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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".

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    $\begingroup$ 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. $\endgroup$
    – François G. Dorais
    Commented Nov 14, 2017 at 22:09
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    $\begingroup$ 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? $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 14, 2017 at 23:41
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    $\begingroup$ (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. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 15, 2017 at 0:06
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    $\begingroup$ see A review of sharps $\endgroup$
    – Mohammad Golshani
    Commented Nov 15, 2017 at 5:10

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In one sense, closure under sharps is itself a standard point in the hierarchy of consistency strengths. Just like the exact consistency strength of "ZFC + measurable" is "ZFC + measurable", so is the case for closure under sharps. However, the value of the consistency strength hierarchy is in terms of the connections it offers, and here we can say more.

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).

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  • $\begingroup$ I thought that $\omega_1$-Erdos only implies that $\forall x\in \mathbb{R}$, $x^{\#}$ exists, and not the full closure under sharps. $\endgroup$
    – Yair Hayut
    Commented Nov 26, 2017 at 11:01
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    $\begingroup$ @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. $\endgroup$
    – Dmytro Taranovsky
    Commented Nov 26, 2017 at 16:35

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