This set is due Monday, October 18.

## 507 – Homework 2

September 29, 2010## Set theory seminar – Marion Scheepers: Coding strategies (III)

September 28, 2010For the second talk (and a link to the first one), see here. The third talk took place on September 28.

In the second case, we fix an with . We can clearly assume that is infinite, and it easily follows that . This is because any can be coded by the pair , and there are only many possible values for the second coordinate.

In particular, is infinite, and we can fix a partition of into countably many pieces, each of size . Recall that we are assuming that and have fixed a set cofinal in of smallest possible size. We have also fixed a perfect information winning strategy for II, and an with for all .

For each , fix a surjection .

We define as follows:

- Given , let
,

and

such that ,

and set .

- Suppose now that , that , and that there is an such that , , and . Let
be such that ,

and

and set .

- Define in other cases.

A straightforward induction shows that is winning. The point is that in a run of the game where player II follows :

- Player II’s moves code the part that lies outside of of player II’s moves in a run of the game following where I plays sets covering the sets in the original run. For this, note that at any inning there is a unique index such that player II’s move covers , is disjoint from , and meets in a set that is neither empty nor all of , and this codes the inning of the game, and the piece of player II’s move in codes the history of the run played so far.
- is eventually covered completely, so in particular the parts inside of player II’s responses in the run are covered as well.

This completes the proof of Theorem 1.

By way of illustration, consider the case where is the ideal of finite sets of some set . Then whether II has a winning coding strategy turns into the question of when it is that . This certainly holds if or if . However, it fails if .

This example illustrates how player II really obtains an additional advantage when playing in rather than just in . To see that this is the case, consider the same as above with . This is an instance of the *countable-finite game*. We claim that II has a winning coding strategy in this case. To see this, consider a partition of into countably many sets with . For each , pick a winning coding strategy for the countable-finite game on , and define a strategy in so that for each it simulates a run of the game with II following , as follows: In inning , II plays on for ; player I’s moves in the “-th board” are the intersection with of I’s moves in , and I’s first move occurred at inning . (II can keep track of in several ways, for example, noticing that, following the proof of Theorem 1 produces coding strategies that never play the empty set.)

Note that this strategy is not winning in , the difference being that there is no guarantee that (for any ) the first moves of I in the -st board are going to be covered by II’s responses. On the other hand, the strategy is winning in , since, no matter how late one starts to play on the -th board, player I’s first move covers I’s prior moves there (and so, II having a winning coding strategy for the game that starts with this move, will also cover those prior moves).

The first place where this argument cannot be continues is when . However, suffices to see that player I has a winning strategy in in this case, and so we can continue. This illustrates the corollary stated in the first talk, that suffices to guarantee that II always has a winning coding strategy in .

The natural question is therefore how much one can weaken the assumption, and trying to address it leads to Theorem 2, which will be the subject of the next (and last) talk.

## Set theory seminar – Marion Scheepers: Coding strategies (II)

September 27, 2010For the first talk, see here. The second talk took place on September 21.

We want to prove of Theorem 1, that if , then II has a winning coding strategy in .

The argument makes essential use of the following:

Coding Lemma.Let be a poset such that for all ,

Suppose that . Then there is a map such that

**Proof.** Note that is infinite. We may then identify it with some infinite cardinal . It suffices to show that for any partial ordering on as in the hypothesis, there is a map such that for any , there is a with such that .

Well-order in type , and call this ordering. We define by transfinite recursion through . Given , let be the set of its -predecessors,

.

Our inductive assumption is that for any pair , we have chosen some with , and defined . Let us denote by the domain of the partial function we have defined so far. Note that . Since has size , it must meet . Take to be least in this intersection, and set , thus completing the stage of this recursion.

At the end, the resulting map can be extended to a map with domain in an arbitrary way, and this function clearly is as required.

Back to the proof of . Fix a perfect information winning strategy for II in , and a set cofinal in of least possible size. Pick a such that for all we have .

Given , let . Now we consider two cases, depending on whether for some we have or not.

Suppose first that for all . Then the Coding Lemma applies with in the role of , and as chosen. Let be as in the lemma.

We define as follows:

- Given , let be such that , and set .
- Given with , let be such that , and set .

Clearly, is winning: In any run of the game with II following , player II’s moves cover their responses following , and we are done since is winning.

The second case, when there is some with , will be dealt with in the next talk.

## Set theory seminar – Marion Scheepers: Coding strategies (I)

September 25, 2010This semester, the seminar started with a series of talks by Marion. The first talk happened on September 14.

We consider two games relative to a (proper) ideal for some set . The ideal is not assumed to be -complete; we denote by its -closure, i.e., the collection of countable unions of elements of . Note that is a -ideal iff is an ideal iff .

The two games we concentrate on are the *Random Game* on , , and the *Weakly Monotonic* game on , .

In both games, players I and II alternate for many innings, with I moving first, moving as follows:

In we do not require that the relate to one another in any particular manner (thus “random”), while in we require that (thus “weakly”, since we allow equality to occur).

In both games, player II wins iff . Obviously, II has a (perfect information) winning strategy, with rather than the weaker .

However, we are interested in an apparently very restrictive kind of strategy, and so we will give some leeway to player II by allowing its moves to over-spill if needed. The strategies for II we want to consider we call *coding strategies*. In these strategies, II only has access to player I’s latest move, and to its own most recent move. So, if is a coding strategy, and II follows it in a run of the game, then we have that for every ,

,

with .

The underlying goal is to understand under which circumstances player II has a winning coding strategy in . Obviously, this is the case if II has a winning coding strategy in .

Theorem 1.For an ideal , the following are equivalent:

- II has a winning coding strategy in .
- .

Corollary.implies that for any ideal , II has a winning strategy in .

We can reformulate our goal as asking how much one can weaken in the corollary.

Let’s denote by , *the weak singular cardinals hypothesis*, the statement that if is singular strong limit of uncountable cofinality, then for no cardinal of countable cofinality, we have .

By work of Gitik and Mitchell, we know that the negation of is equiconsistent with the existence of a of Mitchell order .

Theorem 2.The following are equivalent:

- .
- For each ideal on a singular strong limit of uncountable cofinality, II has a winning strategy in .

We now begin the proof of Theorem 1.

Suppose II has a winning coding strategy in . We want to show that . For this, we will define a map with -cofinal range, as follows: Given , let and for all . Now set

.

To see that is cofinal, given , let , so that the are II’s responses using in a run of the game where player I first plays and then plays in all its subsequent moves. Since is winning, we must have .

## 507 – Problem list (I)

September 21, 2010This is the list of “problems of the day” mentioned through the course.

(Thanks Nick Davidson and Summer Hansen.)

- Frankl’s union-closed sets problem: If a finite collection of finite non-empty sets is closed under unions, must there be an element that belongs to at least half of the members of the collection?
- The inverse Galois problem: Is every finite group a Galois group over ?
- 1. Are there infinitely many Mersenne primes? 2. Are there infinitely many Fermat primes?
- For every positive , is there a prime between and ?
- Does the dual Schroeder-Bernstein theorem imply the axiom of choice?
- The Schinzel–Sierpiński conjecture: Is every positive rational of the form for some primes and ? (The links require a BSU account to access MathSciNet.)
- Are there infinitely many twin primes?
- Are there any odd perfect numbers?
- Is the Euler-Mascheroni constant irrational?
- Is a primitive root modulo for infinitely many primes ? More generally, does Artin’s conjecture hold?

## 170- Quiz 4

September 20, 2010Quiz 4 is here. Please remember that the first midterm is this Wednesday.

Solutions follow.