Let me begin with a couple of comments that may help clarify some of the results from last lecture.
First, I want to show a different proof of Lemma 21.2, that I think is cleaner than the argument I gave before. (The argument from last lecture, however, will be useful below, in the proof of Kunen’s theorem.)
Lemma 1 If is measurable, is a -complete nonprincipal ultrafilter over and is the corresponding ultrapower embedding, then
Proof: Recall that if is Mostowski’s collapsing function and denotes classes in then To ease notation, write for
Let Pick such that for all
Lemma 2 With notation as above, for any
Proof: For a set let denote the function constantly equal to Since is an isomorphism, ‘s lemma gives us that the required equality holds iff
but this last set is just
From the nice representation just showed, we conclude that for all But for any such because by Lemma 21 from last lecture. Hence, which is obviously in being definable from and
The following was shown in the proof of Lemma 20, but it deserves to be isolated.
Lemma 3 If is a normal nonprincipal -complete ultrafilter over the measurable cardinal then i.e., we get back when we compute the normal measure derived from the embedding induced by
Finally, the construction in Lemma 10 and preceeding remarks is a particular case of a much more general result.
Definition 4 Given and an ultrafilter over the projection of over is the set of such that
Clearly, is an ultrafilter over
Notice that if is a partition of into sets not in and is given by the unique such that then is a -complete nonprincipal ultrafilter over (Of course, is possible.)
For a different example, let be a -complete nonprincipal ultrafilter over the measurable cardinal and let represent the identity in the ultrapower by Then is the normal ultrafilter over derived from the embedding induced by
Definition 5 Given ultrafilters and (not necessarily over the same set), say that is Rudin-Keisler below in symbols, iff there are sets and a function such that
Theorem 6 Let be an ultrafilter over a set and an ultrafilter over a set Suppose that Then there is an elementary embedding such that
Proof: Fix and for which there is a map such that Clearly, as witnessed by the map and similarly so it suffices to assume that and
Given let be given by Then is well-defined, elementary, and
In effect, iff iff iff where the second equivalence holds by assumption, and it follows that is well-defined.
If denotes the function with domain and constantly equal to then for any since by definition of the map This shows that
Elementarity is a straightforward modification of the proof of Lemma 10 from last lecture.
One can show that Theorem 6 “very nearly” characterizes the Rudin-Keisler ordering, see for example Proposition 0.3.2 in Jussi Ketonen, Strong compactness and other cardinal sins, Annals of Mathematical Logic 5 (1972), 47–76.
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