Update. Now due Wednesday, April 30 at 2:30 pm.
Corrections. (Thanks to Fedor Manin for noticing these.)
Update. Here is a quick sketch of the proof of the Milner-Rado paradox.
First notice that the result is clear if , since we can write any as a countable union of singletons. So we may assume that is uncountable.
Notice that . This can be checked either by induction on , or by using the characterization of ordinal exponentiation in terms of functions of finite support.
Notice that the function is normal. By the above, it follows that for all uncountable cardinals . In particular, it suffices to prove the result for ordinals that are an ordinal power of , since these ordinals are cofinal in , and a representation as desired for an ordinal gives (by restriction) such a representation for any smaller ordinal.
By the above, for any . It is easy to see that for any , if , then the interval is order isomorphic to ; this can be proved by a straightforward induction on .
For any , we can write , where so it has order type .
Also, if is a limit ordinal below , then we can write for some strictly increasing continuous sequence cofinal in . Let and for . Then and each has order type .
[That the sequence is continuous (at limits) ensures that the cover . That they have the claimed order type follows from the “straightforward inductive argument” three paragraphs above.]
So we have written each as an increasing union of many intervals whose order types are ordinal powers of , and is either or . Now proceed by induction. We may assume that each ordinal below can be written as claimed in the paradox. In particular, each , having order type an ordinal smaller than , can be written that way, say where .
If , this immediately gives the result for : Take and . Clearly their union is and they have small order type as required.
If , take and . Again, their union is , and is at most the order type of concatenating many copies of [it is here that we use that for ], so .