Update. Now due Wednesday, April 30 at 2:30 pm.
Corrections. (Thanks to Fedor Manin for noticing these.)
- On Exercise 2.(c), assume in addition that satisfies the conditions of in item 2.(a); this should really be all that is needed of 2.(c) for later parts of the exercise.
- On Exercise 2.(f), we also need .
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 .
It seems that the Milner-Rado paradox, as stated, is false, because if is successor, the sequence of has to be eventually constant, so for some ; but this is not the case for .
Is there a typo in the problem?
Fedor, the exercise is not requiring that the sets are intervals or that they appear one after another in order type . For , for example, you can even let each (for ) be a singleton. For , the sets corresponding to some cannot be intervals.
This time, something I’m pretty sure is a trivial error: in 2(f), we have to take , since .
Yes, you are right. Thanks for spotting this!
Uh… I think we found a counterexample for 2(c):
Then we have and , but .
If the condition in part (a) holds for and , this problem is simple. I think this would be sufficient to do the later parts of 2.
8) You are right, I was a bit suspicious of the general statement. Yes, if is like in 2.(a), the result holds, and this should be all that is needed for later items.
That set was pretty fun. I don’t feel like I really understood what was going on in 1(d) and 1(e), however. (And maybe that means I only tricked myself into thinking that I understood parts (a)–(c).) Would you be willing to post a solution or something for problem 1?