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-RELATED CONSEQUENCES ON LIMIT SUPERIORS

Daniel Walker, B.S.

Dissertation Prepared for the Degree of

DOCTOR OF PHILOSOPHY

UNIVERSITY OF NORTH TEXAS

May 2013

APPROVED:

Stephen Jackson, Major Professor Douglas Brozovic, Committee Member Su Gao, Committee Member and Chair of the Department of Mark Wardell, Dean of the Toulouse Graduate School Walker, Daniel. Determinacy-Related Consequences on Limit Superiors. Doctor

of Philosophy (Mathematics), May 2013, 26 pp., bibliography, 13 titles.

Laczkovich proved from ZF that, given a countable of Borel sets on a

perfect Polish space, if the limit superior along every subsequence was uncountable, then

there was a particular subsequence whose intersection actually contained a perfect

. Komjath later expanded the result to hold for analytic sets. In this paper, by

adding AD and sometimes V=L(R) to our assumptions, we will extend the result

further. This generalization will include the increasing of the length of the sequence to certain uncountable regular cardinals as well as removing any descriptive requirements

on the sets.

Copyright 2013

by

Daniel Walker

ii TABLE OF CONTENTS

Page

CHAPTER 1. INTRODUCTORY REMARKS ...... 1

CHAPTER 2. THE COUNTABLE CASE ...... 9

CHAPTER 3. THE GENERAL CASE ...... 16

CHAPTER 4. GENERALIZING THE SELECTION LEMMA ...... 19

BIBLIOGRAPHY ...... 25

iii CHAPTER 1 INTRODUCTORY REMARKS

Throughout this paper we are assuming the of ZF. We also equivocate between

R and ωω.

Definition 1. Given a limit ordinal λ, an unbounded S ⊆ λ, and a sequence of sets

0 0 hAαiα∈S, define the limit superior of hAαiα∈S by lim sup Aα = {x : ∀α < λ ∃α > α (α ∈ S α∈S ∧ x ∈ Aα0 )}. That is, the limit superior is simply all those objects that are in cofinally many

Aα’s.

Definition 2. Let X be a of ordinals and κ a cardinal. Define [X]κ = {f : κ → X : ∀α < β f(α) < f(β)}. The elements of [X]κ can also be identified with their ranges. We are very abusive of this ambiguity. When viewing the elements as of X we denote them with capital letters. When viewing them as functions we use standard

0 κ functions variables, such as f, g, etc. We say two functions f, f ∈ [X] are E0-equivalent

κ if their ranges are the same for some point in X on. A set C ⊆ [X] is E0-invariant if it is closed under E0 equivalence.

We first start with the already established results on limit superiors of sets of reals.

Theorem 1 (Laczkovich, 1977). [1] If hAnin∈ω is a sequence of Borel sets of reals

ω ω T such that ∀f ∈ [ω] (lim sup Af(n) is uncountable), then for some f ∈ [ω] ( Af(n) has a n∈ω n∈ω perfect subset).

In 1982, Komjath extended Laczkovich’s result to hold for analytic sets (as well as to all sets when assuming Martin’s ) [2]. Our main goal in this paper is, by strength- ening the axiomatic assumptions, to generalize this phenomenon. Before we state our goal, however, we must first introduce some more preliminary concepts.

1 Definition 3. Consider a game between two players, traditionally referred to as I and II, with the following rules: Before the game starts, a set A ⊆ ωω, known as the payoff set, is fixed. I goes first, playing an integer n0. II then plays an integer n1, which is followed by I playing an integer n2, etc. This continues for ω-many turns. At the end of time, the two players have together built up a real x = hn0, n1, n2, ...i. We say that I wins the game if x ∈ A and II wins otherwise. We say that the game is determined, or more often we refer to the set A itself being determined, if one of the players has a winning strategy. To be clear, a winning strategy for I would be a σ from the set of all even length, finite

ω of integers into ω such that ∀hn0, n1, n2, ...i ∈ ω if ∀k σ(hn0, ..., n2k−1i) = n2k, then hn0, n1, n2, ...i ∈ A. The notion of a winning strategy for II is defined similarly. The , abbreviated AD, is the statement that every set of reals is determined. A winning strategy for one of the players can also be viewed as a continuous function. That is, if σ is a winning strategy for I and y ∈ ωω, let σ(y) = the real that σ demands I play when II plays y. A similar notion is defined for winning strategies for II.

Definition 4. Define the ordinal (actually, the cardinal) Θ as the supremum of the ranks of the prewellorderings on subsets of R. If AC and CH hold, then Θ = ω2. Once we assume AD, however, Θ becomes quite a .

Theorem 2. Assume AD.

(i) Every set of reals has the . That is, if A ⊆ ωω is uncountable, then A has a perfect subset. [3] (ii) The well-ordered of countable sets of reals is countable. (iii) There is no uncountable, well-ordered sequence of distinct, countable sets of reals.

(iv) ω1 and ω2 are regular, but if n ≥ 3, then cof(ωn) = ω2. [4]

(v) If κ < Θ has uncountable cofinality, then there is some S ⊆ R and a prewellordering 1 ≺ on S with length κ that has the Σ1-boundedness property. That is, if A ⊆ S is 1 Σ1, then ∃α < κ ∀y ∈ A |y| < α, where |y| is y’s rank in ≺. [5]

2 (vi) Suppose X , Y are perect Polish spaces and R ⊆ X × Y is such that ∀x ∈ X ∃y ∈ Y R(x, y). Then R has a continuous uniformizing function on some comeager subset of X . (vii) The Axiom of Countable Choice holds for sets of reals.

Definition 5. Informally, Godel’s , denoted by L, is the smallest inner model of . That is, it is the smallest model of set theory that contains ON, the of ordinal . There is another model that is of great inerest to descriptive set theorists, namely L(R), which is the smallest inner model that contains all of the reals.

Definition 6. A predicate R ⊆ ωn1 × (ωω)n2 × (P(ωω))n3 is projective if is of the ~ ω ω ω ~ form R(~n,~x, A) ↔ ∃f1 ∈ ω ∀f2 ∈ ω ... ∃fm ∈ ω ∀k ∈ ω ϕ(~n,~x, A, f1, ..., fm, k), where ϕ is recursive with respect to some oracle machine. To make the definition clear, the predicate

n1 ω n2 ω n3 2 ψ(x, A) ↔ x ∈ A is considered recursive. A predicate R ⊆ ω × (ω ) × (P(ω )) is Σ1 if it is of the form R(~n,~x, A~) ↔ ∃B ∈ P(ωω) ϕ(~n,~x, A,~ B), where ϕ is projective. A predicate

2 2 is ∆1 if both it and its are Σ1.

The model L(R) has some fascinating descriptive properties:

Theorem 3. Assume AD and V=L(R).

(i) Let R ⊆ P(R) be a projective relation (that is, a relation that only allows for real 2 quantification). If R is a non-empty relation, then there is some ∆1 set A ⊆ R such that R(A).

2 2 (ii) Let X , Y be perfect Polish spaces and R ⊆ X × Y a ∆1 (Σ1) relation such that 2 2 ∀x ∈ X ∃y ∈ Y R(x, y). Then R possesses a ∆1 (Σ1) uniformizing function. [6] ω 2 (iii) A ⊆ ω is Σ1 iff A is κ-Suslin for some κ < Θ. [6] (iv) The Axiom of Dependent Choice, DC, is true. [4]

3 Definition 7. One of the topological spaces that we will make use of in chapter 2 is

ω ω the Ellentuck space, h[ω] , εi. The basic open sets are of the form [t, Z] = {X ∈ [ω] : X `(t) = t ∧ X\t ⊆ Z}, where t ∈ 2<ω and Z ∈ [ω]ω.

Definition 8. A ⊆ [ω]ω is completely Ramsey if for every [t, Z] there is some Y ∈ [Z]ω such that either [t, Y ] ⊆ A or [t, Y ] ∩ A = ∅.

The following foundational facts are known about this space:

Theorem 4. (i)[ ω]ω is a Baire space. [7] (ii) A ⊆ [ω]ω is ε-Baire iff it is completely Ramsey. [7]

1 (iii) (AD + V=L(R)) All subsets of [ω]ω are completely Ramsey. [8]

1 1 Definition 9. For n ≥ 1 let δn = sup{| ≺ | :≺ is a ∆n prewellordering on a subset of R}.

Definition 10. A regular cardinal κ has the strong partition property if, given any function h :[κ]κ → {0, 1} there is an H ∈ [κ]κ and an i ∈ {0, 1} such that ∀f ∈ [H]κ h(f) = i.

Definition 11. For this purposes of this paper, when referring to a measure µ on a regular, uncountable cardinal κ we will assume that µ only takes values in {0, 1} and that its domain is P(κ). If ϕ represents some unary formula on κ, then the statement µ({α <

∗ 0 0 κ : ϕ(α)}) = 1 will be written as ∀ α ϕ(α). We write f =µ f if µ({α : f(α) = f (α)}) = 1. Let [κ]κ/µ be the naturally induced ultrapower, which will be another cardinal. A measure is normal if, given any function f : κ → κ such that ∀∗α f(α) < α (a ”pressing down” function), there is some γ < κ such that ∀∗α f(α) = γ.

A few introductory known results:

Theorem 5. Suppose AD is true. For n ≥ 0,

1It is still an open problem as to whether this fact follows from AD alone.

4 1 1 + (i) δ2n+2 = (δ2n+1) [9] 1 (ii) δ2n+1 has the strong partition property. [10] (iii) If we define the ordinal ω(n) recursively by ω(0) = 1 and ω(n + 1) = ωω(n), then

1 + 1 1 1 ω δ2n+1 = (ℵω(2n−1)) . In particular, δ1 = ω1, δ3 = ωω+1, and δ5 = ωωω +1. [10] (v) The cardinals with the strong partition property go cofinally up into Θ. [4]

Theorem 6 (Steel). (AD+V=L(R)) Let κ be an uncountable, regular cardinal. De-

fine µω : P(κ) → {0, 1} by µω(A) = 1 iff A possesses an unbounded subset that is closed

under supremums of length ω. µω is a normal measure. [11]

Definition 12. Given an uncountable cardinal κ, a subset F ⊆ κ is called closed and unbounded, or a club, if it is unbounded in κ and is closed under supremums less than κ.

ω1 Definition 13. A function f ∈ [ω1] has uniform cofinality ω if there is a

function g : ω1 × ω → ω1 such that ∀α < ω1 (∀n ∈ ω g(α, n) < g(α, n + 1) and f(α) = sup g(α, n)). A function f has uniform cofinality α if there is a function g : {hα, βi : β < n∈ω 0 0 α < ω1} → ω1 such that ∀β < β < α < ω1 (g(α, β) < g(α, β ) and f(α) = sup g(α, γ)).A γ<α function f is said to be of the correct type if it is everywhere discontinuous (i.e. f(α) > sup f(β)) and has uniform cofinality ω. A function f is said to be of the other type if it β<α ω1 is everywhere discontinuous and has uniform cofinality α. Given a club F ⊆ ω1, let [F ]c =

ω1 ω1 ω1 {f ∈ [F ] : f is of the correct type} and [F ]o = {f ∈ [F ] : f is of the other type}. In

ω1 Chapter 4 we will rely heavily upon the fact that all of the functions in [ω1] are essentially one of the two types, which we will now show.

ω1 Theorem 7 (Kunen). [13](AD) There is a tree T on ω × ω1 such that ∀f ∈ [ω]

ω ∃x ∈ ω such that Tx is well-founded and for all infinite α < ω1 f(α) ≤ |Tx  α|. T is called the Kunen tree.

1 Proof. Let S be a recursive tree on ω × ω such that p[S] is a Σ1-complete set. Define a tree U on ω × ω1 by hha0, ..., an−1i, hα0, ..., αn−1ii ∈ U iff ∀i, j, k < n (ahi,ji = 1 → αi <

5 2 αj) ∧ [(ahi,ji = 1 ∧ ahj,ki = 1) → ahi,ji = 1], where h, i : ω → ω is some canonical recursive . Therefore, p[U] =WF, the set of all reals coding well-founded relations on ω. Now ~ ~ define a tree V on ω × ω × ω1 × ω × ω by h~s,~a,~α, b,~ci ∈ V iff h~a,~αi ∈ U, hb,~ci ∈ S, and there is some τ ∈ ωω extending s such that τ(~a) = ~b (here viewing τ as a strategy for player II in some game). We claim that, after coding all the integer coordinates into a single integer, V will

ω1 be our T . Consider any f ∈ [ω1] . We look at the game where I plays x ∈WF and II plays

ω y ∈ ω . II will win iff Sy is well-founded and |Sy| > sup{f(β): β ≤ |x|}.

ω 1 Suppose first that I has a winning strategy, σ. σ[ω ] is a Σ1 subset of WF and thus the 1 ranks of its elements are bounded by some γ < ω1. Since p[S] is Σ1-complete, it is not Borel.

Therefore, sup {|Sy| : Sy is well-founded} = ω1. II can then defeat I’s use of σ by playing y∈ωω any y with |Sy| sufficiently high.

Therefore, II must have a winning strategy, τ. Vτ must be well-founded since, being

a winning strategy for II, τ guarantees that Sy is well-founded.

ω Fix some α < ω1 and let x ∈WF have rank α. We can find some ~α ∈ α such that

hx, ~αi ∈ [U]. |Vτ  α| ≥ |Sτ(x)| > sup{f(β): β ≤ |x|} ≥ f(|x|) = f(α). Now letting x = τ (i.e. the x in the statement of the theorem) gives us our result. 

ω1 Lemma 8. If f ∈ [ω1] , then {α : sup f(β) = α} is a club. β<α Proof. First note that, since f(α) < f(β) ↔ α < β, if f(α) 6= α, then f(β) 6= β for all β >

α. If f(α) = α for all α, then we are done. Assume not. Let F = {α < ω1 : sup f(β) = α}. β<α To show that F is non-empty and unbounded, let a0 be some ordinal arbitrarily high such that f(a0) 6= a0. Define recursively an+1 = f(an) and let α = sup an. Note that an < an+1 < α. n<ω sup f(β) = sup f(an) = sup an+1 = α. So, α ∈ F . Now let a0 < a1 < ... all be members of β<α n<ω n<ω F and let α = sup an. Since an+1 ∈ F , f(an) ≤ an+1. sup f(β) = sup f(an) ≤ sup an+1 = α. n<ω β<α n<ω n<ω Since sup f(β) ≥ α, α ∈ F . Therefore, F is closed.  β<α

ω1 Theorem 9. (AD) If L = {α < ω1 : α is a limit ordinal} and f ∈ [L] is such that ∀∗α f(α) > α, then f is almost everywhere one of the two types.

6 Proof. Note first that, via the previous lemma, since f is not almost everywhere equal to the identity function it must be almost everywhere discontinuous. Let T be the Kunen tree. We fix some canonical procedure h, i such that h, i : α<ω → α is a bijection for all α. Let

ω x ∈ ω be such that ∀α f(α) < |Tx  α|. Therefore, ∀β < f(α) ∃u ∈ Tx  α |u|Txα = β. Let

λα = the least γ ≤ α such that sup{|u|Txα : hui ≤ γ} = f(α). ∗ First suppose that ∀ α λα = α. f has unform cofinality α almost everywhere via

g(α, β) = |u|Txα where hui = β. ∗ Now suppose that ∀ α λα < α. The map α 7−→ λα is then a pressing down function,

∗ so ∃γ < ω1 ∀ α λα = γ. γ has countable cofinality, so fix some hγnin∈ω cofinal sequence in γ

such that sup{|un|Txα : n ∈ ω} = f(α), where huni = γn. f has uniform cofinality ω almost

everywhere via g(α, n) = |un|Txα. 

ω1 ω1 Lemma 10. (AD) If we define γf as the ordinal in [ω1] /µω repesenting f ∈ [ω1] ,

then γf is a limit ordinal whenever f is one of the two types.

Proof. Let f be, without loss of generality, of the correct type and let g be the function

ω1 ∗ that witnesses its uniform cofinality. Suppose h ∈ [ω1] is such that γh < γf . Then ∀ α

0 ω1 0 h(α) < f(α). Define h ∈ [ω1] by h (α) = g(α, N), where N is the least integer such that g(α, N) > h(α). N must exist by the definition of the uniform cofinality (in actuality ∀∗α N exists, which is all we really care about). Now we have ∀∗α h(α) < h0(α) < f(α), so

γh < γh0 < γf , which shows that γf cannot be a successor ordinal. A similar but simpler

shows that γf 6= 0. 

ω1 Lemma 11. (AD) Given any partition h :[ω1] → {0, 1} there is a club F ⊆ ω1 and

ω1 ∗ an i ∈ {0, 1} such that ∀f ∈ [F ]c h(f) = i. This is also written as ∀ f f ∈ Ck. The result would still hold if we were to replace ”the correct type” with ”the other type”.

Definition 14. Let κ be a cardinal and Γ a pointclass. Define the Laczkovich-

Komjath property by LK(κ, Γ) iff whenever hAαiα<κ is a sequence of sets in Γ such that

≤κ ≤κ lim sup Af(α) is uncountable whenever f ∈ [κ] is cofinal, there exists a particular f ∈ [κ] α<κ

7 T cofinal such that Af(α) has a perfect subset. That is, Laczkovich proved LK(ω, B) and α<κ 1 Komjath proved LK(ω, Σ1), where B is the pointclass of Borel sets.

We finally come to the main result.

Theorem 12. (AD + V=L(R)) ∀κ LK(κ, P(ωω))

Note: As we show in chapter 3, if we only assume AD we still obtain an impressive result. That is, our main theorem will still hold whenever κ has the strong partition property.

8 CHAPTER 2 THE COUNTABLE CASE

For the most part, we are following Komjath’s original proof for the analytic case. The only problem is what we are calling the Selection Lemma (see below). In Komjath’s situation, the Selection Lemma (he did not use this name) was quite easy to prove. In our situation, it takes a bit more effort.

Definition 15. If f, g ∈ [ω]ω, define f ≤ g iff ran(f) ⊆ ran(g). Let f ≤∗ g iff

ω ∃n ∈ ω f[n, ∞) ≤ g. When viewing elements of [ω] as subsets of ω, we sometimes use the notation G ≤ H to mean G ∈ [H]ω.

Definition 16. Until we complete the proof of the Selection Lemma, let ε be the Ellentuck topology on [ω]ω.

Lemma 13. Let A ⊆ [ω]ω be ε-non-meager and ε-Baire. A has non-empty ε-interior.

Proof. A must be ε-comeager in some ε-basic open set, [t, Z]. Since A is completely Ramsey, for some Y ∈ [Z]ω either [t, Y ] ⊆ A or [t, Y ] ∩ A = ∅. But [t, Z]\A is ε-meager and

[t, Y ] ⊆ [t, Z], so [t, Y ] ⊆ A. 

Theorem 14 (Well-Ordered Additivity of Ellentuck-Meager Sets). (AD+V=L(R)) ω Let hAαiα<κ be a well-ordered sequence of subsets of [ω] . If Aα is ε-meager for every α, S then Aα is also ε-meager. α<κ

Proof. 2 Suppose not and let κ be the least such cardinal. By the σ-additivity of category

we must have κ ≥ ω1. κ must be regular. If not, then let λ = cof(κ) < κ and let htβiβ<λ S be a cofinal sequence in κ. Let Cβ = Aα. By the minimality of κ, Cβ must be ε- α

2Thanks goes to [12] for the inspiration for this proof. They were able to prove a similar assertion arguing from AC, interestingly.

9 a contradiction.

Without loss of generality we may assume that the Aα’s are disjoint. κ < Θ, so there must exist some S∗ ⊆ R and some prewellordering ≺∗ on S∗ with rank κ that has the 1 ∗ ω ∗ Σ1-boundedness property. Let R = {hx, yi|x ∈ A|y|} ⊂ [ω] × S , where |y| is y’s rank in ≺∗. Therefore, the triplet hS∗, ≺∗,R∗i can be considered a counterexample to the theorem. Now define the ternary relation F(S, ≺,R) ↔ (hS, ≺,Ri is a counterexample to the theorem and

1 ≺ has the Σ1-boundedness property). That is, F(S, ≺,R) iff

(1) S ⊆ R 1 (2) ≺ is a prewellordering on S with the Σ1-boundedness property. (3) R ⊂ [ω]ω × R (4) ∀y ∈ S ∃X ∈ [ω]ω R(X, y) (Non-triviality of sections) (5) ∀X ∈ [ω]ω, y, y0 ∈ S [R(X, y) → (R(X, y0) ↔ |y| = |y0|)] (Invariance and disjointness of sections) (5) R has meager sections on the left coordinate but the union of those sections is non-meager.

It can be checked that F is a projective relation. By the existence of hS∗, ≺∗,R∗i, F is non-

2 empty. Therefore, F must be satisfied by some ∆1 triplet. Let hS, ≺,Ri be such a triplet. 2 Since R is at worst ∆1, it must have a uniformizing funcion, F .

Enumerate the left-coordinate sections of R by hDαiα<τ . By possibly switching to a subsequence of the Dα’s, we assume without loss of generality that τ is minimal. That is, given any λ < τ any sub-union of the Dα’s of length λ is meager. S By the previous lemma Dα must contain some [t, W ]. Let hUnin∈ω be an enu- α<τ meration of a basis for R (with regards to the standard topology) with U0 = R. We now ω recursively define a sequence hXnin∈ω in [ω] in the following way: Let X0 = W . Recursively assume that for each k ≤ n

ω (1) Xk+1 ∈ [Xk] ∩ [t, ω]

10 −1 (2) For every s ⊆ {min(Xj\t): j ≤ k} either [t ∪ s, Xk] ⊆ F [Uk] or [t ∪ s, Xk] ∩

−1 F [Uk] = ∅.

By 2n+1 applications of the completely Ramsey property, we now choose an

Xn+1 ∈ [t, Xn\{min(Xn\t)}] that satisfies the recursive requirements. Now let Y = t ∪

{min(Xn\t): n ∈ ω}. It follows that F[t, Y ] is continuous. It is important to note that this is continuity in the topology that [t, Y ] inherits from the Cantor space, not the Ellentuck topology. Let K be the of [t, Y ] under F .[t, Y ] is Borel (once again, in the inherited

1 topology), so K must be analytic. But since ≺ has the Σ1-boundedness property, K must S be ≺-bounded. By the minimality of τ, D|y| must be ε-meager, where |y| is y’s rank in y∈K ≺. But that set contains [t, Y ], which is a contradiction since no non-empty ε-open set can be ε-meager. 

This give us the following lemma.

Lemma 15 (Selection Lemma). (AD+V=L(R)) Let hAnin∈ω be a sequence of sets of S reals and V = α<κ Vα a set of reals with κ < Θ. If V ∩ lim sup Af(n) is uncountable for n∈ω ω ω every f ∈ [ω] , then for some α < κ and H ∈ [ω] Vα ∩ lim sup Af(n) is uncountable for n∈ω every f ∈ [H]ω.

Proof. Let R(α, f) ↔ Vα ∩ lim sup Af(n) is uncountable. Since lim sup Af(n) only depends n∈ω n∈ω ω on a tail end of f, the sets Rα = {f ∈ [ω] : R(α, f)} are E0-invariant. Consider some

ω S S f ∈ [ω] . By assumption Vα ∩ lim sup Af(n) = (Vα ∩ lim sup Af(n)) is uncountable. α<κ n∈ω α<κ n∈ω S ω Therefore, there must be some α < κ such that R(α, f). This means that Rα = [ω] . α<κ ω Since [ω] is ε-non-meager, by the the previous theorem Rα must be ε-non-meager for some

ω α. Rα must contain some [t, H]. Rα is also E0-invariant, so in fact [H] = [∅,H] ⊆ Rα. 

The main theorem will follow as a corollary of the following theorem.

Theorem 16 (Penultimate Claim). (AD+V=L(R)) ∀κ LK(ω, Sκ), where ω Sκ = {A ⊆ ω : A is κ-Suslin}.

11 Note: When the Selection Lemma will be used several times in this chapter, the κ mentioned will not necessarily be the κ of the Penultimate Claim.

Until we prove the Penultimate Claim we will consider hAnin∈ω to be fixed. We will

ω also assume without loss of generality that the An’s are subsets of ω .

Definition 17. Let V be a set of reals and f ∈ [ω]ω. We say f is good for V if

∗ V ∩ lim sup Ag(n) is uncountable for every g ≤ f. Obviously, ≤ can be replaced with ≤ . n∈ω

Lemma 17. (AD+V=L(R)) Let F,A ⊆ ωω, f ∈ [ω]ω with F closed. If f is good for F ∩ A, then for some x ∈ F and f 0 ≤ f f 0 is good for F ∩ A ∩ U for every open neighborhood U of x.

ω S Proof. Since ω is a separable metric space, F can be written as n∈ω Fhni, where each

Fhni is closed and diam(Fhni) < 1. By the Selection Lemma, there is an n0 ∈ ω and f0 ≤ f S such that f0 is good for Fhn0i ∩ A. We now write Fhn0i = Fhn0,ni, where each Fhn0,ni is n∈ω 1 closed and diam(Fhn0,ni) < 2 . Once again, by the Selection Lemma there is an n1 ∈ ω and

f1 ≤ f0 such that f1 is good for Fhn0,n1i ∩A. We recursively extend this procedure to produce

ω n0, n1, n2, ... ∈ ω and f0 ≥ f1 ≥ f2 ≥ .... By the completeness of ω , let x be the unique T of Fhn0,n1,..,nki. k∈ω Let a0 be the least element of f0. Recursively, let an+1 be the least element of fn+1

0 0 that is greater than an. Let f = {an : n ∈ ω}. We claim that f works. Let U be any open

neighborhood of x. By the decreasing diameters, Fhn0,...,nki ⊆ U for some k. Since fk is good 0 ∗ 0 for Fhn0,...,nki ∩ A, it is also good for F ∩ A ∩ U. f ≤ fk, so f is good for F ∩ A ∩ U. 

Definition 18. Let P,Q ⊆ ωω be closed and non-empty. δ∗(P,Q) = min({ρ(x, y): x ∈ P, y ∈ Q}, where ρ is the standard metric on ωω.

Corollary 1. (AD+V=L(R)) Let F,A ⊆ ωω, f ∈ [ω]ω, and ε > 0 with F closed (we are no longer using ε for the Ellentuck topology). If f is good for F ∩ A, then there is

0 ∗ 0 an f ≤ f and perfect sets F0,F1 ⊂ F such that diam(Fi) < ε, δ (F0,F1) > 0 and f is good

for both F0 ∩ A and F1 ∩ A.

12 Proof. By the previous lemma, there is some x ∈ F and f1 ≤ f such that f1 is good for

F ∩ A ∩ U for every open neighborhood U of x. For each n ∈ ω let Qn = {y ∈ F : ρ(x, y) ≥

ε S ω 2n+1 }. Since F = {x} ∪ Qn and no element of [ω] can be good for a , the n∈ω Selection Lemma says that there is a k ∈ ω and an f2 ≤ f1 such that f2 is good for Qk. S Write Qk as Qk = Pn, where each Pn is closed and diam(Pn) < ε. Once again, we use n∈ω 0 0 the Selection Lemma to get an n ∈ ω and an f ≤ f2 such that f is good for Pn. Now we let

ε F0 = {y ∈ F : ρ(x, y) ≤ 2k+2 } and F1 = Pn. Note finally that Fi must be uncountable and closed. Under AD every set has the perfect set property, so without loss of generality we

may reduce Fi to its non-empty perfect part. Since we are only removing countably many

0 elements the goodness of f has not been affected. 

Now,

S T zk Proof. (Penultimate Claim) Since An is κ-Suslin we can write it as An = Fn , z∈κω k∈ω zk zk 1 where Fn is a closed set with diam(Fn ) < 2n . This requires Countable Choice to pick

<ω t S T zk out the Suslin representations for each An. Given a t ∈ κ define An = Fn . z∈κω,tvz k∈ω ω Without loss of generality assume that A0 = ω . We now recursively define for each n ∈ ω the following objects:

(1) an ∈ ω

n (2) A perfect set Ps for each s ∈ 2 (3) A sequence t(k, s) ∈ κn for each k ≤ n and s ∈ 2n

ω (4) Hn ∈ [ω]

They will be defined so as to satisfy the following conditions:

(i) an < an+1 ∈ Hn

ω (ii) Hn+1 ∈ [Hn]

n ∗ (iii) If s ∈ 2 , then Ps∗h0i,Ps∗h1i ⊂ Ps and δ (Ps∗h0i,Ps∗h1i) > 0.

n 1 (vi) If s ∈ 2 , then diam(Ps) < 2n . n n T t(k,s) (v) If s ∈ 2 , then Hn is good for Ps ∩ Aak . k=0

13 n n T t(k,s) (vi) If s ∈ 2 , then Ps ⊆ Fak . k=0 (vii) If s ∈ 2n, k ≤ n, and i ∈ {0, 1}, then t(k, s) v t(k, s ∗ hii).

ω The construction goes as follows: Let a0 = 0, P∅ = ω , and H0 = ω. Obviously, t(0, ∅) = ∅. Now assume that all the objects have been chosen up to stage n. We first claim that there n n T t(k,s) exists b ∈ Hn,G ≤ Hn such that b > an and, for every s ∈ 2 , G is good for Ps∩ Aak ∩Ab. k=0 Suppose not. Then we can recursively construct b0 < b1 < ... and Hn = G0 ≥ G1 ≥ ... such

n that bn ∈ Gn and for each m ∈ ω there is an sm ∈ 2 such that n \ (1) P ∩ At(k,sm) ∩ A ∩ lim sup A sm ak bm t t∈G k=0 m+1 n is countable. Since sm can only take at most 2 different values there must be a particular

n s ∈ 2 such that S = {bm : sm = s} is infinite. Therefore, n \ [ (2) P ∩ At(k,sm) ∩ A ∩ lim sup A sm ak m t t∈S k=0 m∈S is countable, which is impossible by (v) and the fact that S ≤ Hn. Therefore, b and G must exist.

n n Let an+1 = b. By 2 applications of Corollary 1 we get, for each s ∈ 2 , perfect sets

0 0 0 0 1 ∗ 0 0 Ps∗h0i,Ps∗h1i ⊂ Ps and G ≤ G such that diam(Ps∗hii) < 2n+1 , δ (Ps∗h0i,Ps∗h1i) > 0, and for n n 0 0 T t(k,s) every s ∈ 2 G is good for Ps∗hii ∩ Aan ∩ Aan+1 . k=0 Now,

n n+1 \ [ [ [ \ (3) At(k,s) ∩ A ⊆ ... ( Azk ) an an+1 ak k=0 z0 z1 zn+1 k=0

n+1 where z0, ..., zn+1 ∈ κ and t(k, s) v zk for each k ≤ n. That union has length κ, so

n+1 by 2 applications of the Selection Lemma, we can select a particular hz0, ..., zn+1i and a n+1 H ≤ G0 such that H is good for P 0 ∩ T Azk . Now, for each k ≤ n+1 let t(k, s∗hii) n+1 n+1 s∗hii ak k=0 be the corresponding zk. n+1 0 T t(k,s∗hii Finally, let Qs∗hii = Ps∗hii ∩ Fak . Since Hn+1 is good for it, Qs∗hii must k=0 be uncountable. It is also closed, so we let Ps∗hii be its perfect part. This finishes the construction.

14 T S Now we can finish the proof. Let P = Ps. P is perfect. Let x ∈ P . For n∈ω s∈2n n each n ∈ ω there is a unique sn ∈ 2 with x ∈ Psn . Not only this, but s0 v s1 v .... By

t(m,sn) (vi) and (vii), if n ≥ m, then x ∈ Fam and t(m, sm) v t(m, sm+1) v .... Therefore,

t(m,sm) T x ∈ Aam ⊆ Aam and so P ⊆ Aan .  n∈ω We finally complete our goal.

Theorem 18. (AD+V=L(R)) LK(ω, P(ωω))

Proof. Suppose not. Define a unary relation F by F(hBnin∈ω) iff hBnin∈ω is a sequence of subsets of ωω that is a counterexample to the Main Theorem. F is a projective relation

2 and, by assumption, non-empty. Therefore, we can select a particular ∆1 sequence hCnin∈ω

that is a counterexample. The Cn’s must be uniformly κ-Suslin for some κ < Θ and the

Penultimate Claim yields a contradiction. 

15 CHAPTER 3 THE GENERAL CASE

In this chapter we prove the general case. That is, we consider uncountable sequences of sets of reals. It is a bit striking that the countable case takes considerably more effor than the, supposedly, more complicated uncountable case.

Theorem 19. (AD) If κ has the strong partition property, then LK(κ, P(ωω)).

T κ Proof. Suppose not. If we define Cf = Af(α), then ∀f ∈ [κ] Cf is countable. For each α<κ f and β < κ let fβ be the function that enumerates {f(α): f(α) > f(β)} and let f−1 = f.

0 Note that if α < α , then Cfα ⊆ Cfα0 . Since, under AD, there cannot be a strictly increasing 0 uncountable union of countable sets of reals, ∀f ∃α ∀α > α Cfα0 = Cfα . Let αf be the κ least such α. Define a partition on [κ] by C0 = {f : αf = −1} and C1 = {f : αf > −1}.

κ Suppose first that C1 has a homogeneous set, H. Let f = HαH . Since f ∈ [H] , Cf0 6= Cf . But C = C = C = C , a contradiction. f0 HαH +1 HαH f

Therefore, C0 has a homogenenous set, H. Suppose that x ∈ lim sup Aα. By a simple α∈H 0 κ 0 induction, it follows from the choice of H that, given any f, f ∈ [H] , if f and f are E0-

κ T equivalent, then Cf = Cf 0 . From there we get ∀f ∈ [H] ∀β Cf ⊆ Aα. Now x ∈ Cf α

Theorem 20. (AD + V=L(R)) ∀κ > ω LK(κ, P(ωω))

Proof. We may first assume that κ is regular. If not, then we switch to some regular cofinal subsequence and start again. Let P = lim sup Aα. By assumption, P is uncountable. For α<κ each x ∈ P define gx : κ → κ by gx(α) = the least β > α such that x ∈ Aβ. µω is a normal

κ κ measure, so we let λ = [κ] /µω and γf be ordinal in the ultrapower representing f ∈ [κ] .

For each β < λ, let Gβ = {x ∈ P : γf = β}. Under AD we cannot have an uncountable, well-ordered union of distinct countable sets, so there must be some β < λ such that Gβ is

16 0 uncountable. Let P be some perfect subset of Gβ and pick some g such that γg = β. That

0 means for each x ∈ P gx will equal g on some measure 1 set (with regards to µω, of course). This gives us the statement

0 ∗ (4) ∀x ∈ P ∀ α x ∈ Ag(α)

The goal is to switch the quantifiers by switching to an uncountable P 00 ⊆ P 0. Note that a subset of κ will have measure 1 if it has an unbounded subset that is closed under limits of length ω. Every function f : κ → κ such that ∀α f(α) > α defines an ω-closed set. That is, the closed set is {α < κ : α has countable cofinality and ∀β < α f(β) < α}. We

0 will now show that each x ∈ P has an ω-closed set that works for (4) (that is, x ∈ Ag(α) for each α in that set) such that the increasing function that defines the closed set can actually be coded into a real. First consider some x ∈ P 0, some ω-closed set C that works for (4),

1 and some prewellordering of S ⊆ R of length κ with the Σ1-boundedness property (the prewellordering is, of course, chosen independently from x). Now we consider a game G(C)

where players I and II play reals y1, y2 ∈ R against each other and II wins iff y1 ∈ S → (y2 ∈

S ∧ |y1| < |y2| ∈ C. We claim that II has a winning strategy in G(C). Suppose not and let

1 σ be a winning strategy for I. σ[R] ⊆ S is Σ1 and hence bounded by some α < κ. II can

then defeat I’s use of σ by playing any y2 with α ≤ |y2| ∈ C. Now define the relation R(x, τ) ↔

(5) x ∈ P 0 ∧ τ is a winning strategy for II in G(C) for some C that works for (4)

Since P 0 has a perfect subset, we can use AD to uniformize R continuously on a comeager

0 subset. That is, there is some uncountable Q ⊆ P and a continuous function x 7−→ τx on

00 00 Q such that R(x, τx). Let P ⊆ Q be a perfect set. For each x ∈ P , let Cx be the ω-closed

set defined by the hx(α) = inf{τx(y): |y| = α}. Now we will define a master ω-closed

set that will allow us to switch the quantifiers in (4). For each y ∈ S the map x 7−→ τx(y)

is continuous and so as before must be bounded. So, define h(α) = sup hx(α) and let C x∈P 00 be the ω-closed set defined by h. Let x ∈ P 00 and α ∈ C. By the definition of C, α has

17 countable cofinality and ∀β < α h(β) < α. Since hx(β) ≤ h(β), we have α ∈ Cx. Therefore, T C ⊆ Cx. x∈P 00 Finally we have switched the quantifiers in (4) as desired. That is,

∗ 00 (6) ∀ α ∀x ∈ P x ∈ Ag(α)

∗ 00 T where the ”∀ α” means ”∀α ∈ C”. We now have an uncountable P ⊆ Ag(α), which α∈C completes the proof. 

18 CHAPTER 4 GENERALIZING THE SELECTION LEMMA

In this chapter, we will develop some generalizations of the Selection Lemma that was used in chapter 2. It was originally thought that the generalization of the main theorem to uncountably long sequences of sets of reals would follow a similar line of argumentation to the countable case. Even though the proof ended up going in a completely different direction, the following chapter is an interesting discussion on its own. We will be assuming AD for this chapter.

Definition 19. For this chapter we will adopt the following notation: Given any

ordinal β < ω2, there is some equivalence class of functions on ω1 that β represents in the ultrapower. Let π(β) be a function in that equivalence class. There is an abuse here, since there is no way to actually pick out a representative function. However, all functions in the equivalence class will agree almost everywhere and so we will use this notation when the ambiguity won’t cause any problems.

Definition 20. Given a limit ordinal λ define temporarily Cλ = {f : ξ → λ : ξ ≤ λ ∧ f is increasing and unbounded}. Given two infinite cardinals λ, κ < θ define Sel(λ, κ) iff S given any sequence of sets of reals hAαiα<λ and a set of reals V = Vβ, if V ∩ lim sup Af(α) β<κ α∈dom(f) is uncountable for every f ∈ Cλ, then for some g ∈ Cλ and β < κ Vβ ∩ lim sup A(g◦f)(α) α∈dom(f) is uncountable for every f ∈ Cdom(g). This is an unpleasantly complicated definition, but note that it is simply a generalization of the property seen in the second chapter’s Selection Lemma. That is, the Selection Lemma simply states that Sel(ω, κ) holds for every κ < θ.

Lemma 21. (i) Sel(cof(λ), κ) → Sel(λ, κ) (ii) If κ is singular, then (∀ξ < κ Sel(λ, ξ)) → Sel(λ, κ).

Proof. (i) Let λ be a cardinal and let ξ = cof(λ). Fix a sequence of reals hAαiα<λ that satisfies the premise of Sel(λ, κ). Let h : ξ → λ be a cofinal function. Since

19 ξ is regular and hAh(α)iα<ξ satisfies the premise of Sel(ξ, κ), there must be some f ∈ [ξ]ξ and β < κ granted by the conclusion of Sel(ξ, κ). h ◦ f and β now satisfy the conclusion of Sel(λ, κ).

(ii) Let ξ = cof(κ) and let htβiβ<ξ be some cofinal sequence. For each β < ξ let S S S Wβ = Vτ . Wβ = Vβ and Sel(λ, ξ), so there exists some f ∈ Cλ and some τ

0 0 β < κ, so Sel(λ, β ) holds. Therefore, there exists some g ∈ Cdom(f) and some 0 τ < β such that f ◦ g and τ satisfy the conclusion of Sel(λ, β ) (using Vτ , τ < β). But f ◦ g and τ also satisfy the conclusion of Sel(λ, κ).



Therefore, we will now restrict our attention to regular λ’s and also realize that, given a λ, the first κ for which Sel(λ, κ) can fail must be regular. This gives us a much cleaner definition of Sel that allows us to forget about the definition of Cλ.

Theorem 22. Sel(ω1, ω)

Proof. Suppose not and let hhAαiα<ω1 , hVnin<ωi be a counterexample.

0 We will first prove that there is a club F ⊆ ω1 and an n1 < ω such that for every

ω1 ω1 f ∈ [F ]c Vn1 ∩ lim sup Af(α) is uncountable. Suppose not. Let Cn = {f ∈ [ω1]c : α<ω1 ω1 ω1 Vn ∩ lim sup Af(α) is countable}. Cn and [ω1]c \Cn obviously partition [ω1]c . Since we α<ω1 ω1 are supposing our claim is false, [ω1]c \Cn cannot have a homogeneous club. Therefore,

Cn must have a homogeneous club Fn. We can select an Fn for each n using countable

T ω1 choice. Let F = Fn. F is a club, so ∀f ∈ [F ] (Vn ∩ lim sup Af(α) is countable). But n<ω α<ω1 S (Vn ∩ lim sup Af(α)) = V ∩ lim sup Af(α), which is uncountable. Since the countable union n<ω α<ω1 α<ω1 of countable sets is countable, we have a contradiction. This proves our claim.

0 By a similar argument, we construct a club F ⊆ F and an n2 < ω such that for

ω1 every f ∈ [F ]o Vn2 ∩ lim sup Af(α) is uncountable. Since our theorem fails, there must be α<ω1

20 ω1 some f ∈ [F ] such that (Vn1 ∪ Vn2 ) ∩ lim sup Af(α) is countable. By possibly thinning f α<ω1 0 0 ω1 out, we may assume that f =µω f for some f ∈ [F ] of one of the types. Without loss of

0 ω1 0 generality, we assume that f ∈ [F ]c . Let G ⊆ ω1 be some club such that f(α) = f (α) for

ω1 0 0 ω1 every α ∈ G. Let k ∈ [ω1] enumerate G and let h = f ◦ g. Since f is in[F ]c , so is h. But

0 Vn1 ∩ lim sup Ah(α) = Vn1 ∩ lim sup Af (α) = Vn1 ∩ lim sup Af(α) ⊆ Vn1 ∩ lim sup Af(α), which α<ω1 α∈G α∈G α<ω1

is a contradiction since Vn1 ∩ lim sup Ah(α) must be uncountable.  α<ω1

Theorem 23. Sel(ω1, ω1)

ω1 Proof. Consider an appropriate hhAαiα<ω1 , hVβiβ<ω1 i. For each f ∈ [ω1] let βf be the

ω1 least ordinal such that Vβf ∩ lim sup Af(α) is uncountable. Let C0 = {f ∈ [ω1] : βf < f(0)} α<ω1 ω1 and C1 = {f ∈ [ω1] : βf ≥ f(0)}. Suppose that C1 has a homogeneous set G. Let

0 0 0 G = {α ∈ G : α > βG}. G E0G, so βG0 = βG. But βG0 < G (0), a contradiction. Therefore,

ω1 0 let H be a homogeneous set for C0. Let f ∈ [H] . Let f = f ∪ {H(0)}. βf 0 = βf , so

βf < H(0). Since H(0) is a countable ordinal, the pair hhAαia∈H , hVβiβ

conditions of Sel(ω1, ω), which we know to be true. 

Theorem 24. Sel(ω1, ω2)

Proof. We will assume without loss of generality that the Vβ’s are increasing and continuous

at limit stages. By the same argument given in the proof of Sel(ω1, ω), we will first limit

ω1 ourselves to only those functions of the correct type. For each f ∈ [ω1] let βf be as before.

Claim 1. Each βf is a successor ordinal.

S Proof. Suppose βf is a limit ordinal. Vβf = Vβ. First suppose that βf has countable β<βf S cofinality, say with some cofinal sequence hγni. Vβf ∩lim sup Af(α) = (Vγn ∩lim sup Af(α)), α<ω1 n<ω α<ω1 which is a countable union of countable sets, violating the definition of βf . Therefore, βf has uncountable cofinality. Let

[ (7) C = {β < βf :(Vβ\ Vγ) ∩ lim sup Af(α) 6= ∅} α<ω γ<β 1

21 By the definition of βf , C must be cofinal in βf and therefore uncountable. But then S h(Vβ\ Vγ)∩lim sup Af(α)iβ∈C is an uncountable sequence of distinct non-empty countable γ<β α<ω1 sets, which is impossible under AD. 

Since γf must be a limit ordinal when f is of the correct type, βf 6= γf 0 for every f, f 0.

∗ Claim 2. ∀ f βf < γf

∗ Proof. Suppose not. Since βf 6= γf for every f, ∀ f γf < βf . By the definition of βf ,

Vγf ∩ lim sup Af(α) is countable for all f’s of the correct type within some club F . That α<ω1 ω1 means that, given any f ∈ [F ]c , all but countably many x ∈ Vγf are only in countably

many Af(α). That of reals will most likely vary between f’s, even those f’s that agree with each other almost everywhere. Nevertheless, we have the following:

ω1 Sub-Claim. ∀f ∈ [F ]c Vγf ∩ lim sup Aα is countable. α∈F

Proof. Suppose not and fix a counterexample, f. For each x ∈ Vγf ∩ lim sup Aα let gx α∈F

enumerate {α ∈ F : x ∈ Aα}. By possibly thinning out gx we may assume that γgx > γf

(note that gx may not be of the correct type, which is okay). Since the well-ordered union of countable sets is countable, there must be some g ∈ [F ]ω1 (once again, not necessarily of the

correct type) such that {x ∈ Vγf ∩ lim sup Aα : γgx = γg} is uncountable. Fix such a g and α∈F ∗ call that uncountable set W . Since γg > γf , ∀ α f(α) < g(α). Define recursively a function f 0 in the following way:

(i) f 0(0) = f(0) (ii) If α is an ”even” ordinal with f 0(α) = f(α0), then f 0(α + 1) = g(α0). (iii) If α is an ”odd” ordinal, then f 0(α + 1) is the least value of f greater than f 0(α). (iv) If λ is a limit ordinal, then f 0(λ) is the least value of f greater than sup f 0(α). α<λ Given the fact ∀∗λ sup f 0(α) = λ = sup f(α), we have ∀∗α f(α) = f 0(α). Furthermore, since α<λ α<λ ∀∗α (f(α) is in the range of f 0), ∀∗α (g(α) is in the range of f 0). Now, given any x ∈ W ,

22 ∗ ∗ 0 ∀ α gx(α) = g(α), which means that ∀ α (gx(α) is in the range of f ). But then

0 0 (8) x ∈ Vγf ∩ lim sup Af (α) = Vγf0 ∩ lim sup Af (α) α<ω1 α<ω1

0 which means that W ⊆ Vγf0 ∩ lim sup Af (α), contradicting the fact that the latter set is α<ω1 countable. 

Let Cf = Vγf ∩ lim sup Aα. The assignment f 7−→ Cf is µω invariant, so, for some α∈F ∗ particular countable set C, ∀ f (relative to F ) Cf = C. By removing those countably many reals, we may assume that C = ∅. Call the club that witnesses this fact F 0 (so, F 0 ⊆ F ). In

0 ω1 other words, ∀f ∈ [F ]c ∀x ∈ Vγf x 6∈ lim sup Aα. Since every x ∈ V is eventually in some α∈F 0

Vγf , we have ∀x ∈ V x 6∈ lim sup Aα, which violates the assumption of the theorem.  α∈F 0 Consider pairs hh, fi of functions of the correct type such that sup f(β) < h(α) < β<α f(α).

∗ Claim 3. ∀ hh, fi βf < γh

∗ Proof. Suppose not. Then ∀ hh, fi βf > γh. So,

∗ ∗ (9) ∀ hh, fi ∀ α h(α) = π(γh)(α) < π(βf )(α)

Let f be a function of the correct type and let g : ω1 × ω → ω1 be a function that witnesses the fact that f has uniform cofinality ω. By the definition of g,

∗ (10) ∀ α ∃n < ω π(βf )(α) < g(α, n) > sup f(β) β<α

µω is countably complete, so

∗ (11) ∃N < ω ∀ α π(βf )(α) < g(α, N) > sup f(β) β<α Define h by h(α) = g(α, N). hh, fi is therefore one of the pairs we are considering. Therefore,

∗ (12) ∀ α h(α) < π(βf )(α) < g(α, N) = h(α) which is a contradiction. 

23 ω1 Let F ⊆ ω1 be some club such that ∀hh, fi ∈ [F ]c βf < γh. By renaming we may

ω1 assume that F = ω1. Consider some f ∈ [ω1]c . Define hf by hf (α) = sup f(β) + 1. hhf , fi β<α ∗ is one of the pairs we are considering, so βf < γhf . But ∀f ∀ α hf (α) = α + 1. Therefore,

if we define h by h(α) = α + 1, then βf < γh for every f. But this says that the βf ’s are

bounded and therefore there are only ω1 of them. Since we know that Sel(ω1, ω1) is true, we are done. This concludes the consideration of functions of the correct type. We will now consider functions of the other type. The argument actually follows exactly the same as before, with one exception: In the last claim the function h must be defined differently. The argument goes as follows: Fix a function f of the other type and let

g be a function that witnesses the fact that f has uniform cofinality α. Since βf < γf ,

∗ (13) ∀ α ∃β < α π(βf )(α) < g(α, β) > sup f(γ) γ<α

Define a function k : ω1 → ω1 by k(α) = the least β < α such that π(βf )(α) < g(α, β) > sup f(γ) (to be fair, this definition won’t make sense for all α, but it will for γ<α ∗ of them). Since µω is a normal measure, ∃γ < ω1 ∀ α k(α) = γ. We then define h by

h(α) = g(α, γ) and the rest of the proof continues as before. 

Corollary 2. Sel(ω1, ωω)

Proof. All the cardinals from ω3 until ωω are singular. 

24 BIBLIOGRAPHY

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[6 ]Donald Martin and John Steel, The Extent of Scales in L(R), Games, Scales, and Suslin Cardinals (Lecture Notes in Logic, Volume 31), pp. 110-120, Cambridge University Press, New York, 2008. [7 ]Alexander Kechris, Classical Descriptive Set Theory, Springer-Verlag, New York, 1995. [8 ]Adrian Mathias, Happy families, Annals of , Volume 12 (1977), no. 1, pp. 59-111. [9 ]Alexander Kechris, AD and projective ordinals, Wadge Degrees and Projective Or- dinals (Lecture Notes in Logic, Volume 37), pp. 304-345, Cambridge University Press, Cambridge, 2012.

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