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Is it consistent that there is a model of $\mathsf{ZFC}$ (or $\mathsf{ZF}$) with the following properties:

(1) For all $x \in {}^\omega 2$, $x^\sharp$ exists (or $\mathbf{\Sigma}_1^1$ determinacy)

(2) For all sets $A$, there exists $x \in {}^\omega 2$ such that $A \in L[x]$, i.e. every set is constructible from a real.

If one takes an arbitrary model $V$ of $\mathsf{ZFC}$ satisfying (1), will $L[({}^\omega2)^V]$ have this property or does it necessarily add a set that is not constructible from any single real?

Thanks for any information or clarification.

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  • $\begingroup$ You would be probably be interested in the consistency strength of the theory ZFC+ every real has a sharp+ every $\Gamma$ subset of $\aleph_1$ is contructible from a real. There is work of Kechris and Woodin on this question in the Cabal volumes. $\endgroup$ Feb 4, 2015 at 17:41

1 Answer 1

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The theory is not consistent, since $\mathbb{R}$ is a set, but if $\mathbb{R}\in L[x]$, then every real is in $L[x]$, and this contradicts the existence of $x^\sharp$.

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