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Let $\lambda$ be a singular strong limit cardinal and let $G \subset \text{Col}(\omega,\mathord{<}\lambda)$ be a $V$-generic filter. Let $\mathbb{R}^*_G = \bigcup_{\alpha < \lambda} \mathbb{R}^{V[G \restriction \alpha]}$. Then countable choice fails in $V(\mathbb{R}^*_G)$, even for reals: $\omega_1$ is $\lambda$, which is singular. On the other hand, $\mathsf{CC}_\mathbb{R}$ can hold in $L(\mathbb{R}^*_G)$: for example if $\lambda$ is a limit of Woodin cardinals then $L(\mathbb{R}^*_G)$ satisfies the Axiom of Determinacy, which implies $\mathsf{CC}_\mathbb{R}$. In fact we then get $\mathsf{CC}$ in $L(\mathbb{R}^*_G)$ because everything is $\text{OD}$ from a real. But the large cardinal hypothesis seems like it may be overkill. So my question is:

What is the consistency strength of the statement "$\lambda$ is a singular strong limit cardinal and Countable Choice holds in $L(\mathbb{R}^*_G)$"?

As far as I know, it could be consistent relative to $\mathsf{ZFC}$, but I only see how to get it from infinitely many Woodin cardinals.

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    $\begingroup$ It must be at least $0^\sharp$ because if $\lambda$ is a singular cardinal and Jensen's covering holds, then $\lambda$ is singular in $L$. $\endgroup$
    – Monroe Eskew
    Commented Oct 24, 2014 at 2:06
  • $\begingroup$ I always thought that the trick was to collapse the Woodin cardinals, not everything. You might think there is no difference, but there is. Compare the Feferman-Levy model to Truss models. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 24, 2014 at 2:16
  • $\begingroup$ @Monroe Oh, quite right. And it looks like every bounded subset of $\lambda$, and hence every real in $\mathbb{R}^*_G$, has a sharp for the same reason. To go further and get an inner model with a Woodin cardinal, we could try the same thing with $K$ instead of $L$. But I'm not sure how that would work, because we used (full) covering for $L$ to get sharps and $K$ may only satisfy weak covering (as far as I know.) $\endgroup$
    – Trevor Wilson
    Commented Oct 24, 2014 at 2:53
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    $\begingroup$ Also, the first thing I'd try is to shoot a Prikry to a measurable and collapse. $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 24, 2014 at 2:56
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    $\begingroup$ @Asaf Aha, that does seem to show that it follows from a measurable. Prikry forcing doesn't change $\mathbb{R}^*_G$, so it doesn't change $L(\mathbb{R}^*_G)$. And because $\lambda$ was measurable (hence inaccessible) to begin with, $L(\mathbb{R}^*_G) \models \mathsf{DC}$. So together with Monroe's comment this shows that the consistency strength is between a measurable and "every real has a sharp." $\endgroup$
    – Trevor Wilson
    Commented Oct 24, 2014 at 3:38

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