sign in sign up
<<
Home , Analytical hierarchy, Analytic set, Computable ordinal, Pointclass

Math 117c notes: Effective DST

Forte Shinko May 27, 2019

1 Overview of effective DST

Effective descriptive theory is a field at the intersection of and descriptive . It can often help us prove results in which are not easily handled with the classical machinery. First of all, recall the on N:

0 0 0 Σ1 Σ2 Σ3 0 0 0 ∆1 ∆2 ∆3 ··· 0 0 0 Π1 Π2 Π3

The insight of effective DST is relate this hierarchy to the :

0 0 0 Σ1 Σ2 Σ3 0 0 0 ∆1 ∆2 ∆3 ··· 0 0 0 Π1 Π2 Π3

These are defined as follows (on any topological space):

0 1. A set is Σ1 if it is open. 0 0 2. A set is Πn if its complement is Σn. 0 0 3. A set is Σn+1 if it is the union of countably many Πn sets. 0 0 0 4. A set is ∆n if it is both Σn and Πn. Here is an outline of the corresponding notions between the fields:

Computability theory Effective DST Classical DST N N := Nω arithmetical hierarchy first ω levels of Borel hierarchy 0 0 0 0 0 0 Σα, Πα, ∆α Σα, Πα, ∆α set effectively open set hyperarithmetical set computable ordinal countable ordinal CK ω1 ω1 1 1 1 1 1 1 Σn, Πn, ∆n Σn, Πn, ∆n

1 2 Effective open sets

ω <ω Recall that the Baire space N := N has a basic open set Ns for every s ∈ N , defined by

Ns := {x ∈ N : x s}

The basic open sets in N are of the form {n} for n ∈ N. We will work on spaces of the form X = Nk × N l. One reason is that we’d like to recover as a special case the usual computability theory on N. A better reason is that our of X should be thought of as sets definable in second order arithmetic.

Definition 1. Let X = Nk × N l. 0 k <ω l 1. A subset A ⊂ X is Σ1 (or effectively open) if there is a c.e. B ⊂ N × (N ) such that [ A = {n1} × · · · × {nk} × Ns1 × · · · Nsl

(n1,...,nk,s1,...,sl)∈T

(ie. A is an “effective union” of basic open sets).

0 0 2. A subset A ⊂ X is Π1 (or effectively closed) if its complement is Σ1. 0 Recall that a Σ1 subset of a is defined to be an open subset, so the notation reflects the fact that this is its effective analogue. Remark 1. 1. There are only countably many effectively open sets, since there are only countably many c.e. sets. So there are many open sets which are not effectively open.

2. A subset of N is effectively open iff it is c.e. 0 S <ω 3. If X = N , then A ⊂ N is Σ1 iff A = s∈B Ns for some c.e. B ⊂ N . We can choose B to be closed under extensions (ie. if s ∈ B and s ≺ t then t ∈ B), and it is an exercise to show that we can choose 0 B to be computable. So the complement of B is computable tree, and thus a subset of N is Π1 iff A = [T ] for some computable tree on N (where [T ] is the set of infinite branches of T ). k l 0 k+l Proposition 1. Let X = N × N and let A ⊂ X. Then A is Σ1 iff there is a computable relation R ⊂ N such that A(x1, . . . , xk, n1, . . . , nk) ⇐⇒ ∃mR(x1  m, . . . , xk  m, n1, . . . , nk) S <ω Proof. We’ll do the case X = N . Let A = s∈B Ns for some c.e. B ⊂ N . Let B = im f for some computable f : N → N<ω (this doesn’t cover when B is empty, but this case is trivial). Then we have

A(x) ⇐⇒ ∃s ∈ B[s ≺ x] ⇐⇒ ∃n[f(n) ≺ x] ⇐⇒ ∃m∃n[f(n) = x  m] The converse is immediate. The analog of a continuous function is a computable function.

Definition 2. Let X = Nk × N l. 1. A function f : X → N is computable if the relation f(x) = n is effectively open, ie. the set {(x, n) ∈ X × N : f(x) = n} is effectively open.

2. A function f : X → N is computable if the relation f(x) ∈ Ns is effectively open, ie. the set <ω {(x, s) ∈ X × N : f(x) ∈ Ns} is effectively open. 3. A function X → Nm × N n is computable if each of its projections is computable. Remark 2. In the case of a function f : N → N, this agrees with the usual notion of a computable function.

2 Proposition 2.

0 1. Σ1 is closed under the following: finite union, finite intersection, bounded quantification, projection along N (ie. ∃N), projection along N (ie. ∃N ), computable preimage (ie. preimage under a computable function).

0 2. Π1 is closed under the following: finite union, finite intersection, bounded quantification, coprojection along N (ie. ∀N), coprojection along N (ie. ∀N ), computable preimage. One should think of projection as “effective union”. 0 A useful tool is that of a universal Σ1 set. k l 0 0 Proposition 3. Let X = N × N . Then there is a universal Σ1 set for X, that is, a Σ1 subset U ⊂ N × X 0 such that for every Σ1 subset A ⊂ X, there is some n ∈ N such that Un = A.

Proof. We will do the case X = N . Fix a computable enumeration (We)e of the c.e. sets, then let U(e, x) := ∃nWe(x  n). We can relativize all of the above notions to an oracle x ∈ N .

k l 0 0 Definition 3. Let X = N × N and let x ∈ N . A subset A ⊂ X is Σ1(x) if there is a Σ1(x) subset B ⊂ Nk × (N<ω)l such that [ A = {n1} × · · · × {nk} × Ns1 × · · · Nsl

(n1,...,nk,s1,...,sl)∈B The importance of relativization is that it relates the effective notion of an open set to the classical one.

Proposition 4. Let A ⊂ Nk × N l. Then TFAE: 0 1. A is Σ1 (ie. open). 0 2. A is Σ1(x) for some x ∈ N . 0 S <ω 0 Proof. We will do the case X = N . If A is Σ1, then A = s∈B Ns for some B ⊂ N . Then A is Σ1(B). This allows us to prove theorems in descriptive set theory (boldface pointclasses) by using the effective theory (lightface pointclasses).

3 The arithmetical hierarchy

We now define the rest of the arithmetical hierarchy.

0 0 0 k l Definition 4. The pointclasses Σn, Πn, ∆n for n ≥ 1 for spaces of the form N × N are defined as follows: 0 1. A set is Σ1 if it is effectively open. 0 0 2. A set is Πn if its complement is Σn. 0 0 3. A set is Σn+1 if it is the projection along N of a Πn set. 0 0 0 4. A set is ∆n if it is both Σn and Πn. A set in one of these pointclasses is called an . The arithmetical hierarchy looks like the following:

0 0 0 Σ1 Σ2 Σn 0 0 0 ∆1 ∆2 ··· ∆n ··· 0 0 0 Π1 Π2 Πn

3 Proposition 5. Let X = Nk × N l.

0 N 1. Σn is closed under bounded quantification, ∃ -quantification, and computable preimage.

0 N 2. Πn is closed under bounded quantification, ∀ -quantification and computable preimage. 0 3. ∆n is closed under bounded quantification and computable preimage. 0 CK Remark 3. We can define Σα sets for α < ω1 by taking “effective unions” at limit stages, giving what is known as the hyperarithmetical hierarchy. Again, we can relativize to any oracle, and this lets us relate hyperarithmetical sets to Borel sets:

0 0 0 Proposition 6. Let Γ be one of Σα, Πα, ∆α, and let Γ be the corresponding boldface . Then TFAE for a subset A of Nk × N l: 1. A is Γ. 2. A is Γ(x) for some x ∈ N .

4 The analytical hierarchy

1 1 1 k l Definition 5. The pointclasses Σn, Πn, ∆n for n ≥ 1 for spaces of the form N × N are defined as follows: 1 0 1. A set is Σ1 if it is the projection along N of a Π1 set. 1 1 2. A set is Πn if its complement is Σn. 1 1 3. A set is Σn+1 if it is the projection along N of a Πn set. 1 1 1 4. A set is ∆n if it is both Σn and Πn. A set in one these pointclasses is called an analytical set.

1 Remark 4. “Analytical set” should not be confused with “”, which means Σ1. The analytical hierarchy extends the hyperarithmetical hierarchy:

0 0 0 1 1 1 Σ1 Σ2 Σα Σ1 Σ2 Σn 0 0 0 1 1 1 ∆1 ∆2 ··· ∆α ··· ∆1 ∆2 ··· ∆n ··· 0 0 0 1 1 1 Π1 Π2 Πα Π1 Π2 Πn

Proposition 7. Let X = Nk × N l.

1 N N N 1. Σn is closed under ∃ , ∀ , ∃ , computable preimage.

1 N N N 2. Πn is closed under ∃ , ∀ , ∀ , computable preimage.

1 N N 3. ∆n is closed under ∃ , ∀ , computable preimage. 1 1 We saw above that every arithmetical set is ∆1. More generally, every hyperarithmetical set is ∆1, and the converse is also true.

1 Theorem 1 (Suslin-Kleene). A set is hyperarithmetical iff it is ∆1. As always, we can relativize these notions to relate them to the boldface ones.

1 1 1 Proposition 8. Let Γ be one of Σn, Πn, ∆n, and let Γ be the corresponding boldface pointclass. Then TFAE for a subset A of Nk × N l: 1. A is Γ. 2. A is Γ(x) for some x ∈ N .

4 1 5 Structural properties of the pointclass Π1

1 Proposition 9. There is a universal Π1 set. k l 0 Proof. Fix X = N × N . Let U ⊂ N × N × X be a universal Π1 set for N × X (take the complement of a 0 N 1 1 universal Σ1 set). Then ∃ U ⊂ N × X is a universal Σ1 set for X, and its complement is universal Π1. 1 Next we introduce two very important Π1 sets. Definition 6.

<ω <ω 1. Let WF ⊂ NN be the set of codes of well-founded trees; that is, x ∈ WF iff Tx := {s ∈ N : x(s) = 1} is a well-founded tree (ie. it is a tree with no infinite branches).

× 2. Let WO ⊂ NN N be the set of codes of well-orders; that is, x ∈ WO iff the relation ≤x defined by

n ≤x m ⇐⇒ x(n, m) = 1 is a well-order.

1 Proposition 10. WF and WO are Π1. Proof. Note that x ∈ WF iff the conjunction of the following formulas holds:

1. ∀s, t[(s ∈ Tx ∧ t ≺ s) → t ∈ Tx](Tx is a tree).

N 2. ¬∃ y∀n[y  n ∈ Tx](Tx has no infinite branches). 1 Thus WF is Π1. Now for WO, note that x ∈ WO iff the conjunction of the following formulas holds:

1. ∀n, m[(n ≤x m ∧ m ≤x n) → n = m](≤x is antisymmetric).

2. ∀n, m, p[(n ≤x m ∧ m ≤x p) → n ≤x p](≤x is transitive).

3. ∀n, m[n ≤x m ∨ m ≤x n](≤x is a linear order).

N 4. ∀ y∃n[y(n) ≤x y(n + 1)] (≤x is a well-order). 1 Thus WO is Π1. k l 1 1 1 Definition 7. Let X = N × N . A set A ⊂ X is Π1-complete if it is Π1 and if for any Π1 set B ⊂ Y (for any Y ), there is a computable function f : Y → X such that f −1(A) = B (we say f reduces B to A).

1 1 AΠ1-complete set should be thought of as “the hardest Π1 set”. 1 Theorem 2. WF and WO are Π1-complete. 1 1 Proof. First we show that WF is Π1-complete. Let B ⊂ N be Π1 (it’s not hard to show that we only need to 1 consider Π1 subsets of N ). Fix a computable tree T on N × N such that x∈ / B ⇐⇒ ∃y((x, y) ∈ [T ]). For every x ∈ N , let T (x) = {(s, t) ∈ T : s ≺ x}. Then T (x) is a subtree of T which is well-founded iff x ∈ B. So the map x 7→ T (x) reduces B to WF (we should really turn T (x) into a code). 1 Next we show that WO is Π1-complete. It suffices to show that WF reduces to WO. T Recall the Kleene-Brouwer ordering: given a tree T on N, there is a linear order

k l 1 Definition 8. Let X = N × N and let A ⊂ X.AΠ1-rank on A is a function φ : A → ORD such that the ∗ ∗ 1 relations x <φ y and x ≤φ y defined below are Π1:

∗ x <φ y ⇐⇒ x ∈ A ∧ (y ∈ A → φ(x) < φ(y)) ∗ x ≤φ y ⇐⇒ x ∈ A ∧ (y ∈ A → φ(x) ≤ φ(y))

5 ∗ One can think of ≤φ as follows: extend φ to all of X by setting φ(x) = ∞ for x∈ / A, and take the “prewellorder” on X induced by φ, except where ∞ cannot be compared to itself.

1 1 1 Theorem 3. Π1 is ranked; that is, every Π1 set admits a Π1-rank. 1 1 Proof. Since WO is Π1-complete, it suffices to find a Π1-rank on WO. We claim that the map φ : WO → ORD 1 sending x to the of ≤x is a Π1-rank. By definition, we have

∗ x <φ y ⇐⇒ x ∈ WO ∧ (y ∈ WO → ≤y does not embed in ≤x)

1 1 Note that “≤y does not embed in ≤x” is Π1, since it can be expressed via the following Π1 formula:

N ¬∃ z∀n, m[n

1 1 However, the entire formula above is not Π1, since “y ∈ WO” is Π1. To fix this, we have the following equivalent fact: ∗ x <φ y ⇐⇒ x ∈ WO ∧ (y ∈ LO → ≤y does not embed in ≤x) 0 where LO is the set of codes for linear orders, which is a Π1 set. This holds since if a linear order embeds ∗ 1 into a well-order, then it is a fortiori also a well-order. Thus x <φ y is Π1. ∗ For x ≤φ y, we just need to add an extra condition:

∗ x ≤φ y ⇐⇒ x ∈ WO ∧ ((y ∈ LO → ≤y does not embed in ≤x)

∨ (y ∈ WO ∧ ≤y does not embed into a proper initial segment of ≤x))

1 Similarly to above, we can express “≤y does not embed into a proper initial segment of ≤x” via a Π1 formula:

N ¬∃ z∃k∀n, m[n

∗ 1 Thus x ≤φ y is Π1.

1 1 1 ∗ Theorem 4. Π1 has the reduction property; that is, for any Π1 sets A, B ⊂ X, there are Π1 sets A ⊂ A and B∗ ⊂ B such that A∗ ∪ B∗ = A ∪ B and A∗ ∩ B∗ = ∅. 1 1 1 Proof. Let A, B ⊂ X be Π1. Define the Π1 set C := A × {0} ∪ B × {1} ⊂ X × N. SinceΠ 1 is ranked, there 1 ∗ ∗ is a Π1-rank φ : C → ORD. Then we can define A and B as follows:

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

6 Effective theorem

Recall the classical perfect set theorem from descriptive set theory:

1 Theorem 5 (Souslin). Let A ⊂ N be Σ1. Then exactly one of the following holds: 1. A is countable. 2. A contains a perfect set. To state the effective version, we need a definition.

1 1 Definition 9. Let x ∈ N . We say that x is ∆1 (or x is a hyperarithmetical real) if its graph is ∆1. We denote by HYP the set of hyperarithmetical reals. Proposition 11. Let x ∈ N . Then TFAE:

1 1. x is ∆1.

6 1 1 2. x is Σ1 (ie. its graph is Σ1). 1 1 3. x is Π1 (ie. its graph is Π1). 1 1 4. {x} is ∆1 (ie. x is a ∆1 ). 1 1 5. {x} is Σ1 (ie. x is a Σ1 singleton). 1 1 1 Proof. The following shows that x is Σ1 iff x is Π1 iff x is ∆1: x(n) = m ⇐⇒ ∀k[x(n) = k → k = m]

1 1 If x is ∆1, then the following shows that {x} is ∆1: y ∈ {x} ⇐⇒ ∀n, m[y(n) = m ↔ x(n) = m]

1 1 If {x} is Σ1, then the following shows that x is Σ1: x(n) = m ⇐⇒ ∃N y[y ∈ {x} ∧ y(n) = m]

The effective perfect set theorem is the following statement:

1 Theorem 6 (Harrison, 1967). Let A ⊂ N be Σ1. Then exactly one of the following holds: 1. A ⊂ HYP. 2. A contains a perfect set. Remark 5. Note that HYP is countable, so the effective perfect set theorem implies the classical one, but with a much stronger conclusion. Before we prove the effective perfect set theorem, we need to establish some preliminary facts. 1 1 First of all, since we’re dealing with the set of ∆1 reals, we’ll need something like a universal ∆1 set. Such an object cannot literally exist (by the usual diagonalization argument), but the following will be sufficient:

k l 1 1 Proposition 12. Let X = N × N . Then there are Π1 sets U, V ⊂ N × X and a Π1 set C ⊂ N such that:

1. For every n ∈ C, we have Un = X \ Vn.

1 2. For every ∆1 subset A ⊂ X, there is some n ∈ C such that A = Un = X \ Vn. 1 1 Proof. Let W ⊂ N × X be a universal Π1 set for X. Then we can define Π1 subsets W0,W1 of N × N × X as follows:

W0(n0, n1, x) ⇐⇒ W (n0, x)

W1(n0, n1, x) ⇐⇒ W (n1, x)

1 1 Since Π1 has the reduction property, there are Π1 sets U ⊂ W0 and V ⊂ W1 such that W0 ∪ W1 = U ∪ V and U ∩ V = ∅. Now define C ⊂ N × N as follows:

(n0, n1) ∈ C ⇐⇒ Wn0 ∪ Wn1 = X N ⇐⇒ ∀ x[W (n0, x) ∨ W (n1, x)]

1 We can see from this definition that C is Π1. 1 Corollary 1. HYP is Π1. 2 1 2 Proof. Set X = N in ?? to obtain Π1 sets U, V ⊂ N × N and C ⊂ N. Then x ∈ HYP iff the following holds:

∃n ∈ C∀k∀l[(x(k) = l → Un(k, l)) ∧ (x(k) 6= l → Vn(k, l))]

1 So HYP is Π1.

7 We now turn to the proof of the effective perfect set theorem: Proof of ??. Suppose that A 6⊂ HYP. We must show that A contains a perfect set. Note that A \ HYP is 1 also Σ1 by ??, so by replacing A with A \ HYP, we can assume that A is nonempty and disjoint from HYP. 1 Since A is Σ1, there is a computable tree T on N × N such that A = p[T ], where p is the projection p(x, y) = x. In other words: x ∈ A ⇐⇒ ∃y((x, y) ∈ [T ]) Let T ∗ be the pruning of T :

(s, t) ∈ T ∗ ⇐⇒ ∃(x, y) ∈ [T ]((s, t) ≺ (x, y))

∗ ∗ We claim that for any (s, t) ∈ T , there are extensions (s0, t0), (s1, t1) ∈ T of (s, t) such that s0 ⊥ s1. In 1 other words, we claim that p[T(s,t)] is not a singleton. Note that by ??, every Σ1 singleton is contained in 1 HYP. So p[T(s,t)] cannot be a singleton, since it is Σ1 and disjoint from HYP. ∗ <ω Thus we can construct a Cantor scheme (su, tu)u∈2<ω ⊂ T such that su0 ⊥ su1 for every u ∈ 2 . This ω gives an embedding from 2 → N defined by x 7→ limn sxn.

8