Existence of k-complete uniform ultrafilter over a regular cardinal, k is strongly compact

Question

This is a question about set theory. Let $\kappa\leq \lambda$ be infinite cardinals such that $\kappa$ is strongly compact and $\lambda$ is regular. My question is: how to construct a $\kappa$-complete uniform ultrafilter over $\lambda$?

I try to construct such an ultrafilter as follows. $\lbrace X\subseteq\lambda\mid\vert\lambda\setminus X\vert<\lambda\rbrace$ is an uniform and $\lambda$-complete filter, and hence is also $\kappa$-complete. Since $\kappa$ is strongly compact, the filter can be extended into a $\kappa$-complete ultrafilter over $\lambda$. But is the new filter also uniform? Or, is there any other method to solve this problem? Thanks for any comments or suggestions.

Answer 1

Joel has given an answer based on a modern view of strong compactness in terms of elementary embeddings. I'd like to complement that with an answer based on the oldest view of strong compactness of $\kappa$: The infinitary logic $L_{\kappa,\kappa}$ has the compactness property that, for any set $S$ of sentences, if each subset of $S$ of cardinality $<\kappa$ is satisfiable, then $S$ itself is satisfiable. In fact, this compactness property is needed only for the propositional fragment of $L_{\kappa,\kappa}$ (which I suppose should be called $L_{\kappa,1}$).

Work with a vocabulary that has propositional symbols (= 0-ary predicate symbols) $\hat X$, one for each subset $X$ of $\lambda$. Let $S$ consist of the following sentences of $L_{\kappa,1}$: $$\widehat{\bigcap_{i\in I}X_i}\iff\bigwedge_{i\in I}\widehat{X_i}$$ for each family $\{X_i:i\in I\}$ of subsets of $\lambda$ where the index set $I$ has cardinality $<\kappa$, $$ \widehat{\lambda-X}\iff\neg\hat X$$ for each $X\subseteq\lambda$, and $$\neg\hat Y$$ for every subset of $\lambda$ of cardinality $<\lambda$.

I claim that, if this set $S$ is satisfiable, say by a truth assignment $v$, then $$ \mathcal U=\{X\subseteq\lambda:v(\hat X)=\mathtt{true}\} $$ is a $\kappa$-complete uniform ultrafilter on $\lambda$. Indeed, $\mathcal U$ contains $\lambda$ by the first batch of sentences in $S$ (take $I=\varnothing$), is closed under supersets (take $I=\{0,1\}$ with $X_0\subseteq X_1$), and is closed under intersections of $<\kappa$ sets (take the $X_i$ to be the sets you want to intersect). So $\mathcal U$ is a $\kappa$-complete filter on $\lambda$. It is an ultrafilter by the second batch of sentences in $S$. Finally, it is uniform by the third batch of sentences. So, if $S$ is satisfiable then $\mathcal U$ is as desired.

By strong compactness of $\kappa$, it suffices to show that each subset $S_0\subseteq S$ of cardinality $<\kappa$ is satisfiable. Fix such an $S_0$ and let $B$ be the union of all the $Y$'s such that $\neg\hat Y$ is in $S_0$. Since each such $Y$ has cardinality $<\lambda$, since there are $<\kappa\leq\lambda$ such $Y$'s, and since $\lambda$ is regular, $B$ cannot be all of $\lambda$. Let $\alpha$ be any member of $\lambda-B$ and define a truth assignment $v$ by setting, for each $X\subseteq\lambda$, $$ v(\hat X)=\mathtt{true} \iff \alpha\in X. $$ The first two batches of sentences in $S_0$ (in fact in all of $S$) are trivially satisfied by $v$. The third batch of sentences in $S_0$ are satisfied because $\alpha\notin B$, i.e., $\alpha\notin Y$ for any $Y$ such that $(\neg\hat Y)\in S_0$.

Answer 2

Suppose that $\kappa$ is $\lambda$-strongly compact and $\lambda$ is regular. Let $j:V\to M$ be the ultrapower by a fine measure $\mu$ on $P_\kappa\lambda$. The object $s=[\text{id}]_\mu$ in $M$ is a subset of $j(\lambda)$ of size less than $j(\kappa)$, hence bounded, and $j"\lambda\subseteq s$ by fineness. So $\sup j"\lambda<j(\lambda)$. Let $\delta$ be any ordinal in that interval, and define $U$ for $X\subseteq\lambda$ by $$X\in U\quad\iff\quad \delta\in j(X).$$ This is a $\kappa$ complete ultrafilter on $\lambda$ in $V$. It is uniform, since every small subset of $\lambda$ is bounded and therefore its image is contained in some $j(\alpha)$, whereas $\delta$ exceeds $\sup j"\lambda$. So every set in $U$ is unbounded in $\lambda$.

By the way, the truly interesting question is the converse direction. Namely, it is a theorem of Ketonen that if a cardinal $\kappa$ admits $\kappa$-complete uniform ultrafilters on every regular $\lambda\geq\kappa$, then $\kappa$ is strongly compact. I've given above the easy direction of the biconditional, for it is the converse direction that is Ketonen's fundamental contribution.