# Inaccessible cardinals and well-founded models of ZFC

This was a problem in an exam I took last semester, but I never got the chance to ask my professor how to solve it. Here goes:

Let $\kappa$ be an inaccessible cardinal. Then $V_\kappa$ is a transitive model of ZFC (with the usual $\in$). Let $U$ be a non-principal ultrafilter over $\kappa$ and let $\tilde{M} = (V_\kappa)^\kappa/U$ be the corresponding ultrapower of $V_\kappa$.

We say that an ultrafilter $U$ has the $\sigma$-property if given sets $A_i \subset \kappa, i \in \mathbb{N}$, we have that if $A_i \in U$ for every $i \in \mathbb{N}$ then $\bigcap_{i \in \mathbb{N}} A_i \in U$.

(a) Show that $\tilde{M}$ is well-founded iff $U$ has the $\sigma$-property.

(b) Suppose $\tilde{M}$ is well-founded, and let $M$ be obtained from $\tilde{M}$ through the Mostowski collapse (so that $M$ is a transitive model of ZFC). Show that $M$ is elementarily equivalent to $V_\kappa$ but $M \neq V_\kappa$.

IDEAS: One of the definitions of well-founded (and I believe the best one for this problem) is: A model of ZFC $(M, \varepsilon)$ is NOT well-founded iff there exists a sequence $x_1, x_2, \ldots$ in $M$ (note that we're talking about a ''true'' sequence and not a sequence of the model) with $x_{n+1}$ $\varepsilon$ $x_n$ for all $n \in \mathbb{N}$.

I managed to prove the easier direction in part (a): Suppose $\tilde{M}$ is not well-founded. Then there exists a sequence such that $\ldots$ $\varepsilon$ $x_2$ $\varepsilon$ $x_1$. Let $f_i$ be an element in the equivalence class $x_i$. By definition, $x_{i+1}$ $\varepsilon$ $x_{i} \Leftrightarrow A_i := \{k \in V_\kappa | f_{i+1}(k) \in f_i(k) \} \in U$.

Since $V_\kappa$ is well-founded, $\bigcap_{i \in \mathbb{N}} A_i = \emptyset \notin U,$ thus $U$ does not have the $\sigma$-property.

Also, in part (b), $M$ is clearly elementary equivalent to $V_\kappa$ since it is isomorphic to $\tilde{M}$ (through the Mostowski collapse), which is elementary equivalent to $V_\kappa$.

Any thoughts on the other parts?

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## 1 Answer

The $\sigma$-property is also commonly known as saying that the ultrafilter is countably complete, and the existence of an ultrafilter with this property is equivalent to the existence of a measurable cardinal, since any countably complete ultrafilter is actually $\delta$-complete for some measurable cardinal $\delta$.

Question (a) is treated in any of the standard accounts of large cardinals, such as Kanamori's book The Higher Infinite. There are two directions. You have shown that if the ultrafilter is countably complete, then the ultrapower is wellfounded. Conversely, suppose that the ultrafilter is not countably complete. Then there is a sequence $X_n\subset\kappa$ with each $X_n\in U$ but $\bigcap_n X_n\notin U$. We may assume that $\bigcap X_n=\emptyset$, by throwing out a measure zero set, and we may assume that $X_{n+1}\subset X_n$. For each $\alpha\lt\kappa$, let $n_\alpha$ be least such that $\alpha\notin X_{n_\alpha}$. Let $f_k(\alpha)=n_\alpha-k$, if $\alpha\in X_k$. Note that as $k$ increases, these functions decrease on their domain, which have measure $1$, and so we have a violation of the well-foundedness of the ultapower. Indeed, the ultrapower has a nonstandard $\omega$, since our functions have values in $\omega$.

For part (b), you've observed that $V_\kappa$ is elementary equivalent to its ultrapower by the Los theorem. To see that $M\neq V_\kappa$, one could argue directly from the Kunen inconsistency, which implies it as an immediate consequence. Alternatively, we may argue as follows. First, if every set in $U$ has size $\kappa$, then the identity function $id:\kappa\to\kappa$ is larger than the constant functions on a measure one set, and hence represents an ordinal in the ultrapower at least as large as $\kappa$ itself, which ensures that the height of $M$ is strictly taller than $\kappa$, and so $M\neq V_\kappa$. So we may assume that $U$ concentrates on some $\gamma\lt\kappa$, and we may correspondingly restrict the functions to $\gamma$. Again it follows that $j(\gamma)\gt\gamma$, where $j$ is the map $V_\kappa\to M$, since the identity function exceeds the constant functions. Now let $\delta\lt\kappa$ be the $\gamma^{\rm th}$ cardinal with $\delta^\gamma=\delta$. It follows that $j(\delta)\gt\delta$, but observe that every element of $j(\delta)$ is represented by a function $f:\gamma\to\delta$, of which there are only $\delta$ many. This violates elementary if $V_\kappa=M$, since $j(\delta)$ is supposed to be a cardinal. QED

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Part (a) is clear, although I'd like to add the following comment that made all the difference to me: note that $\{\alpha : f_{k+1}(\alpha) \in f_k(\alpha)\} \supset X_{k+2} \in U$. As for part (b), the part with the cardinal $\delta$ got me confused. Could you provide more detail? –  Pedro M. Dec 27 '11 at 14:02
Ok, I almost understood it. One question though: why does $\delta$ have to be the $\gamma^{th}$ cardinal with such property? –  Pedro M. Dec 27 '11 at 14:12
We need that $j(\delta)\gt\delta$ to get the contradiction, and this follows if $\delta$ is the $\gamma$-th with a property, since we already know $j(\gamma)\gt\gamma$. –  JDH Dec 27 '11 at 14:36
I confess I didn't understand all the details, but the proof seems correct. Besides, the Kunen inconsistency does kill the rabit, so I'll accept the answer. Thanks! –  Pedro M. Dec 28 '11 at 10:37