10
$\begingroup$

I am not 100% certain this question is appropriate for MO; I may just be missing something obvious. Also, I vaguely recall a similar question being asked here a while ago, but I can't find it; if it turns out this is a duplicate, I'll delete this question. Anyways, apologies in advance if this is too easy or is a duplicate. Note that e.g. Can measures be added by forcing? prevents the obvious nuke from working.

Also, the "descriptive-set-theory" tag is purely a guess on my part, based on the surprising ubiquity of descriptive set theory in similar-sounding questions.

Suppose I have a transitive model $M$ of $ZFC$, and - in $M$ - $U$ is a measure on $\kappa$. Then the transitive collapse of the ultrapower of $M$ along $U$ is an inner model, $N\subset M$.

My question is:

Can we ever have $M$ be a generic extension of $N$ (either by set or class forcing in $M$)?

As mentioned above, I am almost certain the answer is "no", even if $M$ has loads of large cardinals, but I don't see how to prove this.

EDIT: Danielle Ulrich answered the question for set forcing; I've asked the class forcing version as a separate question.

A small observation:

Say (inside a model $W$) a cardinal $\mu$ is

  • potentially measurable if $\mu$ is measurable in some forcing extension; and

  • reversibly measurable if $\mu$ is measurable, and $W$ is a forcing extension of the transitive collapse of the ultrapower of $W$ by a measure on $\mu$ (that is, if $\mu$ is as above).

Then suppose we had such an $M, N, U, \kappa$, with $j$ the elementary embedding. Then $N$ satisfies "There is a potential measurable below $j(\kappa)$," so - pulling back along $j$ - $M$ satisfies "There is a potential measurable below $\kappa$." This shows that - in $M$ - the least potentially measurable is strictly less than the least reversibly measurable (otherwise we get a descending chain of measurable cardinals).

Now, it feels plausible to me that there's a clever trick that can be done here to outright build a descending sequence of reversibly measurables from a single reversibly measurable; but I don't see it.

Improve this question
$\endgroup$
4
  • $\begingroup$ The term "$\kappa$ is generically measurable" means that you can find a generic ultrapower embedding [into a transitive class] with $\kappa$ as a critical point. It just seemed relevant here. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 26, 2015 at 21:29
  • $\begingroup$ @AsafKaragila It's not obvious to me, though, what the exact connection is; for instance, $\omega_1$ can be generically measurable, clearly never potentially measurable. (I suspect there is a connection in consistency strengths, but I don't quite see it.) $\endgroup$
    – Noah Schweber
    Commented Dec 26, 2015 at 22:18
  • $\begingroup$ I think there may be a mistake in the url of your link for the class forcing question. Probably you meant this one: mathoverflow.net/q/259628/1946. $\endgroup$
    – Joel David Hamkins
    Commented Jan 15, 2017 at 11:14
  • $\begingroup$ @JoelDavidHamkins Whoops, just saw this - fixed, thanks! $\endgroup$
    – Noah Schweber
    Commented Sep 28, 2022 at 18:46

1 Answer 1

Reset to default
16
$\begingroup$

For set-forcing, the answer is no, see the following article

Joel David Hamkins, Greg Kirmayer, and Norman Lewis Perlmutter, Generalizations of the Kunen inconsistency, Ann. Pure Appl. Logic 163 (2012), no. 12, 1872--1890. (see also arxiv.org/abs/1106.1951 and Hamkins's blog post)

The "Kunen Inconsistency" is the theorem that there is no nontrivial elementary embedding $j: V \to V$. The above article shows (among several other things) that if $V[G]$ is any set-forcing extension of $V$ then there is no nontrivial elementary embedding $j: V[G] \to V$. So even if you replaced ultrapowers by extenders, for example, the answer remains no.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Excellent! I'm still interested in the situation with class forcing (and I'm kicking myself a bit; I read that paper recently, and I just didn't make the connection :P). $\endgroup$
    – Noah Schweber
    Commented Dec 26, 2015 at 22:46
  • $\begingroup$ Thanks for mentioning our paper! I added fuller reference information. $\endgroup$
    – Joel David Hamkins
    Commented Dec 26, 2015 at 23:51
  • 3
    $\begingroup$ And incidentally, theorem 10 of the paper applies to many class forcing extensions. So you cannot "undo" an ultrapower by forcing that is eventually stationarily correct at sufficiently large regular cardinals. $\endgroup$
    – Joel David Hamkins
    Commented Dec 27, 2015 at 0:25

You must log in to answer this question.

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