Descriptive Set theory

by George Barmpalias

Institute for Language, Logic and Computation

Lectures 1–8

Last updated on the 10th of November 2009

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 definedI Definability 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 definedI Definability theory

What (meta) mathematics are involved in it?

Answer

I Analysis, topologyI Set theoryI Computability / recursion theory

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

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

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

Polish spaces: complete, separable metric spaces with noisolated points (perfect).

What (meta) mathematics are involved in it?Answer

I Analysis, topologyI Set theoryI Computability / recursion theory

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

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

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

Polish spaces: complete, separable metric spaces with noisolated points (perfect).

Birth of Descriptive set theory

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

This was not immediately appreciated.

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

. . . the classification and description of pointsets in classicalanalysis

It was then that Cantor’s ideas received general acceptance

Birth of Descriptive set theory

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

This was not immediately appreciated.

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

. . . the classification and description of pointsets in classicalanalysis

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 acomplete 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 inthe 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 withoutisolated points) in connection to a problem in Fourier series

Later he studied them in order to settle the continuumhypothesis.

He hoped to show that every uncountable set contains a perfectset.

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’sprogram fails.

Non-uniqness of the real line

In the 1960s Cohen’s method of forcing showed a number ofindependence results illustrating the non-uniqness of the realline.

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

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

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

Effective theory

Independently of the classical developments, logician StephenKleene studied the definability of sets of natural numbers in the1940s and 1950s

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

Classical Effective

Reals integerscontinuous functions computable functions

Borel sets Hyperarithmetical setsanalytic sets Σ1

1 setsprojective 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 theoremsI Games and determinacyI Hyperarithmetic theoryI 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, byR. 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 naturalnumbers.

It has a natural topology generated by the basic open setsNσ = {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 densesubset.

Show that the Baire space is not compact.

The Cantor space is compact (see below).

The Baire space N is homeomorphic to the irrationals viacontinued fractions.

A homeomorphism is a continuous function between twotopological spaces with a continuous inverse function.

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

Note: Every closed set in the Baire or Cantor space can berepresented 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 iffit 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 thatβ � n is defined and has the property that there areinfinitely many strings on the tree which extend it.

I Since the tree is finite branching, there is m ∈ N such thatthere 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.

CorollaryThe 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 pathsI 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 .

CorollaryThe clopen sets in the Cantor space are exactly the finiteunions 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 = ∪iNσi and C − A = ∪iNτi for suitable families of

strings.I Hence C = (∪iNσi )∪ (∪iNτi ) and the union of these families

is a cover of C.I By compactness C = (∪i<kNσi ) ∪ (∪i<kNτi ) for some k ∈ ωI Obviously ∪i<kNσi ⊆ A.I But A ⊆ ∪i<kNσi because if α ∈ A then α 6∈ ∪i<kNτi , soα ∈ ∪i<kNσi .

Task

Show that there is a clopen set in the Baire space which is nota finite union of basic open sets.

The language we use to talk about N and ω hasI Variables for integersI Variables for reals (Greek letters)I constants for natural numbersI +, ·, expI function applications e.g. α(x + 3)

I equalities, inequalitiesI quantifiers (first order, second order), negations,

conjunctions, disjunctions

Examples

∀α∃n [α(n + m + 1) < 5] This is a Π11 formula.

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

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

So we only count quantifier alternations.

∃β∀α∀n∃m (m > n ∧ β(α(n)) = Y (m + 1)) This is a Σ12

formula.

∀n (n < 2n) is a Π01 formula.

The negation of a Π11 formula is Σ1

1; and vice-versa.

A set A is Σ12 if it can be defined by a Σ1

2 formula.

That is, A = {X | P(X )} where P is a Σ12 formula with free

variable X .

The sets that are both in Σ1m and Π1

m are called ∆1m.

Similarly, the sets that are both in Σ0m and Π0

m are called ∆0m.

The arithmetical hierarchy consists of the classes Σ0m, Π0

m, ∆0m.

The analytical hierarchy consists of the classes Σ1m, Π1

m, ∆1m.

Formal definition

A formula in arithmetic is Σ0n if it can be written as a sentence

with no quantifiers, prefixed by an alternating string of nfirst-order quantifiers starting with ∃.

A formula is arithmetical if it is in the arithmetical hierarchy, i.e.in ∪n∈ωΣ0

n.

A formula in second order arithmetic is Σ1n if it can be written as

an arithmetical sentence, prefixed by an alternating string of nsecond-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 recursiverelations are exactly the ∆0

1 relations.

They showed also that they are exactly the ones that arerepresentable in (the formal system of Peano) arithmetic.

Recall that R((x)) is representable in PA (Peano arithmetic) iffthere 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 withnumerals.

You can see the computable functions as a limit class i.e. thesmallest class containing some basic functions and closedunder certain schemata.

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

Discussion

Such definitions are characterized by a ‘top-down’ approach asno stratification of the class is given.

No explicit constructive means are given in order to reach themost distant element in the class, starting from the simplerones.

With such definitions the task is often to stratify the class andclassify the members according to complexity.

We will do this for the Borel sets.For the class ∆0

1 of the computable functions/relations no suchcomplete 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 ofnumbers) in our programming language and allow expressionsα(n) = m then the decidable sets are called computable(recursive) in α.

Here ‘programming language’ is any universal language like:I Fortran, CI λ-calculusI language of recursive functionsI Turing machine languageI etc.

Extended languages and relative definability

Similarly, if we extend the language of arithmetic withparameter α, we get ‘relativized’ classes like ∆0

1(α), Π01(α) etc.

The sets that are computable in α are the ones in ∆01(α).

CodingThere are many way we can code information into numbers orinfinite sequences.

Prime decomposition: t = 〈s0, . . . , sn〉 = ps0+10 · · · · · psn+1

nwhere 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:〈n,m〉 = (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 theodd or the even digits.

Given α an infinite sequence of numbers in ω, let (α)0 be thesequence of the even digits and (α)1 be the sequence of theodd 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 incompletenesstheorems.

Task

Show that the coding functions and their inverses are ∆01.

Using coding, show that Σ0n is closed under (first order) ∃ and

Π0n is closed under (first order) ∀.

Show that Σ0n, Π0

n are closed under ∨,∧.

Show that Σ0n, Π0

n are closed under ∆01 substitution: when a

function f is ∆01 then the relations P(n,m) and P(f (n),m) have

the same arithmetical complexity.

Show that Σ11 consists of the formulas which can be written as a

Π01 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 asP).

And that a formula of the form ∃α∃mP(α,m) (where P isarithmetical) 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 thearithmetical 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 samearithmetical complexity as P (since we got it by ∆0

1 substitutioninto 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 samearithmetical 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 iffsequence 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.

Hence WF is Π11.

König: A finite branching tree is well-founded iff it is finite.

TheoremA tree is well-founded iff it can be mapped into an ordinal by amap 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 aninfinite descending chain of ordinals.

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

I The restriction T/σ consists of the strings in T whichextend 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

h(t) =

{α � n, if t = ∅hk (t ′), if t = k ∗ t ′

then h : T/α � n→ σ is order preserving. Contradiction.

TheoremThe 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 inthe 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 acountable collection of countable ordinals is countable.

DefinitionThe 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 wellfounded tree is∞.

DefinitionIf 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 moreaxiomatic.

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 toformalize 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 twoincompleteness theorems.

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

(and cannot prove its own consistency)

The birth of computability theory came with the arithmetizationintroduced in Gödel’s theorems

Incompleteness

ϕ(e,n): ‘machine e does not halt on input n’ is a Π01 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 primitivenotion are the relations in the language of arithmetic withoutquantifiers.

Equivalently, the computable relations.

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

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

We denote the corresponding classes as before, only usingboldface letters this time.

It is not hard to see that existential quantifiers correspond tounions, universal quantifiers correspond to intersections.

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

For example, the set {α | P(α) ∨Q(α)} (where P,Q arerelations 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 Nwe 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) hierarchiescomputable or basic arithmetical relations existed in any powerof 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 areopen.

The topology of N is the product topology of infinitely manycopies of ω.

Similarly the topology on any space ωn ×Nm is given by theproduct topology.

So we can speak about open sets on the product spaces.

From now on the sets we consider will be subsets of any finiteCartesian 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, ishomeomorphic to the Cantor space.

Boldface classes of finite order

Σ01 are the countable unions of basic open sets.

If P ⊆ ω ×N is clopen, then {α | ∃n P(n, α)} is Σ01.

Π01 consists of the complements of Σ0

1 relations

Π01 consists of the closed sets (in any of the product spaces).

If P ⊆ ω2 ×N is clopen, then {(m, α) | ∀n P(n,m, α)} is Π01.

Similarly Π02 are the countable intersections of open sets.

Σ02 consists of the countable unions of closed sets.

Recall other notations for these classes

The analysts call the Σ02 sets Fσ sets.

and Π02 are called Gδ sets.

Π03 are the Gδσ, Σ0

3 are the Fσδ sets etc.

We let ∆0n = Σ0

n ∩Π0n.

Formally, it is an inductive definition: Σ01 consists of the open

sets.

Π0n consists of the complements of Σ0

n.

Σ0n+1 consists of the countable unions of sets in Π0

n.

Notice that ∆01 consists of the clopen sets.

Closure properties

DefinitionWe say that a class of D of relations is closed under continuoussubstitution if, whenever P ∈ D (and P ⊆ Y for some space Y),and f : X → Y is a continuous function, the relation f−1[P] isalso 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 allcontinuous.

TheoremThe classes Σ0

n,Π0n,∆

0n are closed under continuous

substitution, ∨ and ∧ and bounded quantifiers. Moreover Σ0n

are closed under existential (first order) quantifiers and Π0n 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 ∀.I This quantifier contraction shows the ∃-closure of Σ0

n andthe ∀-closure of Π0

n.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 = 〈x0, . . . , xm−1〉 we have ¬P(n, (u)n) for eachn < m.

I Hence the equivalenceI 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 areequivalent.

I But this follows by the previous clause.I the bounded quantifier closure now follows from the

inductive definition of the classes.

Projective hierarchy

In the same way that we got the classes Σ1n by allowing second

order quantification, we can get the boldface analogues:

Σ11 is the class of sets we get by an application of a second

order existential quantifier in front of a Π01 relation.

Π11 are the complements of Σ1

1 sets.

Equivalently, it is the class of sets we get by an application of asecond order universal quantifier in front of a Σ0

1 relation.

Σ1n+1 is the class of sets we get by an application of a second

order existential quantifier in front of a Π1n

Π1n consists of the compliments of sets in Σ1

n.

∆1n = Σ1

n ∩Π1n.

Task

Show that in the above definition of Σ11, ‘Π0

1 relation’ can bereplaced by ‘Π0

n relation’.

This involves some coding, manipulation of quantifiers.

The general theme of manipulating a formula and producing asimplest 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 firstorder 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) functionf (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) andg1(α)(n) = α(2n + 1).

Task Solution

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

A second order ∃ quantifier in front of a Π0n sentence is

equivalent to a second order ∃ quantifier in front of a Π01

sentence.

Similarly for ∀.

The same proof shows quantifier contraction in the analyticalhierarchy, since f ,g0,g1 are ∆0

1.

Hence we have that a sentence is Σ11 iff it can be written as a

second order ∃ quantifier in front of a Π01 sentence or

equivalently, as a second order ∃ quantifier in front of anarithmetical sentence or equivalently, as a second order ∃

quantifier in front of a Π0n sentence, for some n ∈ N.

Borel sets

DefinitionThe class of Borel sets is the smallest class that contains theopen sets and is closed under complements and countableunions.There is a natural way to stratify the class of Borel sets.

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

It is not hard to see that for all n ∈ N the classes Σ0n,Π

0n,∆

0n 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 asurjection π : I → Y.

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

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

In fact we can include in this mix any Polish space, and thetheory 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 thanindividual pointsets.

For example, the Borel pointclasses Σ0n.

Given two spaces X ,Y and P ⊆ Y ×X for any y ∈ Y define they -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 mapy → Gy

is a parametrization of Γ � X on Y.

That is, for each P ⊆ X ,

P ∈ Γ ⇐⇒ P = Gy , for some y ∈ Y.

DefinitionA pointclass Γ is Y-parametrized if for every product space Xthere is some G ⊆ Y × X which is universal for Γ � X .Recall that C is the Cantor space.

Theorem (Parametrization theorem for Σ01)

The pointclass Σ01 is C-parametrized.

Take any product space X . It has a countable set of basic opensets.

Countable sequences of basic open sets can be coded intoinfinite binary sequences.

Therefore there is some G ⊆ C × X which is universal for theopen 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 binarystring 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.

Theorem (Parametrization theorem for Σ01)

For every perfect product space Y the pointclass Σ01 is

Y-parametrized.

Tasks

I Is Σ01 is ω-parametrized?

I Show that there are 2ℵ0 clopen sets in N .I Is ∆0

1 is ω-parametrized?

Projections

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

∃YP = {x ∈ X | ∃y P(x , y)}.

Also for Q ⊆ X define

¬Q = X −Q = {x | ¬Q(x)}.

TheoremIf a pointclass Γ is Y-parametrized then ∃ZΓ and ¬Γ are alsoY-parametrized, for all spaces Z.

Proof.If G ⊆ Y × X is universal for Γ � X then ¬G = Y × X −G isuniversal for ¬Γ � X .

If G ⊆ Y × X × Z is universal for Γ � (X × Z) then the followingH ⊆ Y × X ,

H(y , x) ⇐⇒ ∃z G(y , x , z)

is universal for ∃ZΓ � X .

Theorem (Hierarchy Lemma)Let Γ be a pointclass such that for every product space X andevery P ⊆ X ×X in Γ the diagonal {x | P(x , x)} is also in Γ. If Γis 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 evenCantor’s diagonal argument.

Theorem (Hierarchy thm for Borel classes of fin. order)Let X be any perfect product space. The class ∆0

n � X is is aproper subclass of Σ0

n � X and the latter is a proper subclass of∆0

n+1 � X .

Proof.The class Σ0

n satisfies the hypothesis of the hierarchy lemma.

Hence there is some P ⊆ X which is in Σ0n � X but not in

Π0n � X .

Moreover if Σ0n = ∆0

n+1 then Π0n = ∆0

n+1 so Π0n ⊇ Σ0

n.

This contradicts P ∈ Σ0n −Π0

n.

Boldface classes of finite order don’t go very farNow we have all that is needed in order to show that there is aBorel set which is not in ∪nΣ

0n.

Let Pt (x , y) be universal for the Σ0t relations on N .

Notice that for each t0 ∈ N the relationQ(t , x , y) ⇐⇒ t = t0 ∧ Pt0(x , y) is in Σ0

t0 .

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

But if it was Σ0k for some k ∈ ω then Q(t , x , y) ⇐⇒ Pk (x , y).

This cannot happen as Σ0k+1 6⊆ Σ0

k .

Extending boldface classes into the transfinite

We need to make a higher hierarchy where suchdiagonalizations 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 exhausta class which is closed under countable operations.

DefinitionFor any ordinal ξ < ω1 we define Σ0

ξ the class of (coutable)unions of sets in ¬(∪η<ξΣ0

η). We denote:

Σ0ξ =

ω∨¬(∪η<ξΣ0

η).

As before,Π0ξ consists of the complements of sets in Σ0

ξ , namely

Π0ξ = ¬Σ0

ξ .

Finally ∆0ξ = Σ0

ξ ∩Π0ξ .

It is not hard to show that the finite levels of this hierarchycoincide with the original definition of the boldface classes.

TheoremThe class of Borel sets is ∪ξ<ω1Σ

0ξ .

Proof:I Let D = ∪ξ<ω1Σ

0ξ .

I The class D contains the open sets and is closed undercomplements.

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 whereDi 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.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 ⊆ D′.

TheoremThe class of Borel sets is closed under continuous inverseimages.

Proof.It is one of the exercises. Use induction on the transfinitehierarchy of Borel sets, starting from the bottom.

Task

The definition of the transfinite Borel hierarchy could be givenby transfinite induction on all ordinals (not only the countableones). Show that in that case, the hierarchy collapses at levelω1. In other words, that B = Σ0

σ for all σ ≥ ω1.

The projective sets

The classes in the projective hierarchy are also called Lusinclasses.

Notice the use of projections in the definition.

For each pointclass Λ let ∃NΛ be the set of projections of setsin Λ across N :

∃NΛ = {∃NP | P ∈ Λ � (X ×N ) for some X}.

Lusin classes

Now define:

Σ11 = ∃NΠ0

1

Σ1n+1 = ∃N¬Σ1

n+1

Π1n = ¬Σ1

n

∆1n = Σ1

n ∩Π1n

Theorem (Hierarchy thm for projective hierarchy)Let X be any perfect product space. The class ∆1

n � X is is aproper subclass of Σ1

n � X and the latter is a proper subclass of∆1

n+1 � X .

Proof.It follows from the parameterization theorem for Σ0

1, the generalparameterization theorems and the hierarchy lemma.

Hierarchy theorems for lightface classes

TheoremThere is a Σ0

1 relation U which is universal, i.e. every Σ01

relation P is a section of U:

P(n) ⇐⇒ U(〈e,n〉)

for some e ∈ N.Proof:

Every Σ01 (say, of one variable) is represented by number e via

standard coding (it is a finite sequence of arithmetic symbolsafter all).

U(〈e,n〉) is the relation ‘there exists n which satisfies therelation that is represented by code e’.

It remains to show that U is Σ01. It suffices to show that the

relation ‘n satisfies the relation that is represented by code e’ isΣ0

1.

This involves some recursion theory, so we omit it. It goes asfollows: primitive recursive relations are ∆0

1. By arithmetizationand a recursive definition we can express the relation ‘nsatisfies the relation that is represented by code e’ in a primitiverecursive way. So it is ∆0

1.

Note: Hilbert’s 10th problem

The work of Davis, Matiyachevich and Robinson (’60s and ’70s)on the undecidability of diophantine equations (polynomialequations with integer coefficients) gave more informationabout Σ0

1.

They showed that a Σ01 relation can be written as the set of

values for which a diophantine equation has solutions.

So if Q(x) is Σ01 then there exists some polynomial P such that

P(x) ⇐⇒ ∃nP(x , n) = 0.

Polynomial equations is a specific form of a quantifier-freearithmetic formula.

Their argument involves a fair amount of number-theory (andcomputability theory) and can be found on a recent edition ofMartin Davis’ ‘Computability and Unsolvability’.

TheoremFor every n, there is a Σ0

n relation which is universal for theclass of Σ0

n relations. Also, there is a Σ1n relation which is

universal for the class of Σ0n relations. And the same for the Π

classes.

Proof.The argument is the same as for the boldface classes. Byinduction, starting with Σ0

1.

TheoremA universal Σ0

n relation is not Π0n. The same for Σ1

n and Π1n.

Proof.Exactly as with the boldface classes. Let P be a Σ0

n relation oftwo arguments which is universal for the Σ0

n relations of oneargument. If it was also Π0

n then ¬P(n,n) would be Σ0n. 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 appliesto Σ1

n, Π1n.

TheoremThe class of ∆0

1 relations/functions on ωn consists of thecomputable relations/functions on ωn.Examples: division, multiplication, bounded sums and products(e.g.

∑ni=0(7i + m)) inverse functions of computable functions,

etc.

Thus the arithmetical hierarchy can be equivalently built,starting with computable relations and applying quantifierprefixes.

Since the classes Σ0n, Π0

n are closed under ∆01 substitution, we

have the following.

TheoremThe classes Σ0

n, Π0n are closed under computable substitution.

Hence, if P(n,m) is Σ0n and f is computable, then P(f (n), f (m))

is Σ0n.

Universal relations

Universal relations are very useful in undecidability problems.

They are the prototypes of undecidability. For example thereare Σ0

1 relations that are not ∆01 (i.e. decidable).

When we want to show that a problem is undecidable, we oftenembed a universal relation into the problem.

TheoremThe class ∆0

n is is a proper subclass of Σ0n and the latter is a

proper subclass of ∆0n+1. The same holds for Σ0

n and ∆0n.

Proof.It follows from the previous theorem. This also shows that allthe classes in the arithmetical and analytical hierarchies aredistinct.

By basic logic (prenex normal forms) we know that anysentence in the first order or second order arithmetic can bewritten in the form of a quantifier prefix in front of aquantifier-free formula.

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

This shows that the arithmetical and analytical hierarchiesstratify all sentences, of the first/second order arithmeticrespectively.

We wish to have a constructive approach to the class of Borelsets.

A bottom-up approach connecting the language with the waythat 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.

PropositionThe topological closure of A is TA.

Proof.

α ∈ A iff every basic neighborhood of α intersects Aiff ∀n [Nα�n ∩ A 6= ∅]iff ∀n α � n ∈ TA

iff α ∈ [TA].

We wish to use codes to represent how a Borel set isrepresented (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 , orwell-founded trees.

DefinitionThe relation α is a code for the Borel set A is definedinductively as follows:

(a) If α(0) = 1, A is closed and for β(n) = α(n + 1), β is acode for TA then α codes A.

(b) If α(0) = 2, β codes A and α(n + 1) = β(n), then αcodes N − A.

(c) If α(0) = 3 and for each n the functionm→ α(〈n,m〉+ 1) codes An then α codes ∪nAn.

In other words, the set of pairs (α,A) such that α is a code for Ais 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 forα(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 setA’, 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 ofunions 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 isα(〈k ,n〉).

The tree Tα of a code α

Every α ∈ N is associated with a tree Tα with a real numberattached 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 ofTα 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 inTα and Fα(s ∗ 0)(n) = Fα(s)(n + 1).

(d) If Fα(s)(0) = 3 then s ∗ k ∈ Tα for all k ∈ ω andFα(s ∗ k)(n) = Fα(〈k ,n〉+ 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 andthe subtree below it is described by the digits > 0ofFα(s).

(d) If Fα(s)(0) = 3 then all successors of s survive the treebelow 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 ,Twhich 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 Tand F (s ∗ 0)(n) = F (s)(n + 1).

(d1) If F (s)(0) = 3 then s ∗ k ∈ T for all k ∈ ω andF (s ∗ k)(n) = F (〈k ,n〉+ 1)

Coding Borel sets into reals

TheoremA 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 codefor Borel set A’ and Tα is well founded.

Proof

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

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

Now it suffices to prove the converse.

This is done by transfinite induction along the possible lengthsof 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 ofthe ω subtrees.

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

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

Task

Draw the tree Tα for the case where α codes a Σ0n or Π0

n set forn ≤ 3.

CorollaryThe relation ‘α is the code of a Borel set’ is Π1

1.

Proof.

The set Tα is computable in α i.e. ∆01(α).

The relation P(β, α): ‘β is a path through Tα’ is arithmetical(Π0

1).

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

This is Π11.

By the same argument we have the following.

PropositionThe relation φ(α,T ,F ) : ‘α,T ,F satisfy (a1)-(d1)’ is Π0

1.

Proof.Notice that (a’)-(d’) are computable conditions with free variablethe node s of T .

So we need to say ‘for all s, (a1)-(d1) which is a Π01

condition.

PropositionThe relation ψ(α,T ,F ) : φ(α,T ,F ) ∧ T is well-founded’ is Π1

1.

Proof.The relation ‘T is well-founded’ means that T does not haveinfinite paths. Hence it can be written as:

∀β ∃n β � n 6∈ T .

This is a Π11 relation. Also, φ is Π0

1.

So ψ is a conjunction of a Π11 and a Π0

1 formula. The normalform of such sentences is Π1

1.

TheoremLet α code a Borel set B. The predicate ‘β is in the set codedby α’ is ∆1

1(α) (i.e. ∆11 with parameter α).

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

Indeed, if we determine membership (or not) for all of thepredecessors of a given node, we can easily determine it forthe set corresponding to the given node.

If we determine membership of β with respect to a set A, wecan 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 withrespect to the tope node α, i.e. the set we are given.

Of course membership with respect to the leaves (i.e. closedsets) is easy, it is a Π0

1 relation (stay on the corresponding treefor ever).

The predicate P below is defined inductively using exactly thisidea.

Given T ,F as before, we define a predicate P (i.e. function withbinary 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.

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

Proof.The proof follows by a simultaneous induction on the length ofT . During the steps of the induction we show that β is in the setrepresented by a node s of T iff P[β](s) = 1 (otherwise thevalue is 0).

Notice that the conditions (I)-(III), although arithmetical, theyare not computable. Therefore the complexity of P[β](s) can bequite high, since it depends on the length of T . If the length isinfinite, there is a possibility that P[β](s) is not arithmetical.However we can overcome these problems by through asecond order definition.

PropositionThe sentence ρ(β,T ,F ,P, β): ‘P[β] satisfies (I)-(III)’ is Π0

2. Inparticular, it is arithmetical.

Proof.It suffices to say that for all nodes s of T , (I)-(III) hold. Since allof these sentences are either Σ0

1 or Π01, 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].

This is a ∆11(α) definition, since ψ and ρ are arithmetical.

CorollaryEvery Borel set is ∆1

1 in a real parameter.

TheoremFor each n ∈ N, every Σ0

n set is Σ0n in a real parameter, and

every Π0n set is Π0

n.

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.I Thus every open set is Σ0

1 in the real parameter whichdescribes the countable union.

I The negation of a Σ01(α) set is Π0

1(α).I The union of the a countable collection of sets Ai which are

Π0n(αi) is Σ0

n+1(α), where α is a parameter which codes thecountable collection of reals αi , i ∈ N.

TheoremI Every Σ1

1 set is Σ11 in a real parameter.

I Every Π11 set is Π1

1 in a real parameter.I Every ∆1

1 set is ∆11 in a real parameter.

Similar relations hold for the higher levels of the hierarchy (byinduction).

Proof

A Σ11 set is definable by a formula like ∃ωαP(α,n), where P is a

Π01 formula.

But by previous theorem, P is Π01 in a real parameter γ.

So the formula ∃ωαP(α,n) is Σ11 in γ and so is the set it defines.

The same holds for Π11 and thus for ∆1

1.

PropositionFor all parameters α we haveI Σ0

1(α) ⊆ Σ01 and Σ1

1(α) ⊆ Σ11

I Π01(α) ⊆ Π0

1 and Π11(α) ⊆ Π1

1I ∆0

1(α) ⊆ ∆01 and ∆1

1(α) ⊆ ∆11

Analogous relations hold for higher classes of the hierarchies.

ProofIt suffices to show that every Σ0

1(α) (or every ∆01(α)) predicate

is open.

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

Suppose that α satisfies the above sentence P.

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

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

As long as another real β agrees with α on those digits, it willalso 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 theproof.

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 thesame decision of any β in place of α, which agrees with α in thefinitely 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 complexityof classes and sets.

For example the various hierarchies of complexity.

To conclude the episode of characterizing the complexity of theclass 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 spaceis countable.

Denote by ||T || the length of a tree T .

TheoremLet S,T be trees. Then ||S|| ≤ ||T || iff there is an orderpreserving map f : S → T .Proof:

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

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 onthe length of ||T ||.

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

||S � sn|| < ||S|| ≤ ||T || = supt 6=∅(||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 mapfn : S � sn → T � tn.

For all s ∈ S define

f (s) =

{∅ if 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: theset

{〈α, β〉 | α codes tree S and β codes tree T with ||S|| ≤ ||T ||}

is Σ11.

Identifying codes with the objects that they code, we have:

CorollaryThe relation ‘||S|| ≤ ||T ||’ is Σ1

1.

CorollaryIf T is a well founded tree, then the set {S | ||S|| ≤ ||T ||} is ∆1

1in parameter T .

Proof.Since we have shown that ||S|| ≤ ||T || is Σ1

1, it suffices to showthat its negation, ||T || < ||S|| is also Σ1

1.

Indeed, if T is well-founded, then

||T || < ||S|| ⇐⇒ ∃n ||T || ≤ ||S � 〈n〉||.

TheoremFor every countable ordinal σ the set WFσ is Borel.

ProofBy 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 thefull subtree T � s of T below node s.

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

WFσ = ∩s ∪τ<σ {T | T � s ∈WFτ} = ∩s ∩τ<σ f−1s (WFτ ).

Since the Borel class is closed under continuous inverseimages, countable unions and intersections this shows thatWFσ is Borel.

Trees in product spacesIn the same way that we used trees to represent closed sets inN 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 〈s, t〉of strings (finite sequences) of numbers, such that |s| = |t |.

Here ‘downward closed’ means that if 〈s, t〉 is in the tree ands′ ⊆ s, t ′ ⊆ t and |s′| = |t ′| then 〈s′, t ′〉 is also in the tree. Using

the topology of N ×N you can verify (as in the case of N ) thata set in that product space is closed iff it is the set of infinitepaths 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 factsdiscussed above.

Theorem (Normal form theorem for Π11)

A set A is Π11 iff there is a tree on ω × ω (i.e. in the space

N ×N ) such that α ∈ A ⇐⇒ ∀β ∃n 〈α � n, β � n〉 6∈ T .

If Tα = {σ | 〈α � |σ|, σ〉 ∈ T} is the α-section of thetwo-dimentional tree T , then we have:

α ∈ A ⇐⇒ Tα ∈WF .

Notice that Tα is a continuous function on α (exercise).

Theorem (Normal form theorem for Π01)

A set A ⊆ N ×N is Π01 iff there is a ∆0

1 (i.e. computable) tree onω × ω (i.e. in the space N ×N ) such that

〈α, β〉 ∈ A ⇐⇒ ∀n 〈α � n, β � n〉 ∈ T .

Theorem (Normal form theorem for Π11)

A set A is Π11 iff there is a ∆0

1 (i.e. computable) tree on ω × ω(i.e. in the space N ×N ) such that

α ∈ A ⇐⇒ ∀β ∃n 〈α � n, β � n〉 6∈ T .

TheoremFor each n ∈ N and i = 0,1 we have Σi

n 6⊆ Πin and Πi

n 6⊆ Σin.

Proof.This is an exercise. First use continuous substitution to showthat if e.g. P(α, β,n) ∈ Πi

n then for any constant α0 the relationQ(β,n) ≡ P(α0, β, n) is also in Πi

n.

To prove our hierarchy theorems (show that the hierarchies donot collapse) we had to devise somewhat artificial universalrelations.

Here is a natural example:

TheoremWF is Π1

1 but not Σ11.

Proof:

We already know that it is Π11.

Assume that it was Σ11 and let A ∈ Σ1

1 −Π11.

Proof (cont.)

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

α ∈ A ⇐⇒ Tα 6∈WF

where Tα is the α-column of T , i.e. {s | 〈α � |s|, s〉 ∈ T}.

But Tα 6∈WF is Π11 since its complement is Σ1

1.

Then A is Π11, a contradiction.

Theorem (Boundedness theorem)If A is a Σ1

1 subset of WF then there is a countable ordinal σsuch 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 ||].

But we have shown that ‘||S|| ≥ ||T ||’ is Σ11, so this is a Σ1

1definition of WF .

This is impossible !

TheoremA set is Borel iff it is ∆1

1 iff it is ∆11 in a real parameter.

For the proof it remains to show that every ∆11 set is Borel.

Proof:

Let A ∈ ∆11. As a Π1

1 set

α ∈ A ⇐⇒ Tα ∈WF

for some tree T .

Proof (cont.)

Since A is Σ11 the set {Tα | α ∈ A} is a Σ1

1 subset of WF .By the boundedness theorem it is contained in some WFσhence

α ∈ A ⇐⇒ Tα ∈WFσ.

This gives A as a continuous inverse image of a Borel set,hence A is Borel.

DefinitionA perfect set A is a closed set with no isolated point, i.e. forevery basic open set Ns, Ns ∩ A is either empty or containsmore than one element.

TheoremEvery 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 everynode 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 letf (t ∗ 0) = u and f (t ∗ 1) = v where u, v are two incompatibleextensions of f (t).

Proof of theorem:

By induction for any infinite path α in T we haveF (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, βdiffer on digint n − 1, then f (α � n), f (β � n) will be incompatiblestrings, 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 asthe disjoint union of a perfect (closed) set and a countable set.

DefinitionA point x is a condensation point of a set F , if Ns ∩ F isuncountable for every Ns such that x ∈ Ns.

Proof of C.-B. theoremPropositionIf P is the set of condensation points of X , then X − P iscountable.

Proof.For every point in X − P there is a a basic open set Ns whichcontains 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 imagef−1(Ns) is contained in Ns ∩ X , hence it is countable.

Since X −P is the union of all f−1(Ns), where Ns is an image off , it is a countable union of countable sets, hencecountable.

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 thenon-condensation points of it.

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

The set of non-condensation points of F is countable by aboveproposition.

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 thatx ∈ N.

Then N ∩ P is closed. Also, perfect because an isolated path init 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 thatN ∩ P = ∅, since P is closed.

Since S is countable, N ∩ F is countable. So x is not acondensation point.

Cantor-Bendixson derivative

DefinitionThe Cantor-Bendixson derivative A′ of a closed set A is the setof limit points of A. An element of the space is a limit point if it isnot isolated in A. That is, if there is no basic open Ns such thatNs ∩ A 6= ∅ consists exactly of that element.

Let A(0) = A and for limit ordinals σ let A(σ) = ∩τ<σA(τ).

So we have the following.

PropositionA closed set [T ] is perfect iff [T ]′ = [T ].

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

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

Clearly A(σ) ⊆ A(τ) for ordinals τ ≤ σ.

TheoremI The set ∩ξA(ξ) (where ξ runs through all ordinals) is perfect

(perhaps empty).

I A− A(ξ) is countable for all ξ.

I There is a countable ordinal λ such that ∩ξA(ξ) = A(λ).

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).

Hence ∩ξA(ξ) is closed.

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

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

Since A(σ) ⊆ A(τ) for ordinals τ ≤ σ we have Nσ 6= Nτ for τ ≤ σ.

Since there are only countably many neighborhoods this meansthat the σ such that A(ξ) − A(λ) 6= ∅ for all λ < ξ ≤ σ form a acountable initial segment of the ordinals.

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

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

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

would too.

So A(τ) 6= A(τ+1) (contradiction).

PropositionIf P is the set of condensation points of a closed set X , thenP ⊆ 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) iscountable.

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 leastcountable σ such that F (σ) = F (σ+1).

We call this the Cantor-Bendixson rank of F .

CorollaryEvery closed set has cardinality either ℵ0 or 2ℵ0 .

Proof.If a close set is not countable, then by the Cantor-Bendixsontheorem it contains a perfect (closed) set. The latter is hascardinality 2ℵ0 as we have shown. Therefore any superset of it(subset of the space) has the same cardinality.

TheoremThe class B of Borel sets is the smallest class containing theclosed sets and closed under countable intersections andcountable disjoint unions.

For the proof, let D be the class defined in the theorem, and ¬Dthe 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 setsand is closed under complementation and countable unions.

The closed sets are in D so it suffices to show that the opensets 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 itremains to show that it is closed under countable unions.

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

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

¬∩n An = ∪n¬An = ∪n(¬An−(∪k<n¬Ak )) = ∪n(¬An∩(∩k<nAk ))

The last expression shows that the set is a disjoint union of setsin D, hence it is in D.

TheoremEvery Borel set is a 1–1 continuous image of a closed subset ofthe 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 countableintersections and disjoint unions.

First disjoint unions. Suppose that for each n, fn is a continuous1–1 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(〈n,m〉).

The map α→ αn is continuous.

Let F = ∩n{α | αn ∈ [Tn]}.

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

CorollaryEvery uncountable Borel set has cardinality 2ℵ0 .

Proof:

Given an uncountable Borel set, it is a 1–1 continuous image ofan 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 infinitepaths 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 compactset.

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

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

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

This finishes the proof.

Although the continuum problem can be solved for Borel setsvia perfect sets, the same does not hold in general.

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

TheoremThere is a subset of the Baire space which neither contains noris disjoint from any perfect set.

Proof

We construct it. Let 〈Pσ | σ < 2ℵ0〉 be a well-ordering of allperfect trees.

We define two sequences of sets (Aσ)σ<2ℵ0 , (Bσ)σ<2ℵ0 bytransfinite 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 cardinalitybelow 2ℵ0 .

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

Such α, β exist because [Pσ] has cardinality 2ℵ0 and each ofthe 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 fromany perfect set.