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
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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
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
![]()
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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.