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What strengthenings of measurability does the Mostowski collapse of the ultrapowers possess?

Ok, I already posted this question, but a couple of notational errors and assumptions were made in the previous version. Hopefully, this new equivalent version will be better understood.

Let $U$ be a nonprinciple $\lambda$-complete ultrafilter over $\lambda$. Let $\pi_U(f)$ for a function $f$ with domain $\lambda$ be defined as follows: $$\pi_U(f)=\{\pi_U(g):\{\alpha<\lambda:g(\alpha)\in f(\alpha)\}\in U\}$$

Let $M$ be $\{\pi_U(f):\mathrm{Dom}(f)=\lambda\}$. Finally, let $\lambda_0=\lambda$ and: $$\lambda_{n+1}=|\{\pi_U(f):\{\alpha<\lambda:g(\alpha)\in \lambda_n\}\in U\}|$$

Then, which of the following are always true:

  • If $\lambda$ is $\theta$-strong, then $V_\theta\subset M$.
  • If $\lambda$ is $\theta$-supercompact, then $M^\theta\subset M$.
  • If $\lambda$ is $n$-superstrong, then $V_{\lambda_n}\subset M$.
  • If $\lambda$ is $n$-huge, then $M^{\lambda_n}\subset M$.
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  • $\begingroup$ This question is oddly phrased. How could $V_\alpha$ be a subset of the ultrapower anyway? Since the ultrapower is a collection of equivalence classes. Most sets are not equivalence classes. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 6, 2017 at 15:33
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    $\begingroup$ I'm confused by the question. When we say $M^\theta\subset M$ for an ultrapower $M$, we mean that $M$ is closed under $\theta$ sequences in a way that makes sense with thinking about $\langle M,\in^M\rangle$ as a structure in the language of set theory, which is of course not literally the same as saying it is closed under $\theta$-sequences. When $M$ is well-founded, as it is with complete ultrafilters, then this amounts to saying the corresponding thing about the Mostowski collapse of $M$, which trivializes the question. Set theorists usually identify the ultrapower with its collapse. $\endgroup$
    – Joel David Hamkins
    Commented Nov 6, 2017 at 15:39
  • $\begingroup$ @AsafKaragila $MO$ is not the ultrapower, rather the Mostowski collapse of it. $\endgroup$
    – Keith Millar
    Commented Nov 6, 2017 at 15:55
  • $\begingroup$ @JoelDavidHamkins $M$ is a transitive inner model such that $j:V\rightarrow M$ is an elementary embedding with critical point $\kappa$. $M$ has no relation to $MO$. Am I misinterpreting your comment? $\endgroup$
    – Keith Millar
    Commented Nov 6, 2017 at 15:58
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    $\begingroup$ I think what you are asking is whether many (or all?) of the reflection properties captured by large cardinal embeddings can be realized by ultrapowers by ultrafilters. Is this correct? And if so, do you want to restrict the carrier set of the ultrafilter ($\kappa$-complete ultrafilters on $\kappa$? Or on $\mathcal{P}_\kappa(\lambda)$? Or something else?) $\endgroup$
    – Miha Habič
    Commented Nov 6, 2017 at 16:09

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Let's adopt Miha's interpretation of your question: which kinds of large cardinal properties can be witnessed by ultrapower embeddings?

Here, there are a variety of things one can say.

If one allows $\kappa$-complete measures on arbitrary sets, then basically all the usual large cardinal properties can be witnessed with suitably chosen ultrapower embeddings. For example, supercompactness embeddings $j:V\to M$ are ultrapowers by normal fine measures on $P_\kappa\theta$, and these kinds of embeddings are also $\alpha$-strong, when $\beth_\alpha\leq\theta$.

Hugeness and almost hugeness are also witnessed by ultrapower embeddings on a suitable set, so this means that (assuming appropriate consistency strength), we can realize all the properties on your list with ultrapower embeddings.

But meanwhile, one cannot expect to define the various extender-based large cardinal notions in ZFC using only ultrapowers. For example, it is relatively consistent that $\kappa$ is strong, but for large enough $\alpha$, there is no ultrapower embedding $j:V\to M$ by a measure on any set for which $V_\alpha\subset M$.

If you mean to allow only ultrapowers by measures on $\kappa$ itself, then much of those embeddings are not possible.

  • No ultrapower embedding by a measure on $\kappa$ can realize $\kappa+2$ strongness, since $M_{j(\kappa)}$ will have size $2^\kappa$ in $V$ and therefore cannot contain all of $V_{\kappa+2}$.

  • In particular, no ultrapower embedding by a measure on $\kappa$ can realize $2^\kappa$-supercompactness, for then it would realize $\kappa+2$-strongness.

  • Meanwhile, there is a kind of border right at that level. There is an interesting theorem of Woodin showing how to make the least measurable cardinal also a little supercompact, and in that model, we get that the $\kappa^+$-supercompactness of $\kappa$ is witnessed by a normal ultrapower on $\kappa$. Necessarily, by the previous observations, $\kappa^+<2^\kappa$ in this model. (Arthur Apter also has some work on this.)

That last example would be perhaps one of the most interesting situations fitting under the umbrella of this interpretation of your question.

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  • $\begingroup$ Although $\theta$-supercompactness can be witnessed by an ultrapower as you said, full supercompactness seems to need a lot of utrapowers --- or am I missing something here. $\endgroup$
    – Andreas Blass
    Commented Nov 6, 2017 at 16:27
  • $\begingroup$ @AndreasBlass Yes, being supercompact is properly a $\Pi_3$ statement. It cannot be $\Sigma_2$, because then the existence of a supercompact cardinal would reflect below the least supercompact cardinal, leading to big problems. $\endgroup$
    – Miha Habič
    Commented Nov 6, 2017 at 16:32
  • $\begingroup$ I was interpreting the question differently, and more simply: which large cardinal properties can be realized by ultrapower embeddings, in place of the arbitrary embeddings one might state. If you define $\kappa$ is supercompact to be: for every $\theta$ there is $j:V\to M$ with critical point $\kappa$ and $M^\theta\subset M$, then indeed it is equivalent to also insist that $j$ is an ultrapower embedding. In the case of strong, etc., this is not an equivalence, but meanwhile it is relatively consistent that strongness is witnessed by ultrapowers, e.g., by supercompactness ultrapowers. $\endgroup$
    – Joel David Hamkins
    Commented Nov 6, 2017 at 16:35
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    $\begingroup$ Aha, @JoelDavidHamkins, this new comment that you have provided is precisely my question. Thank you for taking the time to understand my poorly written question. $\endgroup$
    – Keith Millar
    Commented Nov 6, 2017 at 23:52

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