# Consistency of “the sharp of every set exists”

If there exists a Jónsson cardinal $\kappa$, then $x^\#$ exists for every $x\in V_\kappa$ (in particular $V\neq L[x]$). It follows that if there is a proper class of Jónsson cardinals, then the sharp of every set should exist (this happens if for example if there is a proper class of Ramsey or measurable cardinals).

So the existence of a proper class of Jónsson cardinals is an upper bound for the consistency of "for every $x$, $x^\#$ exists". $\mathbf{\Pi}^1_1$-determinacy, being equivalent to the existence of the sharp of every real, is a lower bound.

Is the exact consistency strength of "for every $x$, $x^\#$ exists" known?

Edit: as François G. Dorais pointed out in a comment, if $\kappa$ is Jónsson then $V_\kappa\models$ "for every $x$, $x^\#$ exists". (we aren't guarenteed that $V_\kappa\models ZFC$ but that doesn't matter consistency-strength-wise), so "there is a Jónsson cardinal" is an upper bound for the consistency strength of "for every $x$, $x^\#$ exists".

• If $\kappa$ is Jónsson then $V_\kappa \vDash$ all sharps exist, so the existence of a single Jónsson cardinal is strictly stronger than the existence of all sharps in consistency strength. – François G. Dorais Nov 14 '17 at 22:09
• How do you define $\Bbb R^\#$? Is it by asserting the existence of an embedding $j\colon L(\Bbb R)\to L(\Bbb R)$? Or the existence of indiscernibles? Or a real which codes the truth predicate and some clever requirements? – Asaf Karagila Nov 14 '17 at 23:41
• (1) It's a bit odd to ask about a technical concept without understanding the odds and ends of that technical concept. (2) I know there are ways to define sharps, but I am interested in how you think about sharps, because that would matter for the final answer, at least to some extent. – Asaf Karagila Nov 15 '17 at 0:06
• – Mohammad Golshani Nov 15 '17 at 5:10

Closure under sharps is equivalent to the failure of the covering lemma (alternatively, weak covering lemma) for every $L[x]$. It also equivalent to $\mathbf{\Pi}^1_1$-determinacy in every generic extension of V. The nature of the assertion "for every $x$, $x^\#$ exists" is such that equiconsistent statements tend to be actually equivalent to it.
Jónsson cardinals are equiconsistent with Ramsey. A consistency-wise weaker assertion that is consistency-wise stronger than closure under sharps is existence of $ω_1$-Erdos cardinals. A still weaker assertion (that remains consistency-wise stronger than closure under sharps) is determinacy in level $ω^2$ of the difference hierarchy of analytic sets. Some other (somewhat technical) notions can be found in "On unfoldable cardinals, ω-closed cardinals, and the beginning of the inner model hierarchy" (P.D. Welch, 2004).
• I thought that $\omega_1$-Erdos only implies that $\forall x\in \mathbb{R}$, $x^{\#}$ exists, and not the full closure under sharps. – Yair Hayut Nov 26 '17 at 11:01
• @YairHayut $ω_1$-Erdős cardinal does not imply closure under sharps above it, but I think it implies closure under sharps below it, and in any case has a higher consistency strength. – Dmytro Taranovsky Nov 26 '17 at 16:35