0
$\begingroup$

or (contrariwise) that $NGB$ + "There exists a Reinhardt cardinal" is consistent?

The question is partially in the title. $NGB$ is used for the reasons stated in the Hamkins, Kirmayer, and Perlmutter paper, "Generalizations of the Kunen inconsistency", Annals of Pure and Applied Logic, 163 (2012) pp. 1873-1876 (Section 1, "A few metamathematical preliminaries").

In order that this question should not be deemed overbroad, answers must be limited in scope to specific mathematical difficulties (eg. the Erdos-Hajnal theorem used in in proving the Kunen inconsistency in $NGBC$ or $NBC$ + $Choice$ has no known proof in $NGB$ or is known not to be provable without choice--however, those who would say that the Erdos-Hajnal theorem cannot be proven without choice must show how this fact allows for a non-trivial elementary embedding $j$: $V$ $\rightarrow$ $V$ and the existence of the requisite critical point $\kappa$ in $NGB$). Those who are so inclined might make mention of Rupert McCallum's recent attempt in showing that "the existence of a non-trivial elementary embedding $j$: $V_{\lambda + 2}$ $\rightarrow$ $V_{\lambda + 2}$ for some ordinal $\lambda$...is not consistent with $ZF$" (quote from McCallum's withdrawn short preprint, The Choiceless Cardinals are Inconsistent, arXiv:1712.09768, which is discussed on Joel David Hamlin's blog — I would be especially interested in getting a nice explanation (i.e. analysis) of the gap in this argument and the relavance of the Laver paper which Gabriel Goldberg directed him to).

Though at first glance the answers to these questions might seem "primarily opinion-based", the mathoverflow community should realize the value of informed opinion and (yes) informed speculation regarding this topic, and use this question as an opportunity to provide these 'clues' (think of the proof of the 4-color theorem) to both the mathoverflow community and the mathematical community as a whole.

$\endgroup$
6
  • 1
    $\begingroup$ Although the title question is clear enough (and answered by the dependence of the proof on the Erdös-Hajnal theorem), the parenthetical addendum in the question itself, "must show how this fact allows for a non-trivial elementary embedding $j:V\to V$," seems to require a proof of consistency (relative to large cardinals) of NBG + existence of such $j$. $\endgroup$ Commented Oct 13, 2018 at 17:02
  • 3
    $\begingroup$ To elaborate on Andreas' comment: since at present we don't know a definitive answer, we can't possibly show that a given point is a necessary stumbling block. The best that can be done is to point out the key step which "no known argument can get around," and observe that this step doesn't go through without choice. So I don't understand what you're looking for in an answer, short of a new major result in set theory, if that doesn't satisfy you. $\endgroup$ Commented Oct 13, 2018 at 23:26
  • $\begingroup$ @AndreasBlass: Well, at least a proof of relative consistency.... $\endgroup$ Commented Oct 15, 2018 at 13:49
  • $\begingroup$ @NoahSchweber: While one may not have a definitive answer, one might have relevant facts that could lead to a definitive answer. An easy example, there might be, say, in some obscure journal, a proof of the relevant Erdos-Hajnal theorem that doesn't use choice, or a choice principle weaker than $AC$ (say, the Axiom of Dependent Choices). The relevant Erdos-Hajnal theorem has been around a long time so such a theorem might be out there. A proof that the relevant Erdos-Hajnal theorem cannot do without $AC$ may suggest that one might be able to construct a model of $NGB$ that has a $\endgroup$ Commented Oct 15, 2018 at 14:06
  • $\begingroup$ (cont.) Reinhardt cardinal, but then again there might be some structural peculiarities (mimicking the 'weirdness' of $ZF$ relative to $ZFC$ Asaf Karagila speaks of in his blog or his answers and comments in mathoverflow and mathstackexchange) of models $NGB$ relative to models of$NGBC$ or $NGB$ + $Choice$ (but if one could localize $Choice$ in $NGB$ + $Choice$ to somwhere above the critical point of an $I0$ embedding, and above that, $Choice$ would fail-- where would the point of demarcation be?), how would those peculiarities relate to the success or failure of the Kunen inconsistency $\endgroup$ Commented Oct 15, 2018 at 14:29

1 Answer 1

10
$\begingroup$

Here is the strategy behind McCallum's attempted proof that there is no nontrivial elementary embedding $j :V\to V$.

Step 1 is to cite the theorem that if a $j :V\to V$ is consistent, then a $j : V\to V$ is consistent with $V_\lambda\vDash \text{ZFC}$ where $\lambda = \sup_{n < \omega} j^n(\text{crt}(j))$. This is proved by a minor modification of Woodin's technique for forcing choice assuming a supercompact. So it suffices to show this stronger assumption is inconsistent.

Step 2 is to try to reflect the failure of AC given by Kunen's theorem into $V_\lambda$. There is a technique for reflecting certain properties of tame substructures of $V_{\lambda+2}$ to cardinals below $V_\lambda$. This is called inverse limit reflection, originally invented by Laver and recently vastly extended by Scott Cramer. For example, a theorem of Cramer implies that in this situation, and assuming DC, there is some $\bar \lambda < \lambda$ such that there is an elementary $\pi : L(V_{\bar \lambda+1})\to L(V_{\lambda+1})$ and even a $\pi : L(V_{\bar \lambda + 1},V_{\bar \lambda + 1}^\#)\to L(V_{\lambda + 1},V_{\lambda + 1}^\#)$.

[Here is a successful application of this proof strategy. Note that $I_0$ holds at $\bar \lambda$ since $I_0$ at $\lambda$ is a first-order fact in $(V_{\lambda+1},V_{\lambda+1}^\#)$. Thus $V_\lambda$ is a model of ZFC + $I_0$. This is how I proved NBG + DC + $j :V\to V$ implies Con(ZFC + $I_0$).]

McCallum had rediscovered the inverse limit reflection technique, and he thought he could reflect the existence of an $\omega$-Jonsson cardinal into $V_\lambda$ using it. The issue, however, was that the statement that $\lambda$ is $\omega$-Jonsson is a fact about $V_{\lambda+2}$, and inverse limit reflection does not seem to be able to reflect statements about the whole structure $V_{\lambda+2}$ into $V_\lambda$. One can only reflect properties of canonical inner models over $V_{\lambda+1}$, like $L(V_{\lambda+1})$ and $L(V_{\lambda+1},V_{\lambda+1}^\#)$.

So maybe the obstruction to carrying out this proof idea is, very vaguely, that one must first extend inverse limit reflection / inner model theory to "reach" $V_{\lambda+2}$.

$\endgroup$
19
  • 1
    $\begingroup$ Cramer showed that if $I^\#_0$ holds (i.e., there is a $j :L(V_{\lambda+1}^\#)\to L(V_{\lambda+1}^\#)$ with sup of critical sequence $\lambda$) then $I_0$ holds in $V_\lambda$. If you do the AC forcing and carry out Cramer's proof in DC then you get that $\text{ZF + DC + }I_0^\#$ implies $\text{Con(ZFC + }I_0)$. The fact that a Reinhardt implies $I_0^\#$ requires proving $V_{\lambda+1}^\#$ exists from a Reinhardt. In fact a Reinhardt implies every set has a sharp, another result of mine from that paper (which is still unpublished). $\endgroup$ Commented Oct 23, 2018 at 23:08
  • 1
    $\begingroup$ It is in "Inverse limit reflection and the structure of $L(V_{\lambda+1})$." The result is Theorem 3.9. $\endgroup$ Commented Oct 24, 2018 at 15:04
  • 1
    $\begingroup$ Yes $V_{\lambda+1}^\#$ can be viewed as a subset of $V_{\lambda+1}$. It is the theory of $\omega$-many Silver indiscernibles of $L(V_{\lambda+1})$ with parameters from $V_{\lambda+1}$. You should look at Caicedo's "A review of sharps." $\endgroup$ Commented Oct 30, 2018 at 12:29
  • 1
    $\begingroup$ andrescaicedo.files.wordpress.com/2008/04/sharps.pdf $\endgroup$ Commented Oct 30, 2018 at 12:48
  • 1
    $\begingroup$ $I_0^\#$ at $\lambda$ is precisely the statement that $V^\#_{\lambda+1}$ exists and is an Icarus set. But if $\lambda$ is least such that $I_0$ holds at $\lambda$ and $V_{\lambda+1}^\#$ exists, then $I_0^\#$ fails at $\lambda$ by Cramer's results. $\endgroup$ Commented Oct 30, 2018 at 14:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .