In Jech's SET THEORY (a very early edition to which I have access), it is shown that the existence of 0sharp implies the existence of a truth definition for the constructible universe L. Does the converse hold? I ask this question because in Koepke's paper "Turing Computations on Ordinals", he defines an ordinal computable "bounded truth predicate" (at least as I understand it, and I possibly don't understand it correctly) for each stage L_i of the constructible universe L. Are these two notions (0sharp and Koepke's bounded truth predicate) at all related, and if so, in what way? At first glance (to me, at least) they seem to be because in Jech's text, the satisfaction relation is defined for each stage L_i of the hierarchy for L, just as the satisfaction relation seems so defined for Koepke's bounded truth predicate.

$\begingroup$ Comment: It is possible to define a truth predicate for $V$ (and so for $L$) in GödelBernays + $\Sigma^1_1$Induction, which consistency wise is much weaker than MK and so an inaccessible. $\endgroup$– Philip WelchAug 2, 2013 at 8:30
1 Answer
The answer is no, one can have models of ZFC set theory with a definable truth predicate for firstorder truth in $L$, but without having $0^\sharp$.
One way to build such a model is like this. In KellyMorse KM set theory, you can prove the existence of a truth predicate for firstorder truth for the whole universe $V$, and then by forcing you can code this class into the GCH pattern, for example, in order to make it definable. The result is a forcing extension $V[G]$ which has a firstorder definable class predicate for firstorder truth in the ground model $V$. From this, one can easily define a truth predicate for firstorder truth in $L$.
But meanwhile, KM is weaker than $0^\sharp$ in consistency strength, and so we can find such a model $V$ and hence also $V[G]$ without $0^\sharp$. The theory KM is weaker than $0^\sharp$ because its consistency follows from the existence of a single inaccessible cardinal: if $\kappa$ is inaccessible, then $V_\kappa$ is a model of KM when equipped with its full secondorder part $V_{\kappa+1}$. In contrast, $0^\sharp$ implies the consistency of a proper class of inaccessible cardinals (and more), since under $0^\sharp$ the Silver indiscernibles are all inaccesible in $L$.
This kind of example shows another direct way to build the desired model. Start with $\kappa$ inaccessible. So $V_\kappa$ is a model of ZFC, and remains a model of ZFC(S) even when we add the satisfaction class S for firstorder truth in $V_\kappa$. Now we may force to make S definable in a forcing extension $V_\kappa[G]$. In $V_\kappa[G]$, we have a definable class for firstorder truth in $V_\kappa$, from which we can define satisfaction in its $L$. But we needn't have $0^\sharp$, since in fact we could have started with $V=L$.

$\begingroup$ @Professor Hamkins: Thanks for the counterexamplesthey are very nice! Regarding forcing: since Cohen used forcing to 'create' (would 'create' be the proper term?) nonconstructible sets, could one use forcing to 'create' nonconstructible sets that are in some sense 'simpler' than 0sharp (I guess for want of a better definition of 'simpler', simpler in this case would mean not implying the consistency of a proper class of inaccessible cardinals)? Also, given Koepke's main theorem: a set S is constructible iff S is ordinal computable from finitely many ordinal parameters, what would $\endgroup$ Jun 7, 2013 at 7:14

$\begingroup$ (ramified) forcing 'look like' from the perspective of ordinal computability? $\endgroup$ Jun 7, 2013 at 7:18

$\begingroup$ In fact, what would 0sharp 'look like' from the perspective of ordinal computability? $\endgroup$ Jun 7, 2013 at 7:31

$\begingroup$ With $0^\sharp$ as an oracle, you can compute $L$generic Cohen reals, since $0^\sharp$ allows you to enumerate the dense subsets of this forcing in a countable sequence, which you can then descend through by diagonalization, building the generic real. Similarly, any other definable forcing notion in $L$, including the forcing notion I had mentioned (using the least inaccessible cardinal of $L$, which will be countable in $L[0^\sharp]$, admit $L$generic filters that are Turing computable from $0^\sharp$. $\endgroup$ Jun 7, 2013 at 10:36

1$\begingroup$ I'm curious about a related question: What is the simplest nonconstructible set of integers (say, in the analytical hierarchy) that is compatible with the nonexistence of $0^\sharp$? In particular, can there still be a nonconstructible $\Delta^1_3$ set of integers in the absence of $0^\sharp$? $\endgroup$ Nov 29, 2014 at 2:21