Due Wednesday, May 7 at 2:30 pm.

**Update**. In exercise 2, , of course. In exercise 3, in the right hand side must be .

**Hint for exercise 5**: Consider the *smallest* such that (ordinal addition).

**Update**. Here is a quick sketch of the solutions:

- 1.(a). Write for some (not necessarily strictly) increasing -sequence of cardinals . (If is regular, take for all . If is singular, let the be increasing and cofinal in .) Then but also and the result follows. If is strong limit, then , and we get .
- 1.(b). Under the assumption, we may choose the as in 1.(a) such that and for all . Then .

- 2. Write as a disjoint union of sets , , each of size . This is possible since . Since each has size , it is necessarily cofinal in . We then have .

- 3. If , then .
- If then any function is bounded, so , and ; the other inequality is clear.
- Suppose that , , and is eventually constant as approaches . Choose a strictly increasing sequence cofinal in such that for . Then . The other inequality is clear.
- 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.

- 5. 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 exercise 1.(b), since is singular (it has cofinality ). This is a contradiction, since by assumption, .

- 4. We want to show that for all limit ordinals . Write where or a limit ordinal, and proceed by induction on , noticing that the case follows from exercise 2.
- 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 holds by Hausdorff’s result that . Finally, as wanted, where the previous to last equality is by exercise 2.
- 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 exercise 3 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 . Then , by induction (it is here that we used the assumption that is a limit of limit ordinals). And .
- Finally, if and is not eventually constant as approaches , then and , by exercise 2, but , and we are done.

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