Due Wednesday, May 28 at 2:30 pm.
Update. I present here a quick sketch of the solution of Exercise 3.(b). See Lecture 18, where it is shown that the result actually holds in , although the proof uses choice.
Let and be two well-orderings of a set . We want to find a subset of of the same size as where the two well-orderings coincide. Let . By combining with an isomorphism between and its order type, we may assume that is an ordinal and . By restricting attention to the subset of , we may assume is a well-ordering of . By further restricting to the subset of of order type under , we may assume that as well.
Assume first that is regular. The result follows easily. The desired set can be built by a straightforward recursion: Given and a sequence of elements of increasing under both well-orderings, regularity ensures that the sequence is bounded under both well-orderings, and we can find which is larger than all the previous under both orderings.
The argument for singular is slightly more delicate. Namely, we may not be able to carry out the construction above since the sequence could be unbounded in one of the orderings when . We circumvent the problem by only considering ordinals whose cofinality is larger than the cofinality of . Notice that if an increasing sequence of order type is unbounded in an ordinal of cofinality , then .
To implement this idea, let be an increasing sequence of regular cardinals cofinal in , with . Consider the subset . It must contain a subset of size where coincides with . By the remark above, this subset is bounded in . Let denote the shortest initial segment of containing . By removing from the set , we are left with a set of size , and any ordinal there is larger than the elements of under both orderings. The induction continues this way, by considering at stage a set of size .