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Axiom of Determinacy Axiom of Determinacy William Gunther March 21, 2010 Abstract The Axiom of Determinacy (AD) states that every subset of the Baire Space, !!, is determined. It is simple to see this is inconsistent with the Axiom of Choice. But, a weaker version of choice called Dependent Choice (DC) is both consistent with AD and strong enough to prove basic theorems of analysis. We shall look at consequences of AD, with a focus on the connections with measurable cardinals. Definition 1. The Axiom of Determinacy (AD) is the statement Det(}(!!)) That is, it states all subsets of the Baire Space are determined. Proposition 1. The Axiom of Choice (AC) is inconsistent with AD. Proof. AC implies there exists a subset of !! which is not Lebesgue measurable. But AD says everything everything is Lebesgue measurable. Remark 1. As you can imagine, models of ZF+AD can be very strange because they can strongly violate choice. Because of this, we will use a weakening of the axiom of choice that is sufficent for some basic theorems of anaylsis called dependent choice (DC). To give you some intuition over the power of DC, under ZF, DC is equivilent to every pruned tree having a branch. Theorem 1 (ZFC). !1 is not measurable. Proof. !! PROOF PAGE 131 OF JECH !! 1 ⊆ 1 Lemma 1 (Σ 1 Boundedness Lemma). If B WO is Σ 1 then there is α < !1 such that type(x) < α for all x 2 B. e e Theorem 2 (ZF+AD+DC). The club filter on !1 is an σ-complete ultrafilter. In particular, !1 is measur- able. Proof. Let S ⊆ !1. Our goal is to show either it, or it's compliment contains a club. We first define the ordinal ordinal game GS . In GS , two players alternate turns playing elements of !1. I α0 α1 ··· αi; βi 2 !1; 8i 2 ! II β0 β1 Let γ = sup(f αi j i < ! g [ f βi j i < ! g). Player I wins if γ 2 S. Otherwise Player II wins. Claim 1. If S contains a club, then Player I has a winning strategy. If !1 n S is contains a club club, then Player II has a winning strategy. Proof. The strategy would be just to play on your club in the obvious way. ! Setting this game aside for a moment, we will play GS on !. Here two players alternate playing natural numbers. I x(0) x(1) ··· x(i); y(i) 2 !; 8i 2 ! II y(0) y(1) 1 And let x = hx(0); x(1);:::i and y = hy(0); y(1);:::i be the corresponding elements of the Baire Space. In each of these we will decode (in the natural way) elements of !×!×!2, which we will view as of !-many functions mapping ! × ! ! 2. That is M M x = xi y = yi i<! i<! xi; yi : ! × ! ! 2 We're viewing the xi's and yi's as characteristic functions of a binary relation on !. Then Player I wins if either of the following hold:- • There is some i < ! such that yi 2= WO • If γ = sup(f type(xi) j i < ! g [ f type(yi) j i < ! g) is an element of S. Player II wins otherwise. Let F = f S ⊆ !1 j Player I has a winning strategy in GS g Claim 2. F is a σ-complete non-principal ultrafilter. Proof. Non-principal is easy; if S is finite then it has a maximal element α Player II can ignore Player I, and just play whatever sequence codes x0 = α + 1 and xi 2 WO; 8i 2 !. Then he wins. If A 2 F then we get a strategy σ for GA. If A ⊆ B then we claim that σ also wins GB. ! Play according to the strategy, and get x; y 2 !. If i < ! such that yi 2= WO then we are done. Otherwise, look at γ = sup(f type(xi) j i < ! g [ f type(yi) j i < ! g). By definition of σ as a winning strategy we have γ 2 A ⊆ B. Thus Player I has a winning strategy for in GB, so B 2 F . To show it is ultra, look at S ⊆ !1. By AD, GS is determined. If Player I has a winning strategy, we're done. Otherwise Player II has a winning strategy, σ. Then, play the game G!1nS with auxillary game GS. In GS, play anything for Player I. Then, we get according to σ a play that Player I will play in G!1nS. Then we continue, plugging in Player II's plays in GS for Player I's. !! COMPLETE!! THIS IS A BIT OF A PROBLEM, AS PICKING ANY RANDOM NUMBER WILL MESS THINGS UP. BUT THE IDEA IS RIGHT; IF PLAYER II HAS A W.S. IN S THEN PLAYER I PLAYING THAT STRATEGY IN !1 n S WILL WORK !! !! HAVE TO DEAL WITH SIGMA COMPLETE !! Claim 3. If a player has a winning strategy in GS then he also has a winning strategy in ordinal GS Proof. Suppose that Player I had a winning strategy σ in GS. We would like to define a strategy ordinal τ in GS . Consider the set ! ! A0 = f x0 2 ! j 9y 2 ! (x; y) = σ ∗ y g 1 1 ⊆ 1 Note A0 is Σ1(σ); in particular it is Σ 1. Also, A0 WO. Then by Lemma 1 (Σ 1 Boundedness), there is some α0 < !1 such that fore every x 2 A0, α0 > type(x). In particular,e we can assume ordinal that this is the smallest such. So, in GS we play α0, in which II answer β0. We continue. Define [ ] ! ! Ai = f xi 2 ! j 9y 2 ! 8j < i[yj = βj] and (x; y) = σ ∗ y g From this, we get αi minimal such that !1 > αi > type(x) for all x 2 Ai. We play αi. ordinal ⊆ Claim that this strategy wins GS . Suppose not; then there is some ζ !1 that wins against τ. Let f αi j i < ! g be the ordinals players by τ against ζ. Then as Player II wins γ = sup(f ζi j i < ! g [ f αi j i < ! g) 2= S L In the auxillary game GS play according to σ against y, chosen such that y = i ζi. As σ is a winning strategy for Player I on S, we have sup(f ζi j i < ! g [ f type(xi) j i < ! g) 2 S 2 Claim 4. F is the club filter on !1. Proof. 3 Let's do another proof focused more on recursion theory. To begin with, we need to know more ramifi- cations of AD. ! Lemma 2 (ZF+DC+AD). There is no injection from !1 into 2 . ! Proof. For sake of contradiction, suppose that f : !1 ! 2 was an injection, with image A. From AD, we know that A has the perfect set property. As jAj = j!1j A is not countable, and therefore has a perfect subset. Let P ⊆ A be closed and perfect. Then it contains a perfect tree, and therefore there is a homeomorphism ! π : 2 ! P . As P ⊆ im(f) there are f αi j i < !1 g such that f f(αi) j i < !1 g = P . Then define ! g : !1 ! 2 −1 i 7! π (f(αi)) Clearly this is a bijection. Then we can define the relation R ⊆ 2! × 2! by iRj () g−1(i) < g−1(j) I claim this is not Lebesgue measurable (as a subset of 2! × 2!). To see this, just note that for any fixed y 2 2! f x 2 2! j xRy g is countable, and thus has measure 0, whereas f y 2 2! j yRx g is co-countable, and thus has measure 1. This contradicts Fubini's Theorem, thus R is not measurable. Definition 2. Recall, for A; B ⊆ !, that A is recursive in B if membership of A can be decided by using B as an oracle. We also say that A is Turing Reducible to B, and denote it A ≤T B. If A ≤T B and B ≤T A then they are Turing equivilent denoted A ≡T B. This is an equivilence relation on }(!); the equivilences classes are called Turing degrees. Theorem 3. The club filter on !1 is a σ-complete ultrafilter; in particular, !1 is measurable. Alternate Proof. Let D = f [X]T j X ⊆ ! g be the set of Turing Degrees. We defined F ⊆ }(D) by F = f S ⊆ D j 9X ⊆ ! . 8Y ⊆ ! . X ≤T Y ! [Y ]T 2 S g A better definition for F is to first to define a cone. For x ⊆ !, we define the cone of x to be cone(x) = f y ⊆ ! j x ≤T y g Then F ⊆ }(D) where F 2 F () F contains a cone. Claim 1. F is a σ-complete filter. Proof. We will first show that F is a σ-complete filter. D is clearly in F , by taking X = ;, or analogously observing that cone(;) = D (or any recursive set). F is clearly upward closed, for if A 2 F we get a corresponding XA; then if A ⊆ B then for any Y ⊆ ! such that XA ≤T Y we have [Y ]T 2 A ⊆ B. F 2 f j g ⊆ F Thus we needT only show is σ-complete. Fix β !1. Take Aα α < β . We will F ⊆ ⊆ show that A = α<β Aα is in , ie. there is some x ! such that cone(x) A. For each α < β, ! we get xα ⊆ ! such that cone(xα) ⊆ Aα.
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