3. The Galvin-Hajnal theorems.
In this section I want to present two theorems of Galvin and Hajnal that greatly generalize Silver’s theorem. I focus on a “pointwise” (or everywhere) result, that gives us information beyond the pointwise theorems from last lecture, like Corollary 23. Then I state a result where the hypotheses, as in Silver’s theorem, are required to hold stationarily rather than everywhere. From this result, the full version of Silver’s result can be recovered.
Both results appear in the paper Fred Galvin, András Hajnal, Inequalities for Cardinal Powers, The Annals of Mathematics, Second Series, 101 (3), (May, 1975), 491–498, available from JSTOR, that I will follow closely. For the notion of -inaccessibility, see Definition II.2.20 from last lecture.
Theorem 1. Let be uncountable regular cardinals, and suppose that is -inaccessible. Let be a sequence of cardinals such that for all Then also
The second theorem will be stated next lecture. Theorem 1 is a rather general result; here are some corollaries that illustrate its reach:
Corollary 2. Suppose that are uncountable regular cardinals, and that is -inaccessible. Let be a cardinal, and suppose that for all cardinals Then also
Proof. Apply Theorem 1 with for all
Corollary 3. Suppose that are uncountable regular cardinals, and that is -inaccessible. Let be a cardinal of cofinality and suppose that for all cardinals Then also
Proof. Let be a sequence of cardinals smaller than such that and set for all Then for all by assumption. By Theorem 1, as well.
Corollary 4. Let be cardinals, with and regular and uncountable. Suppose that for all cardinals Then also
Proof. This follows directly from Corollary 2, since is regular and -inaccessible.
Corollary 5. Let be cardinals, with and of uncountable cofinality Suppose that for all cardinals Then also
Proof. This follows directly from Corollary 3 with
Corollary 6. Let be an ordinal of uncountable cofinality, and suppose that for all Then also
Proof. This follows from Corollary 5 with and
Corollary 7. Let be an ordinal of uncountable cofinality, and suppose that for all cardinals and all Then also
Proof. This follows from Corollary 4: If , then by Theorem II.1.10 from lecture II.2. But so both and are strictly smaller than
Corollary 8. If for all then also
Proof. By Corollary 5.
Corollary 9. If for all then also
Proof. By Corollary 7.
Notice that, as general as these results are, they do not provide us with a bound for the size of for the first cardinal of uncountable cofinality that is a fixed point of the aleph sequence, not even under the assumption that is a strong limit cardinal.
Now I proceed to the proof of Theorem 1.
Definition 10. Let be a cardinal and let be a sequence of nonempty sets. A set is an almost disjoint transversal system (a.d.t.) for iff whenever
As in the proof of Silver’s theorem, Theorem 1 follows from an analysis of the possible sizes of a.d.t.s for a particular sequence
Theorem 11. Let be uncountable regular cardinals, and suppose that is -inaccessible. Let be a sequence of nonempty sets such that for all Let be an a.d.t. for Then
Before proving Theorem 11, I show how it implies Theorem 1:
Lemma 12. Let be a cardinal and let be a sequence of cardinals. For let be the set product and set Then there is an a.d.t. for with
Proof of Theorem 1. Let and be as in the statement of Theorem 1. Let where for all By Lemma 12 there is an a.d.t. for of size By assumption, for all and it follows from Theorem 11 that as well.
Proof of Lemma 12. Let be the cardinal Consider an injective enumeration of the set product For let be given by for all Clearly, implies that is an ordinal below namely, the least such that It follows that is an a.d.t. for
All that remains is to show Theorem 11.
Proof. The argument is organized as an induction. For this, we need a rank on which to induct.
Definition 13. Let be a cardinal. Let be the partial ordering on given by iff (The subindex stands for “bounded.”)
Lemma 14. If is a regular, uncountable cardinal, then is well-founded.
The result is fairly general. For example, instead of the ideal of bounded sets, we could have used the ideal of nonstationary sets in Definition 13 (i.e., requiring that for club many values of ), and Lemma 14 would still hold; all we need is a -complete ideal. Below I follow the usual convention that a -complete ideal means one closed under unions of any length less that
Proof. Let be the filter of cobounded subsets of Notice that is -complete, by regularity of In particular, is indeed a partial order: If and then because contains the intersection of two sets in so it is also in
To prove well-foundedness, consider a -decreasing sequence and let so for all But then In particular, If then and is a decreasing sequence of ordinals, contradiction.
Whenever we are given a well-founded partial order on a set we can assign a rank (an ordinal) to the elements of as follows:
Definition 15. Let be well-founded. The rank of a set is defined by transfinite recursion as
It follows from standard arguments that, in the situation of Definition 15, is well-defined. For example, considering some for which is not defined (or unbounded), it follows that for some either is not defined or it is also unbounded. This easily gives us a strictly -decreasing sequence of members of contradicting that is well-founded. It is immediate from the definition that if then
Definition 16. Given a cardinal and a sequence of nonempty sets, let
- for i.e., only depends on the cardinalities of the terms of the sequence
- (Monotonicity). If and are sequences of nonempty sets with for all then
For let be where By monotonicity, Theorem 11 will follow if I can show that for all since if is finite for some one can just replace it with It is this reformulation that is proved by induction on the -rank of
We argue by contradiction: Assume the claim above fails, and let be a counterexample of least rank. It is our goal to show that there must be another counterexample of strictly smaller rank. For this, consider the following set:
All we need from Lemma 17 below is that is a nontrivial proper ideal (i.e., it is closed under finite unions, does not contain and contains the bounded sets), but proving -completeness requires about the same amount of work.
Lemma 17. is a nontrivial -complete proper ideal on
Proof. We prove this in four steps.
- Clearly, is -downward closed.
Suppose as witnessed by so and for all or
By -minimality of has size
If is an a.d.t. for then for we have that is bounded, so for many values of and they are finite! Let be a subset of of size The number of functions from to is and there are at most many such sets so the number of elements of is at most and contradiction.
It follows that i.e., is proper.
- If is bounded, then
For bounded, let be given by if and otherwise.
I claim that witnesses that Since it suffices to show that
Let be an a.d.t. for and define as where so and is injective since is bounded and for all but boundedly many values of Thus, is an a.d.t. for and it follows that
- It remains to argue that is closed under unions of fewer than sets.
For this, let be sets in for where as witnessed by functions Let Let be given by for
Let We argue that there is an a.d.t. for of size at least Since this holds for all such it follows that and witnesses that
To build fix an a.d.t. for of size for each and let be an injective enumeration of
Partition into (possibly empty) sets such that for all Set for so if and then Let
Notice that since each is an a.d.t., if then is bounded in so is bounded in by regularity of so is an a.d.t. for
Split as where and As argued in the proof of Lemma 17, so the following holds:
Proof. Since is regular and then is bounded, and we can find such that for all Let and Then
For each fix an a.d.t. for with For and let so is an a.d.t. for Moreover, since is a limit ordinal for all we have that For sufficiently large so we have that and therefore there must be some such that
Since is regular, there must be some fixed such that for many ordinals But then and witnesses that as we wanted to show.
It follows that
For define so if and otherwise. If it follows that for some
Since there is some such that and for all Let be an a.d.t. for of size strictly larger than
Given and let It follows that is an a.d.t. for where if and otherwise. By definition of we have that for all and it follows that there is an a.d.t. for of size But then
Let so If has size there is then some and some
We then have that and It follows that since it is in fact a bounded subset of Also, since otherwise it is a set such that is defined and contradiction. Similarly,
This is a contradiction, because then is the union of three sets in and is therefore also in