It is easy to solve negatively the question immediately following Homework problem 5 on lecture II.1. I asked whether if is Dedekind-finite but is Dedekind-infinite, then it followed that there is an infinite Dedekind-finite set such that .
To exhibit a counterexample, it is enough to know that it is consistent to have an infinite Dedekind finite set that is the countable union of finite sets (in fact, sets of size 2). Notice that is a surjective image of so is Dedekind-infinite. Suppose that Then certainly so is a countable union of finite sets If is infinite then for infinitely many values of But then is also a surjective image of , so (and in fact ) injects into and therefore into contradiction.
At the end of last lecture we showed Theorem 10, a general result that allows us to compute products for infinite cardinals namely:
Let and be infinite cardinals. Let Then
The following easy consequence is particularly useful:
Corollary 11. (Hausdorff).
Proof. The result is immediate from the formula above if If, on the other hand, then certainly and it is easy to see that both sides of the equation we want to prove equal
For example, for any
The following more general result is most easily established by induction on
Homework problem 8. (Tarski). If then
For example, Hence if then
Now we begin to explore the behavior of infinitary products, mainly as they relate to the gimel function but the results to follow are more general.
First, an easy observation.
Fact 12. Suppose for all Then
Proof. There is an obvious injection from the product set into It follows that from which the result is immediate.
Theorem 13. Suppose that is a cardinal and is a strictly increasing sequence of nonzero cardinals with supremum Then
Note that the weaker claim that is an obvious consequence of Facts 1 (see lecture II.1) and 12.
Proof. Write as a disjoint union of sets for each of size This is possible since Since each has size it is necessarily cofinal in We then have
For example, for any infinite cardinal We will prove a more general result below, in Theorem 15, but notice for now that, in particular, One of the most celebrated results of Shelah’s pcf theory (established by a careful analysis of this product) is the fact that In particular, if is strong limit, so by the Bukovský-Hechler Theorem 9.2, then In fact, the inequality is strict, by König’s lemma, since
The following is another consequence of Theorem 9:
Corollary 14. If there is an ordinal such that for all we have that then
Proof. Suppose that and for all Let be least such that exists since On the other hand, is a limit ordinal, since and so for any and any and we would have a contradiction to the minimality of
Let Then and By the Bukovský-Hechler theorem, since is singular (it has cofinality ). This is a contradiction since, by assumption,
Remark. On the other hand, for any nonzero Woodin proved that it is consistent with that The case is of course the (For this is a joint result of Foreman and Woodin.) However, for this situation is incompatible with the existence of strongly compact cardinals, as we will see in Section 4.
Theorem 15. (Tarski).
- For any limit ordinal
- If is zero or a limit ordinal then, for and any
Remark. Theorem 15 does not follow from the proof of Theorem 13. The same argument of Theorem 13 actually shows that whenever is a strictly increasing sequence of nonzero cardinals and is a countable limit ordinal, then where However, the argument used to prove Theorem 13 does not apply if since there is no way of splitting into many cofinal sets.
Proof. Notice that item 2. is trivial if and an immediate consequence of 1. and Hausdorff’s formula, Corollary 11, if is an infinite limit ordinal.
Now we prove 1.: Write where or a limit ordinal, and proceed by induction on noticing that the case follows from Theorem 13.
Notice that any limit ordinal can be written as for some nonzero ordinal this is easily established by induction. If is a limit ordinal, then any sequence converging to gives rise in a natural way to a sequence of limit ordinals converging to On the other hand, if is a successor, then for some or limit. In summary: Any limit ordinal is either a limit of limit ordinals, or else it has the form for or limit.
We divide our induction on in two cases. Suppose first as above. Then by induction. Since we have where the last equality follows from Hausdorff’s formula. Finally, as wanted, where the previous to last equality is by Theorem 13.
Now suppose that is a limit of limit ordinals. Choose a strictly increasing sequence of limit ordinals cofinal in We argue according to which of the four possibilities described in Theorem 10 holds. If then
Notice that and so the second possibility does not occur.
Suppose now that we are in the third possibility, i.e., and is eventually constant as approaches . Call this eventual value. Then , by induction (it is here that we use the assumption that is a limit of limit ordinals). And .
Finally, if and is not eventually constant as approaches , then and , by Theorem 13, but , and we are done.
Remark. Tarski conjectured that Theorem 13 also holds in this generality. This is independent of It follows, for example, from Its failure is a strong refutation of the Singular Cardinals Hypothesis, Indeed , there are counterexamples in models of Magidor where fails. Notice that the proof of Theorem 15 shows that the first time we run into trouble trying to generalize Theorem 13 has to be at some of the form for some or limit. Indeed, the inductive proof above in the case that is a limit of limit ordinals generalizes without difficulty. However, the proof in the case for or limit uses Hausdorff’s formula, and does not apply in general. In Thomas Jech, Saharon Shelah, On a conjecture of Tarski on products of cardinals, Proceedings of the American Mathematical Society 112, (1991), 1117–1124, it is shown that there is a counterexample if and only if there is one of shortest possible length,