Let me begin with a remark related to the question of whether . We showed that this is the case if for some , or if is Dedekind-finite.
Theorem. The axiom of choice is equivalent to the statement that any Dedekind-infinite cardinal is a square.
Proof. Let be a set. Assuming that every D-infinite cardinal is a square, we need to show that is well-orderable. We may assume that . Otherwise, replace with . Let . Assume that is a square, say . Then . By Homework problem 2, , so for some , and .
Lemma. Suppose are D-infinite sets and is an (infinite) initial ordinal. If then either or .
Proof. Let be an injection. If there is some such that we are done, so we may assume that for all there is some such that . Letting be the least such , the map is an injection of into .
By the lemma, it must be that either or else . The former is impossible since , so is well-orderable, and thus so is , and since , then is well-orderable as well.
This is a very recent result, see David Feldman, Mehmet Orhon, Generalizing Hartogs’s trichotomy theorem, preprint (2008), available at the ArXiv.
Definition. Trichotomy is the statement that any two sets are comparable.
Theorem. (Hartogs). Trichotomy is equivalent to choice.
Proof. If choice holds, any set is equipotent with an ordinal, and any two ordinals are comparable.
Conversely, if any two sets are comparable, in particular and are. Since , we must have , and we are done.
Definition. Let . -trichotomy is the statement that given any sets, at least two are comparable.
Theorem. For any , -trichotomy is equivalent to choice.
Proof. (Blass). Assume -trichotomy. Let be a set. We want to show that is well-orderable. Let .
Lemma. Let be an initial ordinal. If , then is well-orderable.
Here, is to be understood as disjoint union.
Proof. Let be 1-1. For each , since does not inject into , it must be the case that the range of meets . Let be the least member of this intersection. Then is an injection of into .
Remark. If , then , and it follows that . However, it is not necessarily the case that for infinite. For example, this fails if is D-finite.
Question. If for some initial ordinal , does it follow that is well-orderable?
Define a sequence of cardinals, as follows:
We claim that if is not well-orderable, the set contradicts -trichotomy. Or, positively, -trichotomy applied to this set implies the well-orderability of . Because if and there is an injection , then in particular . If , then . Otherwise, is defined, and (since is the aleph of the set that injects into). Since the sequence of cardinals is increasing, we must have , so and (since ) in fact .
Let . Then , using that . Also, , by assumption. But , and we have that . By the lemma, is well-orderable and we are done, since injects into .
Open question. With the obvious definition, does -trichotomy imply choice?
Let -trichotomy be the statement that any infinite family of sets contains two that are comparable.
Open question. Does -trichotomy imply -trichotomy? Does -trichotomy imply choice?
10. The generalized continuum hypothesis.
The last topic of this part of the course is Specker’s theorem that implies .
Definition. Let be a cardinal. The continuum hypothesis for , , is the statement that for any cardinal , if , then either or else .
Definition. The Generalized Continuum Hypothesis, , is the statement that holds for all infinite cardinals .
Theorem. (Specker). implies , the axiom of choice.
Remark. In fact, it suffices that holds for all infinite initial ordinals. I am not sure who first proved this; it is a consequence of a result of Kanamori and Pincus in the paper mentioned last lecture.
Specker’s result is in fact a local argument. What one proves is the following:
Theorem. Assume and . Then is well-orderable and, in fact, .
Next lecture will be devoted to the proof of this result.