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 .

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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 .

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.

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?

The only reference I know for precisely these matters is the handbook chapter MR2768702. Koellner, Peter; Woodin, W. Hugh. Large cardinals from determinacy. In Handbook of set theory. Vols. 1, 2, 3, 1951–2119, Springer, Dordrecht, 2010. (Particularly, section 7.) For closely related topics, see also the work of Yong Cheng (and of Cheng and Schindler) on Harr […]

As other answers point out, yes, one needs choice. The popular/natural examples of models of ZF+DC where all sets of reals are measurable are models of determinacy, and Solovay's model. They are related in deep ways, actually, through large cardinals. (Under enough large cardinals, $L({\mathbb R})$ of $V$ is a model of determinacy and (something stronge […]

Throughout the question, we only consider primes of the form $3k+1$. A reference for cubic reciprocity is Ireland & Rosen's A Classical Introduction to Modern Number Theory. How can I count the relative density of those $p$ (of the form $3k+1$) such that the equation $2=3x^3$ has no solutions modulo $p$? Really, even pointers on how to say anything […]

(1) Patrick Dehornoy gave a nice talk at the Séminaire Bourbaki explaining Hugh Woodin's approach. It omits many technical details, so you may want to look at it before looking again at the Notices papers. I think looking at those slides and then at the Notices articles gives a reasonable picture of what the approach is and what kind of problems remain […]

It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurable. (This result can be found in a variety of places, it is a standard exercise in measure theory books. As of this writing, there is a short proof here (Intern […]

Let $s$ be the supremum of the $\mu$-measures of members of $\mathcal G$. By definition of supremum, for each $n$, there is $G_n\in\mathcal G$ with $\mu(G_n)>s-1/n$. Letting $G=\bigcup_n G_n$, then $G\in \mathcal G$ since $\mathcal G$ is closed under countable unions, and $\mu(G)=s$, since it is at least $\sup_n\mu(G_n)$ but it is at most $s$ (by definiti […]

The result you are trying to prove is false. For example, if $a=\omega+1$ and $b=\omega+\omega$, then $a+b=\omega\cdot 3>b$. Here is what is true: first, the key result you should establish (by induction) is that An ordinal $\alpha>0$ has the property that for all $\beta

Very briefly: Yes, there are several programs being developed that can be understood as pursuing new axioms for set theory. For the question itself of whether pursuing new axioms is a reasonably line of inquiry, see the following (in particular, the paper by John Steel): MR1814122 (2002a:03007). Feferman, Solomon; Friedman, Harvey M.; Maddy, Penelope; Steel, […]

This is a very interesting question and the subject of current research in set theory. There are, however, some caveats. Say that a set of reals is $\aleph_1$-dense if and only if it meets each interval in exactly $\aleph_1$-many points. It is easy to see that such sets exist, have size $\aleph_1$, and in fact, if $A$ is $\aleph_1$-dense, then between any tw […]

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?