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What is the simplest non-constructible set of integers (say, in the analytical hierarchy) that is compatible with the nonexistence of $0^\sharp$? In particular, can there still be a non-constructible $\Delta^1_3$ set of integers if one assumes the nonexistence of $0^\sharp$?

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It is consistent, relative to just $ZFC$, that there exists a non-constructible $\Delta_3^1$ set of integers. Such a result was first produced in the paper ``Some allpications of almost disjoint sets'' by Jensen-Solovay, using almost disjoint forcing.

In fact we can prove the following stronger result which is due to Jensen ``definable sets of minimal degree'':

It is consistent that $V=L[R],$ where $R$ is minimal (in the degree of constructibility) and it is the unique solution of a $\Pi_2^1$ predicate (hence it is $\Delta_3^1$).

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  • $\begingroup$ By "compatible with the nonexistence of $0^\sharp$" I mean consistent with ZFC+"$0^\sharp$ does not exist." Even if "there is a non-constructible $\Delta^1_3$ set of integers" is consistent with ZFC, I'm wondering if it is also consistent with ZFC+"$0^\sharp$ does not exist." $\endgroup$
    – Jesse Elliott
    Commented Nov 29, 2014 at 6:27
  • $\begingroup$ It is certainly not compatible with ZF+$V = L$, and the thought is that ZFC+"$0^\sharp$ does not exist" is perhaps not too far away from ZF+$V = L$. $\endgroup$
    – Jesse Elliott
    Commented Nov 29, 2014 at 6:34
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    $\begingroup$ As the results are equiconsistent with $ZFC$, you just need to start from L, and force the resulting model in the extension. In the extension $0^\sharp$ does not exist. $\endgroup$
    – Mohammad Golshani
    Commented Nov 29, 2014 at 6:36
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    $\begingroup$ The fact that in the absence of $0^\sharp,$ $V$ is close to $L$ is a consequence of Jensen's covering lemma, which says if $0^\sharp$ does not exist, then any uncountable set of ordinals is covered by a constructible set of ordinals of the same size. $\endgroup$
    – Mohammad Golshani
    Commented Nov 29, 2014 at 6:43
  • $\begingroup$ I see. If the existence of a non-constructible $\Delta^1_3$ set of integers implied $0^\sharp$ exists, then its consistency strength would be greater than that of ZFC, which it is not. I misread your meaning of "consistent, relative to just ZFC." I haven't been able to locate the two papers yet but I'll take your word for it and accept your answer! $\endgroup$
    – Jesse Elliott
    Commented Nov 29, 2014 at 7:50

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