# Is the critical point of an embedding of a model of set theory inaccessible in it?

Can we find an elementary embedding $j:M\to N$ with $M,N$ transitive $ZFC$-models, $\kappa$ being the critical point, so that $\kappa$ is not inaccessible in $M$ ? ($\kappa$ is regular in $M$.) I would tend towards "probably yes", but at the moment I don't know where to start looking.

To complement Asaf's answer, let me point out that you don't need any large cardinals at all to get embeddings like this; they exist as long as there is an uncountable transitive model of ZFC+"there are no inaccessible cardinals".

If $N$ is such an uncountable model we can get an embedding by taking a countable elementary substructure $X\prec N$ and collapsing it to a transitive $M$. The inverse collapse map gives us the embedding $j\colon M\to N$. Since the critical point $\kappa$ has the same properties in $M$ as $j(\kappa)$ has in $N$ and $N$ had no inaccessibles, $\kappa$ cannot be inaccessible in $M$.

It is less clear to me what happens if there are no uncountable transitive models of this theory lying around. First of all, there might not be any uncountable transitive models of ZFC at all. And secondly, it doesn't seem impossible that every uncountable transitive model believes that there is an inaccessible (and when you chop it off at that inaccessible the model becomes countable). On the other hand, we are still ok in this second case, at least consistency-wise, since we can pass to $L$ where this cannot happen.

• Right. You can also use a weakly compact cardinal, as noted by Camilo under my answer. Without uncountable transitive models of $\sf ZFC$ you can probably still get this to happen. Implication-wise it's easy, assume there is a largest cardinal for transitive models and collapse it to be countable. – Asaf Karagila Aug 31 '15 at 11:15
• This is the simplest example to me. Thank you. – user35359 Aug 31 '15 at 12:10
• @user35359: I think that the reason you got three very different answers is that it wasn't clear if you want $M$ and $N$ to be set models, or class models or have the same ordinals, or one of them to be a subclass of the other, or if $j$ should be definable internally, or so on and so forth. But at this point I think that we covered most bases, I guess. – Asaf Karagila Aug 31 '15 at 12:40
• In understand, but I didn't know yet if the most general form of question has an example, although I thought there is probably one, I didn't know how difficult it is, or if restrictions wil make it harder to answer. I also thought about asking : What if $M$ is a model of ZFC minus powerset and $\kappa$ is the largest cardinal in $M$. Anyway, I got some nice answers, so I'm happy. – user35359 Aug 31 '15 at 13:02
• @user35359: Yes, you got two very nice answers, and a suggestion to learn Hebrew and master time travel. All in all, not a bad deal. :-D – Asaf Karagila Aug 31 '15 at 15:59

No, not necessarily.

Suppose $\kappa$ is measurable in $W$, let $M$ be $W[G]$ the model obtained after forcing with $\operatorname{Col}(\omega,<\kappa)$. If $j\colon W\to W'$ is an ultrapower embedding with critical point $\kappa$, then we can force over $M$ to obtain a model in which $j$ can be extended from $W$ to $M$.

Another important example is the stationary tower forcing, which can produce from a Woodin cardinal in $V$, an embedding [definable in $V[G]$] from $V\to M$, with critical point $\omega_1$.

Somewhere to look for this could be Matt Foreman's chapter in the Handbook "Ideals and Generic Elementary Embeddings". Although admittedly I just learned these facts from my advisor (I don't have notes to distribute, though, sorry).

• What I'm suddenly not sure about is whether $\omega_1$ in the case of the stationary tower forcing is measurable in an inner model. But it is inaccessible in $L$, that's for sure. – Asaf Karagila Aug 30 '15 at 4:49
• Actually, for your argument above you need only a weakly compact $\kappa$ cardinal, as for any transitive $M$ that is a model of a decent chunk of $\mathsf{ZFC}$ with $|M|=\kappa\in M$, you can get a embedding $j:M\rightarrow N$ with $crit(j)=\kappa$ – Camilo Arosemena-Serrato Aug 31 '15 at 4:10
• Sure, but this opens to interpretation what do you mean by models. Sets or classes. – Asaf Karagila Aug 31 '15 at 8:33
• Thank you. I'll have to read into these 3 (with Camilo's) examples. It'll be fun I guess. About the first example, which forcing do you mean when you say "we can force over M"? Or is that better left as an exercise to figure out? – user35359 Aug 31 '15 at 12:23
• @user35359: It's a nontrivial thing, but if you read about generic ultrapowers, this is really the first example. I mean, to have an extension you'd need to get it from $W[G]$ to $W'[H]$ where $H$ is a suitable generic filter, but a generic filter for what? Well, $j(\operatorname{Col}(\omega,<\kappa))$. So you really just need to add this part to $W[G]$ in order to be able and extend the embedding. In the case of this simple example a lot of things really tie neatly together and you get something nice. But it should probably be the first example you'll run into. – Asaf Karagila Aug 31 '15 at 12:37

Note that in Asaf's answer, the elementary embedding $j: M\rightarrow N$ does not "live" (that is, is not definable in) $M$.

By contrast, if we have an elementary embedding $j: M\rightarrow N$ which is definable in $M$ (from parameters in $M$), then $crit(j)$ is inaccessible, in fact measurable, in $M$: letting $\kappa=crit(j)$, we form the ultrafilter $$\mathcal{U}=\{S\subseteq \kappa: \kappa\in j(S)\},$$ and note that this ultrafilter is $<\kappa$-closed (since $j(\bigcup_{\alpha<\mu}A_\alpha)=\bigcup_{\alpha<\mu} j(A_\alpha)$ for $\mu<\kappa$, since $\kappa=crit(j)$ so $j(\mu)=\mu$); since $\mathcal{U}$ exists in $M$ (since $j$ is definable), we have $\kappa$ is measurable in $M$.

Even if $j$ is not definable in $M$, however, we may form the ultrafilter $\mathcal{U}$ described above; it might not live in $M$, is all that changes. In inner model theory, we are often interested in embeddings which, though maybe not definable over their domain model, nonetheless yield reasonably tame ("amenable") ultrafilters.

• That's a good point. – Asaf Karagila Aug 30 '15 at 8:22
• For inaccessibility, we don't need that $N$ is definable in $M$, just that $N\subseteq M$. Right? – GME Aug 30 '15 at 10:12
• @GME: If $j$ is definable in $M$, then $N$ is definable in $M$. – Asaf Karagila Aug 30 '15 at 10:17
• @AsafKaragila Sure. But $N\subseteq M$ is weaker than definability of $j$ or $N$ in $M$ in general. Or am I missing something? – GME Aug 30 '15 at 10:18
• @GME In fact, it should be enough that $M$ and $N$ agree on the power set of $\kappa$. – Miha Habič Aug 30 '15 at 12:15