10
$\begingroup$

Let $\alpha\not= 0$ be such that for every $\beta<\alpha$ there is $\beta<\gamma<\alpha$, where $V_\gamma$ is an elementary substructure of $V_\alpha$. In other words, $V_\alpha$ is a limit of its $V_\beta$ elementary substructures. Then it is a simple result that $V_\alpha$ models replacement.

My question: Let $\alpha\not= 0$ be such that for every $\beta<\alpha$ there is $\beta<\gamma<\alpha$ and an elementary embedding from $V_\gamma$ to $V_\alpha$. Does it follow that $V_\alpha$ models replacement?

$\endgroup$
8
  • $\begingroup$ Oh, you changed the question. I was about to post an answer to the earlier one. $\endgroup$ – Joel David Hamkins May 13 at 11:36
  • $\begingroup$ @JoelDavidHamkins I’m so sorry! The original question is the one I’m interested in ultimately, but I thought the new question would be more accessible. If you have an answer, though, should I just reinstate that question? $\endgroup$ – Sam Roberts May 13 at 11:49
  • $\begingroup$ The questions are similar, since you get $\Sigma_n$-elementary embeddings in the old version by using a universal $\Sigma_n$-truth predicate. I'll try to post an answer a bit later. $\endgroup$ – Joel David Hamkins May 13 at 11:50
  • $\begingroup$ @JoelDavidHamkins Right. But I thought the new question might be more tractable. In fact, the new question is the one I was focusing my attention on when I was thinking this all through. $\endgroup$ – Sam Roberts May 13 at 11:53
  • 2
    $\begingroup$ @AsafKaragila That’s right, the assumption is weaker. But getting an elementary embedding between ranks that isn’t the identity requires a lot of large cardinal strength. To show that we can have the assumption without replacement, we need a bunch of elementary embeddings that aren’t the identity and thus a lot of large cardinal strength. (I was working with extendables and Vopkenka’s principle, e.g. to try and get such a model.) $\endgroup$ – Sam Roberts May 13 at 15:13
11
$\begingroup$

If $\alpha$ is a limit of $2^\alpha$-supercompact cardinals, then by the Magidor characterization of supercompactness, for each $2^\alpha$-supercompact cardinal $\kappa < \alpha$, for some $\gamma < \kappa$, there is an elementary embedding $j : V_{\gamma}\to V_\alpha$ with critical point arbitrarily large below $\kappa$. Thus for each $\beta < \gamma$, there is an elementary embedding $j : V_{\gamma}\to V_\alpha$ such that $\gamma$ is between $\beta$ and the next $\alpha$-supercompact cardinal, which yields the property you asked about.

But if $\alpha$ is the least cardinal that is a limit of $\alpha$-supercompact cardinals, then $V_\alpha$ does not model replacement: from the perspective of $V_\alpha$, there are $\omega$-many supercompact cardinals that are cofinal in the ordinals. The reason is that the $2^{\alpha}$-supercompacts of $V$ are precisely the supercompacts of $V_\alpha$. Clearly the forwards implication holds, but conversely if $\delta$ is supercompact in $V_\alpha$, then $\delta$ is supercompact up to a cardinal $\kappa < \alpha$ that is $2^\alpha$-supercompact, and as a consequence, $\delta$ itself is $2^\alpha$-supercompact.

The optimal hypothesis is the existence of an ordinal $\alpha$ that is a limit of $\alpha$-Magidor supercompact cardinals, where a cardinal $\kappa$ is $\alpha$-Magidor supercompact if for some $\gamma < \kappa$, there is an elementary embedding $j : V_\gamma\to V_\alpha$ such that $\kappa = j(\text{crit}(j))$. Let $\alpha$ be the least ordinal as in your question. Let $\beta$ be the supremum of the $\alpha$-Magidor supercompact cardinals. If $\beta < \alpha$, then fix an elementary embedding $j : V_\gamma\to V_\alpha$ with $\beta < \gamma < \alpha$, and note that $j(\text{crit}(j)) \leq \beta < \gamma$ and hence $j$ witnesses that $\text{crit}(j)$ is huge, contrary to the minimality of $\alpha$. So $\alpha = \beta$ and hence $\alpha$ is a limit of $\alpha$-Magidor supercompact cardinals.

$\endgroup$
3
  • $\begingroup$ Thanks, Gabe! That's just what I was looking for. I take it the easiest way to implement your idea is to let $\alpha$ be the limit of the first $\omega$ supercompacts. Then we get that $\beta$ is supercompact in $V_\alpha$ just in case it is supercompact simpliciter and thus that they are the first $\omega$ supercompacts from the perpsective of $V_\alpha$. $\endgroup$ – Sam Roberts May 14 at 7:06
  • $\begingroup$ Yes, that works. And after all the theory "ZFC + infinitely many supercompacts" is not really much stronger than "ZFC + the existence of an ordinal $\alpha$ that is a limit of $\alpha$-Magidor supercompacts," which already implies the consistency of Zermelo set theory + infinitely many supercompacts. $\endgroup$ – Gabe Goldberg May 15 at 0:19
  • $\begingroup$ Exactly! And it makes obtaining the relevant limit trivial. $\endgroup$ – Sam Roberts May 15 at 4:49

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.