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.

9. -trichotomy.

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:

- .
- .

(Then )

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.**

[…] , so, if has size , then is equipotent with , as shown on lecture 4, section 9 (remark after the proof of the first […]