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

For 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.

43.614000-116.202000

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4 Responses to Set theory seminar – Marion Scheepers: Coding strategies (II)

(As I pointed out in a comment) yes, partial Woodinness is common in arguments in inner model theory. Accordingly, you obtain determinacy results addressing specific pointclasses (typically, well beyond projective). To illustrate this, let me "randomly" highlight two examples: See here for $\Sigma^1_2$-Woodin cardinals and, more generally, the noti […]

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Craig: For a while, there was some research on improving bounds on the number of variables or degree of unsolvable Diophantine equations. Unfortunately, I never got around to cataloging the known results in any systematic way, so all I can offer is some pointers to relevant references, but I am not sure of what the current records are. Perhaps the first pape […]

Yes. Consider, for instance, Conway's base 13 function $c$, or any function that is everywhere discontinuous and has range $\mathbb R$ in every interval. Pick continuous bijections $f_n:\mathbb R\to(-1/n,1/n)$ for $n\in\mathbb N^+$. Pick a strictly decreasing sequence $(x_n)_{n\ge1}$ converging to $0$. Define $f$ by setting $f(x)=0$ if $x=0$ or $\pm x_n […]

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One way we formalize this "limitation" idea is via interpretative power. John Steel describes this approach carefully in several places, so you may want to read what he says, in particular at Solomon Feferman, Harvey M. Friedman, Penelope Maddy, and John R. Steel. Does mathematics need new axioms?, The Bulletin of Symbolic Logic, 6 (4), (2000), 401 […]

"There are" examples of discontinuous homomorphisms between Banach algebras. However, the quotes are there because the question is independent of the usual axioms of set theory. I quote from the introduction to W. Hugh Woodin, "A discontinuous homomorphism from $C(X)$ without CH", J. London Math. Soc. (2) 48 (1993), no. 2, 299-315, MR1231 […]

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Fix a model $M$ of a theory for which it makes sense to talk about $\omega$ ($M$ does not need to be a model of set theory, it could even be simply an ordered set with a minimum in which every element has an immediate successor and every element other than the minimum has an immediate predecessor; in this case we could identify $\omega^M$ with $M$ itself). W […]

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