Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose $j:V_\lambda \rightarrow V_\eta$ is (elementary and) cofinal. Can $j$ be extended to all of $V$?

(Subsidiary question: What conditions are there on an ultrafilter/extender/whatever so that the induced $j:V \rightarrow M$ has $V_\eta$ as an initial segment of $M$? Note that $j(\kappa)$ might well be less than $\eta$, where $\kappa$ is the critical point.)

share|improve this question
6  
Welcome to MathOverflow! –  Joel David Hamkins Oct 31 '13 at 23:57

1 Answer 1

up vote 11 down vote accepted

At successor ordinals, the answer is no, not necessarily, assuming the consistency of a nontrivial instance of your hypothesis.

For a counterexample, let $\kappa$ be the least $1$-extendible cardinal. So there is $j:V_{\kappa+1}\to V_{\eta+1}$, and this map is elementary and cofinal, mapping $\kappa$ to $j(\kappa)=\eta$.

I claim that this map cannot extend to $j:V\to M$ for a transitive set $M$, with $V_{\eta+1}\subset M$. The reason is that the size of $j\upharpoonright V_{\kappa+1}$ is $|V_{\kappa+1}|=2^\kappa$, which is therefore coded by a size $2^\kappa$ subset of $V_{\eta+1}$, which is therefore an element of $V_{\eta+1}$. So $M$ will see that $\kappa$ is $1$-extendible, and so by elementarity there must be a $1$-extendible cardinal below $\kappa$, contradicting minimality.

Similarly, if $\kappa$ is the least $(\theta+1)$-extendible cardinal, then there is $j:V_{\kappa+\theta+1}\to V_{\eta+1}$. But since $j\upharpoonright V_{\kappa+\theta+1}$ is a size $|V_{\kappa+\theta+1}|$ subset of $V_{\eta+1}$, it is coded by an element of $V_{\eta+1}$, and so if $j:V\to M$ extends $j$, then $M$ would see that $\kappa$ is $(\theta+1)$-extendible, violating minimality.

Update. But at limit ordinals $\lambda$, the answer is yes, all such elementary cofinal embeddings $j:V_\lambda\to V_\eta$ lift to $j:V\to M$ for some transitive class with $V_\eta\subset M$. To see this, for any $a\in V_\eta$, let $\mu_a$ be the measure generated via $j$ by $a$, so that $X\in \mu_a\iff a\in j(X)$. This measure will concentrate on any set $D_a\in V_\lambda$ for which $a\in j(D_a)$, and such a set exists since the map was cofinal. Let $j_a$ be the ultrapower of $V$ by $\mu_a$. Elementary seed theory shows that $j_a\upharpoonright V_\lambda$ is a factor embedding of $j$, and furthermore, the range of $j_a\upharpoonright V_\lambda$ is precisely the seed hull $X_a=\{\ j(f)(a)\mid f:D_a\to V_\lambda,\ f\in V_\lambda\ \}\prec V_\eta$. These elementary substructures form a directed system, and the map $j$ is the direct limit of the maps $j_a\upharpoonright V_\lambda$. But now the point is that the maps $j_a$ are defined on all of $V$, since they are simply ultrapower maps by the measures $\mu_a$. And so one may take the corresponding direct limit of the full maps $j_a:V\to M_a=\text{Ult}(V,\mu_a)$. The result is an elementary embedding $j:V\to M$ into a transitive class (after the collapse), which extends the given map $j:V_\lambda\to V_\eta$.

Basically, what is happening here is that we use the original map $j:V_\lambda\to V_\eta$ to define an extender, which is then applied to the whole of $V$. This strategy didn't work fully at succcessor ordinals $j:V_{\theta+1}\to V_{\eta+1}$, since there was no way to ensure that all of $V_{\eta+1}$ was picked up in the range of the induced extender, as those seeds $a$ are on the top level of $V_{\eta+1}$ and not covered by any element of $V_{\theta+1}$ (and perhaps this could be considered a violation of cofinality of $j$, even though $j(\theta)=\eta$). But even in the successor ordinal case, we can still define the induced extender, and the resulting $j:V\to M$ will have $V_\eta\subset M$ and it will agree with the original $j:V_{\theta+1}\to V_{\eta+1}$, and in particular include the image of $V_{\theta+1}$ under $j$, although it may not have all of $V_{\eta+1}$ contained in $M$.

share|improve this answer
1  
Joel, Does your observation have any bearing on my question: mathoverflow.net/questions/118689/…? –  Everett Piper Nov 1 '13 at 1:51
1  
@EverettPiper, yes, indeed, the same argument works there, and I have now posted an answer. –  Joel David Hamkins Nov 1 '13 at 2:02
    
Joel, I am so grateful to you. First, for your kind welcome to me on Mathoverflow. Second, for your correct answer about $\kappa$ being the least 1-extendible. Although correct, it left me unsatisfied, so I wanted to post an additional question about the situation when $\lambda$ is a limit. But before I even got to do that, you answered that question too. This helps me with my research with Norman. I hope you will be pleased with the result when it's done. –  Robert Lubarsky Nov 1 '13 at 14:27
    
I'll look forward to hearing all about it... –  Joel David Hamkins Nov 1 '13 at 17:45

Your Answer

 
discard

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.