At the end of last lecture, we showed Theorem 7, König’s lemma, stating that if for all then We begin by looking at some corollaries:

**Corollary 8. **

*If is a limit ordinal and is a strictly increasing sequence of nonzero cardinals, then**If is an -indexed sequence of nonzero cardinals and for all then**(Cantor)**For any infinite , one has*

**Proof. **1. Notice that for all , so by König’s lemma. But, by an obvious injection,

2. Let If for all then König’s lemma gives us that

3. Use König’s lemma with , and for all Then

4. Let be a cofinal function. Let and for all Note that and

5. but, by 4.,

Note that once one decodes the argument, the proof of Cantor’s theorem above is exactly the same as the usual proof. On the other hand, Corollary 8.5 gives us more information and begins to show that the notion of cofinality is quite relevant in the study of cardinal arithmetic; much more dramatic illustrations of this claim are shown below. König’s lemma is the first genuinely new result in cardinal arithmetic (with choice) past Cantor’s theorem. For example, Specker’s result in Section I.5 and the Halbeisen-Shelah theorem in Section I.6 are both trivial once is assumed.

Some of the results that follow are available elsewhere in this blog, but I include them here to make these notes reasonably self-contained.

**Definition. **A cardinal is *strong limit *iff for all cardinals

Notice that the name is justified, since any strong limit is in particular ( or) a limit cardinal, i.e., an with a limit ordinal.

Of course, is a strong limit cardinal and, if holds below then is also strong limit. Without this assumption, examples of strong limit cardinals can be found by iterating the power set operation. More precisely, define the sequence of *beth* cardinals by and for a limit ordinal. Then the strong limit cardinals are precisely the with or a limit ordinal.

**Theorem 9. (Bukovský-Hechler). **

*If is infinite then**In particular, if is strong limit, then**Let be singular, and assume that the exponential is eventually constant below ; i.e., there is some and some such that for all cardinals with one has Then also*

The theorem illustrates that the exponential map must satisfy certain restrictions, at least on its values on the singular cardinals. It is well known since the beginnings of forcing, from the work of Easton, that the exponential is `essentially’ arbitrary on the regular cardinals except that, of course, it is monotonic: If then and must obey König’s lemma, However, once large cardinals are considered, these are not the only restrictions. We will show this in Section 4.

**Proof.** 1. Let be infinite and let Let be strictly increasing, and let for all so Then as the obvious bijection verifies. But On the other hand, clearly for all so and therefore The result follows from these two inequalities.

2. If is strong limit and is as above, then On the other hand, for any so , and 2. follows from 1.

3. Assume is singular. With as above, if is sufficiently large, . But then , since we can take The result follows from 1.

For example, if for all (this situation is easily achieved by forcing) then also

The next result provides an easy `algorithm’ to compute any power, provided we know the values of the exponential function, the cofinality map, and the *gimel *function I’ll expand on this remark after the proof of the theorem.

**Theorem 10.** *Let and be infinite cardinals. Let Then *

**Proof. **1. If then

2. If then any function is bounded, so and

3. Suppose that and is eventually constant (and equal to ) as approaches

Since is a singular cardinal and we can choose a strictly increasing sequence of cardinals cofinal in such that for all In particular, .

Note that for each so also [Of course, it follows from König’s lemma that in fact we have a strict inequality, but we only need this weaker estimate.]

We now have The other inequality is clear.

4. Finally, suppose that and is not eventually constant as approaches

Notice that if then Otherwise, for any , and the map would be eventually constant below after all. Hence,

Choose an increasing sequence of cardinals cofinal in Then The other inequality is clear.

**Remark. **By induction, it follows from Theorem 10 that the computation of the function `reduces’ to computations of the functions and for a cardinal, and the function for an ordinal. More precisely, if and are two models of set theory (with choice) with the same ordinals, and the values of any of these three functions are the same whether they are computed in or in , then also all the powers are the same, whether they are computed in or in

Forcing has shown that there is not much one can say about the exponential function when restricted to successor cardinals and small *large cardinals*. This indicates that to understand cardinal arithmetic, one’s efforts must concentrate on the exponential function on singular and large cardinals, and on the gimel function.

In fact, the gimel function suffices to compute the exponential by Theorem 9: First, for regular. Assume now that is singular. Let If is not eventually constant below then clearly , and by Theorem 9.1, while if is eventually constant below then by Theorem 9.3.

This observation explains why the bulk of research in cardinal arithmetic concentrates on understanding the gimel function. As we will see next time, this naturally leads to the study of certain infinite products. This latter study has proved quite fruitful. We will study two outcomes: Silver’s theorem, and one of the Galvin-Hajnal results. Then, we will briefly mention how Shelah has extended and generalized these theorems with his development of pcf theory.

Thank you so much professor Caicedo for these online classes.

In the paragraph right before the last one, I think you meant to write (if I am not mistaken) (the formula right after “by theorem 9”).

Hi,

What I meant is: First, and second, (trivially) if is regular, in which case it then just happens that

There was a typo at the end of that paragraph, though; the second “is not eventually constant” should have been “is eventually constant.” It is now fixed.

Oh I see, indeed is a regular cardinal so we can write Thank you professor Caicedo.

I don’t want to take to much of your time but I would like to mention a theorem in Jech’s Set Theory whose demonstration seems quite hard to understand. It is Pospisil Theorem in page 75 which states that for every infinite cardinal there exist uniform ultrafilters on First Jech proves a lemma stating that there exists an independent family of subsets of of cardinality

He defines an independent family: a family of subsets of is independent if for any distinct the intersection

has cardinality

Now in the proof of Pospisil’s theorem, Jech states: Let be an independent family of subsets of

For every function consider this family of subsets of :

This family is independent and thus has the finite intersection property (he does this to prove that this family is included in an ultrafilter because thanks to a previous lemma if a family has the finite intersection property then it is included in a filter, and then we extend it to an ultrafilter).

The problem is that I can’t see why this family independent and has the finite intersection property.

Any hints?

Thank you.

[…] the end of last lecture we showed Theorem 10, a general result that allows us to compute products for infinite cardinals […]

Hi hurburble,

I don’t think you need (or want) to prove that is independent. This is false, just look at sets in the first of the three members of the union that make up What you want is that

1. has the finite intersection property. This is easy since is independent. I’ll let you think about it on your own.

2. Any ultrafilter extending is uniform. This is easy from the sets in the first of the three members of the union that make up

3. If then does

nothave the finite intersection property. You want this to guarantee that any ultrafilter extending is different from any ultrafilter extending This is easy from the fact that if then for some we have but4. There are at least many functions so there are at least that many collections and therefore at least that many ultrafilters. This is easy from the size of

Hi Professor Caicedo,

Thank you so much for your explanation

1) Indeed, you are right, is not independent, it is which is independent.

I think I figured out why has the finite intersection property.

if is independent then none of its elements can be the null set. If none of the elements of this family can be the null set then has the finite intersection property because none of the elements of can be the null set since is constructed with the help of elements of In other words the elements of are pairwise disjoint.

2) Any ultrafilter extending is uniform because for all the cardinality of is

3) I now see why we don’t want to have the finite intersection property. It is to guarantee that we have distinct ultrafilters in the end.

4) Indeed, the cardinality of is

Again, thank you!

Hi again,

One last pass at it:

1) What you say is not enough to guarantee that has the finite intersection property. I also assume at the end you meant “not pairwise disjoint,” but what you say does not imply this. You actually need to use the fact that is independent, and the way that is defined.

2) What you say is not enough. You need to use the way that is defined.

3) and 4) are fine.

Hi again,

1) Except for the fact that any finite subset of has nonempty intersection because is constructed with elements of I am not able to find another argument to make it accurate.

2) It seems to me that the sets that make up are large enough as to have an impact on the cardinality of the set resulting from taking them out of which means that they have infinite cardinality cofinal in Except from that argument, I can’t see the full justification.

I think the source of all my trouble comes from me being in difficulty understanding the way an independent family is defined.

It seems to me that in order for the condition to be fulfiled (i.e. for the intersection to have cardinality ) we need to choose the sets large enough, with a cardinality and the sets small enough (with a cardinality that can’t be mapped cofinally in ) so we can end up with an intersection of cardinality

Does has the finite intersection property because it is a filter (it looks like a principal filter because of the condition ) and it is filter because of being independent?

I am afraid that’s all I can make out of all of this.

[…] any power can be computed from the cofinality and gimel functions (see the Remark at the end of lecture II.2). What we can say about the numbers varies greatly depending on whether is regular or not. If is […]

[…] Before stating Silver’s result, we need one additional result. Recall that was introduced last lecture in Definition 4, it is the statement that for all infinite cardinals We break the result into two parts, Lemmas 17 and 18, corresponding respectively to Theorems II.1.9 (describing the behavior of the exponential ) and II.1.10 (describing the computation of powers ) from lecture II.2. […]

[…] This follows from Corollary 4: If , then by Theorem II.1.10 from lecture 2. But so both and are strictly smaller than […]

[…] 2, recall from Corollary 8 in lecture II.2 […]

[…] , then . I have addressed both these computations in my lecture notes for Topics in Set Theory, see here and […]

[…] , then . I have addressed both these computations in my lecture notes for Topics in Set Theory, see here and […]

[…] Before stating Silver’s result, we need one additional result. Recall that was introduced last lecture in Definition 4, it is the statement that for all infinite cardinals We break the result into two parts, Lemmas 17 and 18, corresponding respectively to Theorems II.1.9 (describing the behavior of the exponential ) and II.1.10 (describing the computation of powers ) from lecture II.2. […]

[…] the end of last lecture we showed Theorem 10, a general result that allows us to compute products for infinite cardinals […]