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".

  • $\begingroup$ Note that even superstrong cardinals have "upward reflection." $\endgroup$ – Monroe Eskew Jun 4 '19 at 22:29

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ If $\lambda$ is worldly, then doesn't $V_\lambda\vDash ZFC+j\,is\,a\,nontrivial\,elementary\,embedding$, and so providing a model for $ZFC+There\,exists\,a\,Reinhardt\,cardinal$? $\endgroup$ – Master Jun 7 '19 at 3:42
  • $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Jun 7 '19 at 8:27
  • $\begingroup$ Ah, so the image of $\omega$ under $j$ is not in $V_\lambda$. $\endgroup$ – Master Jun 7 '19 at 15:58
  • $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Jun 7 '19 at 16:01
  • 1
    $\begingroup$ I think I understand now. The problem is the maping $n\mapsto \kappa_n$. $\endgroup$ – Master Jun 7 '19 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.