Descriptive Set theory

by George Barmpalias

Institute for Language, Logic and Computation

Lectures 1–8

Last updated on the 10th of November 2009 Answer

I The study of sets of real numbers that can be described.

I . . . sets of reals that can be defined

I Definability theory

What is Descriptive set theory? What is Descriptive set theory?

Answer

I The study of sets of real numbers that can be described.

I . . . sets of reals that can be defined

I Definability theory Answer

I Analysis, topology

I Set theory

I Computability / recursion theory

Although the continuum is our primary concern, a lot of the results hold for more general spaces.

Also the study of the continuum often requires the study of more general spaces.

This is why descriptive set theory studies abstract spaces that are ‘similar’ to the continuum.

Polish spaces: complete, separable metric spaces with no isolated points (perfect).

What (meta) mathematics are involved in it? What (meta) mathematics are involved in it? Answer

I Analysis, topology

I Set theory

I Computability / recursion theory

Although the continuum is our primary concern, a lot of the results hold for more general spaces.

Also the study of the continuum often requires the study of more general spaces.

This is why descriptive set theory studies abstract spaces that are ‘similar’ to the continuum.

Polish spaces: complete, separable metric spaces with no isolated points (perfect). In early 1900s analysts Lebesgue, Suslin, Lusin, Borel (and others) applied his ideas to analysis. . .

. . . the classification and description of pointsets in classical analysis

It was then that Cantor’s ideas received general acceptance

Birth of Descriptive set theory

Cantor discovered higher infinities (late 1800s) in an attempt to solve a problem on zeros of a Fourier series.

This was not immediately appreciated. Birth of Descriptive set theory

Cantor discovered higher infinities (late 1800s) in an attempt to solve a problem on zeros of a Fourier series.

This was not immediately appreciated.

In early 1900s analysts Lebesgue, Suslin, Lusin, Borel (and others) applied his ideas to analysis. . .

. . . the classification and description of pointsets in classical analysis

It was then that Cantor’s ideas received general acceptance Fathers of Descriptive set theory (early 1900s) Timeline: complexity of the continuum

Pythagoreans discovered the irrational numbers.

Only in the 19th century the continuum was conceived as a complete ordered field.

The axiom of completeness supplied the concept of limits, intermediate value theorem, mean value theorem etc.

. . . essentially the ingredients for the success of the calculus in the 19th century.

This axiom also opened a whole Pandora’s box of set- theoretical difficulties in the 20th century. Cantor’s program

Cantor started studying perfect sets (closed sets without isolated points) in connection to a problem in Fourier series

Later he studied them in order to settle the continuum hypothesis.

He hoped to show that every uncountable set contains a perfect set. Cantor’s program

It was known that every perfect set has cardinality 2ℵ0 .

Also, every uncountable Borel set contains a perfect subset.

Lusin (1921) showed that under the axiom of choice, Cantor’s program fails. Non-uniqness of the real line

In the 1960s Cohen’s method of forcing showed a number of independence results illustrating the non-uniqness of the real line.

There are natural question about the reals whose answer depends on additional axioms in set theory.

These questions arise because the completeness axiom of the continuum forces us to introduce non-absolute concepts into analysis. . . like the concept of power sets.

It became a task of descriptive set theory to discover just where in the various hierarchies of complexity the disease occurs. Effective theory

Independently of the classical developments, logician Stephen Kleene studied the definability of sets of natural numbers in the 1940s and 1950s

This theory presented many similarities with the classical theory of definability of sets of reals.

Classical Effective

Reals integers continuous functions computable functions Borel sets Hyperarithmetical sets 1 analytic sets Σ1 sets projective sets analytical sets In this course we will study the two theories in parallel.

Among the topics will be:

I Definability hierarchies (Borel, Baire, Projective etc.)

I Separation theorems

I Games and determinacy

I Hyperarithmetic theory

I Equivalence relations

. . . with lots of examples, history and exercises. I Assessment: 80% assignments + 20% exam

I Methodology: mathematical

I Course webpage: http://www.barmpalias.net/Descr.html Bibliography

I Recursive aspects of descriptive set theory, by R. Mansfield and G. Weitkamp (Oxford logic guides)

I Descriptive set theory, by Y. Moschovakis (Studies in logic and the FOM)

I Classical descriptive set theory, by A. Kechris (Graduate texts in Mathematics)

I Descriptive set theory notes, by David Marker: http://www.math.uic.edu/ marker/math512/dst.pdf The Baire space N is the set of all sequences of natural numbers.

It has a natural topology generated by the basic open sets Nσ = {α | σ ⊂ α}.

These are clopen (both closed and open) sets in N .

There is a natural metric for this space:

d(α, β) = 1/the least position where they differ and d(α, β) = 0 if α = β. The Baire space is separable i.e. it has a countable dense subset.

Show that the Baire space is not compact.

The Cantor space is compact (see below). The Baire space N is homeomorphic to the irrationals via continued fractions.

A homeomorphism is a continuous function between two topological spaces with a continuous inverse function.

A tree in N is a set of finite sequences which is closed downward under initial segments.

Note: Every closed set in the Baire or Cantor space can be represented as the set of infinite paths through a tree. Theorem (König’s Lemma) In the Baire space, a finite branching tree has an infinite path iff it is infinite. Proof.

I If the tree is finite, obviously it does not have infinite paths.

I If it is infinite, we construct an infinite path by induction:

I Starting with the empty sequence, inductively assume that β  n is defined and has the property that there are infinitely many strings on the tree which extend it.

I Since the tree is finite branching, there is m ∈ N such that there are infinitely many strings on the tree which extend (β  n) ∗ m. I Pick such n and let β  n + 1 = (β  n) ∗ m. I By induction, β is an infinite path in the tree. Corollary The Cantor space C is compact. Proof.

I Let {σi } be an open cover of the Cantor space (assume ∀i σi 6= ∅).

I Consider the set of binary strings τ such that ∀i, σi 6⊆ τ.

I This set of strings is a tree T . . . without infinite paths

I Therefore it is finite and it has finitely many ‘leafs’ (maximal nodes).

I the immediate successors of those ‘leafs’ are also finitely many and belong to the original cover of C.

I The set of their immediate successors form the required finite sub-cover of the space.

I It covers C: any point in its complement would have all of its initial segments prefixed by some σi . Corollary The clopen sets in the Cantor space are exactly the finite unions of the basic open sets Nσ. Proof.

I Each Nσ is both open and closed (easy to show).

I Therefore their finite unions are both open and closed.

I For the other direction, let A be a clopen set.

I Then A = ∪i Nσi and C − A = ∪i Nτi for suitable families of strings.

I Hence C = (∪i Nσi ) ∪ (∪i Nτi ) and the union of these families is a cover of C.

I By compactness C = (∪i

I Obviously ∪i

I But A ⊆ ∪i

α ∈ ∪i

Show that there is a clopen set in the Baire space which is not a finite union of basic open sets. The language we use to talk about N and ω has

I Variables for integers

I Variables for reals (Greek letters)

I constants for natural numbers

I +, ·, exp

I function applications e.g. α(x + 3)

I equalities, inequalities

I quantifiers (first order, second order), negations, conjunctions, disjunctions Examples

1 ∀α∃n [α(n + m + 1) < 5] This is a Π1 formula.

It has a second order universal quantifier in front of a first order existential quantifier.

Two universal quantifiers amount to one, through standard coding. Same for existential..

So we only count quantifier alternations. 1 ∃β∀α∀n∃m (m > n ∧ β(α(n)) = Y (m + 1)) This is a Σ2 formula.

n 0 ∀n (n < 2 ) is a Π1 formula.

1 1 The negation of a Π1 formula is Σ1; and vice-versa. 1 1 A set A is Σ2 if it can be defined by a Σ2 formula.

1 That is, A = {X | P(X)} where P is a Σ2 formula with free variable X.

1 1 1 The sets that are both in Σm and Πm are called ∆m.

0 0 0 Similarly, the sets that are both in Σm and Πm are called ∆m.

0 0 0 The arithmetical hierarchy consists of the classes Σm, Πm, ∆m.

1 1 1 The analytical hierarchy consists of the classes Σm, Πm, ∆m. Formal definition

0 A formula in arithmetic is Σn if it can be written as a sentence with no quantifiers, prefixed by an alternating string of n first-order quantifiers starting with ∃.

A formula is arithmetical if it is in the arithmetical hierarchy, i.e. 0 in ∪n∈ωΣn.

1 A formula in second order arithmetic is Σn if it can be written as an arithmetical sentence, prefixed by an alternating string of n second-order quantifiers starting with ∃. Computability and definability

A subset of ω is computable (or recursive) if there is a program (in a programming language) that decides it.

That is, on input n outputs 0 if n 6∈ A and 1 otherwise.

The fathers of mathematical logic showed that the recursive 0 relations are exactly the ∆1 relations.

They showed also that they are exactly the ones that are representable in (the formal system of Peano) arithmetic. Recall that R(¯(x)) is representable in PA (Peano arithmetic) iff there is a formula ϕ in the language of arithmetic such that:

I If R(x¯) then PA ` ϕ(x¯)

I If ¬R(x¯) then PA ` ¬ϕ(x¯) Here we identified (used the same symbols for) numbers with numerals.

You can see the computable functions as a limit class i.e. the smallest class containing some basic functions and closed under certain schemata.

This can be seen as a miniaturization of the classes we study in descriptive set theory, like the Borel sets. Discussion

Such definitions are characterized by a ‘top-down’ approach as no stratification of the class is given.

No explicit constructive means are given in order to reach the most distant element in the class, starting from the simpler ones.

With such definitions the task is often to stratify the class and classify the members according to complexity. We will do this for the Borel sets. 0 For the class ∆1 of the computable functions/relations no such complete stratifications has succeeded.

The biggest problem of the theory of ‘subrecursive hierarchies’ is to find such a complete stratification.

Their hierarchies of recursive functions do not exhaust the ‘limit’ class of recursive/computable functions. Extended languages and relative computability

If we add a fixed parameter α ∈ ωω (infinite sequence of numbers) in our programming language and allow expressions α(n) = m then the decidable sets are called computable (recursive) in α.

Here ‘programming language’ is any universal language like:

I Fortran, C

I λ-calculus

I language of recursive functions

I Turing machine language

I etc. Extended languages and relative definability

Similarly, if we extend the language of arithmetic with 0 0 parameter α, we get ‘relativized’ classes like ∆1(α), Π1(α) etc.

0 The sets that are computable in α are the ones in ∆1(α). Coding There are many way we can code information into numbers or infinite sequences.

s0+1 sn+1 Prime decomposition: t = hs0,..., sni = p0 ····· pn where pn is the nth prime.

The inverses are given by (t)i = si . Numbers code finite

sequences.

Standard 1–1 pairing function from N to N × N: hn, mi = (n + m)(n + m + 1)/2 + n

Its two inverses are denoted by (x)i , i = 0, 1 Each number can

be seen as a pair of numbers. One infinite sequence can be seen as two, by looking at the odd or the even digits.

Given α an infinite sequence of numbers in ω, let (α)0 be the sequence of the even digits and (α)1 be the sequence of the odd digits of α. These are the inverses of this coding. ω can be seen as ω × ω.

Coding and decoding are computable/effective.

These are the devices that Gödel used in his incompleteness theorems. Task

0 Show that the coding functions and their inverses are ∆1.

0 Using coding, show that Σn is closed under (first order) ∃ and 0 Πn is closed under (first order) ∀.

0 0 Show that Σn, Πn are closed under ∨, ∧.

0 0 0 Show that Σn, Πn are closed under ∆1 substitution: when a 0 function f is ∆1 then the relations P(n, m) and P(f (n), m) have the same arithmetical complexity.

1 Show that Σ1 consists of the formulas which can be written as a 0 Π1 formula, prefixed by a second-order ∃-quantifier. We show the last claim, leaving the others as exercise:

It suffices to show that a formula of the form ∃α∀n∃mP(α, n, m) (where P is arithmetical) is equivalent to a formula of the form ∃α∀nQ(α, n) (where Q has the same arithmetical complexity as P).

And that a formula of the form ∃α∃mP(α, m) (where P is arithmetical) is equivalent to a formula of the form ∃αQ(α) (where Q has the same arithmetical complexity as P).

Indeed, in that case a straightforward induction across the arithmetical hierarchy shows the whole claim. ∃α∀n∃mP(α, n, m) ⇐⇒ ∃β∀nP((β)0, n, ((β1))n)

This shows the claim as P((β)0, n, ((β1))n) has the same 0 arithmetical complexity as P (since we got it by ∆1 substitution into P).

For the second claim we have: ∃α∃mP(α, m) ⇐⇒ ∃βP((β)0, (β)1(0)).

Again this suffices as P((β)0, (β)1(0)) has the same arithmetical complexity as P. Trees

A tree can be coded by an infinite sequence of numbers.

The particular way of coding is not important, as long as it is 0 ∆1.

A tree is well-founded if it has no infinite paths.

There is f : ω → ω computable such that α(f (k)) = 1 iff sequence k belongs to the tree coded by α (and 0 otherwise). WF = {α | α codes a tree and ∀β∃n [α(β  n) = 0]}

WF is the set of codes of well founded trees.

1 Hence WF is Π1.

König: A finite branching tree is well-founded iff it is finite. Theorem A tree is well-founded iff it can be mapped into an ordinal by a map f such that s < t implies f (s) < f (t). Proof:

I If there is such a map f then an infinite path would give an infinite descending chain of ordinals.

I For the other direction we show that if there is no such f then T is not well-founded.

I The restriction T /σ consists of the strings in T which extend σ.

I Let S = {σ ∈ T | there is no such f for T /σ}.

I Define by induction an infinite path α through S. Proof continued

I ∅ ∈ S and S is a subtree of T . I Suppose α  n ∈ S has been defined. I For the induction step it suffices to show that there is k ∈ ω such that (α  n) ∗ k is in S. I Indeed, if not, for each k ∈ ω there is an ordinal σk and an order preserving map hk : T /(α  n) ∗ k → σk .

If we let σ = supk (σk + 1) and ( α n, if t = ∅ ( ) =  h t 0 0 hk (t ), if t = k ∗ t

then h : T /α  n → σ is order preserving. Contradiction. Theorem The previous theorem holds if ‘tree’ is replaced with ‘tree in N ’ and ‘ordinal’ with ‘countable ordinal’. In the terminology of the next definition, well founded trees in the Baire space have countable height. Proof. This is an exercise. Hint: follow the proof of the above theorem, making sure that the ordinals are all countable.

Use the fact that the alphabet is countable and that the sup of a countable collection of countable ordinals is countable. Definition The height/length of a tree T is the smallest ordinal in the class

{f (∅) | f is order preserving from T into an ordinal}.

The height of the empty tree is −1 and the height of a non well founded tree is ∞. Definition If σ is an ordinal, WFσ is the set of trees with height ≤ σ. Task

Draw trees of heights: (i)1 (ii)2 (iii)3 (iv) ω (v) ω + 1 (vi) ω + ω (vii) ω · ω Foundational tales

In the 19th century mathematics started becoming more axiomatic.

People started being concerned about the foundations of maths

Trying to find a good formal foundation of maths

Some took it to the extreme, like Hilbert who had a plan to formalize all mathematics

and then mechanically prove their consistency. One of the most popular formal systems became ZF set theory

Some of the systems proposed turned out to be inconsistent! Gödel (1931) destroyed Hilbert’s plan with his two incompleteness theorems.

Gödel proved that any reasonably strong formal system (like Peano arithmetic) is bound to be incomplete

(and cannot prove its own consistency)

The birth of computability theory came with the arithmetization introduced in Gödel’s theorems Incompleteness

0 ϕ(e, n): ‘machine e does not halt on input n’ is a Π1 statement in the language of arithmetic.

Let e0 be the code of the following machine:

M(n) = 0 if PA ` ϕ(n, n) M(n) ↑ if PA ` ¬ϕ(n, n)

Then the statement ϕ(e0, e0) is true but undecidable in PA. Boldface hierarchies

In the arithmetical and the analytical classes the primitive notion are the relations in the language of arithmetic without quantifiers.

Equivalently, the computable relations.

The classes where defined by applying first order and second order quantifiers to ‘primitive’ (i.e. computable) sentences.

If our primitive notion are the clopen relations, we get much wider classes.

We denote the corresponding classes as before, only using boldface letters this time. It is not hard to see that existential quantifiers correspond to unions, universal quantifiers correspond to intersections.

In the same way that conjunctions correspond to (finite), negation corresponds to complement and disjunctions correspond to (finite) unions.

For example, the set {α | P(α) ∨ Q(α)} (where P, Q are relations on N ) equals

{α | P(α)} ∪ {β | Q(β)}. Also,

{α | P(α) ∧ Q(α)} = {α | P(α)} ∩ {β | Q(β)}.

and if T ⊆ ω × N then

{α | ∃n T (α, n)} = ∪n∈ω{α | T (α, n)}. Product spaces

It is becoming evident that in order to stratify the subsets of N we need to involve the product spaces of N with

(a) ω (for first order quantification) (b) N (itself) for second order quantification.

Recall that in the ‘lightface’ (arithmetical/ analytical) hierarchies computable or basic arithmetical relations existed in any power of ω (ω × ω, ω × ω × ω etc.). Product topologies

In fact N = ωω can be seen as the space ω × ω × · · ·

Here ω is the discrete topological space, where all sets are open.

The topology of N is the product topology of infinitely many copies of ω.

Similarly the topology on any space ωn × N m is given by the product topology.

So we can speak about open sets on the product spaces. From now on the sets we consider will be subsets of any finite Cartesian product of ω and N . Digression: Cantor set

The Cantor set, or ‘middle third’ set is the prototype of a fractal.

What is the relation with our Cantor space?

Our Cantor space is the product of infinitely copies of {0, 1} with the discrete topology.

The Cantor set, as a subspace of the real line, is homeomorphic to the Cantor space. Boldface classes of finite order

0 Σ1 are the countable unions of basic open sets. 0 If P ⊆ ω × N is clopen, then {α | ∃n P(n, α)} is Σ1.

0 0 Π1 consists of the complements of Σ1 relations

0 Π1 consists of the closed sets (in any of the product spaces).

2 0 If P ⊆ ω × N is clopen, then {(m, α) | ∀n P(n, m, α)} is Π1. 0 Similarly Π2 are the countable intersections of open sets.

0 Σ2 consists of the countable unions of closed sets. Recall other notations for these classes

0 The analysts call the Σ2 sets Fσ sets.

0 and Π2 are called Gδ sets.

0 0 Π3 are the Gδσ, Σ3 are the Fσδ sets etc. 0 0 0 We let ∆n = Σn ∩ Πn.

0 Formally, it is an inductive definition: Σ1 consists of the open sets.

0 0 Πn consists of the complements of Σn.

0 0 Σn+1 consists of the countable unions of sets in Πn.

0 Notice that ∆1 consists of the clopen sets. Closure properties

Definition We say that a class of D of relations is closed under continuous substitution if, whenever P ∈ D (and P ⊆ Y for some space Y), and f : X → Y is a continuous function, the relation f −1[P] is also in D. Recall that

x ∈ f −1[P] ⇐⇒ f (x) ∈ P ⇐⇒ P(f (x)). Task

Show that the coding functions that we have seen so far are all continuous. Theorem 0 0 0 The classes Σn, Πn, ∆n are closed under continuous 0 substitution, ∨ and ∧ and bounded quantifiers. Moreover Σn 0 are closed under existential (first order) quantifiers and Πn are closed under universal (first order) quantifiers. Proof

I continuous substitution, ∨ and ∧ are left as exercises.

I Consider the standard coding of finite sequences into numbers (via primes) and let (n)i , be its inverses.

I ∃x∃yP(x, y, z) is equivalent to ∃nP((n)0, (n)1, z).

I Similarly for ∀. 0 I This quantifier contraction shows the ∃-closure of Σn and 0 the ∀-closure of Πn. I For the bounded quantifiers: ∃n < m ∀x P(n, x, z) is equivalent to ∀u∃n ≤ m P(n, (u)n).

I Indeed, the first one obviously implies the second one. Proof continued

I Now suppose that the first one does not hold.

I Hence for each n < m there exists xn such that ¬P(n, xn, z).

I Then for u = hx0,..., xm−1i we have ¬P(n, (u)n) for each n < m.

I Hence the equivalence

I Also, ∀m ≤ n ∃s P(m, s, z) is equivalent to ∃u ∀n ≤ m P(m, (u)m).

I For this it suffices to show that their negations are equivalent.

I But this follows by the previous clause.

I the bounded quantifier closure now follows from the inductive definition of the classes. Projective hierarchy

1 In the same way that we got the classes Σn by allowing second order quantification, we can get the boldface analogues:

1 Σ1 is the class of sets we get by an application of a second 0 order existential quantifier in front of a Π1 relation.

1 1 Π1 are the complements of Σ1 sets.

Equivalently, it is the class of sets we get by an application of a 0 second order universal quantifier in front of a Σ1 relation. 1 Σn+1 is the class of sets we get by an application of a second 1 order existential quantifier in front of a Πn

1 1 Πn consists of the compliments of sets in Σn.

1 1 1 ∆n = Σn ∩ Πn. Task

1 0 Show that in the above definition of Σ1,‘Π1 relation’ can be 0 replaced by ‘Πn relation’.

This involves some coding, manipulation of quantifiers.

The general theme of manipulating a formula and producing a simplest equivalent form of it is called ‘normal forms’. Task solution

It suffices to show that first order quantifiers can be ‘absorbed’ into a second order quantifier, leaving behind only one first order quantifier.

First, it is easy to see that ∃α∃nP(α, n) is equivalent to ∃βP(f (β), β(0)), where f is the (continuous) function f (β)(n) = β(n + 1).

Second, ∃β∀n∃m P(β, n, m) is equivalent to ∃α∀n∃m P(g0(α), n, g1(α)(n)) where g0(α)(n) = α(2n) and g1(α)(n) = α(2n + 1). Task Solution

Now the claim follows by induction on the prefix of quantifiers:

0 A second order ∃ quantifier in front of a Πn sentence is 0 equivalent to a second order ∃ quantifier in front of a Π1 sentence.

Similarly for ∀.

The same proof shows quantifier contraction in the analytical 0 hierarchy, since f , g0, g1 are ∆1. 1 Hence we have that a sentence is Σ1 iff it can be written as a 0 second order ∃ quantifier in front of a Π1 sentence or equivalently, as a second order ∃ quantifier in front of an arithmetical sentence or equivalently, as a second order ∃

0 quantifier in front of a Πn sentence, for some n ∈ N. Borel sets

Definition The class of Borel sets is the smallest class that contains the open sets and is closed under complements and countable unions. There is a natural way to stratify the class of Borel sets.

We can use the boldface classes to classify the Borel sets according to their complexity.

0 0 0 It is not hard to see that for all n ∈ N the classes Σn, Πn, ∆n are contained in the Borel sets.

But do they exhaust them? Parametrization and Universality

In general, a class Y is parametrized by a set I if there is a surjection π : I → Y.

In other words, if we can give names to elements of Y from the pool of elements of I.

Recall that we work simultaneously on the class of products of the spaces ω, N .

In fact we can include in this mix any Polish space, and the theory will apply to this more general case. The elements of these spaces are called points.

A subset of any product of ω, N is called pointset.

A collection of pointsets is called pointclass. In Descriptive set theory we study pointclasses, rather than individual pointsets.

0 For example, the Borel pointclasses Σn. Given two spaces X , Y and P ⊆ Y × X for any y ∈ Y define the y-section Py of P as follows:

Py = {x ∈ X | P(y, x)} Let Γ be a pointclass and

Γ  X = {Q ⊆ X | Q ∈ Γ}.

A pointset G ⊆ Y × X is called universal for Γ  X if G ∈ Γ and the map y → Gy is a parametrization of Γ  X on Y.

That is, for each P ⊆ X ,

P ∈ Γ ⇐⇒ P = Gy , for some y ∈ Y. Definition A pointclass Γ is Y-parametrized if for every product space X there is some G ⊆ Y × X which is universal for Γ  X . Recall that C is the Cantor space.

0 Theorem (Parametrization theorem for Σ1) 0 The pointclass Σ1 is C-parametrized. Take any product space X . It has a countable set of basic open sets.

Countable sequences of basic open sets can be coded into infinite binary sequences.

Therefore there is some G ⊆ C × X which is universal for the open sets of X .

G = {(y, x) | x ∈ X is a point in the open set coded by y}.

Notice that G is the union of all {(y, x) | σ ⊂ y ∧ x ∈ M ∧ σ codes M} where σ is a binary string and M is a basic open set of X .

So G is open. Hence the theorem. General Parametrization

In fact, the following more general statement can be shown. 0 Theorem (Parametrization theorem for Σ1) 0 For every perfect product space Y the pointclass Σ1 is Y-parametrized. Tasks

0 I Is Σ1 is ω-parametrized? ℵ I Show that there are 2 0 clopen sets in N . 0 I Is ∆1 is ω-parametrized? Projections

Given P ⊆ X × Y define the projection of P along Y as:

∃Y P = {x ∈ X | ∃y P(x, y)}.

Also for Q ⊆ X define

¬Q = X − Q = {x | ¬Q(x)}. Theorem If a pointclass Γ is Y-parametrized then ∃Z Γ and ¬Γ are also Y-parametrized, for all spaces Z. Proof. If G ⊆ Y × X is universal for Γ  X then ¬G = Y × X − G is universal for ¬Γ  X .

If G ⊆ Y × X × Z is universal for Γ  (X × Z) then the following H ⊆ Y × X , H(y, x) ⇐⇒ ∃z G(y, x, z) Z is universal for ∃ Γ  X . Theorem (Hierarchy Lemma) Let Γ be a pointclass such that for every product space X and every P ⊆ X × X in Γ the diagonal {x | P(x, x)} is also in Γ. If Γ is Y-parametrized, then some P ⊆ Y is in Γ but not in ¬Γ. Proof. Let G ⊆ Y × Y be universal for Γ  Y and take P = {y | G(y, y)}.

By hypothesis P ∈ Γ. If ¬P ∈ Γ then for some fixed y0 ∈ Y we would have

G(y0, y) ⇐⇒ ¬P(y) ⇐⇒ ¬G(y, y).

This gives a contradiction for y = y0. This argument is a reminiscent of Russell’s paradox or even Cantor’s diagonal argument. Theorem (Hierarchy thm for Borel classes of fin. order) 0 Let X be any perfect product space. The class ∆n  X is is a 0 proper subclass of Σn  X and the latter is a proper subclass of 0 ∆n+1  X . Proof. 0 The class Σn satisfies the hypothesis of the hierarchy lemma.

0 Hence there is some P ⊆ X which is in Σn  X but not in 0 Πn  X .

0 0 0 0 0 0 Moreover if Σn = ∆n+1 then Πn = ∆n+1 so Πn ⊇ Σn.

0 0 This contradicts P ∈ Σn − Πn. Boldface classes of finite order don’t go very far Now we have all that is needed in order to show that there is a 0 Borel set which is not in ∪nΣn.

0 Let Pt (x, y) be universal for the Σt relations on N .

Notice that for each t0 ∈ N the relation Q(t, x, y) ⇐⇒ t = t ∧ P (x, y) is in Σ0 . 0 t0 t0

Now ∪t∈ωQ(t, x, y) is Borel, as a countable union of Borel relations.

0 But if it was Σk for some k ∈ ω then Q(t, x, y) ⇐⇒ Pk (x, y).

0 0 This cannot happen as Σk+1 6⊆ Σk . Extending boldface classes into the transfinite

We need to make a higher hierarchy where such diagonalizations within the class of Borel sets do not work.

The ‘spine’ of our short hierarchy was ω. The ‘spine’ of our

extended hierarchy will be the first uncountable ordinal ω1.

The uncountable length is expected as we are trying to exhaust a class which is closed under countable operations. Definition 0 For any ordinal ξ < ω1 we define Σξ the class of (coutable) 0 unions of sets in ¬(∪η<ξΣη). We denote:

ω 0 _ 0 Σξ = ¬(∪η<ξΣη).

0 0 As before,Πξ consists of the complements of sets in Σξ , namely

0 0 Πξ = ¬Σξ .

0 0 0 Finally ∆ξ = Σξ ∩ Πξ . It is not hard to show that the finite levels of this hierarchy coincide with the original definition of the boldface classes. Theorem 0 The class of Borel sets is ∪ξ<ω1 Σξ . Proof: 0 I Let D = ∪ξ<ω1 Σξ . I The class D contains the open sets and is closed under complements.

I We show that it is closed under countable unions.

I Let (Di )i∈ω be a countable sequence of sets in D. Proof continued

I Let αi be the least level of the transfinite hierarchy where Di occurs.

I Then the supremum of all αi , i ∈ ω is a countable ordinal β.

I By definition of hierarchy ∪i∈ωDi occurs in level β + 1.

I It remains to show that D is the smallest class with this property. 0 I Suppose that D was another class containing the open sets and closed under complement and countable union.

I By transfinite induction through the ordinals up to ω1 it follows that D ⊆ D0. Theorem The class of Borel sets is closed under continuous inverse images. Proof. It is one of the exercises. Use induction on the transfinite hierarchy of Borel sets, starting from the bottom. Task

The definition of the transfinite Borel hierarchy could be given by transfinite induction on all ordinals (not only the countable ones). Show that in that case, the hierarchy collapses at level 0 ω1. In other words, that B = Σσ for all σ ≥ ω1. The projective sets

The classes in the projective hierarchy are also called Lusin classes.

Notice the use of projections in the definition.

For each pointclass Λ let ∃N Λ be the set of projections of sets in Λ across N :

N N ∃ Λ = {∃ P | P ∈ Λ  (X × N ) for some X}. Lusin classes

Now define:

1 N 0 Σ1 = ∃ Π1

1 N 1 Σn+1 = ∃ ¬Σn+1

1 1 Πn = ¬Σn

1 1 1 ∆n = Σn ∩ Πn Theorem (Hierarchy thm for projective hierarchy) 1 Let X be any perfect product space. The class ∆n  X is is a 1 proper subclass of Σn  X and the latter is a proper subclass of 1 ∆n+1  X . Proof. 0 It follows from the parameterization theorem for Σ1, the general parameterization theorems and the hierarchy lemma. Hierarchy theorems for lightface classes

Theorem 0 0 There is a Σ1 relation U which is universal, i.e. every Σ1 relation P is a section of U:

P(n) ⇐⇒ U(he, ni)

for some e ∈ N. Proof:

0 Every Σ1 (say, of one variable) is represented by number e via standard coding (it is a finite sequence of arithmetic symbols after all).

U(he, ni) is the relation ‘there exists n which satisfies the relation that is represented by code e’. 0 It remains to show that U is Σ1. It suffices to show that the relation ‘n satisfies the relation that is represented by code e’ is 0 Σ1.

This involves some recursion theory, so we omit it. It goes as 0 follows: primitive recursive relations are ∆1. By arithmetization and a recursive definition we can express the relation ‘n satisfies the relation that is represented by code e’ in a primitive 0 recursive way. So it is ∆1. Note: Hilbert’s 10th problem

The work of Davis, Matiyachevich and Robinson (’60s and ’70s) on the undecidability of diophantine equations (polynomial equations with integer coefficients) gave more information 0 about Σ1.

0 They showed that a Σ1 relation can be written as the set of values for which a diophantine equation has solutions.

¯ 0 So if Q(x) is Σ1 then there exists some polynomial P such that P(x¯) ⇐⇒ ∃nP¯ (x¯, n¯) = 0. Polynomial equations is a specific form of a quantifier-free arithmetic formula.

Their argument involves a fair amount of number-theory (and computability theory) and can be found on a recent edition of Martin Davis’ ‘Computability and Unsolvability’. Theorem 0 For every n, there is a Σn relation which is universal for the 0 1 class of Σn relations. Also, there is a Σn relation which is 0 universal for the class of Σn relations. And the same for the Π classes. Proof. The argument is the same as for the boldface classes. By 0 induction, starting with Σ1. Theorem 0 0 1 1 A universal Σn relation is not Πn. The same for Σn and Πn. Proof. 0 Exactly as with the boldface classes. Let P be a Σn relation of 0 two arguments which is universal for the Σn relations of one 0 0 argument. If it was also Πn then ¬P(n, n) would be Σn. Then it would be a column of P, so there would be some e0 such that

¬P(n, n) ⇐⇒ P(e0, n) for all n.

This is a contradiction for n = e0. The same argument applies 1 1 to Σn, Πn. Theorem 0 n The class of ∆1 relations/functions on ω consists of the computable relations/functions on ωn. Examples: division, multiplication, bounded sums and products Pn i (e.g. i=0(7 + m)) inverse functions of computable functions, etc.

Thus the arithmetical hierarchy can be equivalently built, starting with computable relations and applying quantifier prefixes. 0 0 0 Since the classes Σn, Πn are closed under ∆1 substitution, we have the following.

Theorem 0 0 The classes Σn, Πn are closed under computable substitution.

0 Hence, if P(n, m) is Σn and f is computable, then P(f (n), f (m)) 0 is Σn. Universal relations

Universal relations are very useful in undecidability problems.

They are the prototypes of undecidability. For example there 0 0 are Σ1 relations that are not ∆1 (i.e. decidable).

When we want to show that a problem is undecidable, we often embed a universal relation into the problem. Theorem 0 0 The class ∆n is is a proper subclass of Σn and the latter is a 0 0 0 proper subclass of ∆n+1. The same holds for Σn and ∆n. Proof. It follows from the previous theorem. This also shows that all the classes in the arithmetical and analytical hierarchies are distinct. By basic logic (prenex normal forms) we know that any sentence in the first order or second order arithmetic can be written in the form of a quantifier prefix in front of a quantifier-free formula.

Moreover we can ensure that in this form, there are no adjacent quantifiers of the same kind.

This shows that the arithmetical and analytical hierarchies stratify all sentences, of the first/second order arithmetic respectively. We wish to have a constructive approach to the class of Borel sets.

A bottom-up approach connecting the language with the way that a Borel set is constructed.

Stratification of this class via a hierarchy of complexity. Coding Borel sets into reals

Given A ⊆ N define TA = {s | Ns ∩ A 6= ∅}.

Clearly TA is a tree. Proposition

The topological closure of A is TA. Proof.

α ∈ A¯ iff every basic neighborhood of α intersects A ∀n [N ∩ A 6= ∅] iff αn iff ∀n α  n ∈ TA iff α ∈ [TA]. We wish to use codes to represent how a Borel set is represented (as unions and complements of open sets).

For Borel sets that only involve finite unions, it is easy to do.

For the general case our codes will be points of N , or well-founded trees. Definition The relation α is a code for the Borel set A is defined inductively as follows: (a) If α(0) = 1, A is closed and for β(n) = α(n + 1), β is a code for TA then α codes A. (b) If α(0) = 2, β codes A and α(n + 1) = β(n), then α codes N − A. (c) If α(0) = 3 and for each n the function m → α(hn, mi + 1) codes An then α codes ∪nAn.

In other words, the set of pairs (α, A) such that α is a code for A is the smallest subset of N × C which contains coding pairs for (basic open) open sets and is closed under the two operations. Without loss of generality we could add a clause for α(0) ∈ ω − {1, 2, 3} to be identical as in the case α(0) = 1.

In that case all points in N are used as ‘codes’.

By the (inductive) definition of ‘Borel set’ and ‘α codes Borel set A’, every Borel set has a code.

Is every point of N a code for a Borel set? Any given Borel set has infinitely many descriptions in terms of unions and complements of open sets.

Every such description corresponds to a different code.

Therefore every Borel set has infinitely many different codes.

The k-column of a real α is the sequence whose n-th digit is α(hk, ni). The tree Tα of a code α

Every α ∈ N is associated with a tree Tα with a real number attached to each node, indicating the sub-tree below that node.

So if β is attached on node s of Tα, then Tβ is the restriction of Tα below node s.

This attachment is represented by a function Fα : Tα → N . Definition (Simultaneous induction for Fα, Tα)

(a) ∅ ∈ Tα and Fα(∅) = α

(b) If Fα(s)(0) 6= 2, 3 then s has no extensions in Tα.

(c) If Fα(s)(0) = 2 then s ∗ 0 is the only successor of s in Tα and Fα(s ∗ 0)(n) = Fα(s)(n + 1).

(d) If Fα(s)(0) = 3 then s ∗ k ∈ Tα for all k ∈ ω and Fα(s ∗ k)(n) = Fα(hk, ni + 1). Recasting the definition

(a) Tα has a root and α is attached to it.

(b) If Fα(s)(0) 6= 2, 3 then no extension of s survives.

(c) If Fα(s)(0) = 2 then only successor s ∗ 0 survives and the subtree below it is described by the digits > 0 ofFα(s).

(d) If Fα(s)(0) = 3 then all successors of s survive the tree below node s ∗ k is dictated by the ‘k-column’ of Fα(s). Equivalent definition

Given a real α, the function Fα and tree Tα are the unique F, T which satisfy the following conditions:

(a1) ∅ ∈ T and F(∅) = α.

(b1) If F(s)(0) 6= 2, 3 then s has no extensions in T .

(c1) If F(s)(0) = 2 then s ∗ 0 is the only successor of s in T and F(s ∗ 0)(n) = F(s)(n + 1).

(d1) If F(s)(0) = 3 then s ∗ k ∈ T for all k ∈ ω and F(s ∗ k)(n) = F(hk, ni + 1) Coding Borel sets into reals

Theorem A set is Borel iff it has a code α such that Tα is well-founded.

That is, a set A is Borel iff there is α ∈ N such that ‘α is a code for Borel set A’ and Tα is well founded. Proof

Notice that the class of sets that have a code α such that Tα is well-founded is closed under complements and countable unions.

Therefore every Borel set has this property, by the definition of the Borel sets.

Now it suffices to prove the converse.

This is done by transfinite induction along the possible lengths of Tα. If the length is 0 then the set is closed, hence Borel.

Now assume that it holds for all lengths < σ.

And let A be a set with code α such that |Tα| = σ.

If the root of Tα is ω-branching the hypothesis holds for each of the ω subtrees.

Therefore A is a countable union of Borel sets, therefore it is Borel.

If the root is 1-branching (so σ is successor ordinal) then similarly A is the negation of a Borel set, therefore it is Borel. Task

0 0 Draw the tree Tα for the case where α codes a Σn or Πn set for n ≤ 3. Corollary 1 The relation ‘α is the code of a Borel set’ is Π1. Proof.

0 The set Tα is computable in α i.e. ∆1(α).

The relation P(β, α):‘β is a path through Tα’ is arithmetical 0 (Π1).

Now α is a code of a Borel set iff ∀β ¬P(β, α).

1 This is Π1. By the same argument we have the following. Proposition 0 The relation φ(α, T , F): ‘α, T , F satisfy (a1)-(d1)’ is Π1. Proof. Notice that (a’)-(d’) are computable conditions with free variable the node s of T .

0 So we need to say ‘for all s,(a1)-(d1) which is a Π1 condition. Proposition 1 The relation ψ(α, T , F): φ(α, T , F) ∧ T is well-founded’ is Π1. Proof. The relation ‘T is well-founded’ means that T does not have infinite paths. Hence it can be written as:

∀β ∃n β  n 6∈ T .

1 0 This is a Π1 relation. Also, φ is Π1.

1 0 So ψ is a conjunction of a Π1 and a Π1 formula. The normal 1 form of such sentences is Π1. Theorem Let α code a Borel set B. The predicate ‘β is in the set coded 1 1 by α’ is ∆1(α) (i.e. ∆1 with parameter α).

Given T , F describing the Borel set we can determine membership of β to each of the sets corresponding to the nodes of T , inductively starting from the leaves.

Indeed, if we determine membership (or not) for all of the predecessors of a given node, we can easily determine it for the set corresponding to the given node. If we determine membership of β with respect to a set A, we can determine its membership with respect to ¬A.

If we determine it with respect to a countable sequence of sets, we can determine it with respect to the union of those sets.

And so on. . . inductively we can determine membership with respect to the tope node α, i.e. the set we are given.

Of course membership with respect to the leaves (i.e. closed 0 sets) is easy, it is a Π1 relation (stay on the corresponding tree for ever).

The predicate P below is defined inductively using exactly this idea. Given T , F as before, we define a predicate P (i.e. function with binary values) on the nodes of T (and reals β) as follows:

(I) If s is an endnode of T , then P[β](s) = 1 iff ∀n F(s)(β  n) = 1. (II) If F(s)(0) = 2 then P[β](s) = 1 iff P[β](s ∗ 0) = 0 (III) If F(s)(0) = 3 then P[β](s) = 1 iff ∃k P[β](s ∗ k) = 1. Proposition

If T , F satisfy (a1)-(d1) for some Borel code α (so T is well founded), there is a unique relation P(s) satisfying (i)-(III) and P(s) is defined for all β and nodes s of T . Moreover, it is a function and β belongs to the set represented by α iff P[β](∅) = 1.

Proof. The proof follows by a simultaneous induction on the length of T . During the steps of the induction we show that β is in the set represented by a node s of T iff P[β](s) = 1 (otherwise the value is 0). Notice that the conditions (I)-(III), although arithmetical, they are not computable. Therefore the complexity of P[β](s) can be quite high, since it depends on the length of T . If the length is infinite, there is a possibility that P[β](s) is not arithmetical. However we can overcome these problems by through a second order definition. Proposition 0 The sentence ρ(β, T , F, P, β): ‘P[β] satisfies (I)-(III)’ is Π2. In particular, it is arithmetical. Proof. It suffices to say that for all nodes s of T , (I)-(III) hold. Since all 0 0 of these sentences are either Σ1 or Π1, the claim follows. Now we can give the following definition.

β ∈ B iff ∃F, T , P [ψ(α, T , F) ∧ ρ(β, T , F, P, β) ∧ P[β](∅) = 1] iff ∀F, T , P [ψ(α, T , F) ∧ ρ(β, T , F, P, β) ⇒ P[β](∅) = 1].

1 This is a ∆1(α) definition, since ψ and ρ are arithmetical.

Corollary 1 Every Borel set is ∆1 in a real parameter. Theorem 0 0 For each n ∈ N, every Σn set is Σn in a real parameter, and 0 0 every Πn set is Πn. Proof. Exercise. Hint:

I Start from the bottom of the hierarchy and use induction.

I A countable collection basic open sets can be coded into a single real. 0 I Thus every open set is Σ1 in the real parameter which describes the countable union. 0 0 I The negation of a Σ1(α) set is Π1(α). I The union of the a countable collection of sets Ai which are 0 0 Πn(αi ) is Σn+1(α), where α is a parameter which codes the countable collection of reals αi , i ∈ N. Theorem 1 1 I Every Σ1 set is Σ1 in a real parameter. 1 1 I Every Π1 set is Π1 in a real parameter. 1 1 I Every ∆1 set is ∆1 in a real parameter.

Similar relations hold for the higher levels of the hierarchy (by induction). Proof

1 ω A Σ1 set is definable by a formula like ∃ αP(α, n), where P is a 0 Π1 formula.

0 But by previous theorem, P is Π1 in a real parameter γ.

ω 1 So the formula ∃ αP(α, n) is Σ1 in γ and so is the set it defines.

1 1 The same holds for Π1 and thus for ∆1. Proposition For all parameters α we have 0 0 1 1 I Σ1(α) ⊆ Σ1 and Σ1(α) ⊆ Σ1 0 0 1 1 I Π1(α) ⊆ Π1 and Π1(α) ⊆ Π1 0 0 1 1 I ∆1(α) ⊆ ∆1 and ∆1(α) ⊆ ∆1

Analogous relations hold for higher classes of the hierarchies. Proof 0 0 It suffices to show that every Σ1(α) (or every ∆1(α)) predicate is open.

So let P(α) ⇐⇒ ∃nQ(α, n) where Q is a finite arithmetic statement with no quantifiers (or, if you like, a computable predicate).

Suppose that α satisfies the above sentence P.

It suffices to find an open neighborhood of α such as all of the reals in there satisfy P.

Fix n provided by the ∃ quantifier. The rest of the sentence Q is finite, thus it involves finitely many digits of α.

As long as another real β agrees with α on those digits, it will also satisfy P (with the same n). Proof (cont.)

Let m be the largest digit involved in Q(α, n).

Then all reals extending α  m + 1 satisfy P. This finishes the proof.

Alternatively, if you like to think of Q as a computable predicate, notice that each computation is finite.

Thus the machine which decides that Q(α, n) will make the same decision of any β in place of α, which agrees with α in the finitely many digits involved n the computation.

Hence as before P holds for all reals in a neighborhood of α. We have described various devices which measure complexity of classes and sets.

For example the various hierarchies of complexity.

To conclude the episode of characterizing the complexity of the class of Borel sets, let us call our old friends WFσ.

Remember, this is the class of well-founded trees of length σ, and WF is the class of all well-founded trees.

Recall that the length of a well-founded tree in the Baire space is countable.

Denote by ||T || the length of a tree T . Theorem Let S, T be trees. Then ||S|| ≤ ||T || iff there is an order preserving map f : S → T. Proof:

One direction is easy. If such f exists, let h : T → ||T || be order preserving.

Then h ◦ f : S → ||T || is order preserving. Proof (cont.)

For the other direction suppose ||S|| ≤ ||T ||. If T has an infinite

path, then S can be mapped into that.

So suppose that T is well-founded. Use transfinite induction on the length of ||T ||.

The case ||T || = 0 is trivial. For each n < ω which is extendible in T , let sn be the one-bit sequence n.

||S  sn|| < ||S|| ≤ ||T || = supt6=∅(||T  t|| + 1). Proof (final)

So for each such n there is some sequence tn such that ||S  sn|| ≤ ||T  tn||.

By induction hypothesis there is an order preserving map fn : S  sn → T  tn.

For all s ∈ S define ( ∅ if s = ∅ f (s) = tn ∗ fn(t) if s = sn ∗ t

Then f : S → T is order preserving. The significance of the previous result lies in the following: the set

{hα, βi | α codes tree S and β codes tree T with ||S|| ≤ ||T ||}

1 is Σ1.

Identifying codes with the objects that they code, we have: Corollary 1 The relation ‘||S|| ≤ ||T ||’ is Σ1. Corollary 1 If T is a well founded tree, then the set {S | ||S|| ≤ ||T ||} is ∆1 in parameter T . Proof. 1 Since we have shown that ||S|| ≤ ||T || is Σ1, it suffices to show 1 that its negation, ||T || < ||S|| is also Σ1.

Indeed, if T is well-founded, then

||T || < ||S|| ⇐⇒ ∃n ||T || ≤ ||S  hni||. Theorem For every countable ordinal σ the set WFσ is Borel. Proof

By induction on σ. For σ = 0, WF0 consists of the empty tree, which is clearly Borel.

Now suppose that Wσ is Borel for all τ < σ.

Given a finite sequence s of integers and a tree let fs(T ) be the full subtree T  s of T below node s.

Notice that fs is continuous (check it). Then:

−1 WFσ = ∩s ∪τ<σ {T | T  s ∈ WFτ } = ∩s ∩τ<σ fs (WFτ ). Since the Borel class is closed under continuous inverse images, countable unions and intersections this shows that WFσ is Borel. Trees in product spaces In the same way that we used trees to represent closed sets in N we can do this in N × N and in general all product spaces.

A tree in N × N is a ‘downward closed’ collection of pairs hs, ti of strings (finite sequences) of numbers, such that |s| = |t|.

Here ‘downward closed’ means that if hs, ti is in the tree and s0 ⊆ s, t0 ⊆ t and |s0| = |t0| then hs0, t0i is also in the tree. Using

the topology of N × N you can verify (as in the case of N ) that a set in that product space is closed iff it is the set of infinite paths through a (two-dimentional) tree in the space.

Using this fact, we can use trees in the Σ and Π classes, instead of open or closed sets. For example, the following is a consequence of the facts discussed above. 1 Theorem (Normal form theorem for Π1) 1 A set A is Π1 iff there is a tree on ω × ω (i.e. in the space N × N ) such that α ∈ A ⇐⇒ ∀β ∃n hα  n, β  ni 6∈ T.

α If T = {σ | hα  |σ|, σi ∈ T } is the α-section of the two-dimentional tree T , then we have:

α ∈ A ⇐⇒ T α ∈ WF.

Notice that T α is a continuous function on α (exercise). 0 Theorem (Normal form theorem for Π1) 0 0 A set A ⊆ N × N is Π1 iff there is a ∆1 (i.e. computable) tree on ω × ω (i.e. in the space N × N ) such that

hα, βi ∈ A ⇐⇒ ∀n hα  n, β  ni ∈ T .

1 Theorem (Normal form theorem for Π1) 1 0 A set A is Π1 iff there is a ∆1 (i.e. computable) tree on ω × ω (i.e. in the space N × N ) such that

α ∈ A ⇐⇒ ∀β ∃n hα  n, β  ni 6∈ T . Theorem i i i i For each n ∈ N and i = 0, 1 we have Σn 6⊆ Πn and Πn 6⊆ Σn. Proof. This is an exercise. First use continuous substitution to show i that if e.g. P(α, β, n) ∈ Πn then for any constant α0 the relation i Q(β, n) ≡ P(α0, β, n) is also in Πn.

To prove our hierarchy theorems (show that the hierarchies do not collapse) we had to devise somewhat artificial universal relations.

Here is a natural example: Theorem 1 1 WF is Π1 but not Σ1.

Proof:

1 We already know that it is Π1.

1 1 1 Assume that it was Σ1 and let A ∈ Σ1 − Π1. Proof (cont.)

By normal form theorem there is a tree T on ω × ω such that

α ∈ A ⇐⇒ T α 6∈ WF

α where T is the α-column of T , i.e. {s | hα  |s|, si ∈ T }.

α 1 1 But T 6∈ WF is Π1 since its complement is Σ1.

1 Then A is Π1, a contradiction. Theorem (Boundedness theorem) 1 If A is a Σ1 subset of WF then there is a countable ordinal σ such that A ⊆ WFσ Proof. Otherwise for all σ < ℵ1 there is S ∈ A such that |S| ≥ σ, so

T ∈ WF ⇐⇒ ∃S [S ∈ A ∧ ||S|| ≥ ||T ||].

1 1 But we have shown that ‘||S|| ≥ ||T ||’ is Σ1, so this is a Σ1 definition of WF.

This is impossible ! Theorem 1 1 A set is Borel iff it is ∆1 iff it is ∆1 in a real parameter.

1 For the proof it remains to show that every ∆1 set is Borel.

Proof:

1 1 Let A ∈ ∆1. As a Π1 set

α ∈ A ⇐⇒ T α ∈ WF for some tree T . Proof (cont.)

1 α 1 Since A is Σ1 the set {T | α ∈ A} is a Σ1 subset of WF. By the boundedness theorem it is contained in some WFσ hence α α ∈ A ⇐⇒ T ∈ WFσ. This gives A as a continuous inverse image of a Borel set, hence A is Borel. Definition A perfect set A is a closed set with no isolated point, i.e. for every basic open set Ns,Ns ∩ A is either empty or contains more than one element.

Theorem Every perfect subset of N has 2ℵ0 elements. Proof of theorem:

Let A be a perfect set.

Then A = [P] for some perfect tree P, i.e. a tree such that every node has two incomparable/incompatible extensions.

Let T be the full binary tree. We construct a map f : T → P.

Let f (∅) = ∅ and inductively, if t is a node of length n + 1 let f (t ∗ 0) = u and f (t ∗ 1) = v where u, v are two incompatible extensions of f (t). Proof of theorem:

By induction for any infinite path α in T we have F(α  n) ⊆ F(α  n + 1) for every n.

Thus we can define: g(α) = ∪n<ωf (α  n).

This map from the Cantor space to A is 1-1 because if α, β differ on digint n − 1, then f (α  n), f (β  n) will be incompatible strings, hence g(α), g(β) will be different.

Hence g is 1-1 and A has to be uncountable. Theorem (Cantor-Bendixson theorem) Every uncountable closed set can be expressed uniquely as the disjoint union of a perfect (closed) set and a countable set. Definition A point x is a condensation point of a set F, if Ns ∩ F is uncountable for every Ns such that x ∈ Ns. Proof of C.-B. theorem Proposition If P is the set of condensation points of X, then X − P is countable. Proof. For every point in X − P there is a a basic open set Ns which contains this point and such that X ∩ Ns is countable.

Consider this map f from X − P to the basic open sets

and notice that for each Ns in the range of f , the inverse image −1 f (Ns) is contained in Ns ∩ X, hence it is countable.

−1 Since X − P is the union of all f (Ns), where Ns is an image of f , it is a countable union of countable sets, hence countable. Proof of C.-B. theorem

Now let F be an uncountable closed set.

Consider F as the union of the condensation points of it and the non-condensation points of it.

The set of condensation points of F is perfect, by definition, and is contained in F since F is closed.

The set of non-condensation points of F is countable by above proposition. Proof of C.-B. theorem (uniqueness)

Now suppose that F = P ∪ S where P ∩ S = ∅, P is perfect (closed) and S is countable.

Take x ∈ P and a basic open set (neighborhood) N such that x ∈ N.

Then N ∩ P is closed. Also, perfect because an isolated path in it would be isolated in P.

Then by previous theorem it is uncountable.

Hence x is a condensation point. Proof of C.-B. theorem (uniqueness) cont.

It suffices to show that no point in S is a condensation point.

If x ∈ S then there exists some neighborhood N of x such that N ∩ P = ∅, since P is closed.

Since S is countable, N ∩ F is countable. So x is not a condensation point. Cantor-Bendixson derivative

Definition The Cantor-Bendixson derivative A0 of a closed set A is the set of limit points of A. An element of the space is a limit point if it is not isolated in A. That is, if there is no basic open Ns such that Ns ∩ A 6= ∅ consists exactly of that element.

(0) (σ) (τ) Let A = A and for limit ordinals σ let A = ∩τ<σA . So we have the following. Proposition A closed set [T ] is perfect iff [T ]0 = [T ].

That [T ]0 ⊆ [T ] follows from the fact that [T ] is closed, thus it contains all of its limit points.

That [T ] ⊆ [T ]0 follows by the definition of derivative and the fact that [T ] is perfect.

Clearly A(σ) ⊆ A(τ) for ordinals τ ≤ σ. Theorem (ξ) I The set ∩ξA (where ξ runs through all ordinals) is perfect (perhaps empty).

(ξ) I A − A is countable for all ξ.

(ξ) (λ) I There is a countable ordinal λ such that ∩ξA = A . To prove the theorem, first notice that all A(ξ) are closed sets (the set of limit points of a closed set is a closed set).

(ξ) Hence ∩ξA is closed.

Suppose for a contradiction that A(ξ) − A(λ) 6= ∅ for some ξ and all λ < ξ.

(ξ) (λ) To each such ξ corresponds a point xξ ∈ A − A and a (λ) neighborhood N such that N ∩ A = {xξ}.

(σ) (τ) Since A ⊆ A for ordinals τ ≤ σ we have Nσ 6= Nτ for τ ≤ σ. Since there are only countably many neighborhoods this means that the σ such that A(ξ) − A(λ) 6= ∅ for all λ < ξ ≤ σ form a a countable initial segment of the ordinals.

Their supremum τ is also countable and A(τ + 1) = A(τ).

(ξ) (τ) (ξ) (τ) So A = A for all ξ ≥ τ and ∩ξA = A .

(ξ) (τ) Also ∩ξA is perfect because if it has an isolated point, A would too.

So A(τ) 6= A(τ+1) (contradiction). Proposition If P is the set of condensation points of a closed set X, then P ⊆ X (σ) for all ordinals σ. Proof. Let τ be an ordinal such that X (ξ) = X (τ) for all ξ ≥ τ. Then X = X τ ∪ (X − X (τ)) where X τ is perfect and X − X (τ) is countable.

By uniqueness in the Cantor-Bendixon theorem P ⊆ X (τ) hence P ⊆ X (σ) for all ordinals σ. Cantor-Bendixson rank

By the corollary above, for every set closed F there is a least countable σ such that F (σ) = F (σ+1).

We call this the Cantor-Bendixson rank of F. Corollary ℵ Every closed set has cardinality either ℵ0 or 2 0 . Proof. If a close set is not countable, then by the Cantor-Bendixson theorem it contains a perfect (closed) set. The latter is has cardinality 2ℵ0 as we have shown. Therefore any superset of it (subset of the space) has the same cardinality. Theorem The class B of Borel sets is the smallest class containing the closed sets and closed under countable intersections and countable disjoint unions.

For the proof, let D be the class defined in the theorem, and ¬D the class of the complements of sets in D. Clearly D ⊆ B.

To show equality, it suffices to show that B ⊆ D ∩ ¬D.

For this it suffices to show that D ∩ ¬D contains all open sets and is closed under complementation and countable unions.

The closed sets are in D so it suffices to show that the open sets are also in D.

But every open set is the disjoint union of basic open sets, which happen to be clopen.

So every open set is the disjoint countable union of closed sets, hence it is in D. D ∩ ¬D is by definition closed under complementation so it remains to show that it is closed under countable unions.

By De Morgan laws, it suffices to show that it is closed under countable intersections.

So suppose An ∈ D ∩ ¬D for each n. Then ∩nAn is in D.

¬∩n An = ∪n¬An = ∪n(¬An −(∪k

The last expression shows that the set is a disjoint union of sets in D, hence it is in D. Theorem Every Borel set is a 1–1 continuous image of a closed subset of the Baire space.

Proof:

Consider the class Θ of such images.

Obviously the closed sets are in Θ.

So it suffices to show that Θ is closed under countable intersections and disjoint unions. First disjoint unions. Suppose that for each n, fn is a continuous 1–1 function from [Tn] onto An.

Let T be the tree-union of Tn, n ∈ ω, i.e. T = {∅} ∪ {k ∗ s | s ∈ Tk }.

− − For each α let α (n) = α(n + 1) and define f (α) = fα(0)(α ).

Then f is 1–1 and continuous with domain [T ] and range ∪nAn.

Now intersections. Let fn, An be as above.

Given α, let αn(m) = α(hn, mi). The map α → αn is continuous.

Let F = ∩n{α | αn ∈ [Tn]}.

F is closed as an intersection of continuous inverse images of closed sets.

Let Fn = {α ∈ F | fn(αn) = f0(α0)}. It is closed.

Let G = ∩nFn and for α ∈ G let f (α) be f0(α0).

f is continuous, 1–1 and has range ∩nAn. f continuous: as composition of cont. functions f0 and α → α0.

f 1–1: Let α 6= β. Then αn 6= βn for some n.

Hence fn(αn) 6= fn(βn) so f (α) = f0(α0) 6= f0(β0) = f (β).

Finally given β ∈ ∩nAn choose α so that fn(αn) = β for each n.

Then α ∈ G and f (α) = β. Corollary Every uncountable Borel set has cardinality 2ℵ0 .

Proof:

Given an uncountable Borel set, it is a 1–1 continuous image of an uncountable closed subset of the Baire space.

But an uncountable closed set contains a perfect subset.

Every perfect subset in the Baire space is the set of infinite paths through a perfect tree (no isolated paths).

Every perfect tree contains a finite branching perfect subtree. By Königs Lemma, the paths through the latter form a compact set.

Hence every perfect set in the Baire space contains a compact perfect set.

A 1–1 continuous image of a compact perfect set is compact and perfect.

Hence every uncountable Borel set contains a compact perfect set, hence has cardinality 2ℵ0 .

This finishes the proof. Although the continuum problem can be solved for Borel sets via perfect sets, the same does not hold in general.

The following theorem of Lusin showed that Cantor’s program for solving the continuum problem (as described above) fails.

Theorem There is a subset of the Baire space which neither contains nor is disjoint from any perfect set. Proof

ℵ We construct it. Let hPσ | σ < 2 0 i be a well-ordering of all perfect trees.

We define two sequences of sets (Aσ)σ<2ℵ0 , (Bσ)σ<2ℵ0 by transfinite induction.

Step 0: Choose two distinct reals α, β in [P0] and let A0 = {α} and B0 = {β}. Step σ: At this stage (Aτ )τ<σ, (Bτ )τ<σ have been defined and:

A0 ⊆ A1 ⊆ · · · ⊆ Aτ . . . τ < σ

B0 ⊆ B1 ⊆ · · · ⊆ Bτ . . . τ < σ and for each τ < σ, Aτ ∩ Bτ = ∅ and Aτ ∪ Bτ has cardinality below 2ℵ0 .

Find α, β ∈ [Pσ] − ∪τ<σAτ − ∪τ<σBτ so that α 6= β.

ℵ Such α, β exist because [Pσ] has cardinality 2 0 and each of the unions has cardinality less than that. Define

Aσ = ∪τ<σAτ ∪ {α}

Bσ = ∪τ<σBτ ∪ {β}

This completes stage σ.

Now let A = ∪σ<2ℵ0 Aσ and B = ∪σ<2ℵ0 Bσ. Clearly A intersects every perfect set, and so does B.

But A, B are disjoint.

Hence A cannot have a perfect subset, and is not disjoint from any perfect set.