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Are Cohen generics (in $2^\omega$) minimal covers?

I'm ultimately interested in this question for some more effective notion of forcing but I realized I wasn't sure how to show this even assuming full set-theoretic forcing (with the reduction being relative constructability). If using full set-theoretic forcing introduces complications I'm fine with just forcing with respect to hyperarithmetic statements (i.e., $\omega_1$ generics in the sense of Fefferman and the reduction becoming is $\Delta^1_1$ in) or in some even more effective setting.

My thought was to show that if $f$ is definable from our generic $g$ then either $f$ is itself definable (e.g. in the ground model) or $g$ is uniquely determined by $f$ plus some finite initial segment of $g$ or there is some kind of factorization of the forcing and that $g$ must be generic relative to some notion of forcing for elements which produce $f$. But maybe this is the wrong way to go about it.

In any case, I'm guessing this question has already been answered in the set theoretic case and rather than bashing my head against the wall on something that probably has a straightforward answer I should look at how it was answered there.

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Indeed, this has been answered very negatively in the literature:

Abraham, Uri; Shore, Richard A., The degrees of constructibility of Cohen reals, Proc. Lond. Math. Soc., III. Ser. 53, 193-208 (1986). ZBL0625.03033.

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  • $\begingroup$ Unfortunately, the part I'm having trouble with is the very first theorem. That if a is cohen generic and 0 < x < a then x is of cohen generic degree. The paper cites to Jech but I only have the 3rd edition and I don't think it preserved the numbers since the cited results don't seem obviously related. $\endgroup$
    – Peter Gerdes
    Commented Nov 17 at 3:52
  • $\begingroup$ That's a pretty common problem when reading papers from 38 years ago. I'm not sure what you're expecting me to do about that. $\endgroup$
    – François G. Dorais
    Commented Nov 17 at 4:02
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    $\begingroup$ I was thinking that if you knew why that claim was true off the top of your head you'd respond and otherwise you'd ignore it. Sorry if it felt like I was expecting you to solve my problem. Anyway, I finally managed to download something from libgen that was labeled right and figured it out. Thanks for the citatoin $\endgroup$
    – Peter Gerdes
    Commented Nov 17 at 4:08
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    $\begingroup$ No worries Peter! All good! $\endgroup$
    – François G. Dorais
    Commented Nov 17 at 4:44
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    $\begingroup$ It's because every subforcing of a countable forcing notion is countable, hence is itself Cohen forcing. $\endgroup$
    – Joel David Hamkins
    Commented Nov 17 at 20:23

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