# Showing that extenders do not have nice closure properties

This is an exercise from Kanamori's "The Higher Infinite" which I solved by strengthening the hypothesis and I would like to know how we can solve it without doing that.

So the exercise is:

Exercise. Suppose that $$\kappa \lt \lambda$$, $$\lambda$$ is a strong limit of cofinality $$\omega$$, and there is a $$j:V\prec M$$ with $$\operatorname{crit}(j) = \kappa$$ and $$V_\lambda \subseteq M$$. Let $$\mathcal{E}$$ be the $$(\kappa, \lambda)$$-extender derived from $$j$$ and $$j_\mathcal{E}:V\prec M_\mathcal{E}$$. Then $$^\omega M_\mathcal{E} \not \subseteq M_\mathcal{E}$$.
Hint. $$V_\lambda \subseteq M_\mathcal{E}$$ by $$(*)$$, and so if $$\kappa \le \beta \lt \lambda$$, the $$(\kappa, \beta)$$-extender $$\mathcal{E}|[\beta]^{\lt \omega} \in M_\mathcal{E}$$. But $$\mathcal{E} \not \in M_\mathcal{E}$$.

The $$(*)$$ that is being referred to is:

$$(*)$$. Suppose we are working with a $$N$$-$$(\kappa, \beta)$$-extender $$\mathcal{E}$$ derived from some $$j:N\prec M$$. Then for any $$\gamma$$ satisfying $$|V_\gamma|^M \le \beta$$, we have $$V_\gamma^M = V_\gamma^{M_\mathcal{E}}$$.

This exercise seems quite straight-forward, especially with hint. But when I tried to show that $$|V_\lambda|^M \le \lambda$$, I ran into cofinality issues.

What I did was, I assumed that $$\lambda$$ was the limit of an $$\omega$$-sequence of inaccessibles and that did the job. Because inaccessibility is downward-absolute for inner models of ZFC, for each $$\alpha \lt \lambda$$ we can bound $$|V_\alpha|^M$$ with the size of the inaccessible cardinal above it, which in turn is less than $$\lambda$$. And we are done.

My question is: Can we say this for general strong limits?

It is true that it may happen that $$\vert V_\lambda\vert^M>\lambda$$. Note that the additional assumption you made is overkill for your argument: One can show that for $$\alpha>\omega$$, $$\vert V_\alpha\vert=\alpha$$ if and only if $$\beth_\alpha=\alpha$$. So you could also assume that $$\lambda$$ is a $$\beth$$-fixed point and then it is quite easy to see that $$\lambda$$ is a $$\beth$$-fixed point in $$M$$ as well (observe that the $$\beth$$-fixed points form a club and that $$(\beth\upharpoonright\lambda)^V=(\beth\upharpoonright\lambda)^M$$). Furthermore, a limit of inaccessibles is always a $$\beth$$-fixed point as all inaccessibles are $$\beth$$-fixed points, so this assumption is really less restrictive.

Anyhow the exercise is true without any additional assumption. To see that, we will code the extender more "rank-efficient". Let $$\gamma$$ be the maximal ordinal $$\alpha$$ with $$\vert V_\alpha\vert^M\leq\lambda$$. I claim that $$\gamma$$ is a limit of countable cofinality in $$V$$ and satisfies $$\vert V_\gamma\vert^M=\lambda$$. First note that the function $$h(\alpha)=\vert V_\alpha\vert^M$$ is strictly increasing and continuous (so normal) map (it is just the function $$(\omega+\alpha\mapsto\beth_{\alpha})^M)$$. So if $$\alpha<\gamma$$ then $$\vert V_\alpha\vert^M<\lambda$$. As $$\lambda$$ is a strong limit in $$V$$ and $$V_\lambda\subseteq M$$, $$\lambda$$ is a strong limit in $$M$$ as well. Thus $$\vert V_{\alpha+1}\vert^M=(2^{\vert V_\alpha\vert})^M<\lambda$$. Hence $$\gamma$$ is a limit. The rest of the claim follows from the normality of $$h$$.

Thus we may take a bijecion $$f:V_\gamma\rightarrow\lambda$$ in $$M$$; we will even take $$f$$ in $$M_{\mathcal E}$$ (note that $$V_\gamma$$ is the same in $$V$$, $$M$$ and $$M_{\mathcal E}$$ and has size $$\lambda$$ everywhere as well). We will moreover assume that whenever $$\alpha<\gamma$$ then $$f\upharpoonright V_\alpha$$ is a bijection between $$V_\alpha$$ and $$\vert V_\alpha\vert$$. We will now code the $$(\kappa, \lambda)$$ extender $$E$$, which is a directed system of ultrafilters indexed by $$a\in[\lambda]^{<\omega}$$ as a "$$(\kappa, V_\gamma)$$"-extender $$F$$ indexed by $$a\in [V_\gamma]^{<\omega}$$. Just put, for $$a\in [V_\gamma]^{<\omega}$$, $$F_a=\{\{(f^{-1}(b_0), \dots, f^{-1}(b_{\vert a\vert-1}))\mid (b_0,\dots, b_{\vert a\vert-1})\in A\}\mid A\in E_{f"a}\}$$ Here, I understand $$E_{f"a}$$ to be an ultrafilter on $$[\xi]^{\vert a\vert}$$ for $$\xi$$ least with $$j(\xi)>\operatorname{max}f"a$$ (so that $$\xi<\lambda$$). By our assumption on $$f$$, $$F_a\subseteq V_\alpha$$ for some $$\alpha<\gamma$$ only depending on and increasing in $$\operatorname{max} f"a$$, so that $$F\upharpoonright[V_\beta]^{<\omega}\in V_\gamma$$ whenever $$\beta<\gamma$$. Finally, let $$(\gamma_n)_{n<\omega}$$ be a sequence in $$V$$, witnessing $$\gamma$$ to have countable cofinality. We have $$F\upharpoonright [V_{\gamma_n}]^{<\omega}\in V_\gamma\subseteq M_\mathcal E$$ for all $$n<\omega$$, however $$F\notin M_\mathcal E$$ as otherwise $$M_\mathcal E$$ could decode $$E$$ from $$F$$ using $$f$$.

Edit: Regarding the questions:

1. We have seen that $$h:\operatorname{Ord}\rightarrow\operatorname{Ord}$$ is a normal map and that $$\gamma$$, the least ordinal with $$h(\gamma)\geq\lambda$$ is a limit ordinal. We have also seen that $$h(\alpha)<\lambda$$ implies $$h(\alpha+1)<\lambda$$. It now follows from normality that $$h(\gamma)=\lambda$$. It is an easy exercise to show that for any normal function $$h^\prime$$ and any two limit ordinals $$\gamma^\prime, \lambda^\prime$$ with $$h^\prime(\gamma^\prime)=\lambda^\prime$$ it follows that $$\operatorname{cof}(\gamma^\prime)=\operatorname{cof}(\lambda^\prime)$$. Thus $$\operatorname{cof}(\gamma)=\operatorname{cof}(\lambda)=\omega$$.
2. $$M_\mathcal E$$ has its own version of $$h$$, the map $$h^\prime(\alpha)=\vert V_\alpha\vert^{M_\mathcal E}$$. It follows from $$(\ast)$$ that $$V_\gamma\subseteq M_\mathcal E$$ and thus $$h\upharpoonright\gamma=h^\prime\upharpoonright\gamma$$. But $$h$$ and $$h^\prime$$ are both normal so that $$\vert V_\gamma\vert^{M_\mathcal E}=h^\prime(\gamma)=h(\gamma)=\lambda$$. Hence we can find this bijection in $$M_\mathcal E$$.
3. There are different essentially equivalent ways of defining what an extender is. The version in Kanamori where he fixes one such $$\xi$$ and lets each $$E_a$$ be an extender on $$[\xi]^{\vert a\vert}$$ is notationally convenient, but in some situations, like the one we are dealing with, not optimal. Associating an $$\xi_a$$ to every $$a$$ that indexes an ultrafilter in the extender is notationally more cumbersome, but that notion of extender has nice properties that are desirable in some situations. For example if you have an elementary embedding $$j:V\rightarrow M$$ with critical point $$\kappa<\nu<\mu$$ and you take $$E$$ the $$(\kappa, \nu)$$ extender derived from $$j$$ and $$E^\prime$$ the $$(\kappa, \mu)$$ extende derived from $$j$$ then for any $$a\in[\nu]^{<\omega}$$ it will be the case that $$E_a=E^\prime_a$$, which is not true in general for the definition of extender Kanamori uses. In any case, both of these two versions of defining extenders convey the same information in the sense that the one is easily translateable into the other so that both induce the exact same directed system of elementary embeddings and ultrapowers (so that the ultrapowers by the extenders are the same as well). This definition of extender is used for example in Ralf Schindlers book "Set Theory: Exploring Independence and Truth". Note that there are still more ways of defining extenders.
• Thanks for the answer. I am kinda okay with the argument as a whole, but I have $3$ minor questions. First is that, how do we prove that $\gamma$ has countable cofinality? Second is that although $V_\gamma$ is the same everywhere, how can we prove that it has also size $\lambda$ in $M_\mathcal{E}$? And lastly, you say that you consider $E_{f"a}$ to be an ultrafilter on $[\xi]^{|a|}$ for the least $\xi$ with $j(\xi) \gt \max f"a$. But in the definition in the book we take a fixed least $\xi$ s.t. $j(\xi) \gt \lambda$, how is this exchange possible? Feb 11, 2020 at 11:49
• Since my last $2$ questions might take a longer answer, I would also appreciate any book or reference that you could point me towards, so I can study by myself. Feb 11, 2020 at 11:51
• I have adressed your questions in an edit. Feb 11, 2020 at 13:28
• Many thanks for answering my questions. Actually after finishing this extender business from Kanamori I was planning to start Ralf Schindlers book! Feb 11, 2020 at 16:00
• Go for it! Keep up the good work. Feb 12, 2020 at 10:56