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* stationaril*y 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 a*lmost 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

Clearly,

- 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:

**Claim 18.**

**Claim 19.**

**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

[…] Last lecture, I covered the first theorem of the Galvin-Hajnal paper and several corollaries. Recall that the result, Theorem 3.1, states that if and are uncountable regular cardinals, and is -inaccessible, then for any sequence of cardinals such that for all […]

[…] Proof: Suppose first that is -complete. If there is a -decreasing sequence let Then for all and therefore (since it is in fact an element of ) If is in this intersection, then is an -decreasing sequence of sets, contradiction. (This is just the proof of Lemma 14 in lecture II.6.) […]

[…] 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. […]

[…] extend and improve the Galvin-Hajnal results on powers of singulars of uncountable cofinality (see lectures II.6 and II.7). His first success was an extension to cardinals of countable cofinality. For example: […]

Hi,

In the proof of claim 19, in the line before the last one, should it be instead of or am I being silly?

Thanks.

Hi! There were a few other typos as well: In the proof of Lemma 17, “partition ” should have been “partition ”, and should have been In the paragraph above, should have been . They are now fixed.

Hi,

Ha yes, it is true, since the double indexed functions go into because is an a.d.t for and that’s where the functions are by definition.

Right after the proof of claim 19, in the third paragraph of the proof of do we need “and otherwise” instead of “and otherwise”?

R.A

You are right, thanks!

[…] notes, I have a proof of a general version of this result, due to Galvin and Hajnal, see Lemma 12 here; essentially, we list all functions , and then replace them with (appropriate codes for) the […]

[…] notes, I have a proof of a general version of this result, due to Galvin and Hajnal, see Lemma 12 here; essentially, we list all functions , and then replace them with (appropriate codes for) the […]

[…] 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. […]

[…] extend and improve the Galvin-Hajnal results on powers of singulars of uncountable cofinality (see lectures II.6 and II.7). His first success was an extension to cardinals of countable cofinality. For […]

[…] Last lecture, I covered the first theorem of the Galvin-Hajnal paper and several corollaries. Recall that the result, Theorem 3.1, states that if and are uncountable regular cardinals, and is -inaccessible, then for any sequence of cardinals such that for all […]

[…] Proof: Suppose first that is -complete. If there is a -decreasing sequence let Then for all and therefore (since it is in fact an element of ) If is in this intersection, then is an -decreasing sequence of sets, contradiction. (This is just the proof of Lemma 14 in lecture II.6.) […]