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?

Let $C$ be the standard Cantor middle-third set. As a consequence of the Baire category theorem, there are numbers $r$ such that $C+r$ consists solely of irrational numbers, see here. What would be an explicit example of a number $r$ with this property? Short of an explicit example, are there any references addressing this question? A natural approach would […]

Suppose $M$ is an inner model (of $\mathsf{ZF}$) with the same reals as $V$, and let $A\subseteq \mathbb R$ be a set of reals in $M$. Suppose further that $A$ is determined in $M$. Under these assumptions, $A$ is also determined in $V$. The point is that since winning strategies are coded by reals, and any possible run of the game for $A$ is coded by a real, […]

Yes. This is obvious if there are no such cardinals. (I assume that the natural numbers of the universe of sets are the true natural numbers. Otherwise, the answer is no, and there is not much else to do.) Assume now that there are such cardinals, and that "large cardinal axiom" is something reasonable (so, provably in $\mathsf{ZFC}$, the relevant […]

Please send an email to mathrev@ams.org, explaining the issue. (This is our all-purpose email address; any mistakes you discover, not just regarding references, you can let us know there.) Give us some time, I promise we'll get to it. However, if it seems as if the request somehow fell through the cracks, you can always contact one of your friendly edit […]

The characterization mentioned by Mohammad in his answer really dates back to Lev Bukovský in the early 70s, and, as Ralf and Fabiana recognize in their note, has nothing to do with $L$ or with reals (in their note, they indicate that after proving their result, they realized they had essentially rediscovered Bukovský's theorem). See MR0332477 (48 #1080 […]

The paper MR1029909 (91b:03090). Mekler, Alan H.; Shelah, Saharon. The consistency strength of "every stationary set reflects". Israel J. Math. 67 (1989), no. 3, 353–366, that you mention in the question actually provides the relevant references and explains the key idea of the argument. Note first that $\kappa$ is assumed regular. They refer to MR […]

Start with Conway's base 13 function $c $ (whose range on any interval is all of $\mathbb R $), which is everywhere discontinuous, and note that if $f $ only takes values $0$ and $1$, then $c+f $ is again everywhere discontinuous (since its range on any interval is unbounded). Now note that there are $2^\mathfrak c $ such functions $f $: the characteris […]

Yes, there are such sets. To describe an example, let's start with simpler tasks. If we just want $P\ne\emptyset$ with $P^1=\emptyset$, take $P$ to be a singleton. If we want $P^1\ne\emptyset$ and $P^2=\emptyset$, take $P$ to be a strictly increasing sequence together with its limit $a$. Then $P^1=\{a\}$. If we want $P^2\ne\emptyset$ and $P^3=\emptyset$ […]

The result was proved by Kenneth J. Falconer. The reference is MR0629593 (82m:05031). Falconer, K. J. The realization of distances in measurable subsets covering $R^n$. J. Combin. Theory Ser. A 31 (1981), no. 2, 184–189. The argument is relatively simple, you need a decent understanding of the Lebesgue density theorem, and some basic properties of Lebesgue m […]

No, not even $\mathsf{DC}$ suffices for this. Here, $\mathsf{DC}$ is the axiom of dependent choice, which is strictly stronger than countable choice. For instance, it is a theorem of $\mathsf{ZF}$ that for any set $X$, the set $\mathcal{WO}(X)$ of subsets of $X$ that are well-orderable has size strictly larger than the size of $X$. This is a result of Tarski […]

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?