MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is exercise 20.5 out of Jech:

Let $\lambda \geq \kappa$ and let $U$ be a normal measure on $P_{\kappa}(\lambda)$. The ultraproduct $\mathrm{Ult} _U \{ (V _{\lambda _x},\in) : x \in P _{\kappa}(\lambda) \}$ is isomorphic to $(V _{\lambda}, \in)$

Here $\lambda _x$ simply denotes the order type of $x$. The function $x \mapsto \lambda _x$ represents $\lambda$ in the ultrapower of $V$ by $U$. Unless I'm mistaken, the ultraproduct mentioned will be $V^M _{\lambda} = V _{\lambda} \cap M$ where $M$ is the ultrapower of $V$ by $U$. I don't see why this would be $V _{\lambda}$ itself, since I don't see why $V _{\lambda} \subset M$. Clearly $H _{\lambda ^+} \subset M$ since $M$ is closed under $\lambda$ sequences, but I sort of doubt that $V _{\lambda} \subset M$ -- I figure if $\lambda$-supercompactness implied $\lambda$-strongness, I would've seen that mentioned somewhere.

So did I make a mistake in computing the ultraproduct, or is there a mistake in the exercise, or does $\lambda$-supercompactness imply $\lambda$-strongness for some reason I'm not seeing?

share|cite|improve this question
To get curly brackets, put a backtick, `, on the outside of your dollar signs. I've edited to fix it up. – David Roberts Oct 30 '10 at 22:47
Thanks David. Is there every any reason to not put back ticks (aside from a couple extra keystrokes), i.e. is there anything you can do without back ticks that you cannot do with them? – Amit Kumar Gupta Oct 31 '10 at 0:19
up vote 5 down vote accepted

In general, $\lambda$-supercompactness, if consistent, does not imply $\lambda$-strongness. One can see this by observing that the smallest cardinal $\kappa$ that is $\kappa^+$-supercompact is never $\kappa^+$-strong, and in fact, cannot be even $(\kappa+3)$-strong. The reason is that $\kappa^+$-supercompactness is witnessed by a measure on $P_\kappa\kappa^+$, which amounts essentially to (is coded by) a subset of $P(\kappa^+)$, and hence is witnessed inside $V_{\kappa+3}$. Thus, if $j:V\to M$ were any embedding with critical point $\kappa$, by minimality it follows that $\kappa$ is not $\kappa^+$-supercompact in $M$, and hence $M$ cannot have the true $V_{\kappa+3}$. Thus, $\kappa$ is not $(\kappa+3)$-strong and thus definitely not $\kappa^+$-strong.

I haven't looked at the context of the exercise, but perhaps he is merely asking you to make the observation that you did in fact make, that the ultrapower will give you $V_\lambda^M$?

For some kinds of $\lambda$, it does follow that $\lambda$-supercompactness implies $\lambda$-strongness. For example, if $\lambda$ is a beth-fixed point, then every $\lambda$-supercompactness embedding is also $\lambda$-strong and even $(\lambda+1)$-strong, since in this case $|V_\lambda|=\lambda$. But if $\lambda$ is not a beth-fixed point, then a version of my argument above will still apply: if $\lambda$ is not a beth-fixed point and $j:V\to M$ is a Mitchell minimal $\lambda$-supercompactness embedding for $\kappa$, then $\kappa$ is not $\lambda$-supercompact in $M$, and this is witnessed inside $V_\lambda$ by the assumption on $\lambda$, and so $j$ cannot be a $\lambda$-strongness embedding.

share|cite|improve this answer
(Joel: "strength" rather than "strongness", no?) – Andrés E. Caicedo Oct 30 '10 at 22:59
This discrepancy between partial supercompactness and strength causes all sorts of problems when trying to pinpoint the optimal large cardinal hypotheses that allow one to force "localized" versions of forcing axioms, by the way. – Andrés E. Caicedo Oct 30 '10 at 23:01
Andres, I always use "strongness" in this situation, since I find "strength" to be overloaded, refering more generally to large cardinal strength, while degrees of strongness seems less ambiguous. – Joel David Hamkins Oct 30 '10 at 23:07
Joel, that makes sense. :-) – Andrés E. Caicedo Oct 30 '10 at 23:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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