Topics in Descriptive Set Theory (Ma 191)

Home , Descriptive set theory, Polish space, Prewellordering

Lecturer: Alexander Kechris, Notes by: Siddharth Prasad

February 21, 2017

1 Lecture 1 (1/5/17)

1.1 Basic Classical Notions Descriptive set theory is the study of “deﬁnable” sets and functions in well behaved spaces.

Deﬁnition 1.1. A Polish space is a topological space given by a complete, separable metric.

n Example. R, C, R , separable Hilbert spaces, separable Banach spaces. Deﬁnition 1.2. For a Polish space X we are interested in the so called Borel sets in X.

We denote by B∼ (X) the class of all Borel subsets of X. These consist of

0 Σ∼ 1(X) = open sets of X( sometimes denoted by G) 0 Π∼ 1(X) = closed sets of X( sometimes denoted by F ) 0 Σ∼ 2(X) = countable unions of closed sets (Fσ) 0 0 Π∼ 2(X) = complements of Σ∼ 2(X) sets = countable intersections of open sets(Gδ) 0 Σ∼ 3(X) = Gδσ sets 0 Π∼ 3(X) = Fσδ sets and so on. This notion can be extended into the transﬁnite as such: for a countable ordinal ξ > 1 deﬁne

0 0 Σ∼ ξ(X) = {∪nAn : An ∈ Π∼ ξn (X), ξn < ξ} 0 0 Π∼ ξ(X) =∼ Σ∼ ξ(X) 0 0 0 ∆∼ ξ(X) = Σ∼ ξ(X) ∩ Π∼ ξ(X)

1 0 0 0 We have B∼ (X) = ∪ξ<ω1 Σ∼ ξ(X) = ∪ξ<ω1 Π∼ ξ(X) = ∪ξ<ω1 ∆∼ ξ(X), which forms the so called Borel hierarchy.

1 Deﬁnition 1.3. We deﬁne the analytic sets (A) Σ∼ 1(X) = {projX (A): A ⊆ X×Y,A Borel}. 1 1 Their complements are the co-analytic sets (CA) Π∼ 1(X) =∼ Σ∼ 1(X). PCA sets are 1 1 1 1 Σ∼ 2(X) = {projX (A): A ⊆ X ×Y,A is Π∼ 1(X ×Y )}, the CPCA sets are Π∼ 2(X) =∼ Σ∼ 2(X).

This hierarchy of sets forms the projective sets or projective hierarchy, denoted by P∼ (X). 1 Remark. A result of Suslin shows that B∼ (X) = ∆∼ 1(X). Deﬁnition 1.4. A function f : X → Y is a Borel (measurable) function if for each open Borel set U ⊆ Y , f −1(U) is borel in X.

Effective descriptive set theory: Introduces concepts and methods of computability theory, is a refinement of the classical theory. We introduce a class of so called “transfer theorems”, that allow us to work primarily in the so called Baire space N = NN (endowed with product topology). A metric on this space of infinite sequences is given by ( 0 if α = β d(α, β) = 1 2n+1 where n is the least number such that α(n) 6= β(n)

The Cantor space is the subset C = 2N ⊆ N . Elements of Baire space are sometimes ∼ referred to irrationals since N = R \ Q. Theorem 1.1 (Transfer theorems). N and C denote Baire space and Cantor space respectively.

• If K is a compact metrizable space, there exists a continuous surjection f : C K.

• If X is a Polish space, there exists a continuous surjection f : N X . Moreover, there exists a closed F ⊆ N and there exists a continuous bijection f : F X.

• If X is a perfect Polish space, there exists a continuous bijection f : N X. • X and Y are Polish spaces with |X| = |Y | ⇐⇒ there exists a Borel bijection f : X Y . 1 Example. Analytic sets satisfy the perfect set property, i.e. if A ⊂ X is Σ∼ 1, either |A| ≤ ℵ0 (A is countable) or there is a continuous bijection f : C A.

2 1.2 Review of Computability Theory k Main objects of study are partial computable (recursive) k-ary functions f : N → N. Recall we can code TMs by numbers, {e}k(x1, . . . , xk) = the function computed by the TM with code e. k ϕ (e, x¯) = {e}k(¯x) is a universal recursive function.

m Theorem 1.2 (S-m-n). There is a total recursive function Sn such that {e}m+n(¯x, y¯) = m {Sn (e, x¯)}(¯y). Theorem 1.3 (Recursion theorem). For each computable function f, there is e such that {e}(¯x) = f(e, x¯).

k m k ∼ ` A function f : N → N is computable if its projections are computable. So N =c N m are “computably isomorphic”. We have the notion of a coding function hm, ni2 = 2 · 2 (2n + 1) − 1 which gives a bijection h·, ·i : N N. Put hm, n, ki3 = hhm, ni2, ki2. By the nature of the coding function we have that m1, . . . , mn << hm1, . . . , mnin. We ﬁnally extend this to a bijective coding function on N given by hi = 0, hmi = h0, m + 1i2, and hm0, . . . , mk−1i = hk − 1, hm0, . . . , mk−1iki2. k Recall that a set A ⊆ N is computable ⇐⇒ the characteristic function χA is computable. Recall the following equivalent characterizations for recursively enumerable sets.

A ⊆ N is recursively enumerable ⇐⇒ A = ∅ or A = f(N), f total computable ⇐⇒ A = f(N) is computable ⇐⇒ A = dom(f), f computable 2 ⇐⇒ A = proj(B),B ⊆ N ,B computable.

k The relation W (e, x¯) ⇐⇒ {e}k(¯x) ↓ is universal of r.e. subsets of N . Given R r.e., there is e such that W (e, x¯) ⇐⇒ R(e, x¯).

Proposition 1.4. A function f is recursive ⇐⇒ graph(f) is r.e.

Proposition 1.5. A set A is recursive ⇐⇒ A and ∼ A are r.e.

1.3 Eﬀective Topology We will be primarily working with product spaces over the Baire space N and the naturals k N. Let s = (s0, . . . , sk−1) ∈ N . The neighborhoods Ns = {α ∈ N : s ⊆ α} are the so called basic open sets of the Baire space. Then, open sets are of the form ∪i∈NNsi for some sequence (si). The basic open sets for N are singleton sets {n}, and likewise the basic open sets in the n m product space N × N are sets of the form Ns1 × · · · × Nsn × {n1} × · · · × {nm}.

3 n m 0 Definition 1.5. A ⊆ N ×N is effectively open or Σ1 iff A = ∅ or there are total recursive functions f1, . . . , fn, g1, . . . , gm such that [ A = Nf1(k) × · · · × Nfn(k) × {g1(k)} · · · × {gm(k)}. k

0 Remark. Note that there are thus only countably many Σ1 sets. n m 0 Proposition 1.6. A ⊆ N × N is Σ1 if and only if there exists a recursive relation n+m R ⊆ N such that

(α1, . . . , αn, n1, . . . , nm) ∈ A ⇐⇒ ∃kR(¯α1(k),..., α¯n(k), n1, . . . , nm), where α¯(k) = hα ki = hα(0), . . . , α(k − 1)i.

2 Lecture 2 (1/10/17)

We prove the proposition from last time:

n m 0 Proposition 2.1. A ⊆ N × N is Σ1 if and only if there exists a recursive relation n+m R ⊆ N such that

(α1, . . . , αn, n1, . . . , nm) ∈ A ⇐⇒ ∃kR(¯α1(k),..., α¯n(k), n1, . . . , nm).

Proof. For simplicity we will take n = m = 1, so A ⊆ N × N. For the forward direction, if 0 A is Σ1 we can write A = ∪pNf(p) × {g(p)} for total recursive functions f and g. Then we have

(α, n) ∈ A ⇐⇒ ∃p(f(p) ⊆ α&n = g(p))

⇐⇒ ∃p∃q(f(p) ⊆ α q&n = g(p))

⇐⇒ ∃k(f((k)0) ⊆ α &n = g((k)0)) (k)1 ∃kR(¯α(k), n) where R is clearly recursive. < For the other direction, suppose (α, n) ∈ A ⇐⇒ ∃kR(¯α(k), n), where R ⊆ N N × N is recursive. Then, there are recursive functions f, g such that R = {(f(p), g(p)) : p ∈ N}, so we can write A = ∪pNf(p) × {g(p)}.

< Definition 2.1. A Tree on N is a set of finite sequences T ⊆ N N that is closed under the subsequence relation, i.e. if t ∈ T and s ⊆ t, then s ∈ T . The set of all infinite branches of

T is called the body of T and is denoted by [T ], i.e. [T ] = {α ∈ N : ∀n(α n ∈ T )}. 0 Remark. {A ⊆ N : A ∈ Π∼ 1} = {[T ]: T a tree on N}.

4 Proof. See page 138 of [1].

0 Proposition 2.2. A ⊆ N is Π1 ⇐⇒ ∃T (T is recursive &A = [T ]). 0 Proof. For the forward direction assume A ⊆ N is Π1. Then, there exists a recursive R such that α ∈ A ⇐⇒ ∀nR(¯α(n)). Then, take T to be the tree such that s ∈ T ⇐⇒ R(hsi). Then A = [T ] and we are done. For the reverse implication take R = hT i. Then, α ∈ A ⇐⇒ ∀nR(¯α(n)).

m m < Deﬁnition 2.2. Trees on N : A tree T on N is a set of (s1, . . . , sm), si ∈ N N, lh(s1) = ··· = lh(sm), such that if (t1, . . . , tm) ⊆ (s1, . . . , sm) (meaning ti ⊆ si∀i), then (t1, . . . , tm) ∈ T .

m 0 So the eﬀective version of the remark is as follows: A ⊆ N is Π1 ⇐⇒ ∃ a recursive m tree T on N such that A = [T ].

0 N N Proposition 2.3. Σ1 is closed under ∧, ∨, ∃ , ∃ , ∀m ≤ n. N 0 Proof. We will prove closure under ∃ . Suppose P ⊆ N × N × N is Σ1. Then, we can write P (α, β, n) ⇐⇒ ∃tR(¯α(t), β¯(t), n), for some recursive R. We have

∃αP (α, β, n) ⇐⇒ ∃α∃tR(¯α(t), β¯(t), n) ⇐⇒ ∃k∃s(R(s, β¯(k), n)&|s| = k) ⇐⇒ ∃uR0(β¯(u), n), where R0 is recursive.

n m Deﬁnition 2.3. Let X = N × N . A function f : X → N is eﬀectively continuous or recursive if the relation Gf ⊆ X × N given by

−1 Gf (x, s) ⇐⇒ f(x) ∈ Ns ⇐⇒ x ∈ f (Ns)

0 is Σ1. A function f : X → N is eﬀectively continuous/recursive if Gf (x, i) ⇐⇒ f(x) = i 0 n is Σ1. If Y = N × N, then f : X → Y is recursive if its components are.

Example. The map (α, k) 7→ α¯(k) from N × N → N is recursive.

Example. The map (α, β) 7→ hα, βi = γ and its inverse maps h(γ)0, (γ)1i = γ are recursive. This gives a recursive isomorphism between N 2 and N . Example. The map (α, n) 7→ (n, α(0), α(1),...) is recursive and also gives a recursive isomorphism N × N ↔ N . 0 −1 0 Proposition 2.4. If A ∈ Σ1 and f is recursive, then f (A) is Σ1.

5 Proof. We prove it for functions f : X → N , A ⊆ N . Set Q(x) ⇐⇒ P (f(x)), so x ∈ Q ⇐⇒ f(x) ∈ P . We have

f(x) ∈ P ⇐⇒ ∃n(f(x) ∈ Nsn ), where the map n 7→ sn is recursive 0 ⇐⇒ ∃nGf (x, sn), where Gf is Σ1 ⇐⇒ ∃n∃m(Gf (x, m) ∧ m = sn)

Proposition 2.5. If f, g are recursive, so is f ◦ g.

X 0 Proposition 2.6. For each space X, there is a set W = W ⊆ N × X such that W ∈ Σ1 0 and {We : e ∈ N} = Σ1(X). W is called a universal set. Proof. Let X = N × N, A ⊆ X. Then, we can write (α, n) ∈ A ⇐⇒ ∃kR(¯α(k), n) for some recursive R. Now, let U be a universal r.e. set, so

(α, n) ∈ A ⇐⇒ ∃kU(e, α¯(k), n), for some e ∈ N. Taking W (e, α, n) ⇐⇒ ∃kU(e, α¯(k), n) gives the desired universal relation.

0 0 0 Remark. In particular note that Σ1 is not closed under complements, so Σ1 6= Π1. This is shown by a diagonal argument.

2.1 Combinatorial description of recursive functions f : N → N < < Definition 2.4. A function ϕ : N N → N N is monotone if s ⊆ t =⇒ ϕ(s) ⊆ ϕ(t). ∗ ∗ Define a partial function ϕ : N → N as follows: ϕ (α) ↓ ⇐⇒ |ϕ(α n)| → ∞, and then ∗ put ϕ (α) = ∪nϕ(α n) ∈ N . Proposition 2.7. f : N → N is recursive iff f = ϕ∗ for some recursive ϕ. Proof. Suppose f = ϕ∗. Then,

Gf (α, s) ⇐⇒ f(α) ∈ Ns ⇐⇒ ∃k(s ⊆ ϕ(α k)) ⇐⇒ ∃k∀i ≤ |s|(s(i) = ϕ(α k)(i)). For the opposite direction, write f(α)(i) = j ⇐⇒ ∃nR(¯α(n), i, j), R recursive. Put

T (s, i) ⇐⇒ R(s , i, (|s|)1), clearly recursive. Finally set U(n) = (n)1, which is (|s|)0 recursive. Then, f(α)(i) = U(µ`T (¯α(`), i)) (this is a sort of analog of the Kleene normal form for N). To deﬁne ϕ(s) put |ϕ(s)| = Mk ≤ |s|(∀i < k)(∃` ≤ |s|)T (¯s(`), i),

6 and set ϕ(s)(i) = U(µ` ≤ |s|T (¯s(`), i)). Then ϕ is recursive and ϕ∗ = f, and moreover ϕ is monotone.

As with the boldface Borel classes, we can define an effective Borel hierarchy. For 0 0 0 0 0 0 0 example, we take ∆1 = Σ1 ∩ Π1,Σ2 = {∃nP (n, x): P ∈ Π1},Π2 =∼ Σ2 = {∀nQ(n, x): 0 Q ∈ Σ1}, and so on. These are called the arithmetical classes. 0 Remark. Note that f : N → N is recursive iff f : N × N → N is recursive, where f 0(α, n) = f(α)(n).

3 Lecture 3 (1/12/17)

1 We have the effective analog of the projective sets: Σ1(X) = {∃αP (x, α): P ⊆ X ×N ,P ∈ 0 1 1 1 1 1 1 Π1},Π1(X) =∼ Σ1(X), Σn+1 sets are the projections of the Πn sets, Πn+1 =∼ Σn+1, and 1 1 1 ∆n = Σn ∩ Πn. These sets are also called the analytical sets. 1 Analogously with the result for arithmetical sets, we have A ⊆ N is Σ1 iff there is a recursive tree T such that A = p[T ], where T is a tree on N × N (A is the projection of T along its first coordinate).

1 Proposition 3.1. Σn sets are closed under ∧, ∨, ∀m, ∃m, ∃α, and recursive pre-images 1 (continuous substitution). Πn sets are similarly closed under the same operations, except it is closed under ∀α instead of ∃α.

1 0 Proof. We write out the (easy) proofs for ∀m and ∃α. If P is Σ1, there is a Q ∈ Π1 such that P (x, n) ⇐⇒ ∃αQ(x, n, α). Then we have

∀nP (x, n) ⇐⇒ ∀n∃αQ(x, n, α)

⇐⇒ ∃α∀nQ(x, n, (α)n)

1 0 Now let P (x, α) ∈ Σ1, so P (x, α) ⇐⇒ ∃βQ(x, α, β), Q ∈ Π1. We have

∃αP (x, α) ⇐⇒ ∃α∃βQ(x, α, β)

⇐⇒ ∃δQ(x, (δ)0, (δ)1)

1 1 Proposition 3.2. Σ1 and Πn have universal sets.

7 3.1 Relativization k Deﬁnition 3.1. Let α ∈ N . A partial function f : N → N is recursive in α if it is in the smallest class of functions that contains the basic functions (erase, successor, projections) and α and is closed under composition, primitive recursion, and minimalization.

Equivalent deﬁnition using register machines (RMs) with oracles: we simply add an instruction of the following form:

Replace the content of R1, say n, with α(n).

Note that this is strictly a formal deﬁnition and in particular does not have to do with what α actually is.

Deﬁnition 3.2. A function is recursive relative to α iﬀ it can be computed by a RM using oracle α.

We can deﬁne all the similar computability notions in a relativized fashion (r.e. in α, etc.), and all the main results for computable functions also relativize. m Remark. Note that the in the S −m−n theorem, however, the Sn functions are absolutely recursive. α Analagous to the usual notation, {e} (x1, . . . , xn) denotes the recursive in α function that is computed by the RM with code e.

3.2 Normal Form Theorem

Deﬁne T (e, s, x¯) ⇐⇒ the RM with code e terminates on inputx ¯ in at most (|s|)0 steps using as an oracle s and the output is (|s|)1.T is recursive, and if U(s) = (s)1, then

{e}α(¯x) = U(µnT (e, α¯(n), x¯)).

Note that U and T do not depend on α. α α k Thus f (e, x¯) = {e} (¯x) is universal of recursive in α functions. So A ⊆ N is r.e. in α iﬀ for some e, x¯ ∈ A ⇐⇒ {e}α(¯x) ↓ ⇐⇒ ∃nT (e, α¯(n), x¯). Put W α(e, x¯) ⇐⇒ ∃nT (e, α¯(n), x¯). Then this is universal for r.e. in α relations. 0 0 0 0 Note that we can relativize the arithmetical sets Σn, Πn to Σn(α), Πn(α) by considering computable in α unions of basic neighborhoods.

0 0 Proposition 3.3. Let A ⊆ X. Then A is Σ1(α) iﬀ ∃B ⊆ N × X, B ∈ Σ1 such that x ∈ A ⇐⇒ (α, x) ∈ B.

8 Proof. The converse is easy: take f(x) = α, then f is recursive in α, and x ∈ A ⇐⇒ 0 (f(x), x) ∈ B. For the forward direction, let X = N be Σ1(α). Then

x ∈ A ⇐⇒ ∃nR(¯x(n)) ⇐⇒ ∃n∃mS(¯x(n), α¯(m)) ⇐⇒ ∃kS0(¯x(k), α¯(k)) ⇐⇒ (α, x) ∈ B

0 0 where R is recursive in α, S, S are recursive, and B is Σ1. 0 0 0 0 1 1 Corollary 3.4. Σ∼ 1 = ∪αΣ1(α), similar for Σ∼ n, Π∼ n, Σ∼ n, Π∼ n. 0 0 0 Proof. Any Σ1(α) set is open, so ∪αΣ1(α) ⊆ Σ∼ 1. Conversely, let X = N , and suppose 0 A ∈ Σ∼ 1. Then either A = ∅ (in which case we are done), or A = ∪nNsn . Then, taking 0 α = (n 7→ sn) we get A ∈ Σ1(α). 0 0 Corollary 3.5. There exists a Σ1 set W ⊆ N × X which is universal for Σ∼ 1. Similar for 0 0 1 1 Σn, Πn, Σn, Πn. 0 0 Proof. If A ⊆ X is Σ∼ 1, ∃α and ∃B ⊆ N × X, B ∈ Σ1 such that x ∈ A ⇐⇒ (α, x) ∈ B. 0 0 Let W (e, α, x) be a universal Σ1 set for Σ1(N × X). Then, x ∈ A ⇐⇒ W (e, α, x) for 0 some e, so W is universal for Σ∼ 1(X). 1 General Note: Suppose we want to show every A ∈ Σ∼ 1 has some property. Then, it 1 1 suﬃces to show that every A ∈ Σ1 has the property, since then every A ∈ Σ1(α) has the 1 1 property for all α, so every A ∈ Σ∼ 1 = ∪αΣ1(α) has the property.

3.3 Eﬀective Version of the Perfect Set Theorem 1 Theorem 3.6 (Suslin). The Perfect Set theorem for Σ∼ 1 sets. If A ⊆ X where X is a 1 Polish space is Σ∼ 1, then either |A| ≤ ℵ0, or else contains a Cantor set. It is enough to prove the theorem for X = N by the transfer theorems.

1 2 1 Deﬁnition 3.3. We say α ∈ N is ∆1 iﬀ graph(α) ⊆ N is ∆1. Proposition 3.7. The following are equivalent:

1 1 1. α is ∆1 (i.e. graph(α) ∈ ∆1) 1 2. graph(α) ∈ Σ1 1 3. graph(α) ∈ Π1 1 4. {α} is ∆1

9 1 1 5. {α} is Σ1 (but not {α} ∈ Π1) 1 Proof. (2 =⇒ 3): If graph(α) is Σ1, then

0 0 0 1 α(n) = m ⇐⇒ ∀m (α(n) = m =⇒ m = m ) ∈ Π1. (3 =⇒ 2): Similar. 1 (1 =⇒ 4): If α is ∆1, β ∈ {α} ⇐⇒ β = α 1 ⇐⇒ ∀n, m(β(n) = m ⇐⇒ α(n) = m) ∈ ∆1

1 (4 =⇒ 1): If {α} is ∆1, then

1 α(n) = m ⇐⇒ ∃β(β ∈ {α} ∧ β(n) = m) ∈ Σ1,

1 so graph(α) is ∆1. 1 (5 =⇒ 4): If {α} is Σ1, then

1 β ∈ {α} ⇐⇒ ∀γ(γ ∈ {α} =⇒ γ = β) ∈ Π1

1 so {α} is ∆1. 1 1 Deﬁnition 3.4. D1 = {α ∈ N : α is ∆1} 1 Theorem 3.8 (Harrison 1967). Eﬀective version of perfect set theorem. If A ⊆ N is Σ1 then exactly one of the following holds:

1 1. A ⊆ D1. 2. A contains a Cantor set.

1 1 Note: D1 = ∪{A ⊆ Σ1 : |A| ≤ ℵ0}. 1 1 Proposition 3.9. D1 ∈ Π1

4 Lecture 4 (1/17/17)

We restate the eﬀective version of the perfect set theorem due to Harrison in 1967 from last time:

1 Theorem 4.1. If A ⊆ N is Σ1, then exactly one of the following holds: 1 1. A ⊆ D1. 2. A contains a Cantor set.

10 1 1 1 where we recall that D1 = {α ∈ N : α is ∆1 (i.e. graph(α) ∈ ∆1)}. 1 1 We will prove this using the proposition that D1 ∈ Π1, which we shall prove later. 1 1 1 We look at B = A \D1, which is a Σ1 set such that B ∩ D1 = ∅. So it is enough to prove:

1 1 Proposition 4.2. If A ∈ Σ1 and A ∩ D1 = ∅, then A contains a Cantor set. 1 1 Proof. Let A ∈ Σ1 be such that A ∩ D1. Then, A = p[T ] is the projection of a recursive 2 tree T on N . So α ∈ A ⇐⇒ ∃β((α, β) ∈ [T ]). We define the “pruned” version of T , T ∗ as T ∗ = {(s, t) ∈ T : ∃α, β(s ⊆ α ∧ t ⊆ β ∧ (α, β) ∈ [T ]}. The first two conditions are 1 ∗ 1 recursive relations, and the third is a Π0 relation, so T ∈ Σ1. Now there are two cases we must consider. ∗ ∗ Case 1: ∀(s, t) ∈ T , ∃(s0, t0), (s1, t1) ∈ T ((s, t) ⊆ (s0, t0) ∧ (s, t) ⊆ (s1, t1) ∧ s0 ⊥ s1), where s0 ⊥ s1 means that the two branches are incompatible, i.e. they differ after some finite index. homeomorphism, so f(2N) is a Cantor space. ∗ ∗ Case 2: ∃(s, t) ∈ T such that ∀(s0, t0), (s1, t1) ∈ T ((s, t) ⊆ (s0, t0)∧(s, t) ⊆ (s1, t1) =⇒ ∗ ∗ s0, s1 are compatible). So if we call the tree below (s, t) by Ts,t, then A ⊇ p[Ts,t] = {α}. 1 1 Then {α} is Σ1, so α ∈ D1, which is a contradiction. Thus we have proved the effective perfect set theorem, modulo the proposition that 1 1 D1 ∈ Π1.

4.1 Brief excursion - some analysis 1 1 A real x ∈ R is ∆1 if {q ∈ Q : q < x} is ∆1 (bijected with N). Consider f : [0, 1] → R. f is analytically expressible if it can be expressed by a formula involving elementary functions x P∞ (polynomials, sin, cos, e , etc.) and inﬁnite series i=0. 0 R 1 If f is a derivative (∃ diﬀerentiable F : [0, 1] → R such that f = F ) we write 0 f = F (1) − F (0).

R 1 1 1 Theorem 4.3. { 0 f : f is analytically expressible} = D1(R) = {x ∈ R : x is ∆1}.

4.2 Back to Eﬀective PST 1 1 1 Proposition 4.4. For each X, there is a Π1 set C ⊆ N, a Π1 set P ⊆ N × X, and a Σ1 1 set S ⊆ N × X such that n ∈ C =⇒ Pn = Sn =: Dn, and {Dn : n ∈ C} = ∆1(X). So this 1 is a result that gives us a coding of ∆1 sets. Using this, we can prove the proposition:

11 1 1 Proposition 4.5. D1 ∈ Π1. 2 Proof. Take X = N and C,P,S as in the previous proposition. Then,

1 1 α ∈ D1 ⇐⇒ ∃n(n ∈ C ∧ ∀k, `(α(k) = ` ⇐⇒ (k, `) ∈ Dn)) ∈ Π1.

In order to prove the coding result, we need the following result:

1 1 Proposition 4.6 (Reduction theorem for Π1 sets). If A, B ⊆ X are Π1, then there exist ∗ ∗ ∗ ∗ 1 ∗ ∗ ∗ ∗ A ⊆ A, B ⊆ B, A ,B ∈ Π1 such that A ∩ B = ∅ and A ∪ B = A ∪ B. Using this we can prove the coding result (which we restate for convenience):

1 1 1 Proposition 4.7. For each X, there is a Π1 set C ⊆ N, a Π1 set P ⊆ N × X, and a Σ1 1 set S ⊆ N × X such that n ∈ C =⇒ Pn = Sn =: Dn, and {Dn : n ∈ C} = ∆1(X). So this 1 is a result that gives us a coding of ∆1 sets. 1 Proof. Let W ⊆ N × X be universal Π1. Let (n, x) ∈ W0 ⇐⇒ ((n)0, x) ∈ W ,(n, x) ∈ 1 W1 ⇐⇒ ((n)1, x) ∈ W , W0,W1 ∈ Π1.(W0,W1) is called a universal pair in the sense 1 that if A, B ⊆ X are Π1, then ∃n such that A = (W0)n and B = (W1)n (A and B are sections of W0 and W1 respectively wrt the same index in N). Use the reduction theorem ∗ ∗ ∗ ∗ 1 ∗ to get W0 and W1 . Let C = {n :(W0 )n ∪ (W1 )n = X}, so C ∈ Π1. Then set P = W0 ∗ and S =∼ W1 . As a corollary to the reduction theorem, we have:

1 1 Corollary 4.8 (Separation theorem for Σ1 sets). If A, B ⊆ X are Σ1 and A ∩ B = ∅, then 1 there is a ∆1 set C such that A ⊆ C and C ∩ B = ∅. Proof. To separate A and B, reduce their complements ∼ A and ∼ B.

Proposition 4.9. Reduction and separation cannot hold at the same time.

1 Proof. We prove it for Π1 sets. Take a universal pair (W0,W1) and reduce them to get ∗ ∗ 1 (W0 ,W1 ). Then, separating them gives a universal ∆1 set, contradiction.

4.3 Review of well-orderings

Well orderings (X, ≤)-every subset of X which is nonempty has a least element, e.g. (N, ≤). Linear preordering - a ≤ b, b ≤ a does not necessarily mean a = b. Prewellordering - the least elements could be sets of items, each maps into a least element of a well ordering. Ordinals - canonical representation of well orderings up to isomorphism (N, ≤) with the usual ordering is assigned the ordinal ω,(N, ≤) with the ordering where all the even numbers are listed before the odd numbers is assigned the ordinal ω + ω.

12 1 Deﬁnition 4.1. A class Γ of sets (e.g. Π1) is called ranked or has the prewellordering property if for every A ∈ Γ there is a map ϕ : A → ORD and relation≤Γ, ≤Γˆ such that ≤Γ∈ Γ, ≤Γˆ∈ Γ,ˆ and

ˆ y ∈ A ⇐⇒ (x ∈ A and ϕ(x) ≤ ϕ(y) ⇐⇒ x ≤Γ y ⇐⇒ x ≤Γ y), where we assume ϕ(x) = ∞ if x∈ / A (Γˆ denotes the dual of Γ). Such a ϕ is called a Γ-norm.

1 Theorem 4.10. Π1 is ranked (with ordinal < ω1). 1 Given this, we show the reduction theorem for Π1 sets. 1 Proof. Let A, B ⊆ X be Π1. Let C ⊆ X × N be deﬁned by C = A × {0} ∪ B × {1}. Then 1 1 C ∈ Π1, so let ϕ : C → ORD be a Π1-norm. Then, take

x ∈ A∗ ⇐⇒ ϕ(x, 0) ≤ ϕ(x, 1) and x ∈ A x ∈ B∗ ⇐⇒ ϕ(x, 1) ≤ ϕ(x, 0) and x ∈ B.

0 0 Note that for Σn, we have that A ∈ Σn can be written as x ∈ A ⇐⇒ ∃m(x, m) ∈ B, 0 where B ∈ Πn−1, so here we would simply take ϕ(x) = µm(x, m) ∈ B.

5 Lecture 5 (1/19/17)

Let A ⊆ X, ϕ : A → ORD be a norm. A class of sets Γ is ranked if for each A ∈ Γ there is ϕ : A → ORD and ≤Γ, ≤Γˆ in Γ, Γˆ respectively such that

ˆ y ∈ A ⇐⇒ [(x ∈ A ∧ ϕ(x) ≤ ϕ(y)) ⇐⇒ x ≤Γ y ⇐⇒ x ≤Γ y], and ϕ is called a Γ-rank. Fact. We have the following equivalent definition (for most Γ): Given a rank ϕ : A → ORD ∗ define x ≤ϕ y ⇐⇒ x ∈ A ∧ ϕ(x) ≤ ϕ(y) (where we let ϕ(y) = ∞ if y∈ / A) and ∗ ∗ ∗ x <ϕ y ⇐⇒ x ∈ A ∧ ϕ(x) < ϕ(y). Then ϕ is a Γ-rank iff ≤ϕ, <ϕ∈ Γ. Proof. We state without proof the formulas that verify the equivalent definition. For the Γ ∗ Γˆ ∗ converse we have x ≤ y ⇐⇒ x ≤ϕ y and x ≤ y ⇐⇒ ¬(y <ϕ x). For the forward ∗ Γ Γˆ ∗ direction, we have x ≤ϕ y ⇐⇒ x ∈ A ∧ (x ≤ y ∨ ¬y ≤ x) and x <ϕ y ⇐⇒ x ∈ A ∧ ¬y ≤Γˆ x.

Proposition 5.1. Γ ranked =⇒ Γ has the reduction property.

13 Proof. Take A, B ⊆ X, A, B ∈ Γ. Let C ∈ X × N be deﬁned as C = A × {0} ∪ B × {1}. ∗ ∗ C is in Γ, so let ϕ : C → ORD be a Γ-rank. Then put x ∈ A ⇐⇒ (x, 0) ≤ϕ (x, 1) and ∗ ∗ x ∈ B ⇐⇒ (x, 1) <ϕ (x, 0). 1 We will now prove that Π1 is ranked. First, we need to deﬁne the Kleene-Browuer ordering on trees.

S Deﬁnition 5.1. Let S be a tree on N. The Kleene-Brouwer ordering on S

Remark.

S Fact. S is well-founded iﬀ KB s1 >KB s2 >KB ··· is an inﬁnite descending chain. Note that if u, v 6= ∅ and u

1 Theorem 5.2. Π1 is ranked. 1 1 Proof. Let A ⊆ N , suppose A ∈ Π1. So ∼ A ∈ Σ1 =⇒ ∼ A = p[T ], T a recursive tree on

LO = {w ∈ N :

14 1 So now it is enough to show that WO admits a Π1-rank. If for a wellordering W on N we let |W | be the unique ordinal isomorphic to W , let ϕ : WO → ω1 be deﬁned by Π1 Σ1 ϕ(w) = |

Σ1 w ≤ 1 v ⇐⇒ w, v ∈ LO&∃f : N → N such that f embeds

Π1 w ≤ 1 v ⇐⇒ w, v ∈ WO&¬∃f : N → N such that f embeds

1 From this we can easily prove that Σ2 is ranked. 1 Theorem 5.3. Σ2 is ranked. 1 1 Proof. Let A ⊆ N , A ∈ Σ2. Then, there is a Π1 set B such that α ∈ A ⇐⇒ ∃β(α, β) ∈ B. 1 Let ϕ : B → ORD be a Π1-rank on B. Then deﬁne ψ : A → ORD by ψ(α) = min{ϕ(α, β): (α, β) ∈ B}. Then, we take

∗ 0 0 ∗ 0 0 1 α ≤ψ α ⇐⇒ ∃β∀β ((α, β) ≤ϕ (α , β )) ∈ Π1, and ∗ 0 0 ∗ 0 0 1 α <ψ α ⇐⇒ ∃β∀β ((α, β) <ϕ (α , β )) ∈ Π1.

6 Lecture 6 (1/24/17)

1 6.1 Another proof that Π1 is ranked 1 We now discuss another method of showing that Π1 is ranked using the idea of games. We 0 will use the fact that Σn sets are ranked. 1 Theorem 6.1. Π1 sets are ranked. 1 2 0 Proof. Suppose A ⊆ N is Π1, so ∃B ⊆ N , B ∈ Σ1 such that x ∈ A ⇐⇒ ∀α(x, α) ∈ B. 0 Let ϕ : B → ORD be a Σ1-rank on B. For x, y ∈ N , we deﬁne the two player game Gx,y as such: player I chooses some number α(0), in response to which player II chooses some number β(0). Player I then chooses α(1), in response to which player II chooses β(0). If α and β are the elements of Baire space players I and II eventually obtain, we say player ∗ II wins means (x, α) ≤ϕ (y, β) (if x, y ∈ A, this means ϕ(x, α) ≤ ϕ(x, β)). This winning condition is open, so the game is determined by the Gale-Stewart theorem, which states that open games are determined.

15 Deﬁne on the set A the relation x y ⇐⇒ II has a winning strategy in Gx,y. We claim that is a prewellordering on A. That x x is obvious (II just copies II’s moves). Transitivity: x y, y z =⇒ x z. Here, basically player II in Gx,z will simulate player Is moves in Gx,y and Gy,z, and based on II’s winning strategies in those games it is trivial to come up with one for Gx,z. Every pair of elements in A is comparable: either x y or y x for all x, y ∈ A. If x y fails, then by determinacy, I has a winning strategy in Gx,y. Here basically player II copies what player I does in Gx,y in Gy,x by copying player I’s moves in Gy,x in Gx,y. Finally we must show there is no inﬁnite descending chain xo x1 x1 · · · in A. Note that x y ⇐⇒ I has a winning strategy in Gx,y. The proof of this is easy: If x y, then x y, so I has a winning strategy by determinacy. Conversely if I has a winning strategy in Gx,y, then as above x y, and also II has no winning strategy in Gx,y, i.e. x y, so x y.

Then basically what we do is look at all the games Gx0,x1 ,Gx1,x2 , and so on. Suppose I plays αk(0) in Gxk,xk+1 . Then player II copies that and plays αk(0) in the game Gxk−1,xk .

Player I responds in game Gxk,xk+1 with some αk(1), and again player II copies that and plays αk(1) in game Gxk−1,xk , and so on. So ﬁnally we would have come up with sequences α0, α1, α1, α2, α2, α3,..., such that ϕ(x0, α0) > ϕ(x1, α1) > ϕ(x2, α2) > ··· , which is a contradiction, since we cannot have an inﬁnite descending chain of ordinals.

We now prove the Gale-Stewart theorem:

Proof. The winning condition here is that I wins iﬀ (α, β) ∈ A, where A is a (classically) open set. Suppose I has no winning strategy in this game. Say the position p is good if I has no winning strategy from then on. Basic observation: if p = (a0, b0, . . . , an, bn) is ∧ good, then ∀an+1∃bn+1p (an+1, bn+1) is good (note that ∅ is a good position). Here is a strategy for II: II plays so that at any stage, p = (a0, b0, . . . , an, bn) is good. Claim: this is winning for II.

6.2 Further applications of ranks 1 Theorem 6.2 (Kreisel). Number uniformization theorem. If P ⊆ X × N, P ∈ Π1, then 1 ∗ ∗ there is a Π1 subset P ⊆ P such that ∃nP (x, n) ⇐⇒ ∃!nP (x, n). Furthermore if 1 1 ∀x∃nP (x, n), then ∃F : X → N such that F is ∆1 (i.e. graph(F ) ∈ ∆1) and ∀x(x, F (x)) ∈ P .

1 Proof. Let ϕ : P → ORD be a Π1-rank. Then, put

∗ ∗ ∗ P (x, n) ⇐⇒ P (x, n) ∧ ∀m((x, n) ≤ϕ (x, m)) ∧ ∀m < n((x, n) <ϕ (x, m)).

This clearly works. Assume now ∀x∃nP (x, n) holds. Then put F (x) = n ⇐⇒ P ∗(x, n) is 1 1 Π1 and F (x) 6= n ⇐⇒ ∃m(F (x) = m ∧ m 6= n), so graph(F ) is ∆1.

16 By relativizing, we get as corollaries the following boldface results:

1 ∗ Corollary 6.3 (Generalized reduction). Let (An), An ∈ Π∼ 1 be given. Then ∃An ⊆ An, ∗ 1 ∗ ∗ ∗ An ∈ Π∼ 1,An ∩ Am = ∅ if n 6= m and ∪nAn = ∪nAn. 1 1 ∗ 1 Proof. Put P (x, n) ⇐⇒ x ∈ An, so P ∈ Π∼ 1. So P ∈ Π1(α) for some α. Let P ∈ Π1(α) ∗ ∗ uniformize P . Then put x ∈ An ⇐⇒ P (x, n). 1 Corollary 6.4 (Generalized Separation). If (An), An ∈ Σ∼ 1 is given and ∩nAn = ∅, then 1 ∃Bn ⊇ An,Bn ∈ ∆∼ 1 such that ∩nBn = ∅.

6.3 Boundedness Theorems We now discuss a class of so-called “boundedness” theorems.

1 1 Theorem 6.5 (Boundedness Theorem). Let A ∈ Π1 \ ∆1 and let ϕ : A → ORD be a 1 1 Π1-rank. Let B ⊆ A, B ∈ Σ1. Then B is bounded, i.e. ∃x0 ∈ A∀x ∈ B(ϕ(x) ≤ ϕ(x0)). Proof. If not, then 1 Σ1 x ∈ A ⇐⇒ ∃y(y ∈ B ∧ x ≤ϕ y), 1 which is Σ1. 1 Corollary 6.6. If A ⊆ WO is Σ1, then |A| = sup{|α| : α ∈ A} < ω1. 1 1 1 Theorem 6.7. D1 ∈ Π1 \ Σ1. 1 1 1 1 1 Proof. Suppose D1 ∈ ∆1. Put D = {A ⊆ N : A ∈ ∆1} ∈ ∆1. Recall the coding of ∆1 1 1 1 2 1 subsets of N. We have a Π1 \ Σ1 subset C ⊆ N, a Π1 subset P ⊆ N and a Σ1 subset 2 S ⊆ N such that n ∈ C =⇒ Pn = Sn =: Dn and {Dn : n ∈ C} = D. We can write this as ∀A(A ∈ D =⇒ ∃n(n ∈ C ∧ A = Dn)), or equivalently ∀A∃n(A ∈ D =⇒ (n ∈ 1 C ∧ A = Dn)). By number uniformization, let F : D → N be a ∆1 function such that 1 ∀A ∈ D(F (A) ∈ C ∧A = DF (A)). Note that F (D) ⊆ C and F (D) is Σ1, so if ϕ : C → ORD 1 is a Π1-rank, ∃k0 ∈ C such that ∀A ∈ D(ϕ(F (A)) ≤ ϕ(k0)) by boundedness. Then, put

(k, n) ∈ U ⇐⇒ [k ∈ C ∧ ϕ(k) ≤ ϕ(k0)] ∧ n ∈ Dk.

1 Then U is a universal ∆1 function, contradiction. CK Deﬁnition 6.1. ω1 = sup{|α| : α ∈ W O, α is recursive} is called the Church-Kleene Ordinal.

1 Theorem 6.8 (Eﬀective version of boundedness theorem). If A ⊆ WO, A ∈ Σ1, then CK sup{|α| : α ∈ A} < ω1 .

17 7 Lecture 7 (1/25/17)

We prove the eﬀective boundedness theorem from last time:

1 Theorem 7.1 (Eﬀective version of boundedness theorem). If B ⊆ WO, B ∈ Σ1, then CK sup{|α| : α ∈ B} < ω1 . 1 1 1 Proof. Let A ⊆ N,A ∈ Π1 \ Σ1. WO is Π1-complete, so let f : N → N be recursive such CK that n ∈ A ⇐⇒ f(n) ∈ WO. If, towards a contradiction, sup{|α| : α ∈ B} ≥ ω1 , then 1 Σ1 1 n ∈ A ⇐⇒ ∃α(α ∈ B ∧ f(n) ≤|·| α). Then A is Σ1, contradiction.

CK 1 Corollary 7.2. ω1 = sup{|α| : α ∈ W O, α ∈ ∆1}. Deﬁnition 7.1. Let A ⊆ X. ϕ : A → ORD is called regular if ϕ(A) is an initial segment of the ordinals. For any rank ψ : A → ORD, there is a regular ϕ : A → ORD such that ≤ψ=≤ϕ.

1 1 1 Corollary 7.3. If A ⊆ N, A ∈ Π1 \ Σ1, and ϕ : A → ORD is a regular Π1-rank, then CK CK ϕ(A) = {ξ : ξ < ω1 }, or equivalently |ϕ| = ω1 . 1 CK Proof. Since any ϕ-initial segment is ∆1, this shows that |ϕ| ≤ ω1 . If, for contradiction, CK |ϕ| < ω1 , let |α0| = |ϕ|, where α0 is a recursive ordinal. Then we have

∗ ∗ n ∈ A ⇐⇒ ∃f : N → N∃k ∈ N∀a, b ∈ N(a <ϕ b <ϕ n =⇒ f(a) <α0 f(b) <α0 k),

1 which is a Σ1 relation, contradiction.

7.1 Review of Wellfounded relations We consider pairs (X, ≺), where ≺⊆ X2.

Definition 7.2. The relation ≺ is called well-founded if every non-empty subset Y ⊆ X has a minimal element (∃y ∈ Y ∀z ∈ Y (z ⊀ y)). Equivalently, there is no infinite descending chain · · · ≺ x2 ≺ x1 ≺ x0. Definition 7.3 (Principle of Induction). If Y ⊆ X is such that ∀y(y ≺ x =⇒ y ∈ Y ) =⇒ x ∈ Y , then Y = X.

Deﬁnition 7.4 (Principle of deﬁnition by recursion). Given a function g there is a unique function f such that f(x) = g(f {y:y≺x}, x).

Deﬁnition 7.5. Let (X, ≺) be a wellfounded relation. Then the map ρ≺ : X → ORD is deﬁned as ρ≺(x) = sup{ρ≺(y) + 1 : y ≺ x}, where ρ≺(x) = 0 if x is minimal. So ρ≺ maps X onto an initial segment of ORD. Finally, we set ρ(≺) := sup{ρ≺(x) + 1 : x ∈ X}.

18 Deﬁnition 7.6. If (X, ≺X ), (Y, ≺Y ) are two spaces equipped with wellfounded relations, we say a map f : X → Y is order-preserving if x ≺X y =⇒ f(x) ≺Y f(y).

So if

Definition 7.7. We define the rank of a well-founded tree T to be ρ(T ) = sup{ρT (s) + 1 : s ∈ T } = ρT (∅) + 1. Definition 7.8. Let S, T be trees on A, B respectively. ϕ : S → T is (strictly) monotone if s ≺ t =⇒ ϕ(s) ≺ ϕ(t).

Proposition 7.4. Let S, T be trees on A, B respectively. Then assuming T is well-founded, TFAE: (1) S is wellfounded and ρ(S) ≤ ρ(T ), and (2) there is a strictly monotone map f : S → T .

Proof. To be written later.

Let (X, ≺) be any relation, and map it to T≺ a tree on X such that (x0, . . . , xn−1) ∈ T ⇐⇒ x0 ≺ · · · ≺ xn−1. So clearly ≺ is wellfounded iﬀ T≺ is wellfounded. If ≺ is wellfounded, then ρ≺(x) = ρT≺ ((x0, . . . , xn−1, x) ). | {z } any sequence in T≺ ending in x

Corollary 7.5. ρ(≺) = ρT≺ (∅).

1 CK Theorem 7.6. If (X, ≺) is a Σ1 well-founded relatin, then ρ(≺) < ω1 . 1 α Remark. This boundedness can be relativized easily: If ≺ is Σ1 in α, then ρ(≺) < ω1 := 1 sup{|ω| : ω ∈ W O, ω recursive in α}. So if ≺ is Σ∼ 1, then ρ(≺) < ω1. The proof of the theorem is due to Kunen:

Proof. To be written later.

19 8 Lecture 8

1 1 Eﬀective boundedness theorem for Σ1 wellfounded relations: If ≺ is a well-founded Σ1 CK relation on X, then ρ(≺) < ω1 . 1 Corollary 8.1. If ≺ is a wellfounded Σ∼ 1 relation, then ρ(≺) < ω1. 1 1 1 CK Recall if A ⊆ N, A ∈ Π1 \ Σ1, ϕ : A → ORD is a regular Π1 rank, then |ϕ| = ω1 . 1 1 1 Corollary 8.2. If A ⊆ N , A ∈ Π∼ 1 \ Σ∼ 1, and ϕ : A → ORD is a regular Π∼ 1-rank, then |ϕ| = ω1.

1 8.1 Recursion theorem for Π1 relations 1 Fix U ⊆ N × (N × X), which is universal for Π1-subsets of N × X. Put W (n, x) ⇐⇒ 1 U((n)0, (n)1, x). W is universal for Π1 subsets of X, and it satisﬁes the recursion theorem, 1 i.e. given a Π1 set P ⊆ N × X, there is a e0 ∈ N such that We0 (x) ⇐⇒ P (e0, x).

Proof. Let e1 be such that U(e1, e, x) ⇐⇒ P (he, ei, x). Let e0 = he1, e1i, so

W (e0, x) ⇐⇒ U(e1, e1, x) ⇐⇒ P (he1, e1i, x) ⇐⇒ P (e0, x).

1 1 Theorem 8.3. Let ≺ be a Σ1 wellfounded relation on X, and let A ⊆ Y be Π1 complete (i.e. 1 for any Π1 relation P on z, there is a recursive function f such that z ∈ P ⇐⇒ f(z) ∈ A), 1 and let ϕ : A → ORD be a Π1 rank on A. Then, there is a recursive function f : X → Y such that f(X) ⊆ A, and x ≺ y =⇒ ϕ(f(x)) < ϕ(f(y)) (f is order preserving). Proof. Let W ⊆ N × X be as in the recursion theorem. Let g be a recursive function such that W (n, x) ⇐⇒ g(n, x) ∈ A. If there is n0 ∈ N such that W (n0, x) holds for all x, and x ≺ y =⇒ ϕ(g(n0, x)) < ϕ(g(n0, y), then f(x) = g(n0, x). Let n0 be such that ∗ W (n0, x) ⇐⇒ ∀y(y ≺ x =⇒ g(n0, y) <ϕ g(n0, x)).

It is enough to show that W (n0, x) holds for all x. If this fails, let x0 be minimal such that ∗ ¬W (n0, x0). So by deﬁnition of n0, there is x1 ≺ x0 such that g(n0, x1) <ϕ g(n0, x0) fails. ∗ But since x1 ≺ x0, W (n0, x1) holds and g(n0, x1) ∈ A. So g(n0, x1) <ϕ g(n0, x0) must hold, contradiction.

1 8.2 Perfect set property for Π1 sets 1 Deﬁnition 8.1. A ⊆ X is thin if it contains no Cantor set. For A ∈ Σ1, A is thin ⇐⇒ 1 A is countable (this is the perfect set theorem for Σ1 sets we proved). 1 Theorem 8.4. There exists a largest thin Π1 subset of N , denoted by C1. Proof. To be written later.

1 1 Deﬁnition 8.2. We deﬁne the notion of ∆ -reducibility. We write α ≤ 1 ⇐⇒ α ∈ ∆ (β). 1 ∆1 1

20 References

[1] Moschovakis, Y. (2006). Notes on set theory. Springer Science & Business Media.