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Structural Dichotomy Theorems in Descriptive Set Theory

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Structural Dichotomy Theorems in Descriptive Set Theory Structural dichotomy theorems in descriptive set theory Benjamin Miller Universit¨atWien Winter Semester, 2019 Introduction The goal of these notes is to provide a succinct introduction to the primary structural dichotomy theorems of descriptive set theory. The only prerequisites are a rudimentary knowledge of point-set topology and set theory. Working in the base theory ZF + DC, we first discuss trees, the corresponding representations of closed, Borel, and Souslin sets, and Baire category. We then consider consequences of the open dihypergraph dichotomy and variants of the G0 dichotomy. While pri- marily focused upon Borel structures, we also note that minimal modi- fications of our arguments can be combined with well-known structural consequences of determinacy (which we take as a black box) to yield generalizations into the projective hierarchy and beyond. iii Contents Introduction iii Chapter 1. Preliminaries 1 1. Closed sets 1 2. Ranks 2 3. Borel sets 3 4. Souslin sets 4 5. Baire category 9 6. Canonical objects 13 Chapter 2. The box-open dihypergraph dichotomy 27 1. Colorings of box-open dihypergraphs 27 2. Partial compactifications 31 3. Separation by unions of closed hyperrectangles 34 Chapter 3. The G0 dichotomy, I: Abstract colorings 39 1. Colorings within cliques 39 2. Discrete perfect sets within cliques 42 3. Scrambled sets 45 Chapter 4. The G0 dichotomy, II: Borel colorings 49 1. Borel colorings 49 2. Index two subequivalence relations 52 3. Perfect antichains 54 4. Parametrization and uniformization 56 Chapter 5. The (G0; H0) dichotomy 61 1. Borel local colorings 61 2. Linearizability of quasi-orders 67 Bibliography 71 Index 73 v CHAPTER 1 Preliminaries 1. Closed sets Given a set I, define I<N = S In and I≤N = I<N [ IN. The n2N length of a sequence t 2 I≤N is given by jtj = n if t 2 In, and jtj = 1 if t 2 IN. Given sequences s; t 2 I≤N, we say that s is an initial segment of t, or t is an extension of s, if jsj ≤ jtj and s = t jsj, in which case we write s v t. In the special case that s 6= t, we say that s is a proper initial segment of t, or t is a proper extension of s, in which case we < write s @ t.A tree on I is a set T ⊆ I N that is closed under initial segments, in the sense that 8t 2 T 8n < jtj t n 2 T .A subtree of T is a tree S ⊆ T on I.A branch through T is a sequence x 2 IN such that 8n 2 N x n 2 T . We use [T ] to denote the set of all branches through T , and we say that T is well-founded if [T ] = ;. Suppose now that I is a discrete topological space. For each se- <N N quence s 2 I , let Ns denote the set of extensions of s in I . These sets form a basis for the product topology on IN. Proposition 1.1.1. Suppose that I is discrete space and T is a tree on I. Then [T ] is closed. Proof. Observe that if x 2 [T ], then Nxn \ [T ] 6= ; for all n 2 N, so x n 2 T for all n 2 N, thus x 2 [T ]. N Given a set X ⊆ I , we use TX to denote the set of proper initial segments of elements of X. Proposition 1.1.2. Suppose that I is a discrete space and X ⊆ IN. Then X = [TX ]. Proof. Clearly X ⊆ [TX ], so X ⊆ [TX ] by Proposition 1.1.1. Con- versely, if x 2 [TX ], then x n 2 TX for all n 2 N, so Nxn \ X 6= ; for all n 2 N, thus x 2 X. We use (i) to denote the singleton sequence given by s(0) = i. The < concatenation of sequences s; t 2 I N is the extension s a t of s given by (s a t)(jsj + n) = t(n) for all n < jtj. 1 2 1. PRELIMINARIES Proposition 1.1.3. Suppose that I is a well-orderable discrete space and C ⊆ IN is a non-empty closed set. Then there is a func- tion β : TC ! C with the property that 8t 2 TC t v β(t). Proof. Fix a well-ordering of I, and define ι: TC ! I by letting ι(t) be the -minimal element of I for which t a (ι(t)) 2 TC . Define n 0 n+1 n n β : TC ! TC by β (t) = t and β (t) = β (t) a ((ι ◦ β )(t)), and set β(t) = S βn(t). n2N A retraction from a set X onto a subset Y is a surjection φ: X Y whose restriction to Y is the identity. Proposition 1.1.4. Suppose that I is a well-orderable discrete space and C ⊆ IN is a non-empty closed set. Then there is a con- tinuous retraction φ: IN C. Proof. Proposition 1.1.2 ensures that for all sequences x 2 ∼C, there is a maximal proper initial segment ι(x) of x in TC , and Proposi- tion 1.1.3 yields a function β : TC ! C such that 8t 2 TC t v β(t). Let φ: IN C be the retraction agreeing with β ◦ ι off of C. To see that φ is continuous, it is enough to show that if n 2 N and x 2 IN, then φ(Nxn) ⊆ Nφ(x)n. But if x n 2 TC then φ(Nxn) ⊆ Nxn = Nφ(x)n, and if x n2 = TC then φ(Nxn) = fφ(x)g ⊆ Nφ(x)n. Now that we have explicitly proven and applied a particular in- stance of the axiom of choice, it should be noted that the axiom of determinacy rules out simply assuming the latter: Theorem 1.1.5 (Solovay). Suppose that AD holds. Then there is N no injective !1-sequence of elements of N . 2. Ranks Suppose that R is a binary relation on X. For all Y ⊆ X, define 0 (0) (α+1) (α) 0 YR = fy 2 Y j 9x 2 Y x R yg, YR = Y , YR = (YR )R for all (λ) T (α) ordinals α, and YR = α<λ YR for all limit ordinals λ. The rank of (ρ(R)) (ρ(R)+1) R is the least ordinal ρ(R) for which XR = XR . 0 The relation R is well-founded if Y 6= YR for all non-empty sets Y ⊆ X. By DC, this is equivalent to the inexistence of a sequence x 2 XN with the property that 8n 2 N x(n + 1) R x(n). Proposition 1.2.1. A binary relation R on a set X is well-founded (ρ(R)) if and only if XR = ;. (ρ(R)) Proof. It is clear that if R is well-founded, then XR = ;. Conversely, if there is a non-empty set Y ⊆ X for which Y = Y 0, then (ρ(R)) a straightforward transfinite induction shows that Y ⊆ XR . 3. BOREL SETS 3 The rank of a point x 2 X with respect to R is the largest ordinal (ρR(x)) ρR(x) for which x 2 XR , or 1 if no such ordinal exists. We adopt the conventions that 1 = 1 + 1 and α < 1 for all ordinals α. Proposition 1.2.2. Suppose that R is a binary relation on a set X. Then 8x 2 X ρR(x) = supfρR(w) + 1 j w R xg. (α) Proof. Note that if α is an ordinal, w R x, and w; x 2 XR , then (α+1) x 2 XR , so ρR(x) ≥ ρR(w) + 1. But if α ≥ supfρR(w) + 1 j w R xg (α+1) is an ordinal, then x2 = XR , so ρR(x) ≤ α. The horizontal sections of a set R ⊆ X × Y are the sets of the form Ry = fx 2 X j x R yg, where y 2 Y . The vertical sections are the sets of the form Rx = fy 2 Y j x R yg, where x 2 X. Proposition 1.2.3. Suppose that X and Y are sets, R and S are binary relations on X and Y , and φ: X ! Y is a function. x φ(x) (1) If 8x 2 X φ(R ) ⊆ S , then 8x 2 X ρR(x) ≤ ρS(φ(x)). φ(x) x (2) If 8x 2 XS ⊆ φ(R ), then 8x 2 X ρR(x) ≥ ρS(φ(x)). Proof. To see (1), note that if α is an ordinal for which ρR(x) ≤ ρS(φ(x))) whenever ρR(x) < α, then Proposition 1.2.2 ensures that x ρR(x) = supfρR(w) + 1 j w 2 R g x ≤ supfρS(φ(w)) + 1 j w 2 R g ≤ ρS(φ(x)) (ρ(R)) whenever ρR(x) = α. Moreover, if ρR(x) = 1, then x 2 XR , and (ρ(R)) (ρ(S)) since φ(XR ) ⊆ YS , if follows that ρS(φ(x)) = 1. To see (2), note that if α is an ordinal for which ρR(x) ≥ ρS(φ(x))) whenever ρR(x) < α, then Proposition 1.2.2 ensures that x ρR(x) = supfρR(w) + 1 j w 2 R g x ≥ supfρS(φ(w)) + 1 j w 2 R g ≥ ρS(φ(x)) whenever ρR(x) = α. 3. Borel sets Suppose that κ is an ordinal. A family of sets is a κ-complete algebra if it is closed under complements and unions of length strictly less than κ.
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