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
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
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.
3. Large cardinals and
With the preliminary technical machinery established last lecture, we are ready to show how the presence of large cardinals in the universe may influence its cardinal arithmetic.
Theorem 7 (Scott) Let be measurable and let be a normal nonprincipal, -complete measure on If and then In particular, a measurable cardinal cannot be the first counterexample to Similarly, if then and if then
Proof: Let be the ultrapower embedding derived from If and then by ‘s lemma. But since and by Lemma 20 from last lecture, since is normal.
This shows that, in By Lemma 21 from last lecture, we know that so and since It follows that in
The other two statements are shown similarly.
The assumption that is normal cannot be eliminated in the previous theorem. For example, Gitik showed that it is consistent to have a measurable cardinal and a stationary set such that for all but On the other hand, if holds in a club below , then it also holds at since any normal ultrafilter extends the club filter.
Definition 8 Denote by the increasing enumeration of the fixed points of the function.
The following is the first result we obtain about fixed points of the aleph sequence:
Theorem 9 Suppose that is measurable and that is strong limit. Then
Proof: Let be a -complete nonprincipal ultrafilter over and let be the corresponding elementary embedding. The result follows from general facts about continuity points of
- If then while if then
- If is strong limit and then while if then
Proof: To prove 1, we need to determine whether there is some function such that
If let be increasing and cofinal in and take for all
If then any is bounded.
If the -completeness of implies that there is some such that
To prove 2, notice first that is closed under : If then where the first inequality follows from the fact that any ordinal below is for some
The result now follows from item 1: If then but always holds.
If note that so, in particular, Also, since is strong limit. Since then If that was shown in Lemma 21 from last lecture. If then and since is strong limit in
Let now and suppose that is strong limit. Then, by item 2 of Lemma 10, But since by Lemma 21 from last lecture.
Before we continue, let me remark that the continuity results above depend in an essential way on the fact that the embedding being considered is an ultrapower embedding, and they do not need to hold for other elementary maps.
As mentioned at the end of last lecture, the embeddings that measurability provides are rather restrictive since, for example, The results above show that measurability influences cardinal arithmetic, so it is reasonable to expect that stronger large cardinal assumptions will provide us with stronger results. This is indeed the case, as the following example illustrates. Next lecture, I will show another result along these lines, deriving a significantly stronger conclusion from the existence of a significantly stronger cardinal.
Definition 11 Let be an ordinal. A cardinal is -strong iff there is an elementary embedding with and
is strong iff it is -strong for all
Clearly, is measurable iff it is -strong iff it is -strong, but already -strength of is more significant than measurability as it implies that for any ultrafilter on so in particular the measurable cardinals below form a stationary set.
The standard presentation of strength adds the extra requirement that This is not necessary; in fact, if is -strong as defined above, then it is -strong under the stronger requirement of this paragraph. This is a consequence of the following famous result of Kunen:
Proof: There are several nice proofs of this result. Kunen’s argument uses that any infinite set admits an -Jónsson function, that is, a map such that whenever has the same size as then Here, as usual, This is a result of and Hajnal.
Zapletal has found a proof that uses a modicum of pcf theory, namely, Shelah’s result that any singular cardinal admits a scale, that is, there is an increasing sequence of regular cardinals cofinal in and a sequence of functions such that for all whenever then and whenever then for some Here, is the partial order induced by the ideal of bounded subsets of see Definition 13 in lecture II.6.
There are a few other arguments as well. I present here a proof due to Woodin. All arguments begin by noticing that if
then Here, and To see this, apply to both sides of the displayed equation:
It follows that because Therefore, is cofinal in
The same argument that shows that is fixed shows that is unbounded: For any consider the sequence Its supremum is in
Also, is -closed. (A set is -closed iff whenever is unbounded in and then ) This is simply because the supremum of an -sequence of fixed points of is necessarily fixed by
But any -closed unbounded subset of an ordinal of uncountable cofinality contains the restriction to of a club, namely, the club of limit points of elements of In particular, this holds of It follows that if is stationary, then
Woodin’s argument now invokes Ulam’s result that every stationary subset of a successor cardinal can be split into many disjoint stationary sets, this is Theorem 13 in lecture II.5. Consequently, we can partition into many stationary pieces. Let be such a partition. By elementarity, is (in ) a partition of into many stationary pieces. The result will follow by deriving a contradiction from the assumption that is stationary in
For suppose that it is. It follows that is stationary for some since the club filter on is -complete and But then must meet Let be in this intersection. Then and But then so contradicting that is a partition.
The argument can actually be tightened to prove the following two stronger results:
Proof: It is easy to code (in an absolute way) subsets of by elements of so Woodin’s proof of Kunen’s Theorem 12 goes through for since the sets it makes reference to are subsets of and we have a contradiction.
To see that there is such a coding, simply notice (as in the appendix in lecture I.5) that there is a definable bijection between and that the map from a well-ordering of to its order type gives via this bijection a (definable) surjection from onto and that the inverse of this surjection is a (definable) injection of into
Theorem 14 (Kunen) If is elementary and is the first fixed point of past then
Proof: Suppose otherwise, and define an ultrafilter over by
Then is a -complete ultrafilter, and we can form the ultrapower embedding
Lemma 15 and
Proof: Begin by noticing that In effect: Let be arbitrary such that This occurs iff which (by definition) holds iff for some But then and We easily conclude that and for all
Letting be given by is now easily checked to be well-defined and elementary, and .
Note now that if and is given by then
It follows that by the definition of and the same argument. Therefore, In particular, and
Finally, let Then since Note that implies that by the proof of Lemma 21.2 from last lecture, so and since But then
It follows from the lemma that again by the proof of Lemma 21.2 from last lecture. But then since any element of has size at most and we have a contradiction by Theorem 13.
Kunen’s result shows that the definition of strong cardinals given above coincides with the standard one.
Proof: Suppose that is -strong as witnessed by By Theorem 13, where is the first fixed point of past so (and therefore also ) is strictly below for some
Extend to proper classes by setting for all classes Define for by and so Define for by and so each is elementary, and
Corollary 17 If is -strong for all and holds below then it holds for all cardinals below Thus if is strong and holds below then it holds everywhere.
Proof: Let be a cardinal below Then for some By Corollary 16, there is an embedding such that and Hence, holds in below in particular it holds at But so in fact in
As mentioned above, it is consistent that fails at a measurable. This is closely related to the failure of Prikry forcing allows us to change the cofinality of a measurable to while preserving all cardinals and without adding any bounded subsets of Hence, if then after Prikry forcing we obtain a singular strong limit of cofinality with i.e., we have violated
Moreover, the consistency strength of both assumptions (the existence of a measurable where fails and the failure of ) is the same: Both statements are equiconsistent over to the existence of a cardinal of Mitchell order This is a notion significantly stronger than measurability, but weaker than -strength.
Definition 18 The Mitchell order of a measurable cardinal is defined as follows: First, for normal ultrafilters and over set iff It is easy to see that this implies that and therefore is well-founded.
We define the Mitchell order of by is a normal nonprincipal -complete ultrafilter over
It follows that iff is measurable. The assumption implies that the measurable cardinals below form a stationary set, but is stronger than this.
Homework problem 12. Assume that is -strong and determine
One final remark is in order: Through this lecture I have used proper classes freely, without worrying about whether this can be formalized in This is indeed the case, since just as the existence of embeddings corresponds to the existence of -complete nonprincipal ultrafilters, the embeddings discussed here can be coded via sequences of ultrafilters (and so, any mention of proper classes can be eliminated by referring to the (definable) ultrapowers from which they arise). Nevertheless, the right setting for Kunen’s theorem is a set theory that allows classes, so it is perhaps more reasonable to think that we are arguing in either or (even better) My personal opinion is that these matters are not mathematical in nature and should therefore not worry us.
Through this lecture, I have made good use of Akihiro Kanamori, The Higher Infinite, Springer (1994). For an introduction to the Rudin-Keisler order, see Wistar Comfort, Stylianos Negrepontis, The theory of ultrafilters, Springer (1974).
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