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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?

Edit: As Joel's answer shows, we can generate a counterexample using class forcing over $L$. To avoid these, we might as well require that the universe is not a class-generic extension of some $W$ for which $\Bbb R^W=\Bbb R^L$.

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  • $\begingroup$ My knowledge of generics is pretty weak, but from the bit of recursion theory I know, and what I know of the analogies, it seems like some version of the usual means for constructing reals of intermediate degree ought to provide such an $x$? $\endgroup$
    – Steven Stadnicki
    Commented Aug 15, 2013 at 20:58
  • $\begingroup$ Hi Asaf - I came across this old question while researching other things, and I'm very intrigued by your side comment in the question above, that the existence of 0# implies the existence of L-Cohen generics. Could you provide a reference? Many thanks! $\endgroup$
    – jonasreitz
    Commented Aug 1, 2018 at 23:03
  • $\begingroup$ @jonasreitz: That's easy. If $0^\#$ exists, then $\omega_1^L$ is countable, therefore $\operatorname{Add}(\omega,1)^L$ has only countably many dense open sets (in $L$), so there is a generic filter meeting them. The same argument holds for any $V$-countable $L$-cardinal. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 1, 2018 at 23:05
  • $\begingroup$ (Also, as a side note, \# produces $\#$, just for future reference. So you wanted to write $0^\#$, @jonasreitz.) $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 1, 2018 at 23:06
  • $\begingroup$ Thanks, Asaf - that’s great! Do you know if it holds for cardinals not countable in $V$? I don’t see immediately how to generalize the argument, but it would be great if $L[0^\#]$ had L-generics for all posets $Add(\kappa,1)^L$. (Ps. Thanks also for the markup advice - I’m posting from my phone so we’ll see how it comes out). $\endgroup$
    – jonasreitz
    Commented Aug 2, 2018 at 0:59

1 Answer 1

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Great question!

In order to formalize the notion, let us understand the phrase "$x$ is $L$-generic" to mean: there is some partial order $\mathbb{P}\in L$ and some filter $G\subset\mathbb{P}$ that is $L$-generic, such that $x\in L[G]$. In particular, this refers to set-sized forcing only.

In this case, we can give a negative answer. Sy Friedman has a way to undertake the coding-the-universe forcing in such a way that the generic extension $V[G]=L[R]$ is minimal: Minimal Coding, Annals of Pure and Applied Logic, 1989, pp. 233-297. In particular, every real in the extension is either in $V$, or generates $R$, which is not set-generic (although it is generic for class forcing). Thus, if you start in $L$ and then undertake this forcing, you get a class forcing extension in which there are no $L$-generic reals for set forcing, and indeed, no $L$-generic sets of any kind.

In the linked paper, Friedman states the following:

Corollary. There is an $L$-definable forcing for producing a real $R$ which is minimal over $L$ but not set-generic over $L$.

It follows that this extension $L[R]$ has no set-generic objects at all not in $L$. That is, it is a strong counterexample, which applies not just to reals, but to sets of any size.

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  • $\begingroup$ That sure answers the question, thanks! Now comes the obvious follow up, what if we allow class-generics? $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 15, 2013 at 21:38
  • $\begingroup$ Well, that seems far more difficult. First, it cannot really be formalized in ZFC. Second, if you forbid class forcing, then our hands are too-much tied for building a counterexample. $\endgroup$
    – Joel David Hamkins
    Commented Aug 15, 2013 at 21:42
  • $\begingroup$ Sy has also studied a concept of hyperclass forcing, but I don't know whether it provides a counterexample for the follow-up question. If it does, then there's an obvious further follow-up. $\endgroup$
    – Andreas Blass
    Commented Aug 15, 2013 at 21:58
  • $\begingroup$ Joel, here's a slightly more sensitive follow up. Can we characterize (except trivially) models of the form $L[x]$ where $x$ is a real number, in which there are no $L$-generic reals? $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 16, 2013 at 0:26
  • $\begingroup$ I seriously doubt there will be a good characterization of such models, since the coding-the-universe argument shows that any model $V$ is a ground model of such an $L[x]$. So our characterization will have to include in some a way an account of all possible $V$'s. $\endgroup$
    – Joel David Hamkins
    Commented Aug 16, 2013 at 0:31

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