 2020 at 12:51
• I can prove this for the Chang model (i.e., replace $\omega^\omega$ with $\rm Ord^\omega$). May 3, 2020 at 14:01
• 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. May 3, 2020 at 17:41

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 $$s$$.
• +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. May 4, 2020 at 0:38