panagio/116c/w6d2.pdf · Analytic determinacy Today we are going to nish the proof of the following result: Theorem (Martin) If measurable cardinals exist then every analytic A NN

Set theory of the continuum

Caltech

Week 6, day 2

(Caltech) Set theory of the continuum Week 6, day 2 1 / 14

Analytic determinacy

Today we are going to finish the proof of the following result:

Theorem (Martin)

If measurable cardinals exist then every analytic A ⊆ NN is determined.

We will rely on the following theorem which we proved last week.

Theorem (Gale-Stewart)

Let T be a non-empty pruned subtree of AN, for some set A. If D ⊆ [T ] iseither open or closed then D is determined in the game G([T ], D).




Theorem (Martin)








Theorem (Martin)






Measurable cardinals

A cardinal κ is measurable if there is µ : P(κ)→ {0, 1} so that:1 µ(∅) = 0 and µ(κ) = 1;2 µ({ξ}) = 0 for all ξ ∈ κ;3 A ⊆ B =⇒ µ(A) ≤ µ(B);4 if λ < κ and (Aξ ⊆ κ | ξ < λ) with µ(Aξ) = 1, then µ(

⋂ξ<λAξ) = 1

Theorem (Rowbottom)

If κ is measurable and (fi | i ∈ N) is a family of functions fi : [κ]mi → N

then there is X ⊆ κ with |X| = ℵ1 so that each fi is constant on [X]mi .

Here, [X]m :={{x1, . . . , xm} ⊆ X | (∀i, j ≤ m)

(i 6= j =⇒ xi 6= xj

)}.


Measurable cardinalsA cardinal κ is measurable if there is µ : P(κ)→ {0, 1} so that:

1 µ(∅) = 0 and µ(κ) = 1;2 µ({ξ}) = 0 for all ξ ∈ κ;3 A ⊆ B =⇒ µ(A) ≤ µ(B);4 if λ < κ and (Aξ ⊆ κ | ξ < λ) with µ(Aξ) = 1, then µ(

⋂ξ<λAξ) = 1

Theorem (Rowbottom)



Here, [X]m :={{x1, . . . , xm} ⊆ X | (∀i, j ≤ m)

(i 6= j =⇒ xi 6= xj

)}.




⋂ξ<λAξ) = 1

Theorem (Rowbottom)



Here, [X]m :={{x1, . . . , xm} ⊆ X | (∀i, j ≤ m)

(i 6= j =⇒ xi 6= xj

)}.




⋂ξ<λAξ) = 1

Theorem (Rowbottom)



Here, [X]m :={{x1, . . . , xm} ⊆ X | (∀i, j ≤ m)

(i 6= j =⇒ xi 6= xj

)}.


The Kleene-Brouwer ordering

Let N<N be the full tree on N. The Kleene-Brouwer ordering <KB onN<N is the “lexicographic/prefix” ordering: if s, t ∈ N<N and(a0, . . . , an−1) is their meet, i.e. the longest element of N<N with:

s = (a0, . . . , an−1, b0, . . . , bk), t = (a0, . . . , an−1, c0, . . . , cl),

then(s <KB t

)if and only if

(b0 < c0

)∨(t = (a0, . . . , an−1)

).

Fact. Let X be any subset of ORD with |X| = ℵ1. Then, the treeT ⊆ N<N is well-founded iff (T,<KB) embedds into X;




s = (a0, . . . , an−1, b0, . . . , bk), t = (a0, . . . , an−1, c0, . . . , cl),

then(s <KB t

)if and only if

(b0 < c0

)∨(t = (a0, . . . , an−1)

).





s = (a0, . . . , an−1, b0, . . . , bk), t = (a0, . . . , an−1, c0, . . . , cl),

then(s <KB t

)if and only if

(b0 < c0

)∨(t = (a0, . . . , an−1)

).



Table of Contents

1 The proof of analytic determinacy

2 Projective determinacy

3 Borel determinacy



Theorem. If κ measurable exists, then analytic D ⊆ NN are determined.Proof. Let D ⊆ NN be analytic and let T ⊆ (N× N)<N be a tree with:

x ∈ D ⇐⇒ ∃y ∈ NN ((x, y) ∈ [T ])⇐⇒ Tx ∈ IF

Reformulating this using the “fact” from the previous slide we have that:

x ∈ Dc ⇐⇒ (Tx, <KB) embedds into κ,

where Tx is the section tree {t ∈ N<N | (x�|t|, t) ∈ T} of T at x.Fix some enumeration t0, t1, t2, . . . of N<N with |ti| ≤ i.Auxiliary to the usual game G(NN, D):

I a0 a2 a4 a6 · · ·II a1 a3 a5 a7

we define below the game Gκ(NN, D), where ηi are ordinals in κ:

I a0 a2 a4 a6 · · ·II a1, η0 a3, η1 a5, η2 a7, η3


Analytic determinacyTheorem. If κ measurable exists, then analytic D ⊆ NN are determined.

Proof. Let D ⊆ NN be analytic and let T ⊆ (N× N)<N be a tree with:

x ∈ D ⇐⇒ ∃y ∈ NN ((x, y) ∈ [T ])⇐⇒ Tx ∈ IF




I a0 a2 a4 a6 · · ·II a1 a3 a5 a7




Analytic determinacyTheorem. If κ measurable exists, then analytic D ⊆ NN are determined.Proof. Let D ⊆ NN be analytic and let T ⊆ (N× N)<N be a tree with:

x ∈ D ⇐⇒ ∃y ∈ NN ((x, y) ∈ [T ])⇐⇒ Tx ∈ IF



where Tx is the section tree {t ∈ N<N | (x�|t|, t) ∈ T} of T at x.

Fix some enumeration t0, t1, t2, . . . of N<N with |ti| ≤ i.Auxiliary to the usual game G(NN, D):

I a0 a2 a4 a6 · · ·II a1 a3 a5 a7





x ∈ D ⇐⇒ ∃y ∈ NN ((x, y) ∈ [T ])⇐⇒ Tx ∈ IF




I a0 a2 a4 a6 · · ·II a1 a3 a5 a7





x ∈ D ⇐⇒ ∃y ∈ NN ((x, y) ∈ [T ])⇐⇒ Tx ∈ IF




I a0 a2 a4 a6 · · ·II a1 a3 a5 a7




Analytic determinacyFix some enumeration t0, t1, t2, . . . of N<N with |ti| ≤ i. Given this fixedenumeration we define the game Gκ(NN, D). Below, ηi ∈ κ:


Player II wins if for x = (a0, a1, a2, . . .) we have that:(i) ηi = 0, if ti 6∈ Tx; (ii) (ηi < ηj) ⇐⇒ ti <KB tj , if ti, tj ∈ Tx.

To finish the proof we will prove three things:1 if Player I has a winning strategy for Gκ(NN, D) then so does for

Player I has a winning strategy for G(NN, D).2 if Player II has a winning strategy for Gκ(NN, D) then so does for

Player II has a winning strategy for G(NN, D).3 the game Gκ(NN, D) is determined.

For (2): in the new game Player II needs not only to win the old gamebut to also produce an explanation as to why they won.For (3): Gκ(NN, D) is closed game with rules on A =

(N⋃(N× κ)

).









(N⋃(N× κ)

).





To finish the proof we will prove three things:

1 if Player I has a winning strategy for Gκ(NN, D) then so does forPlayer I has a winning strategy for G(NN, D).

2 if Player II has a winning strategy for Gκ(NN, D) then so does forPlayer II has a winning strategy for G(NN, D).

3 the game Gκ(NN, D) is determined.


(N⋃(N× κ)

).






Player I has a winning strategy for G(NN, D).

2 if Player II has a winning strategy for Gκ(NN, D) then so does forPlayer II has a winning strategy for G(NN, D).



(N⋃(N× κ)

).







Player II has a winning strategy for G(NN, D).



(N⋃(N× κ)

).









(N⋃(N× κ)

).








For (2): in the new game Player II needs not only to win the old gamebut to also produce an explanation as to why they won.

For (3): Gκ(NN, D) is closed game with rules on A =(N⋃(N× κ)

).









(N⋃(N× κ)

).



Fix some enumeration t0, t1, t2, . . . of N<N with |ti| ≤ i.Below, ηi ∈ κ:



Claim.If Player I has a winning strategy for Gκ(NN, D) then so does for G(NN, D).

Proof. Let ϕI be a winning strategy for Player I in Gκ(NN, D).We will find a winning strategy for I in G(NN, D).

I a0 a2 a4 a6 · · ·II a1 a3 a5 a7

Assume that s = (a0, . . . , a2n−1) ∈ N2n has been played in G(NN, D).Player I needs to decide what a2n to play.We want to consult ϕI but for that we need to accompany (a0, . . . , a2n−1) withthe additional data (η0, . . . , ηn−1) and different choices of (η0, . . . , ηn−1) wouldgive potentially different a2n output.


Analytic determinacyFix some enumeration t0, t1, t2, . . . of N<N with |ti| ≤ i.Below, ηi ∈ κ:





I a0 a2 a4 a6 · · ·II a1 a3 a5 a7







Proof. Let ϕI be a winning strategy for Player I in Gκ(NN, D).

We will find a winning strategy for I in G(NN, D).

I a0 a2 a4 a6 · · ·II a1 a3 a5 a7








I a0 a2 a4 a6 · · ·II a1 a3 a5 a7








I a0 a2 a4 a6 · · ·II a1 a3 a5 a7

Assume that s = (a0, . . . , a2n−1) ∈ N2n has been played in G(NN, D).

Player I needs to decide what a2n to play.We want to consult ϕI but for that we need to accompany (a0, . . . , a2n−1) withthe additional data (η0, . . . , ηn−1) and different choices of (η0, . . . , ηn−1) wouldgive potentially different a2n output.







I a0 a2 a4 a6 · · ·II a1 a3 a5 a7

Assume that s = (a0, . . . , a2n−1) ∈ N2n has been played in G(NN, D).Player I needs to decide what a2n to play.

We want to consult ϕI but for that we need to accompany (a0, . . . , a2n−1) withthe additional data (η0, . . . , ηn−1) and different choices of (η0, . . . , ηn−1) wouldgive potentially different a2n output.







I a0 a2 a4 a6 · · ·II a1 a3 a5 a7








Given: s = (a0, . . . , a2n−1) ∈ N2n in G(NN, D). Decide: what a2n to play.

In Gκ(NN, D), the additional η0, . . . ηn−1 are matched to t0, t1, . . . , tn−1.

Let Ts be the subset of t0, t1, . . . , tn−1 which are in {t ∈ N<N | (s�|t|, t) ∈ T}.Let ms be the size of Ts. There is 1-1 correspondence between H ∈ [κ]ms andlegal assignments (ηH0 , . . . , η

Hn−1) for (t0, . . . , tn−1), since <KB on Ts is given,

and t 7→ 0 for all t ∈ {t0, t1, . . . , tn−1} \ Ts.We have a fs : [κ]

ms → N with fs(H) := ϕI(a0, a1, ηH0 , . . . , a2n−1, η

Hn−1).













Hn−1).









Let Ts be the subset of t0, t1, . . . , tn−1 which are in {t ∈ N<N | (s�|t|, t) ∈ T}.

Let ms be the size of Ts. There is 1-1 correspondence between H ∈ [κ]ms andlegal assignments (ηH0 , . . . , η




Hn−1).









Let Ts be the subset of t0, t1, . . . , tn−1 which are in {t ∈ N<N | (s�|t|, t) ∈ T}.Let ms be the size of Ts.

There is 1-1 correspondence between H ∈ [κ]ms andlegal assignments (ηH0 , . . . , η




Hn−1).











and t 7→ 0 for all t ∈ {t0, t1, . . . , tn−1} \ Ts.

We have a fs : [κ]ms → N with fs(H) := ϕI(a0, a1, η

H0 , . . . , a2n−1, η

Hn−1).













Hn−1).








Consult ϕI : a2n depends on what (η0, . . . ηn−1) you feed fs : [κ]ms → N.

By Rowbottom there is X ⊆ κ of size ℵ1 so that fs�[X]m is constant for all s.

Set ϕnewI (s) = ϕI(a0, a1, ηH0 , . . . , a2n−1, η

Hn−1) for any H ∈ [X]m.

ϕnewI is winning for I in G(NN, D).Otherwise x 6∈ D for the resulting x = (a0, a1, . . .).Fix embedding e : (Tx, <KB)→ (X,<). Extend it by e(t) = 0 for every t 6∈ Tx.But then II would beat ϕI by playing e along with (an): contradiction.




































ϕnewI is winning for I in G(NN, D).

Otherwise x 6∈ D for the resulting x = (a0, a1, . . .).Fix embedding e : (Tx, <KB)→ (X,<). Extend it by e(t) = 0 for every t 6∈ Tx.But then II would beat ϕI by playing e along with (an): contradiction.












ϕnewI is winning for I in G(NN, D).Otherwise x 6∈ D for the resulting x = (a0, a1, . . .).

Fix embedding e : (Tx, <KB)→ (X,<). Extend it by e(t) = 0 for every t 6∈ Tx.But then II would beat ϕI by playing e along with (an): contradiction.












ϕnewI is winning for I in G(NN, D).Otherwise x 6∈ D for the resulting x = (a0, a1, . . .).Fix embedding e : (Tx, <KB)→ (X,<).

Extend it by e(t) = 0 for every t 6∈ Tx.But then II would beat ϕI by playing e along with (an): contradiction.












ϕnewI is winning for I in G(NN, D).Otherwise x 6∈ D for the resulting x = (a0, a1, . . .).Fix embedding e : (Tx, <KB)→ (X,<). Extend it by e(t) = 0 for every t 6∈ Tx.

But then II would beat ϕI by playing e along with (an): contradiction.














Table of Contents



3 Borel determinacy


Determinacy of infinite games ⇐⇒ Large cardinals

(1970) D.A. Martin: assuming existence ofmeasurable cardinals all analytic subsets of NN aredetermined.

(1975) D.A. Martin: all Borel subsets of NN aredetermined.

(1971) H. Friedman: any proof of Boreldeterminacy requires uncountably many applicationsof the powerset operation and the axiom ofreplacement.

(1984) H. Woodin: if all projective subsets of NN

are determined then it is consistent that infinitelymany Woodin cardinals exist.

(1985) D.A. Martin and J.R. Steel: assumingexistence of infinitely many Woodin cardinals allprojective subsets of NN are determined.


































Table of Contents



3 Borel determinacy


Determinacy of Borel gamesWe will briefly describe the ideas behind Martin’s proof that all Borel subsets ofN are determined.

Notice that every Borel D ⊆ N can be “unraveled” by transfinite induction to its“clopen atoms.”We want to follow this unraveling of D with an “unraveling” of G(N , D).

Let T ⊆ AN be a tree. A covering of T is a triple (T , π, ρ) so that:1 T is a pruned tree on some alphabet A;2 π : T → T is a length preserving monotone projection;3 ρ maps strategies of I (resp. II) in T to strategies of I (resp. II) in T ,

in a Lipschitz way;4 if S is a strategy in T then [ρ(S)] ⊆ [π]([S]), where [π] : [T ]→ [T ] is

the continuous map induced by π.

A covering (T , π, ρ) of T unravels D ⊆ [T ] if π−1(D) is clopen in [T ].Borel determinacy follows by Gale-Steward and the following theorem:

Theorem (Martin)

Every Borel D ⊆ N admits a covering (T , π, ρ) of N<N which unravels D.


Determinacy of Borel gamesWe will briefly describe the ideas behind Martin’s proof that all Borel subsets ofN are determined.Notice that every Borel D ⊆ N can be “unraveled” by transfinite induction to its“clopen atoms.”

We want to follow this unraveling of D with an “unraveling” of G(N , D).





Theorem (Martin)



Determinacy of Borel gamesWe will briefly describe the ideas behind Martin’s proof that all Borel subsets ofN are determined.Notice that every Borel D ⊆ N can be “unraveled” by transfinite induction to its“clopen atoms.”We want to follow this unraveling of D with an “unraveling” of G(N , D).





Theorem (Martin)




Let T ⊆ AN be a tree. A covering of T is a triple (T , π, ρ) so that:

1 T is a pruned tree on some alphabet A;2 π : T → T is a length preserving monotone projection;3 ρ maps strategies of I (resp. II) in T to strategies of I (resp. II) in T ,




Theorem (Martin)




Let T ⊆ AN be a tree. A covering of T is a triple (T , π, ρ) so that:1 T is a pruned tree on some alphabet A;

2 π : T → T is a length preserving monotone projection;3 ρ maps strategies of I (resp. II) in T to strategies of I (resp. II) in T ,




Theorem (Martin)




Let T ⊆ AN be a tree. A covering of T is a triple (T , π, ρ) so that:1 T is a pruned tree on some alphabet A;2 π : T → T is a length preserving monotone projection;

3 ρ maps strategies of I (resp. II) in T to strategies of I (resp. II) in T ,in a Lipschitz way;

4 if S is a strategy in T then [ρ(S)] ⊆ [π]([S]), where [π] : [T ]→ [T ] isthe continuous map induced by π.


Theorem (Martin)








Theorem (Martin)





in a Lipschitz way;

4 if S is a strategy in T then [ρ(S)] ⊆ [π]([S]), where [π] : [T ]→ [T ] isthe continuous map induced by π.


Theorem (Martin)








Theorem (Martin)







A covering (T , π, ρ) of T unravels D ⊆ [T ] if π−1(D) is clopen in [T ].

Borel determinacy follows by Gale-Steward and the following theorem:

Theorem (Martin)








Theorem (Martin)








Theorem (Martin)



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panagio/116c/w6d2.pdf · Analytic determinacy Today we are going to nish the proof of the following result: Theorem (Martin) If measurable cardinals exist then every analytic A NN