David (Mumford)

]]>*Thanks — corrected now.*

I think Anon meant that in the sentence “Moreover, if Player I wins, then there will be some stage at which at least n_1 1s have been chosen, and after that it doesn’t make any difference what anyone plays.” n_1 should be n_2.

*Thanks. I’ve corrected that too now.*

*Yes. Thanks — I’ve changed it now.*

*Thanks — that was a slip, which I have now corrected (by turning (2k-1,2k) into (k-1,k)).*

A different description involves using a second order existential quantifier followed by some first order quantifiers followed by a matrix to describe any analytic set. (“Matrix” is just the term for the part of a formula after the quantifiers, for formulas written with a single block of quantifiers.) If the matrix refers to explicit reals (known as the “parameters”) then the analytic set is lightface analytic relative to its parameters. As usual, Cantor allows us to assume there is a single parameter.

]]>There’s another similar typo in the same paragraph (an that should be an ).

]]>“Note, however, that Player II can postpone this moment for as long as he likes, by choosing n_1 sufficiently large.”

should read:

“Note, however, that Player II can postpone this moment for as long as he likes, by choosing n_2 sufficiently large.”

*Thanks — corrected.*

Many thanks (to both of you) for explaining this to me.

]]>(This is a bit technical, apologies in advance. Hope not many typos survived.)

To clarify, the optimal version of the results on iterated power sets, following Martin’s unpublished book on determinacy (and, where relevant, improving the results from Friedman’s 1971 paper “Higher Set Theory and Mathematical Practice”), is roughly as follows: (“Roughly”, since I do not want to state the intricacies in the notion of tree that Martin uses. The case suffices for most of what follows.)

Let be the theory , without replacement, or the power set axiom. Working in the theory -replacement, if is a tree, , and exists, then all games are determined; if is an infinite countable ordinal, and exists, then all games are determined.

This is optimal, in the following sense: For each ordinal , let be if is finite, and if it is infinite. We then have that there is a transitive model of without the power set axiom, such that , and -determinacy for countable trees fails.

(Note that the background theory may be too weak to show that the iterated power sets under discussion actually exist.)

The converse is not a direct implication, but a consistency result. A problem with stating it is that in order to discuss properties of an ordinal we need a way of referring to it. Martin addresses this by assuming is recursive. (Replacing the background theory appropriately, we can circumvent this formality.) Work in -replacement. For recursive, we have that if all games on are determined, then there is a (least) ordinal such that is a model of without the power set axiom, plus " exists".

Note that the theory Martin works on is stronger than Zermelo set theory (in view of Monroe's comments, this is unavoidable). His assumption of -replacement allows us to discuss ordinals (as long as they appear as ranks of countable trees, but this certainly takes us way beyond ); as Martin puts it, "The point of -replacement is that it

gives us cartesian products, enough ordinal numbers, and some simple definitions by transﬁnite recursion." In Friedman's paper, a similar extension of Zermelo set theory is considered as well.

(Recent additional results have been obtained in the realm of second order arithmetic, by Montalbán and Shore. )

]]>That’s true for a non-determined game, yes.

]]>Sorry, my first comment probably does not make sense. If Player I does not have a winning strategy, it means that for all \sigma there exists a sequence n in the complement of A that agrees with \sigma on even-step moves. I guess I just don’t see how this implies that Player II has a winning strategy when A is open, but I will think about this more later.

]]>The terminology you are using is a bit confusing. Since the game is (or can be) infinite and can be non-determined, it is better to start by calling ‘winning strategy’ a ‘non-losing strategy’. Then when you talk about open games, it makes more sense to say that if Player II has a non-losing strategy then it is automatically a winning strategy, so the game is determined. It is not clear what it means that Player I does not have a winning strategy.

Some corrections:

(iii) If not all pairs are legal moves, then k is even and the only exception is the pair (n_{k-1},n_{k}).

Moreover, if Player I wins, then there will be some stage at which at least n_2 1s have been chosen, and after that it doesn’t make any difference what anyone plays. Note, however, that Player II can postpone this moment for as long as he likes, by choosing n_2 sufficiently large.

If A is an open set and \mathbf{n} is an infinite sequence that belongs to A, then there must be a finite sequence s such that \mathbf{n}\in X_s and X_s\subset A.

Let us call a sequence \mathbf{n} consistent with a strategy \sigma …

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