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Rank-into-rank cardinals have the rather intriguing property that they reflect upwards. I would be interested to know how far the upward reflection goes:

1) Does "There exists a rank-into-rank cardinal (of type I3, say)" imply that "There exists an unbounded class of rank-into-rank cardinals (in V) ?"

2) If 1) is false, then can we say something interesting about the supremum of this upward reflecting sequence ? For instance, maybe the supremum has some large cardinal properties of interest ?

3) Finally, existence of a rank-into-rank implies the existence of pretty much all of the other large cardinals. But can we strengthen this along the lines of "If there exists a rank-into-rank cardinal, then there exists an unbounded class of cardinals with property P (in V)" for interesting some property P? By interesting property P, I mean something like "inaccessible" or "Mahlo" or hopefully even stronger like "measurable".

Anindya
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The answer is negative.

The existence of a rank-to-rank cardinal $j:V_\lambda\to V_\lambda$ is $\Sigma_2$ expressible, since it is witnessed inside any sufficiently large $V_\alpha$. Therefore, if one cuts off at any inaccessible or worldly cardinal above $\lambda$, one still has the rank-to-rank cardinal, but there would be no large cardinals above $\lambda$.

So the answers to questions 1 and 3 are strongly negative. Basically, if $\kappa$ is rank-to-rank, witnessed by embedding $j:V_\lambda\to V_\lambda$ with critical point $\kappa$, then $\lambda$ is the limit of the critical sequence, and this gives you $\omega$ many additional rank-to-rank cardinals. But by cutting off above $\lambda$, one can see that $\lambda$ itself is the limit of any upward reflection phenomenon.

In a sense, this perspective shows that with rank-to-rank cardinals, it is not necessarily the critical point $\kappa$ that is important, but the cardinal $\lambda$ that is relevant. And this way of thinking destroys the upward-reflection idea, since $\lambda$ is not reflecting upward at all.

Meanwhile, the cardinal $\lambda$ itself must be worldly of very high order, since it is the union of an elementary chain of rank-to-rank cardinals. This can be seen as a positive answer to question 2.

Joel David Hamkins
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    If $\lambda$ is worldly, then doesn't $V_\lambda\vDash ZFC+j,is,a,nontrivial,elementary,embedding$, and so providing a model for $ZFC+There,exists,a,Reinhardt,cardinal$? – Master Jun 07 '19 at 03:42
  • No, that isn't right. The structure $V_\lambda$ cannot satisfy ZF in the language with $j$, because the critical sequence will be an $\omega$-sequence unbounded in $\lambda$, which would violate the replacement axiom in $\langle V_\lambda,\in,j\rangle$. In particular, $j$ is not definable in $V_\lambda$, and you cannot add it without destroying ZF. – Joel David Hamkins Jun 07 '19 at 08:27
  • Ah, so the image of $\omega$ under $j$ is not in $V_\lambda$. – Master Jun 07 '19 at 15:58
  • The embedding $j$ fixes every set up to the critical point, so the image of $\omega$ under $j$ is just $\omega$ itself. But the critical sequence $\kappa_0=cp(j)$, $\kappa_{n+1}=j(\kappa_n)$, which is definable from $j$, is cofinal in $\lambda$, and this is what causes the problem. – Joel David Hamkins Jun 07 '19 at 16:01
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    I think I understand now. The problem is the maping $n\mapsto \kappa_n$. – Master Jun 07 '19 at 16:20
  • Yes, that's right. – Joel David Hamkins Jun 07 '19 at 16:22