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Asaf Karagila
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If there is a non-constructible real, is there an $L$-generic real?

If we assume that $\Bbb R^V\neq\Bbb R^L$, can we deduce that there is some $x\in\Bbb R^V$ which is $L$-generic?

Of course if $V$ is a generic extension of $L$ this is true, but if $V=L[0^\#]$ this is also true (the existence of $0^\#$ implies the existence of $L$-Cohen generics), but $0^\#$ cannot be added by forcing. So it is possible to have reals which are not generic over $L$, but their existence does imply that a generic exist.

Is this always the case, or is there a counterexample? Namely, can we have $V=L[x]$ for some real number $x$ such that no real number in $L[x]$ is $L$-generic?

Asaf Karagila
  • 39.7k
  • 8
  • 134
  • 282