descriptive set theory - polito.itcalvino.polito.it/~camerlo/caserta2014-1.pdf · 2014-10-09 ·...

Descriptive set theory

Descriptive set theory: classical and effective

I Classical descriptive set theory was founded by people like Baire,Borel, Lebesgue, Luzin, Suslin, Sierpinskpi and others in the first twodecades of the XXth century. It studies the descriptive complexity ofsets of real numbers.

I Effective descriptive set theory was created later by introducing intothe classical theory the new and powerful tools developed fromrecursion theory.

Basic review on topological and metric spaces

Let X be a topological space.

I X is T1 if given x , y ∈ X there is a neighbourhood of x that doesnot contain y (and viceversa); equivalently, every singleton is closed.

I X is regular if X is T1 and whenever x ∈ X ,A ⊆ X , A is closed andx /∈ A, point x and set A have disjoint neighbourhoods; equivalently,x is T1 and for all x and any open neighbourhood U of x , there isan open neighbourhood V of x with V ⊆ U.

I X is normal if X is T1 and any two disjoint closed subsets havedisjoint neighbourhoods.

I X is Baire if the intersection of countably many open dense sets isdense.

I A subset of X is Gδ if it is a countable intersection of open sets. Thecomplement of a Gδ is an Fσ set: a countable union of closed sets.

I If (X , d) is a metric space, then d ′ = d1+d is a metric compatible

with d . Notice that d ′ ≤ 1 and d is complete iff d ′ is complete.




I X is regular if X is T1 and whenever x ∈ X ,A ⊆ X , A is closed andx /∈ A, point x and set A have disjoint neighbourhoods;

equivalently,x is T1 and for all x and any open neighbourhood U of x , there isan open neighbourhood V of x with V ⊆ U.












I A subset of X is Gδ if it is a countable intersection of open sets.

Thecomplement of a Gδ is an Fσ set: a countable union of closed sets.











with d .

Notice that d ′ ≤ 1 and d is complete iff d ′ is complete.


Urysohn’s theorem. Let X be second countable. Then X is metrisableiff it is regular.

Urysohn’s lemma. If X is metrisable and A,B ⊆ X are closed anddisjoint, then ∃f : X → [0, 1] continuous s.t. f (A) ⊆ {0}, f (B) ⊆ {1}.

Tietze’s extension theorem. If X is metrisable, A ⊆ X is closed andf : A→ R is continuous, then ∃g : X → R continuous s.t. f ⊆ g .If moreover sup f ≤ M, then one can find g with sup g ≤ M.




Tietze’s extension theorem. If X is metrisable, A ⊆ X is closed andf : A→ R is continuous, then ∃g : X → R continuous s.t. f ⊆ g .

If moreover sup f ≤ M, then one can find g with sup g ≤ M.




Tietze’s extension theorem. If X is metrisable, A ⊆ X is closed andf : A→ R is continuous, then ∃g : X → R continuous s.t. f ⊆ g .If moreover sup f ≤ M, then one can find g with sup g ≤ M.

Trees

Definition

I T is a (descriptive) tree on set A if T is non-empty, T ⊆ A<ω andT is closed under subsequences.

I If T is a tree on A, then [T ] = {f ∈ AN | ∀n f |n ∈ T} is the body ofT .

Example. [2<ω] = 2N is the Cantor space.

Trees

Definition



Example. [2<ω] = 2N

is the Cantor space.

Trees

Definition



Example. [2<ω] = 2N is the Cantor space.

A metric on AN

d(a, b) =

{0 if a = b12n if n is least with a(n) 6= b(n)

is a ultrametric on AN, i.e. a metric such thatd(a, b) ≤ max(d(a, c), d(c , b)).

Exercise. Check this.

Open balls centered in g ∈ AN are sets of the formNg |n = {f ∈ AN | g |n = f |n} for n ∈ N.

Thus d is compatible with the product topology on AN, where A is giventhe discrete topology.

A metric on AN

d(a, b) =

{0 if a = b

12n if n is least with a(n) 6= b(n)





A metric on AN

d(a, b) =

{0 if a = b12n if n is least with a(n) 6= b(n)





A metric on AN

Propositiond is a complete metric.

Proof.Let fn be a Cauchy sequence in AN. For any m, eventually all fn’s willhave the first n digits fixed. These digits define an element f ∈ AN suchthat limn→∞ fn = f .

Notice that AN is separable if (and only if) A is at most countable.

This is a first example and will consitute the motivating example for thefirst fundamental definition, that of a Polish space.

A metric on AN


Proof.Let fn be a Cauchy sequence in AN.

For any m, eventually all fn’s willhave the first n digits fixed. These digits define an element f ∈ AN suchthat limn→∞ fn = f .



A metric on AN


Proof.Let fn be a Cauchy sequence in AN. For any m, eventually all fn’s willhave the first n digits fixed.

These digits define an element f ∈ AN suchthat limn→∞ fn = f .



A metric on AN


Proof.Let fn be a Cauchy sequence in AN. For any m, eventually all fn’s willhave the first n digits fixed. These digits define an element f ∈ AN

suchthat limn→∞ fn = f .



A metric on AN


Proof.Let fn be a Cauchy sequence in AN. For any m, eventually all fn’s willhave the first n digits fixed. These digits define an element f ∈ AN suchthat limn→∞ fn = f .



Basics on trees

DefinitionA tree T is pruned if it has no terminal nodes: ∀t ∈ T ∃s ∈ T t ⊂ s.

Example. Finite non-empty trees are not pruned.

Exercise. The map T 7→ [T ] is a bijection between pruned trees on Aand closed subsets of AN. Its inverse is F 7→ TF = {x |n | x ∈ F , n ∈ N}.

Some induced trees. If T is a tree on A and s ∈ A<ω, then thefollowing are trees on A:

I Ts = {t ∈ A<ω | st ∈ T}I T[s] = {t ∈ T | t is compatible with s}

Basics on trees

DefinitionA tree T is pruned if it has no terminal nodes:

∀t ∈ T ∃s ∈ T t ⊂ s.





Basics on trees






Basics on trees





I Ts = {t ∈ A<ω | st ∈ T}

I T[s] = {t ∈ T | t is compatible with s}

Basics on trees






Basics on trees

DefinitionLet T a tree on A and S a tree on B.

A function ϕ : T → S is monotoneif

∀t1, t2 ∈ T (t1 ⊆ t2 ⇒ ϕ(t1) ⊆ t2)

Let D(ϕ) = {x ∈ [T ] | limn→∞ length(ϕ(x |n)) = +∞}. Then one candefine

ϕ∗(x) =⋃n∈N

ϕ(x |n) ∈ [S ]

Exercise. D(ϕ) is Gδ in [T ] and ϕ∗ : D(ϕ)→ [S ] is continuous.

Basics on trees

DefinitionLet T a tree on A and S a tree on B. A function ϕ : T → S is monotoneif

∀t1, t2 ∈ T (t1 ⊆ t2 ⇒ ϕ(t1) ⊆ t2)


ϕ∗(x) =⋃n∈N

ϕ(x |n) ∈ [S ]


Basics on trees


∀t1, t2 ∈ T (t1 ⊆ t2 ⇒ ϕ(t1) ⊆ t2)

Let D(ϕ) = {x ∈ [T ] | limn→∞ length(ϕ(x |n)) = +∞}.

Then one candefine

ϕ∗(x) =⋃n∈N

ϕ(x |n) ∈ [S ]


Basics on trees


∀t1, t2 ∈ T (t1 ⊆ t2 ⇒ ϕ(t1) ⊆ t2)


ϕ∗(x) =⋃n∈N

ϕ(x |n)

∈ [S ]


Basics on trees


∀t1, t2 ∈ T (t1 ⊆ t2 ⇒ ϕ(t1) ⊆ t2)


ϕ∗(x) =⋃n∈N

ϕ(x |n) ∈ [S ]


Polish spaces

Definition

I A Polish space is a separable, completely metrisable topologicalspace

I A Polish metric space is a separable metric space (X , d) such that dis complete

Basic properties.

I Any Polish space is second countable and normal

I A finite or countable product of Polish spaces is Polish

I (Baire category theorem) Polish spaces are Baire (in fact everycompletely metrisable space is Baire)

I A quotient of a Polish space and a subspace of a Polish space arenot necessarily Polish

I A closed subspace of a Polish space is Polish

First examples.

I All countable discrete spaces

I Cantor space 2N

Exercise. Prove that is homeomorphic to the 13 -Cantor set

E 13⊆ [0, 1] via x 7→

∑∞n=0

2x(n)3n+1

I Baire space NN

Exercise. Prove that it is homeomorphic to R \Q (equivalently, to(R \Q) ∩ [0, 1]), using continued fractions.

I Rn,RN

I ]0, 1[, {z ∈ C | |z | = 1}, [0, 1]

I Hilbert cube [0, 1]N

I All separable Banach spaces

Exercise. Compact metrisable spaces are Polish.

First examples.


I Cantor space 2N


E 13⊆ [0, 1]

via x 7→∑∞

n=02x(n)3n+1

I Baire space NN


I Rn,RN

I ]0, 1[, {z ∈ C | |z | = 1}, [0, 1]




First examples.


I Cantor space 2N


E 13⊆ [0, 1] via x 7→

∑∞n=0

2x(n)3n+1

I Baire space NN


I Rn,RN

I ]0, 1[, {z ∈ C | |z | = 1}, [0, 1]




First examples.


I Cantor space 2N


E 13⊆ [0, 1] via x 7→

∑∞n=0

2x(n)3n+1

I Baire space NN

Exercise. Prove that it is homeomorphic to R \Q (equivalently, to(R \Q) ∩ [0, 1])

, using continued fractions.

I Rn,RN

I ]0, 1[, {z ∈ C | |z | = 1}, [0, 1]




First examples.


I Cantor space 2N


E 13⊆ [0, 1] via x 7→

∑∞n=0

2x(n)3n+1

I Baire space NN


I Rn,RN

I ]0, 1[, {z ∈ C | |z | = 1}, [0, 1]




Vietoris topology

Let X be a topological space. Let K (X ) be the family of all compactsubsets of X . The Vietoris topology on K (X ) is the one generated bysets of the form:

{K ∈ K (X ) | K ⊆ U} and {K ∈ K (X ) | K ∩ U 6= ∅}

for U open in X . A basis is given by sets of the form

{K ∈ K (X ) | K ⊆ K0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅}

for n ∈ N and U0,U1, . . . ,Un open.

Exercise.

I ∅ is an isolated point in K (X )

I If X is a subspace of Y , then K (X ) is a subspace of K (Y )

Vietoris topology

Let X be a topological space. Let K (X ) be the family of all compactsubsets of X .

The Vietoris topology on K (X ) is the one generated bysets of the form:

{K ∈ K (X ) | K ⊆ U} and {K ∈ K (X ) | K ∩ U 6= ∅}


{K ∈ K (X ) | K ⊆ K0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅}


Exercise.



Vietoris topology


{K ∈ K (X ) | K ⊆ U} and {K ∈ K (X ) | K ∩ U 6= ∅}


{K ∈ K (X ) | K ⊆ K0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅}


Exercise.



Vietoris topology


{K ∈ K (X ) | K ⊆ U}

and {K ∈ K (X ) | K ∩ U 6= ∅}


{K ∈ K (X ) | K ⊆ K0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅}


Exercise.



Vietoris topology


{K ∈ K (X ) | K ⊆ U} and {K ∈ K (X ) | K ∩ U 6= ∅}


{K ∈ K (X ) | K ⊆ K0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅}


Exercise.



Vietoris topology


{K ∈ K (X ) | K ⊆ U} and {K ∈ K (X ) | K ∩ U 6= ∅}

for U open in X .

A basis is given by sets of the form

{K ∈ K (X ) | K ⊆ K0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅}


Exercise.



Vietoris topology


{K ∈ K (X ) | K ⊆ U} and {K ∈ K (X ) | K ∩ U 6= ∅}


{K ∈ K (X ) | K ⊆ K0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅}


Exercise.



Vietoris topology

PropositionIf X is separable, then K (X ) is separable.

Proof.Let D be countable dense in X . Then Kfin(D) = {K ⊆ D | D is finite} iscountable and it is dense in K (X ). Suppose indeed

V = {K ∈ K (X ) | K ⊆ U0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅} 6= ∅

where U0,U1, . . . ,Un are open sets. If n = 0, then ∅ ∈ V ∩ Kfin(D).Otherwise, U0 ∩ Uj 6= ∅ for all j 6= 0. Picking a point in any U0 ∩ Uj

yields a K ∈ V ∩ Kfin(D).

Vietoris topology


Proof.Let D be countable dense in X .

Then Kfin(D) = {K ⊆ D | D is finite} iscountable and it is dense in K (X ). Suppose indeed

V = {K ∈ K (X ) | K ⊆ U0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅} 6= ∅



Vietoris topology


Proof.Let D be countable dense in X . Then Kfin(D) = {K ⊆ D | D is finite} iscountable

and it is dense in K (X ). Suppose indeed

V = {K ∈ K (X ) | K ⊆ U0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅} 6= ∅



Vietoris topology


Proof.Let D be countable dense in X . Then Kfin(D) = {K ⊆ D | D is finite} iscountable and it is dense in K (X ).

Suppose indeed

V = {K ∈ K (X ) | K ⊆ U0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅} 6= ∅



Vietoris topology



V = {K ∈ K (X ) | K ⊆ U0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅} 6= ∅

where U0,U1, . . . ,Un are open sets.

If n = 0, then ∅ ∈ V ∩ Kfin(D).Otherwise, U0 ∩ Uj 6= ∅ for all j 6= 0. Picking a point in any U0 ∩ Uj


Vietoris topology



V = {K ∈ K (X ) | K ⊆ U0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅} 6= ∅

where U0,U1, . . . ,Un are open sets. If n = 0, then ∅ ∈ V ∩ Kfin(D).

Otherwise, U0 ∩ Uj 6= ∅ for all j 6= 0. Picking a point in any U0 ∩ Uj


Vietoris topology



V = {K ∈ K (X ) | K ⊆ U0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅} 6= ∅

where U0,U1, . . . ,Un are open sets. If n = 0, then ∅ ∈ V ∩ Kfin(D).Otherwise, U0 ∩ Uj 6= ∅ for all j 6= 0.

Picking a point in any U0 ∩ Uj


Vietoris topology



V = {K ∈ K (X ) | K ⊆ U0 ∧ K ∩ U1 6= ∅ ∧ . . . ∧ K ∩ Un 6= ∅} 6= ∅



Lower and upper topological limits

DefinitionLet X be a topological space, Kn ∈ K (X ).

I Topological upper limit:

Tlimn→∞Kn = {x | ∀U open nbhd of x ∃∞n U ∩ Kn 6= ∅} =

=⋂

n∈N⋃∞

i=n Ki

I Topological lower limit:

Tlimn→∞Kn = {x | ∀U open nbhd of x ∀∞n U ∩ Kn 6= ∅}

RemarksI Tlimn→∞Kn ⊆ Tlimn→∞Kn. If they are equal, one writes

Tlimn→∞Kn.I If X is first countable,

TlimnKn = {x | ∃ sequence xn (∀n xn ∈ Kn∧∃ subsequence xni → x)}




Tlimn→∞Kn

= {x | ∀U open nbhd of x ∃∞n U ∩ Kn 6= ∅} =

=⋂

n∈N⋃∞

i=n Ki









Tlimn→∞Kn = {x | ∀U open nbhd of x ∃∞n U ∩ Kn 6= ∅}

=

=⋂

n∈N⋃∞

i=n Ki










=⋂

n∈N⋃∞

i=n Ki










=⋂

n∈N⋃∞

i=n Ki


Tlimn→∞Kn

= {x | ∀U open nbhd of x ∀∞n U ∩ Kn 6= ∅}








=⋂

n∈N⋃∞

i=n Ki










=⋂

n∈N⋃∞

i=n Ki



RemarksI Tlimn→∞Kn ⊆ Tlimn→∞Kn.

If they are equal, one writesTlimn→∞Kn.

I If X is first countable,






=⋂

n∈N⋃∞

i=n Ki




Tlimn→∞Kn.

I If X is first countable,






=⋂

n∈N⋃∞

i=n Ki






Hausdorff metric

Let d ≤ 1 be a compatible metric. The Hausdorff metric on K (X ) isdefined by letting

dH (K , L) =

0 if K = L1 if exactly one of K , L is emptymax(δ(K , L), δ(L,K )) in all other cases

where δ(K , L) = max{d(x , L)}x∈K . So, letting B(K , ε) =⋃

x∈K Bε(x), ifK 6= ∅ 6= L then

dH (K , L) < ε⇔ K ⊆ B(L, ε) ∧ L ⊆ B(K , ε)

Exercise. Hausdorff metric is compatible with Vietoris topology.

Hausdorff metric

Let d ≤ 1 be a compatible metric.

The Hausdorff metric on K (X ) isdefined by letting

dH (K , L) =




dH (K , L) < ε⇔ K ⊆ B(L, ε) ∧ L ⊆ B(K , ε)


Hausdorff metric


dH (K , L) =




dH (K , L) < ε⇔ K ⊆ B(L, ε) ∧ L ⊆ B(K , ε)


Hausdorff metric


dH (K , L) =

0 if K = L

1 if exactly one of K , L is emptymax(δ(K , L), δ(L,K )) in all other cases



dH (K , L) < ε⇔ K ⊆ B(L, ε) ∧ L ⊆ B(K , ε)


Hausdorff metric


dH (K , L) =

0 if K = L1 if exactly one of K , L is empty

max(δ(K , L), δ(L,K )) in all other cases



dH (K , L) < ε⇔ K ⊆ B(L, ε) ∧ L ⊆ B(K , ε)


Hausdorff metric


dH (K , L) =


where δ(K , L) = max{d(x , L)}x∈K .

So, letting B(K , ε) =⋃


dH (K , L) < ε⇔ K ⊆ B(L, ε) ∧ L ⊆ B(K , ε)


Hausdorff metric


dH (K , L) =




dH (K , L) < ε⇔ K ⊆ B(L, ε) ∧ L ⊆ B(K , ε)


Hausdorff metric


dH (K , L) =




dH (K , L) < ε⇔

K ⊆ B(L, ε) ∧ L ⊆ B(K , ε)


Hausdorff metric


dH (K , L) =




dH (K , L) < ε⇔ K ⊆ B(L, ε) ∧ L ⊆ B(K , ε)


Completeness of K (X )

TheoremIf X is completely metrisable, then K (X ) is completely metrisable.

ProofLet d ≤ 1 be a compatible complete metric on X . Let dH be Hausfdorffmetric on K (X ) and fix a dH -Cauchy sequence Kn. Since ∅ is an isolatedpoint in K (X ), one can assume ∀n Kn 6= ∅. Set

K = Tlimn→∞Kn =⋂n∈N

∞⋃i=n

Ki



ProofLet d ≤ 1 be a compatible complete metric on X .

Let dH be Hausfdorffmetric on K (X ) and fix a dH -Cauchy sequence Kn. Since ∅ is an isolatedpoint in K (X ), one can assume ∀n Kn 6= ∅. Set


∞⋃i=n

Ki



ProofLet d ≤ 1 be a compatible complete metric on X . Let dH be Hausfdorffmetric on K (X ) and fix a dH -Cauchy sequence Kn.

Since ∅ is an isolatedpoint in K (X ), one can assume ∀n Kn 6= ∅. Set


∞⋃i=n

Ki



ProofLet d ≤ 1 be a compatible complete metric on X . Let dH be Hausfdorffmetric on K (X ) and fix a dH -Cauchy sequence Kn. Since ∅ is an isolatedpoint in K (X ), one can assume ∀n Kn 6= ∅.

Set


∞⋃i=n

Ki



ProofLet d ≤ 1 be a compatible complete metric on X . Let dH be Hausfdorffmetric on K (X ) and fix a dH -Cauchy sequence Kn. Since ∅ is an isolatedpoint in K (X ), one can assume ∀n Kn 6= ∅. Set


∞⋃i=n

Ki

Proof (cont’d)

Claim 1. K is compact.

Proof of claim. Since K is closed, it in enough to prove that it is totallybounded. For this it is enough to prove that L =

⋃i∈N Ki is totally

bounded:∀n ∃F ⊆ X (F is finite ∧ L ⊆

⋃x∈F

B 12n

(x))

Let Fi be finite such that Ki ⊆⋃

x∈FiB 1

2n+1(x). Let also p be such that

∀j ≥ p dh(Kp,Kj ) <1

2n+1 . Set

F =

p⋃i=0

Fi

Proof (cont’d)


Proof of claim. Since K is closed, it in enough to prove that it is totallybounded.

For this it is enough to prove that L =⋃

i∈N Ki is totallybounded:

∀n ∃F ⊆ X (F is finite ∧ L ⊆⋃x∈F

B 12n

(x))


x∈FiB 1



2n+1 . Set

F =

p⋃i=0

Fi

Proof (cont’d)




bounded:

∀n ∃F ⊆ X (F is finite ∧ L ⊆⋃x∈F

B 12n

(x))


x∈FiB 1



2n+1 . Set

F =

p⋃i=0

Fi

Proof (cont’d)





⋃x∈F

B 12n

(x))


x∈FiB 1



2n+1 . Set

F =

p⋃i=0

Fi

Proof (cont’d)





⋃x∈F

B 12n

(x))


x∈FiB 1

2n+1(x).

Let also p be such that


2n+1 . Set

F =

p⋃i=0

Fi

Proof (cont’d)





⋃x∈F

B 12n

(x))


x∈FiB 1



2n+1 .

Set

F =

p⋃i=0

Fi

Proof (cont’d)





⋃x∈F

B 12n

(x))


x∈FiB 1



2n+1 . Set

F =

p⋃i=0

Fi

Proof (cont’d)

Claim 2. limn→∞ Kn = K

Proof of claim. Fix ε > 0 and let N be s.t. ∀i , j ≥ N dH (Ki ,Kj ) < ε. Itwill be proved ∀n > N dH (Kn,K ) < 2ε.

I Let x ∈ K and let xni ∈ Kni with limi→∞ xni = x . Eventually, ni > Nand d(xni , x) < ε. Then pick yn ∈ Kn s.t. d(xni , yn) < ε. It followsd(x , yn) ≤ d(x , xni ) + d(xni , yn) < 2ε, so δ(K ,Kn) < 2ε.

I Let now y ∈ Kn. Let n = k0 < k1 < . . . s.t.∀m ≥ kj dH (Kkj ,Km) < ε

2j . Define{x0 = y ∈ Kn = Kk0

xj+1 ∈ Kkj+1 s.t. d(xj+1, xj ) <ε2j

So xj is Cauchy. Then x = limj→∞ xj ∈ K . Moreover

d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε

Consequently δ(Kn,K ) < 2ε.

Proof (cont’d)


Proof of claim. Fix ε > 0 and let N be s.t. ∀i , j ≥ N dH (Ki ,Kj ) < ε.

Itwill be proved ∀n > N dH (Kn,K ) < 2ε.






d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)








d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)



I Let x ∈ K

and let xni ∈ Kni with limi→∞ xni = x . Eventually, ni > Nand d(xni , x) < ε. Then pick yn ∈ Kn s.t. d(xni , yn) < ε. It followsd(x , yn) ≤ d(x , xni ) + d(xni , yn) < 2ε, so δ(K ,Kn) < 2ε.





d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)



I Let x ∈ K and let xni ∈ Kni with limi→∞ xni = x .

Eventually, ni > Nand d(xni , x) < ε. Then pick yn ∈ Kn s.t. d(xni , yn) < ε. It followsd(x , yn) ≤ d(x , xni ) + d(xni , yn) < 2ε, so δ(K ,Kn) < 2ε.





d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)



I Let x ∈ K and let xni ∈ Kni with limi→∞ xni = x . Eventually, ni > Nand d(xni , x) < ε.

Then pick yn ∈ Kn s.t. d(xni , yn) < ε. It followsd(x , yn) ≤ d(x , xni ) + d(xni , yn) < 2ε, so δ(K ,Kn) < 2ε.





d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)



I Let x ∈ K and let xni ∈ Kni with limi→∞ xni = x . Eventually, ni > Nand d(xni , x) < ε. Then pick yn ∈ Kn s.t. d(xni , yn) < ε.

It followsd(x , yn) ≤ d(x , xni ) + d(xni , yn) < 2ε, so δ(K ,Kn) < 2ε.





d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)



I Let x ∈ K and let xni ∈ Kni with limi→∞ xni = x . Eventually, ni > Nand d(xni , x) < ε. Then pick yn ∈ Kn s.t. d(xni , yn) < ε. It followsd(x , yn)

≤ d(x , xni ) + d(xni , yn) < 2ε, so δ(K ,Kn) < 2ε.





d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)



I Let x ∈ K and let xni ∈ Kni with limi→∞ xni = x . Eventually, ni > Nand d(xni , x) < ε. Then pick yn ∈ Kn s.t. d(xni , yn) < ε. It followsd(x , yn) ≤ d(x , xni ) + d(xni , yn)

< 2ε, so δ(K ,Kn) < 2ε.





d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)



I Let x ∈ K and let xni ∈ Kni with limi→∞ xni = x . Eventually, ni > Nand d(xni , x) < ε. Then pick yn ∈ Kn s.t. d(xni , yn) < ε. It followsd(x , yn) ≤ d(x , xni ) + d(xni , yn) < 2ε

, so δ(K ,Kn) < 2ε.





d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)




I Let now y ∈ Kn.

Let n = k0 < k1 < . . . s.t.∀m ≥ kj dH (Kkj ,Km) < ε




d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)




I Let now y ∈ Kn. Let n = k0 < k1 < . . . s.t.

∀m ≥ kj dH (Kkj ,Km) < ε2j . Define{

x0 = y ∈ Kn = Kk0



d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)





2j .

Define{x0 = y ∈ Kn = Kk0



d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)








d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)






xj+1 ∈ Kkj+1

s.t. d(xj+1, xj ) <ε2j


d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)








d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)







So xj is Cauchy.

Then x = limj→∞ xj ∈ K . Moreover

d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)







So xj is Cauchy. Then x = limj→∞ xj

∈ K . Moreover

d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)







So xj is Cauchy. Then x = limj→∞ xj ∈ K .

Moreover

d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)








d(y , x)

≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Proof (cont’d)








d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . .

= 2ε


Proof (cont’d)








d(y , x) ≤ d(x0, x1) + d(x1, x2) + d(x2, x3) + . . . = 2ε


Vietoris topology on K (X )

CorollaryIf X is Polish, then K (X ) is Polish.

TheoremIf X is compact metrisable (hence Polish), then K (X ) is compactmetrisable.

Proof.Since K (X ) is metrisable, fixing a compatible metric d ≤ 1 on X , it isenough to show that (K (X ), dH ) is totally bounded.Fix ε > 0. Let F ⊆ X be finite such that X =

⋃y∈F Bε(y). Then

K (X ) =⋃

S⊆F Bε(S). Indeed, given a non-empty K ∈ K (X ), every pointof K is less than ε-apart from some member of F . Throwing away theelements of F that are at least ε-apart from all members of K , gives aset S such that dH (K ,S) < ε.




Proof.Since K (X ) is metrisable

, fixing a compatible metric d ≤ 1 on X , it isenough to show that (K (X ), dH ) is totally bounded.Fix ε > 0. Let F ⊆ X be finite such that X =


K (X ) =⋃





Proof.Since K (X ) is metrisable, fixing a compatible metric d ≤ 1 on X

, it isenough to show that (K (X ), dH ) is totally bounded.Fix ε > 0. Let F ⊆ X be finite such that X =


K (X ) =⋃





Proof.Since K (X ) is metrisable, fixing a compatible metric d ≤ 1 on X , it isenough to show that (K (X ), dH ) is totally bounded.

Fix ε > 0. Let F ⊆ X be finite such that X =⋃

y∈F Bε(y). ThenK (X ) =

⋃S⊆F Bε(S). Indeed, given a non-empty K ∈ K (X ), every point

of K is less than ε-apart from some member of F . Throwing away theelements of F that are at least ε-apart from all members of K , gives aset S such that dH (K ,S) < ε.




Proof.Since K (X ) is metrisable, fixing a compatible metric d ≤ 1 on X , it isenough to show that (K (X ), dH ) is totally bounded.Fix ε > 0.

Let F ⊆ X be finite such that X =⋃

y∈F Bε(y). ThenK (X ) =







⋃y∈F Bε(y).

ThenK (X ) =








K (X ) =⋃

S⊆F Bε(S).

Indeed, given a non-empty K ∈ K (X ), every pointof K is less than ε-apart from some member of F . Throwing away theelements of F that are at least ε-apart from all members of K , gives aset S such that dH (K ,S) < ε.






K (X ) =⋃

S⊆F Bε(S). Indeed, given a non-empty K ∈ K (X ), every pointof K is less than ε-apart from some member of F .

Throwing away theelements of F that are at least ε-apart from all members of K , gives aset S such that dH (K ,S) < ε.






K (X ) =⋃

S⊆F Bε(S). Indeed, given a non-empty K ∈ K (X ), every pointof K is less than ε-apart from some member of F . Throwing away theelements of F that are at least ε-apart from all members of K

, gives aset S such that dH (K ,S) < ε.






K (X ) =⋃


Exercises on Vietoris topology

I If (X , d) is metric, with d ≤ 1, then

X → K (X )x 7→ {x}

is an isometry.

I If X is metrisable and Kn ∈ K (X ) with K0 ⊇ K1 ⊇ K2 ⊇ . . ., thenlimn→∞ Kn =

⋂n∈N Kn.

In particular, the Cantor set E 13⊆ [0, 1] is the limit of

K0 = [0, 1],K1 = [0, 13 ] ∪ [ 23 , 1], . . .


I Let X be metrisable.

1. The relation x ∈ K is closed in X × K(X ) (i.e., the set{(x ,K) ∈ X × K(X ) | x ∈ K} is closed). [Hint: use sequences]

2. The relation K ⊆ L is closed in (K(X ))2.3. The relation K ∩ L 6= ∅ is closed in (K(X ))2. [Hint: use sequences]4. The function (K(X ))2 → K(X ), (K , L) 7→ K ∪ L is continuous.5. Compact unions of compact sets is

compact:∀K ∈ K(K(X ))⋃K ∈ K(X ). Moreover, the union

function K(K(X ))→ K(X ),K 7→⋃K is continuous.

6. Let X ,Y be metrisable and f : X → Y be continuous. Then finduces a map K(X )→ K(Y ),K 7→ f (K). This map is continuous.

7. If X ,Y are metrisable, then the mapK(X )× K(Y )→ K(X × Y ), (K , L) 7→ K × L is continuous.

8. Find a compact space X such that the function(K(X ))2 → K(X ), (K , L) 7→ K ∩ L is not continuous (however it isalways Borel).



1. The relation x ∈ K is closed in X × K(X )

(i.e., the set{(x ,K) ∈ X × K(X ) | x ∈ K} is closed). [Hint: use sequences]










2. The relation K ⊆ L is closed in (K(X ))2.

3. The relation K ∩ L 6= ∅ is closed in (K(X ))2. [Hint: use sequences]4. The function (K(X ))2 → K(X ), (K , L) 7→ K ∪ L is continuous.5. Compact unions of compact sets is









2. The relation K ⊆ L is closed in (K(X ))2.3. The relation K ∩ L 6= ∅ is closed in (K(X ))2. [Hint: use sequences]

4. The function (K(X ))2 → K(X ), (K , L) 7→ K ∪ L is continuous.5. Compact unions of compact sets is









2. The relation K ⊆ L is closed in (K(X ))2.3. The relation K ∩ L 6= ∅ is closed in (K(X ))2. [Hint: use sequences]4. The function (K(X ))2 → K(X ), (K , L) 7→ K ∪ L is continuous.

5. Compact unions of compact sets iscompact:∀K ∈ K(K(X ))

⋃K ∈ K(X ). Moreover, the union









compact:

∀K ∈ K(K(X ))⋃K ∈ K(X ). Moreover, the union









compact:∀K ∈ K(K(X ))⋃K ∈ K(X ).

Moreover, the unionfunction K(K(X ))→ K(X ),K 7→

⋃K is continuous.










6. Let X ,Y be metrisable and f : X → Y be continuous.

Then finduces a map K(X )→ K(Y ),K 7→ f (K). This map is continuous.









6. Let X ,Y be metrisable and f : X → Y be continuous. Then finduces a map K(X )→ K(Y ),K 7→ f (K).

This map is continuous.7. If X ,Y are metrisable, then the map

K(X )× K(Y )→ K(X × Y ), (K , L) 7→ K × L is continuous.8. Find a compact space X such that the function

(K(X ))2 → K(X ), (K , L) 7→ K ∩ L is not continuous (however it isalways Borel).









8. Find a compact space X such that the function(K(X ))2 → K(X ), (K , L) 7→ K ∩ L is not continuous

(however it isalways Borel).


I If X is metrisable, then Kfin(X ) = {K ∈ K (X ) | K is finite} is Fσ inK (X ).

DefinitionA subset of a topological space is perfect if it is non-empty, closed andhas no isolated points.

I If X is separable and metrisable, thenKperf (X ) = {K ∈ K (X ) | K is perfect} is Gδ in K (X ).


I A tree T on N is a subset of N<ω,

so it can be identified, via itscharacteristic function, with an element of 2N<ω , which ishomeomorphic with the Cantor space 2N. Let Tr ⊆ 2N<ω be thesubspace of trees and PTr the subspace of pruned trees. Show thatTr is closed and PTr is Gδ.

I Similarly, if Tr2 is the set of trees on 2 = {0, 1}, then Tr2 can be

identified with a subset of 22<ω , which is again homeomorphic to theCantor space. Let PTr2 be the set of pruned trees on 2. ThenTr ,PTr2 are closed sets. (Similarly for PTrk .) To prove this, provethat the function K (2N)→ PTr2,K 7→ TK is a homeomorphism.

I A tree if finite splitting if every element has only finitely manyimmediate successors. Let Trf ,PTrf be the sets of finite splitting,respectively finite splittind and pruned, trees on N. Prove that theyare not Gδ. [Hint: Use Baire’s category theorem]

I Prove that the map K (NN)→ PTrf ,K 7→ TK is continuous but notan homeomorphism.


I A tree T on N is a subset of N<ω, so it can be identified, via itscharacteristic function, with an element of 2N<ω ,

which ishomeomorphic with the Cantor space 2N. Let Tr ⊆ 2N<ω be thesubspace of trees and PTr the subspace of pruned trees. Show thatTr is closed and PTr is Gδ.






I A tree T on N is a subset of N<ω, so it can be identified, via itscharacteristic function, with an element of 2N<ω , which ishomeomorphic with the Cantor space 2N.

Let Tr ⊆ 2N<ω be thesubspace of trees and PTr the subspace of pruned trees. Show thatTr is closed and PTr is Gδ.






I A tree T on N is a subset of N<ω, so it can be identified, via itscharacteristic function, with an element of 2N<ω , which ishomeomorphic with the Cantor space 2N. Let Tr ⊆ 2N<ω be thesubspace of trees and PTr the subspace of pruned trees.

Show thatTr is closed and PTr is Gδ.






I A tree T on N is a subset of N<ω, so it can be identified, via itscharacteristic function, with an element of 2N<ω , which ishomeomorphic with the Cantor space 2N. Let Tr ⊆ 2N<ω be thesubspace of trees and PTr the subspace of pruned trees. Show thatTr is closed and PTr is Gδ.







I Similarly, if Tr2 is the set of trees on 2 = {0, 1},

then Tr2 can be







identified with a subset of 22<ω ,

which is again homeomorphic to theCantor space. Let PTr2 be the set of pruned trees on 2. ThenTr ,PTr2 are closed sets. (Similarly for PTrk .) To prove this, provethat the function K (2N)→ PTr2,K 7→ TK is a homeomorphism.






identified with a subset of 22<ω , which is again homeomorphic to theCantor space.

Let PTr2 be the set of pruned trees on 2. ThenTr ,PTr2 are closed sets. (Similarly for PTrk .) To prove this, provethat the function K (2N)→ PTr2,K 7→ TK is a homeomorphism.






identified with a subset of 22<ω , which is again homeomorphic to theCantor space. Let PTr2 be the set of pruned trees on 2.

ThenTr ,PTr2 are closed sets. (Similarly for PTrk .) To prove this, provethat the function K (2N)→ PTr2,K 7→ TK is a homeomorphism.






identified with a subset of 22<ω , which is again homeomorphic to theCantor space. Let PTr2 be the set of pruned trees on 2. ThenTr ,PTr2 are closed sets. (Similarly for PTrk .)

To prove this, provethat the function K (2N)→ PTr2,K 7→ TK is a homeomorphism.







I A tree if finite splitting if every element has only finitely manyimmediate successors.

Let Trf ,PTrf be the sets of finite splitting,respectively finite splittind and pruned, trees on N. Prove that theyare not Gδ. [Hint: Use Baire’s category theorem]






I A tree if finite splitting if every element has only finitely manyimmediate successors. Let Trf ,PTrf be the sets of finite splitting,respectively finite splittind and pruned, trees on N.

Prove that theyare not Gδ. [Hint: Use Baire’s category theorem]


Lipschitz functions

Let

I X ,Y : separable real Banach spaces

I L(X ,Y ) = {T : X → Y | T linear and bounded}I ||T || = sup{||T (x)|| | ||x || ≤ 1}I L1(X ,Y ) = {T ∈ L(X ,Y ) | ||T || ≤ 1}

The strong topology on L(X ,Y ) is generated by all functions

fx : L(X ,Y ) → YT 7→ T (x) = fx

for x ∈ X . Basic open nbhds of T are of the form

V Tx1,...,xn,ε = {S ∈ L(X ,Y ) | ∀i ||Sx1 − Txi || < ε}

Lipschitz functions

Let


I L(X ,Y ) = {T : X → Y | T linear and bounded}

I ||T || = sup{||T (x)|| | ||x || ≤ 1}I L1(X ,Y ) = {T ∈ L(X ,Y ) | ||T || ≤ 1}


fx : L(X ,Y ) → YT 7→ T (x) = fx



Lipschitz functions

Let


I L(X ,Y ) = {T : X → Y | T linear and bounded}I ||T || = sup{||T (x)|| | ||x || ≤ 1}

I L1(X ,Y ) = {T ∈ L(X ,Y ) | ||T || ≤ 1}The strong topology on L(X ,Y ) is generated by all functions

fx : L(X ,Y ) → YT 7→ T (x) = fx



Lipschitz functions

Let




fx : L(X ,Y ) → YT 7→ T (x) = fx



Lipschitz functions

Let



The strong topology on L(X ,Y )

is generated by all functions

fx : L(X ,Y ) → YT 7→ T (x) = fx



Lipschitz functions

Let




fx : L(X ,Y ) → YT 7→ T (x) = fx

for x ∈ X .

Basic open nbhds of T are of the form


Lipschitz functions

Let




fx : L(X ,Y ) → YT 7→ T (x) = fx



Lipschits functions

Let D countable dense in X and closed under rational linearcombinations.

Then Y D is Polish with the product topology. Thefunction

Φ : L1(X ,Y ) → Y D

T 7→ T|D

is injective and its range is

F = {f | ∀x , y ∈ D ∀p, q ∈ Q f (px + qy) = pf (x) + qf (y)∧∧∀x ∈ D ||f (x)|| ≤ ||x ||}

Exercise. Show that F is closed (thus Polish) and Φ is ahomeomorphism onto F .

Consequently, L1(X ,Y ) with the strong topology is Polish.

Lipschits functions

Let D countable dense in X and closed under rational linearcombinations. Then Y D is Polish with the product topology.

Thefunction

Φ : L1(X ,Y ) → Y D

T 7→ T|D





Lipschits functions

Let D countable dense in X and closed under rational linearcombinations. Then Y D is Polish with the product topology. Thefunction

Φ : L1(X ,Y ) → Y D

T 7→ T|D





Lipschits functions


Φ : L1(X ,Y ) → Y D

T 7→ T|D

is injective

and its range is




Lipschits functions


Φ : L1(X ,Y ) → Y D

T 7→ T|D





Borel sets

Definition

I A measurable space (X ,S) is a non-empty set X endowed with aσ-algebra S of subsets of X

I A Borel space (X ,B(X )) is a topological space (X , τ) endowed withthe σ-algebra B(X ) = B(X , τ) = B(τ) generated by τ . If τ isPolish, then (X ,B(X )) is a standard Borel space.

I A function f : X → Y between measurable spaces (X ,S), (Y ,T ) ismeasurable if ∀W ∈ T f −1(W ) ∈ S

I A function f : X → Y between a measurable space X and atopological space Y is measurable if it is measurable w.r.t. theσ-algebra B(Y ) on Y . If X is a Borel space, then f is Borel.

Remarks.

I Different topologies on X can yield the same Borel space.

I In order for a function f : X → Y to be Borel, it is enough thatpreimages of open sets (or of closed sets, or of sets from a countablebasis) of Y are Borel in X .

Borel sets

Definition


I A Borel space (X ,B(X ))

is a topological space (X , τ) endowed withthe σ-algebra B(X ) = B(X , τ) = B(τ) generated by τ . If τ isPolish, then (X ,B(X )) is a standard Borel space.



Remarks.



Borel sets

Definition


I A Borel space (X ,B(X )) is a topological space (X , τ)

endowed withthe σ-algebra B(X ) = B(X , τ) = B(τ) generated by τ . If τ isPolish, then (X ,B(X )) is a standard Borel space.



Remarks.



Borel sets

Definition


I A Borel space (X ,B(X )) is a topological space (X , τ) endowed withthe σ-algebra B(X ) = B(X , τ) = B(τ) generated by τ .

If τ isPolish, then (X ,B(X )) is a standard Borel space.



Remarks.



Borel sets

Definition





Remarks.



Borel sets

Definition




I A function f : X → Y between a measurable space X and atopological space Y is measurable if it is measurable w.r.t. theσ-algebra B(Y ) on Y .

If X is a Borel space, then f is Borel.

Remarks.



Borel sets

Definition





Remarks.



Borel sets

Basic facts.

I If E is a countable sub-basis for X , then B(X ) is the σ-algebragenerated by E

I If Y is a subspace of X , then B(Y ) = {Y ∩ B}B∈B(X )

I B(X ) is the smallest subfamily of P(X ) containing open and closedsets, and stable under countable unions and countable intersections

I B(X ) is the smallest subfamily of P(X ) containing open sets andstable under complementation and disjoint countable unions.

Question. Given a σ-algebra Σ on a set X , is there always a topology τon X such that Σ = B(τ)? Characterise the σ-algebras that have thisproperty.

The Borel hierarchy

A motivating remark. If X is a metric space, then every closed set isGδ, being the intersection of all of its ε-nbhds, for ε ∈ Q. (Consequently,every open set is Fσ.)

DefinitionLet X be a topological space. By induction on 1 ≤ ξ < ω1, define

I Σ01(X ) = the open sets of X

I Π0ξ(X ) = {X \ A}A∈Σξ(X )

I Σ0ξ(X ) = {

⋃n An | ∀n ∃ξn < ξ An ∈ Π0

ξn(X )}

I ∆0ξ(X ) = Σ0

ξ(X ) ∩Π0ξ(X )

The Borel hierarchy

A motivating remark. If X is a metric space, then every closed set isGδ, being the intersection of all of its ε-nbhds, for ε ∈ Q.

(Consequently,every open set is Fσ.)



I Π0ξ(X ) = {X \ A}A∈Σξ(X )

I Σ0ξ(X ) = {

⋃n An | ∀n ∃ξn < ξ An ∈ Π0

ξn(X )}

I ∆0ξ(X ) = Σ0

ξ(X ) ∩Π0ξ(X )

The Borel hierarchy

A motivating remark. If X is a metric space, then every closed set isGδ, being the intersection of all of its ε-nbhds, for ε ∈ Q. (Consequently,every open set is Fσ.)



I Π0ξ(X ) = {X \ A}A∈Σξ(X )

I Σ0ξ(X ) = {

⋃n An | ∀n ∃ξn < ξ An ∈ Π0

ξn(X )}

I ∆0ξ(X ) = Σ0

ξ(X ) ∩Π0ξ(X )

The Borel hierarchy

If X is metrisable, then by induction

Σ01 Σ0

2 Σ0ξ Σ0

ξ+1

∆01 ∆0

2 . . . ∆0ξ ∆0

ξ+1 . . .

Π01 Π0

2 Π0ξ Π0

ξ+1

each class in the sequence is contained in all the classes at its right.

Inclassical notation:

Σ01 = open sets, Π0

1 = closed sets, ∆01 = clopen sets

Σ02 = Fσ, Π0

2 = GδΣ0

3 = Gδσ, Π03 = Fσδ

. . .

Notice that each of these classes is closed under continuous preimages.

The Borel hierarchy


Σ01 Σ0

2 Σ0ξ Σ0

ξ+1

∆01 ∆0

2 . . . ∆0ξ ∆0

ξ+1 . . .

Π01 Π0

2 Π0ξ Π0

ξ+1

each class in the sequence is contained in all the classes at its right. Inclassical notation:

Σ01 = open sets,

Π01 = closed sets, ∆0

1 = clopen sets

Σ02 = Fσ, Π0

2 = GδΣ0

3 = Gδσ, Π03 = Fσδ

. . .


The Borel hierarchy


Σ01 Σ0

2 Σ0ξ Σ0

ξ+1

∆01 ∆0

2 . . . ∆0ξ ∆0

ξ+1 . . .

Π01 Π0

2 Π0ξ Π0

ξ+1



1 = closed sets,

∆01 = clopen sets

Σ02 = Fσ, Π0

2 = GδΣ0

3 = Gδσ, Π03 = Fσδ

. . .


The Borel hierarchy


Σ01 Σ0

2 Σ0ξ Σ0

ξ+1

∆01 ∆0

2 . . . ∆0ξ ∆0

ξ+1 . . .

Π01 Π0

2 Π0ξ Π0

ξ+1




Σ02 = Fσ, Π0

2 = GδΣ0

3 = Gδσ, Π03 = Fσδ

. . .


The Borel hierarchy


Σ01 Σ0

2 Σ0ξ Σ0

ξ+1

∆01 ∆0

2 . . . ∆0ξ ∆0

ξ+1 . . .

Π01 Π0

2 Π0ξ Π0

ξ+1




Σ02 = Fσ, Π0

2 = Gδ

Σ03 = Gδσ, Π0

3 = Fσδ. . .


The Borel hierarchy


Σ01 Σ0

2 Σ0ξ Σ0

ξ+1

∆01 ∆0

2 . . . ∆0ξ ∆0

ξ+1 . . .

Π01 Π0

2 Π0ξ Π0

ξ+1




Σ02 = Fσ, Π0

2 = GδΣ0

3 = Gδσ, Π03 = Fσδ

. . .


The Borel hierarchy

By regularity of ω1:

B(X ) =⋃

1≤ξ<ω1

Σ0ξ(X ) =

⋃1≤ξ<ω1

Π0ξ(X ) =

⋃1≤ξ<ω1

∆0ξ(X )

An example and Kuratowski algorithm

A number x ∈]0, 1[ is normal in base 2 if, given its binary expansion withinfinitely many 1’s

x = 0.b1b2b3 . . .

one has

limn→∞

|{i ≤ n | bi = 1}|n

=1

2

PropositionThe set N = {x ∈]0, 1[| x is normal} is Borel.

An example and Kuratowski algorithm

Proof.Consider the functions

dn :]0, 1[ → 2x 7→ the n-th digit of x

dn(x) =

{0 if x ∈] 2k

2n ,2k+12n ]

1 if x ∈] 2k+12n , 2k+2

2n ]for some k < 2n−1

so x =∑∞

n=1dn(x)2n .

Proof (cont’d)

Then x ∈ N ⇔ ∀ε ∈ R+ ∃n ∀m ≥ n |∑m

i=1 di (x)m − 1

2 | < ε

or, better,

x ∈ N ⇔ ∀ε ∈ Q+ ∃n ∀m ≥ n |∑m

i=1 di (x)

m− 1

2| < ε

SoN =

⋂ε∈Q+

⋃n∈N

⋂m≥n

Amε

where Amε = {x |∑m

i=1 di (x)m − 1

2 | < ε}. Each di is constant on intervals of

the form ] k2i ,

k+12i ], so Amε is a finite union of intervals of the that kind.

Since these are Π02, each Amε is Π0

2.

Proof (cont’d)

Then x ∈ N ⇔ ∀ε ∈ R+ ∃n ∀m ≥ n |∑m

i=1 di (x)m − 1

2 | < ε or, better,

x ∈ N ⇔ ∀ε ∈ Q+ ∃n ∀m ≥ n |∑m

i=1 di (x)

m− 1

2| < ε

SoN =

⋂ε∈Q+

⋃n∈N

⋂m≥n

Amε


i=1 di (x)m − 1


the form ] k2i ,



2.

Proof (cont’d)

Then x ∈ N ⇔ ∀ε ∈ R+ ∃n ∀m ≥ n |∑m

i=1 di (x)m − 1


x ∈ N ⇔ ∀ε ∈ Q+ ∃n ∀m ≥ n |∑m

i=1 di (x)

m− 1

2| < ε

SoN =

⋂ε∈Q+

⋃n∈N

⋂m≥n

Amε


i=1 di (x)m − 1

2 | < ε}.

Each di is constant on intervals of

the form ] k2i ,



2.

Proof (cont’d)

Then x ∈ N ⇔ ∀ε ∈ R+ ∃n ∀m ≥ n |∑m

i=1 di (x)m − 1


x ∈ N ⇔ ∀ε ∈ Q+ ∃n ∀m ≥ n |∑m

i=1 di (x)

m− 1

2| < ε

SoN =

⋂ε∈Q+

⋃n∈N

⋂m≥n

Amε


i=1 di (x)m − 1


the form ] k2i ,

k+12i ],

so Amε is a finite union of intervals of the that kind.


2.

Proof (cont’d)

Then x ∈ N ⇔ ∀ε ∈ R+ ∃n ∀m ≥ n |∑m

i=1 di (x)m − 1


x ∈ N ⇔ ∀ε ∈ Q+ ∃n ∀m ≥ n |∑m

i=1 di (x)

m− 1

2| < ε

SoN =

⋂ε∈Q+

⋃n∈N

⋂m≥n

Amε


i=1 di (x)m − 1


the form ] k2i ,



2.

Proof (cont’d)

Then x ∈ N ⇔ ∀ε ∈ R+ ∃n ∀m ≥ n |∑m

i=1 di (x)m − 1


x ∈ N ⇔ ∀ε ∈ Q+ ∃n ∀m ≥ n |∑m

i=1 di (x)

m− 1

2| < ε

SoN =

⋂ε∈Q+

⋃n∈N

⋂m≥n

Amε


i=1 di (x)m − 1


the form ] k2i ,


Since these are Π02,

each Amε is Π02.

Proof (cont’d)

Then x ∈ N ⇔ ∀ε ∈ R+ ∃n ∀m ≥ n |∑m

i=1 di (x)m − 1


x ∈ N ⇔ ∀ε ∈ Q+ ∃n ∀m ≥ n |∑m

i=1 di (x)

m− 1

2| < ε

SoN =

⋂ε∈Q+

⋃n∈N

⋂m≥n

Amε


i=1 di (x)m − 1


the form ] k2i ,



2.

Proof (cont’d)

SoN =

⋂ε∈Q+

⋃n∈N

⋂m≥n Amε

Π02

Π02

Σ03

Π04

Thus N is Π04, hence Borel.

(Note: there is a sharper classification.)

Kuratowski algorithm

The computation is usually carried out directly from the definition:

x ∈ N ⇔ ∀ε ∈ Q+ ∃n ∈ N ∀m ≥ n |∑m

i=1 di (x)m − 1

2 | < ε

Π02

Π02

Σ03

Π04

This is the so-called Kuratowski algorithm.


Note that each ] k2i ,

k+12i ], and so each Amε, is also Σ0

2.

Thus one cancompute the complexity of N as:

x ∈ N ⇔ ∀ε ∈ Q+ ∃n ∈ N ∀m ≥ n |∑m

i=1 di (x)m − 1

2 | < ε

Σ02

Π03

Σ04

Π05

But this calculation is coarser that the former.


Note that each ] k2i ,

k+12i ], and so each Amε, is also Σ0

2. Thus one cancompute the complexity of N as:

x ∈ N ⇔ ∀ε ∈ Q+ ∃n ∈ N ∀m ≥ n |∑m

i=1 di (x)m − 1

2 | < ε

Σ02

Π03

Σ04

Π05

But this calculation is coarser that the former.

Further examples

I C0 = {x = (xn) ∈ RN | limn→∞ xn = 0}

x ∈ C0 ⇔ ∀ε ∈ Q+ ∃n ∀m ≥ n xm ≤ ε

Π01

Π01

Σ02

Π03

So C0 ∈ Π03(RN)

Further examples

I Let f ∈ C([0, 1],R).

ThenDf = {x ∈ [0, 1] | f ′(x) exists} ∈ Π0

3(C([0, 1],R)).Indeed, x ∈ Df iff

∀ε ∈ Q+ ∃δ ∈ Q+ ∀p, q ∈ [0, 1] ∩Q (p, q ∈ Bδ(x) \ {x} ⇒⇒ | f (p)−f (x)

p−x − f (q)−f (x)q−x | ≤ ε)

I N2 = {x ∈ 2N | ∃∞n x(n) = 0} ∈ Π02(2N)

x ∈ N2 ⇔ ∀m ∃n ≥ m x(n) = 0

Further examples

I Let f ∈ C([0, 1],R). ThenDf = {x ∈ [0, 1] | f ′(x) exists}

∈ Π03(C([0, 1],R)).

Indeed, x ∈ Df iff


p−x − f (q)−f (x)q−x | ≤ ε)

I N2 = {x ∈ 2N | ∃∞n x(n) = 0} ∈ Π02(2N)

x ∈ N2 ⇔ ∀m ∃n ≥ m x(n) = 0

Further examples

I Let f ∈ C([0, 1],R). ThenDf = {x ∈ [0, 1] | f ′(x) exists} ∈ Π0

3(C([0, 1],R)).

Indeed, x ∈ Df iff


p−x − f (q)−f (x)q−x | ≤ ε)

I N2 = {x ∈ 2N | ∃∞n x(n) = 0} ∈ Π02(2N)

x ∈ N2 ⇔ ∀m ∃n ≥ m x(n) = 0

Further examples




p−x − f (q)−f (x)q−x | ≤ ε)

I N2 = {x ∈ 2N | ∃∞n x(n) = 0} ∈ Π02(2N)

x ∈ N2 ⇔ ∀m ∃n ≥ m x(n) = 0

Further examples




p−x − f (q)−f (x)q−x | ≤ ε)

I N2 = {x ∈ 2N | ∃∞n x(n) = 0}

∈ Π02(2N)

x ∈ N2 ⇔ ∀m ∃n ≥ m x(n) = 0

Further examples




p−x − f (q)−f (x)q−x | ≤ ε)

I N2 = {x ∈ 2N | ∃∞n x(n) = 0} ∈ Π02(2N)

x ∈ N2 ⇔ ∀m ∃n ≥ m x(n) = 0

Exercises

I Let X be a measurable space, Y a metrisable space, fn : X → Ymeasurable functions

and suppose fn → f pointwise. Then f ismeasurable.

I Let f : [0, 1]→ R be a differentiable function. Then f ′ is Borel.

I Let X ,Z be separable metrisable, Y be a topological space, andf : X × Y → Z . In order for f to be Borel it is enough that:

I for all y ∈ Y , the function f y : X → Z , f y (x) = f (x , y) becontinuous; and

I for a dense subset of x in X , the functionfx : Y → Z , fx(y) = f (x , y) be Borel

Exercises

I Let X be a measurable space, Y a metrisable space, fn : X → Ymeasurable functions and suppose fn → f pointwise.

Then f ismeasurable.





Exercises

I Let X be a measurable space, Y a metrisable space, fn : X → Ymeasurable functions and suppose fn → f pointwise. Then f ismeasurable.





Exercises


I Let f : [0, 1]→ R be a differentiable function.

Then f ′ is Borel.




Exercises






Exercises



I Let X ,Z be separable metrisable, Y be a topological space, andf : X × Y → Z .

In order for f to be Borel it is enough that:I for all y ∈ Y , the function f y : X → Z , f y (x) = f (x , y) be

continuous; andI for a dense subset of x in X , the function

fx : Y → Z , fx(y) = f (x , y) be Borel

Exercises






Exercises

I Let X be Polish. Then B(K (X )) is generated by any one of thefollowing families:

I E0 = {{K ∈ K(X ) | K ⊆ U} | U open}I E1 = {{K ∈ K(X ) | K ∩ U 6= ∅} | U open}

I If X is Polish then the function (K (X ))2 → K (X ), (K , L) 7→ K ∩ L isBorel.

I let X be Polish, Y compact metrisable, and let F ⊆ X × Y beclosed. Then the function

X → K (Y )x 7→ Fx = {y ∈ Y | (x , y) ∈ F}

is Borel.

Exercises




I let X be Polish, Y compact metrisable, and let F ⊆ X × Y beclosed.

Then the function

X → K (Y )x 7→ Fx = {y ∈ Y | (x , y) ∈ F}

is Borel.

Exercises





X → K (Y )

x 7→ Fx = {y ∈ Y | (x , y) ∈ F}

is Borel.

Exercises





X → K (Y )x 7→ Fx = {y ∈ Y | (x , y) ∈ F}

is Borel.

Oscillation

DefinitionLet X be a topological space, Y a metric space, A ⊆ X , and f : A→ Y .For x ∈ A, define

oscf (x) = inf{diam(f (U ∩ A)) | U open nbhd of x}

In particular if x ∈ A, then oscf (x) = 0 iff f is continuous at x .

Oscillation

DefinitionLet X be a topological space, Y a metric space, A ⊆ X , and f : A→ Y .

For x ∈ A, define



Oscillation

DefinitionLet X be a topological space, Y a metric space, A ⊆ X , and f : A→ Y .For x ∈ A, define



Continuity points

PropositionLet X be a topological space, Y be a metrisable space, and f : X → Y .Then Z = {x ∈ X | f is continuous at x} is a Gδ subset of X .

Proof.Since

Z = {x ∈ X | oscf (x) = 0} =∞⋂

n=1

{x ∈ X | oscf (x) <1

n}

it is enough to show that each Aε = {x ∈ X | oscf (x) < ε} is open. Ifx ∈ Aε, let U be an open nbhd of x with diam(f (U)) < ε. Then∀y ∈ U oscf (y) < ε, whence U ⊆ Aε.

Continuity points

PropositionLet X be a topological space, Y be a metrisable space, and f : X → Y .

Then Z = {x ∈ X | f is continuous at x} is a Gδ subset of X .

Proof.Since

Z = {x ∈ X | oscf (x) = 0} =∞⋂

n=1

{x ∈ X | oscf (x) <1

n}


Continuity points


Proof.Since

Z = {x ∈ X | oscf (x) = 0} =∞⋂

n=1

{x ∈ X | oscf (x) <1

n}


Continuity points


Proof.Since

Z = {x ∈ X | oscf (x) = 0}

=∞⋂

n=1

{x ∈ X | oscf (x) <1

n}


Continuity points


Proof.Since

Z = {x ∈ X | oscf (x) = 0} =∞⋂

n=1

{x ∈ X | oscf (x) <1

n}


Continuity points


Proof.Since

Z = {x ∈ X | oscf (x) = 0} =∞⋂

n=1

{x ∈ X | oscf (x) <1

n}

it is enough to show that each Aε = {x ∈ X | oscf (x) < ε} is open.

Ifx ∈ Aε, let U be an open nbhd of x with diam(f (U)) < ε. Then∀y ∈ U oscf (y) < ε, whence U ⊆ Aε.

Continuity points


Proof.Since

Z = {x ∈ X | oscf (x) = 0} =∞⋂

n=1

{x ∈ X | oscf (x) <1

n}

it is enough to show that each Aε = {x ∈ X | oscf (x) < ε} is open. Ifx ∈ Aε, let U be an open nbhd of x with diam(f (U)) < ε.

Then∀y ∈ U oscf (y) < ε, whence U ⊆ Aε.

Continuity points


Proof.Since

Z = {x ∈ X | oscf (x) = 0} =∞⋂

n=1

{x ∈ X | oscf (x) <1

n}

it is enough to show that each Aε = {x ∈ X | oscf (x) < ε} is open. Ifx ∈ Aε, let U be an open nbhd of x with diam(f (U)) < ε. Then∀y ∈ U oscf (y) < ε,

whence U ⊆ Aε.

Continuity points


Proof.Since

Z = {x ∈ X | oscf (x) = 0} =∞⋂

n=1

{x ∈ X | oscf (x) <1

n}


Kuratowski’s theorem

TheoremIf X is metrisable, Y completely metrisable, A ⊆ X and f : A→ Y iscontinuous, then there are a Gδ set G ⊆ X with A ⊆ G ⊆ A, and acontinuous g : G → Y extending f .

Proof.Let G = {x ∈ X | oscf (x) = 0}. This is Gδ. Moreover, since f iscontinuous on A, one has A ⊆ G ⊆ A.For x ∈ G and xn → x , with xn ∈ A, define

g(x) = limn→∞

f (xn)

Exercise. Check that this is well defined, by proving that f (xn) is alwaysCauchy, for any choice of the sequence xn.

In particular, g extends f .


TheoremIf X is metrisable, Y completely metrisable, A ⊆ X and f : A→ Y iscontinuous,

then there are a Gδ set G ⊆ X with A ⊆ G ⊆ A, and acontinuous g : G → Y extending f .


g(x) = limn→∞

f (xn)





Proof.Let G = {x ∈ X | oscf (x) = 0}. This is Gδ.

Moreover, since f iscontinuous on A, one has A ⊆ G ⊆ A.For x ∈ G and xn → x , with xn ∈ A, define

g(x) = limn→∞

f (xn)





Proof.Let G = {x ∈ X | oscf (x) = 0}. This is Gδ. Moreover, since f iscontinuous on A, one has A ⊆ G ⊆ A.

For x ∈ G and xn → x , with xn ∈ A, define

g(x) = limn→∞

f (xn)






g(x) = limn→∞

f (xn)



Proof (cont’d)

It remains to show that g is continuous.

Fix x ∈ G . If U is open in Xand contains x , one has g(U ∩ G ) ⊆ f (U ∩ A), sodiam(g(U ∩ G )) ≤ diam(f (U ∩ A)) = diam(f (U ∩ A)). The infimum ofthese latter numbers is 0, as oscf (x) = 0, so oscg (x) = 0 as well, and gis continuous at x .

Proof (cont’d)

It remains to show that g is continuous. Fix x ∈ G .

If U is open in Xand contains x , one has g(U ∩ G ) ⊆ f (U ∩ A), sodiam(g(U ∩ G )) ≤ diam(f (U ∩ A)) = diam(f (U ∩ A)). The infimum ofthese latter numbers is 0, as oscf (x) = 0, so oscg (x) = 0 as well, and gis continuous at x .

Proof (cont’d)

It remains to show that g is continuous. Fix x ∈ G . If U is open in Xand contains x , one has g(U ∩ G ) ⊆ f (U ∩ A),

sodiam(g(U ∩ G )) ≤ diam(f (U ∩ A)) = diam(f (U ∩ A)). The infimum ofthese latter numbers is 0, as oscf (x) = 0, so oscg (x) = 0 as well, and gis continuous at x .

Proof (cont’d)

It remains to show that g is continuous. Fix x ∈ G . If U is open in Xand contains x , one has g(U ∩ G ) ⊆ f (U ∩ A), sodiam(g(U ∩ G )) ≤ diam(f (U ∩ A)) = diam(f (U ∩ A)).

The infimum ofthese latter numbers is 0, as oscf (x) = 0, so oscg (x) = 0 as well, and gis continuous at x .

Proof (cont’d)

It remains to show that g is continuous. Fix x ∈ G . If U is open in Xand contains x , one has g(U ∩ G ) ⊆ f (U ∩ A), sodiam(g(U ∩ G )) ≤ diam(f (U ∩ A)) = diam(f (U ∩ A)). The infimum ofthese latter numbers is 0, as oscf (x) = 0, so oscg (x) = 0 as well, and gis continuous at x .

Lavrentiev’s theorem

TheoremLet X ,Y be completely metrisable, A ⊆ X ,B ⊆ Y and f : A→ B ahomeomorphism.

Then there are Gδ sets G ,H s.t.A ⊆ G ⊆ X ,B ⊆ H ⊆ Y and a homeomorphism f : G → H with f ⊆ f .

Proof.By Kuratowski’s theorem, let G1,H1 be Gδ sets s.t.A ⊆ G1 ⊆ X ,B ⊆ H1 ⊆ Y with f1 : G1 → Y , g1 : H1 → X continuousand extending f , g , resp. Let R be the graph of f1 and S be the inverseof the grapf of g1, i.e., {(x , y) | x = g1(y)}. Consider the projections

G = π1(R ∩ S),H = π2(R ∩ S)

Then A ⊆ G ⊆ G1,B ⊆ H ⊆ H1.


TheoremLet X ,Y be completely metrisable, A ⊆ X ,B ⊆ Y and f : A→ B ahomeomorphism. Then there are Gδ sets G ,H s.t.A ⊆ G ⊆ X ,B ⊆ H ⊆ Y and a homeomorphism f : G → H with f ⊆ f .


G = π1(R ∩ S),H = π2(R ∩ S)

Then A ⊆ G ⊆ G1,B ⊆ H ⊆ H1.



Proof.By Kuratowski’s theorem, let G1,H1 be Gδ sets s.t.A ⊆ G1 ⊆ X ,B ⊆ H1 ⊆ Y with f1 : G1 → Y , g1 : H1 → X continuousand extending f , g , resp.

Let R be the graph of f1 and S be the inverseof the grapf of g1, i.e., {(x , y) | x = g1(y)}. Consider the projections

G = π1(R ∩ S),H = π2(R ∩ S)

Then A ⊆ G ⊆ G1,B ⊆ H ⊆ H1.



Proof.By Kuratowski’s theorem, let G1,H1 be Gδ sets s.t.A ⊆ G1 ⊆ X ,B ⊆ H1 ⊆ Y with f1 : G1 → Y , g1 : H1 → X continuousand extending f , g , resp. Let R be the graph of f1 and S be the inverseof the grapf of g1, i.e., {(x , y) | x = g1(y)}.

Consider the projections

G = π1(R ∩ S),H = π2(R ∩ S)

Then A ⊆ G ⊆ G1,B ⊆ H ⊆ H1.




G = π1(R ∩ S),H = π2(R ∩ S)

Then A ⊆ G ⊆ G1,B ⊆ H ⊆ H1.


Exercise. Check that

x ∈ G ⇔ g1f1(x) = xy ∈ H ⇔ f1g1(y) = y

Let h : G → H, h(x) = f1(x). Then h is a homeomorphism: it is enoughto notice that ran(h) is indeed H. In remains to check that G ,H are Gδ.

I For G : Consider the continuous function

ρ : G1 → X × Yx 7→ (x , f1(x))

Exercise. Check that G = ρ−1(S).

Since S is closed in X ×H1, so Gδ in X ×Y , its preimage G is Gδ inX .

I For H: Similar.



x ∈ G ⇔ g1f1(x) = xy ∈ H ⇔ f1g1(y) = y

Let h : G → H, h(x) = f1(x).

Then h is a homeomorphism: it is enoughto notice that ran(h) is indeed H. In remains to check that G ,H are Gδ.


ρ : G1 → X × Yx 7→ (x , f1(x))



I For H: Similar.



x ∈ G ⇔ g1f1(x) = xy ∈ H ⇔ f1g1(y) = y

Let h : G → H, h(x) = f1(x). Then h is a homeomorphism: it is enoughto notice that ran(h) is indeed H.

In remains to check that G ,H are Gδ.


ρ : G1 → X × Yx 7→ (x , f1(x))



I For H: Similar.



x ∈ G ⇔ g1f1(x) = xy ∈ H ⇔ f1g1(y) = y



ρ : G1 → X × Yx 7→ (x , f1(x))



I For H: Similar.



x ∈ G ⇔ g1f1(x) = xy ∈ H ⇔ f1g1(y) = y



ρ : G1 → X × Yx 7→ (x , f1(x))


Since S is closed in X ×H1,

so Gδ in X ×Y , its preimage G is Gδ inX .

I For H: Similar.



x ∈ G ⇔ g1f1(x) = xy ∈ H ⇔ f1g1(y) = y



ρ : G1 → X × Yx 7→ (x , f1(x))


Since S is closed in X ×H1, so Gδ in X ×Y ,

its preimage G is Gδ inX .

I For H: Similar.



x ∈ G ⇔ g1f1(x) = xy ∈ H ⇔ f1g1(y) = y



ρ : G1 → X × Yx 7→ (x , f1(x))



I For H: Similar.

Subspaces of Polish spaces

Theorem

I If X is metrisable and Y ⊆ X is completely metrisable, then Y is Gδin X .

I If X is completely metrisable and Y ⊆ X is Gδ in X , then Y iscompletely metrisable.

In particular, if X is Polish and Y ⊆ X ,

Y is Polish⇔ Y is Gδ

Proof.Suppose X metrisable, Y completely metrisable. Since the identityfunction i : Y ⊆ X → Y is continuous, by Kuratowski’s thm there is G , aGδ-subset of X , s.t. Y ⊆ G ⊆ Y , with a continuous extensionf : G → Y ⊆ X of i . The only possibility is that f is identity on G , soY = G is Gδ.


Theorem





Proof.Suppose X metrisable, Y completely metrisable.

Since the identityfunction i : Y ⊆ X → Y is continuous, by Kuratowski’s thm there is G , aGδ-subset of X , s.t. Y ⊆ G ⊆ Y , with a continuous extensionf : G → Y ⊆ X of i . The only possibility is that f is identity on G , soY = G is Gδ.


Theorem





Proof.Suppose X metrisable, Y completely metrisable. Since the identityfunction i : Y ⊆ X → Y is continuous,

by Kuratowski’s thm there is G , aGδ-subset of X , s.t. Y ⊆ G ⊆ Y , with a continuous extensionf : G → Y ⊆ X of i . The only possibility is that f is identity on G , soY = G is Gδ.


Theorem





Proof.Suppose X metrisable, Y completely metrisable. Since the identityfunction i : Y ⊆ X → Y is continuous, by Kuratowski’s thm there is G , aGδ-subset of X , s.t. Y ⊆ G ⊆ Y , with a continuous extensionf : G → Y ⊆ X of i .

The only possibility is that f is identity on G , soY = G is Gδ.


Theorem





Proof.Suppose X metrisable, Y completely metrisable. Since the identityfunction i : Y ⊆ X → Y is continuous, by Kuratowski’s thm there is G , aGδ-subset of X , s.t. Y ⊆ G ⊆ Y , with a continuous extensionf : G → Y ⊆ X of i . The only possibility is that f is identity on G , soY = G is Gδ.

Proof (cont’d)

Conversely, suppose X is completely metrisable, d is a compatiblecomplete metric on X , and Y is Gδ in X .

So let Y =⋂

n∈N Un, with Un

open. Denote Fn = X \ Un. For x , y ∈ Y , define

d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)

Exercise. Check that d ′ is a compatible metric on Y .

It remains to prove that d ′ is complete. Let yi be a d ′-Cauchy sequencein Y . Then yi is d-Cauchy. Let limi→∞ yi = y ∈ X . It remains to showthat y ∈ Y . Indeed, fix n. Then for any ε > 0, for i , j sufficiently large,| 1

d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1

d(yi ,Fn)is Cauchy, so it converges is R. This

means that d(yi ,Fn) is bounded away from 0. Consequently, y /∈ Fn, i.e.,y ∈ Un.

Proof (cont’d)

Conversely, suppose X is completely metrisable, d is a compatiblecomplete metric on X , and Y is Gδ in X . So let Y =

⋂n∈N Un, with Un

open.

Denote Fn = X \ Un. For x , y ∈ Y , define

d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)



d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1



Proof (cont’d)



open. Denote Fn = X \ Un.

For x , y ∈ Y , define

d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)



d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1



Proof (cont’d)




d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)



d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1



Proof (cont’d)




d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)


It remains to prove that d ′ is complete.

Let yi be a d ′-Cauchy sequencein Y . Then yi is d-Cauchy. Let limi→∞ yi = y ∈ X . It remains to showthat y ∈ Y . Indeed, fix n. Then for any ε > 0, for i , j sufficiently large,| 1

d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1



Proof (cont’d)




d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)


It remains to prove that d ′ is complete. Let yi be a d ′-Cauchy sequencein Y .

Then yi is d-Cauchy. Let limi→∞ yi = y ∈ X . It remains to showthat y ∈ Y . Indeed, fix n. Then for any ε > 0, for i , j sufficiently large,| 1

d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1



Proof (cont’d)




d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)


It remains to prove that d ′ is complete. Let yi be a d ′-Cauchy sequencein Y . Then yi is d-Cauchy.

Let limi→∞ yi = y ∈ X . It remains to showthat y ∈ Y . Indeed, fix n. Then for any ε > 0, for i , j sufficiently large,| 1

d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1



Proof (cont’d)




d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)


It remains to prove that d ′ is complete. Let yi be a d ′-Cauchy sequencein Y . Then yi is d-Cauchy. Let limi→∞ yi = y ∈ X .

It remains to showthat y ∈ Y . Indeed, fix n. Then for any ε > 0, for i , j sufficiently large,| 1

d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1



Proof (cont’d)




d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)


It remains to prove that d ′ is complete. Let yi be a d ′-Cauchy sequencein Y . Then yi is d-Cauchy. Let limi→∞ yi = y ∈ X . It remains to showthat y ∈ Y .

Indeed, fix n. Then for any ε > 0, for i , j sufficiently large,| 1

d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1



Proof (cont’d)




d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)


It remains to prove that d ′ is complete. Let yi be a d ′-Cauchy sequencein Y . Then yi is d-Cauchy. Let limi→∞ yi = y ∈ X . It remains to showthat y ∈ Y . Indeed, fix n.

Then for any ε > 0, for i , j sufficiently large,| 1

d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1



Proof (cont’d)




d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)



d(yi ,Fn− 1

d(yj ,Fn)| < ε,

i.e., 1d(yi ,Fn)

is Cauchy, so it converges is R. This


Proof (cont’d)




d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)



d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1

d(yi ,Fn)is Cauchy,

so it converges is R. This


Proof (cont’d)




d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)



d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1

d(yi ,Fn)is Cauchy, so it converges is R.

This


Proof (cont’d)




d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)



d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1


means that d(yi ,Fn) is bounded away from 0.

Consequently, y /∈ Fn, i.e.,y ∈ Un.

Proof (cont’d)




d ′(x , y) = d(x , y) +∑

n

min(1

2n+1, | 1

d(x ,Fn)− 1

d(y ,Fn)|)



d(yi ,Fn− 1

d(yj ,Fn)| < ε, i.e., 1



Cantor schemes

DefinitionA Cantor scheme on a set X is a family {As}s∈2<ω of subsets of X s.t.:

I ∀s ∈ 2<ω As0 ∩ As1 = ∅I ∀s ∈ 2<ω ∀i ∈ 2 Asi ⊆ As

Embedding Cantor space in perfect Polish spaces

TheoremLet X be a perfect Polish space. Then there is a continuous injection2N → X .

Proof.First define by induction on length(s) a Cantor scheme {Us}s∈2<ω on Xs.t. for all s ∈ 2<ω:

1. Us is open non-empty

2. diam(Us) ≤ 12length(s)

3. Us i ⊆ Us

Start with any U∅ satisfying (1) and (2). Given Us , let x0, x1 be distinctpoints in Us , which exist since X is perfect; let U0,U1 be sufficientlysmall, disjoint open nbhds of x0, x1, resp.Now, for every x ∈ 2N, by completeness of X , the set⋂

n∈N Ux|n=⋂

n∈N Ux|nis a singleton {f (x)}. Then f : 2N → X is

injective and continuous.






3. Us i ⊆ Us

Start with any U∅ satisfying (1) and (2).

Given Us , let x0, x1 be distinctpoints in Us , which exist since X is perfect; let U0,U1 be sufficientlysmall, disjoint open nbhds of x0, x1, resp.Now, for every x ∈ 2N, by completeness of X , the set⋂

n∈N Ux|n=⋂








3. Us i ⊆ Us

Start with any U∅ satisfying (1) and (2). Given Us , let x0, x1 be distinctpoints in Us , which exist since X is perfect;

let U0,U1 be sufficientlysmall, disjoint open nbhds of x0, x1, resp.Now, for every x ∈ 2N, by completeness of X , the set⋂

n∈N Ux|n=⋂








3. Us i ⊆ Us

Start with any U∅ satisfying (1) and (2). Given Us , let x0, x1 be distinctpoints in Us , which exist since X is perfect; let U0,U1 be sufficientlysmall, disjoint open nbhds of x0, x1, resp.

Now, for every x ∈ 2N, by completeness of X , the set⋂n∈N Ux|n

=⋂








3. Us i ⊆ Us


n∈N Ux|n=⋂

n∈N Ux|nis a singleton

{f (x)}. Then f : 2N → X isinjective and continuous.






3. Us i ⊆ Us


n∈N Ux|n=⋂

n∈N Ux|nis a singleton {f (x)}.

Then f : 2N → X isinjective and continuous.






3. Us i ⊆ Us


n∈N Ux|n=⋂




CorollaryIf X is a perfect Polish space, then card(X ) = 2ℵ0 .

Cantor-Bendixson theorem

TheoremLet X be a Polish space. Then there is a unique decompositionX = P ∪ C with P perfect and C countable open.

Proof.For Z any topological space, letZ∗ = {x ∈ Z | x is a condensation point of Z}.Let P = X ∗,C = X \ P. If {Un}n is a basis of X , thenC =

⋃{Un | Un is countable}. So C is open countable and P is closed.

To show that P is perfect, let x ∈ P and U a nbhd of x in X . Then U isuncountable, so it contains some other condensation points of X .For uniqueness, let X = P1 ∪ C1 be another decomposition. Since P1 isperfect Polish, any open nbhd of x in P1 is perfect Polish as well, souncountable, and x is a condensation point of P1, hence of X . ThusP1 ⊆ P. If x ∈ C1, since C1 is countable open, x ∈ C , so C1 ⊆ C .Consequently, P = P1,C = C1.



Proof.For Z any topological space, letZ∗ = {x ∈ Z | x is a condensation point of Z}.

Let P = X ∗,C = X \ P. If {Un}n is a basis of X , thenC =





Proof.For Z any topological space, letZ∗ = {x ∈ Z | x is a condensation point of Z}.Let P = X ∗,C = X \ P.

If {Un}n is a basis of X , thenC =






⋃{Un | Un is countable}.

So C is open countable and P is closed.To show that P is perfect, let x ∈ P and U a nbhd of x in X . Then U isuncountable, so it contains some other condensation points of X .For uniqueness, let X = P1 ∪ C1 be another decomposition. Since P1 isperfect Polish, any open nbhd of x in P1 is perfect Polish as well, souncountable, and x is a condensation point of P1, hence of X . ThusP1 ⊆ P. If x ∈ C1, since C1 is countable open, x ∈ C , so C1 ⊆ C .Consequently, P = P1,C = C1.





To show that P is perfect, let x ∈ P and U a nbhd of x in X .

Then U isuncountable, so it contains some other condensation points of X .For uniqueness, let X = P1 ∪ C1 be another decomposition. Since P1 isperfect Polish, any open nbhd of x in P1 is perfect Polish as well, souncountable, and x is a condensation point of P1, hence of X . ThusP1 ⊆ P. If x ∈ C1, since C1 is countable open, x ∈ C , so C1 ⊆ C .Consequently, P = P1,C = C1.





To show that P is perfect, let x ∈ P and U a nbhd of x in X . Then U isuncountable,

so it contains some other condensation points of X .For uniqueness, let X = P1 ∪ C1 be another decomposition. Since P1 isperfect Polish, any open nbhd of x in P1 is perfect Polish as well, souncountable, and x is a condensation point of P1, hence of X . ThusP1 ⊆ P. If x ∈ C1, since C1 is countable open, x ∈ C , so C1 ⊆ C .Consequently, P = P1,C = C1.





To show that P is perfect, let x ∈ P and U a nbhd of x in X . Then U isuncountable, so it contains some other condensation points of X .

For uniqueness, let X = P1 ∪ C1 be another decomposition. Since P1 isperfect Polish, any open nbhd of x in P1 is perfect Polish as well, souncountable, and x is a condensation point of P1, hence of X . ThusP1 ⊆ P. If x ∈ C1, since C1 is countable open, x ∈ C , so C1 ⊆ C .Consequently, P = P1,C = C1.





To show that P is perfect, let x ∈ P and U a nbhd of x in X . Then U isuncountable, so it contains some other condensation points of X .For uniqueness, let X = P1 ∪ C1 be another decomposition.

Since P1 isperfect Polish, any open nbhd of x in P1 is perfect Polish as well, souncountable, and x is a condensation point of P1, hence of X . ThusP1 ⊆ P. If x ∈ C1, since C1 is countable open, x ∈ C , so C1 ⊆ C .Consequently, P = P1,C = C1.





To show that P is perfect, let x ∈ P and U a nbhd of x in X . Then U isuncountable, so it contains some other condensation points of X .For uniqueness, let X = P1 ∪ C1 be another decomposition. Since P1 isperfect Polish, any open nbhd of x in P1 is perfect Polish as well,

souncountable, and x is a condensation point of P1, hence of X . ThusP1 ⊆ P. If x ∈ C1, since C1 is countable open, x ∈ C , so C1 ⊆ C .Consequently, P = P1,C = C1.





To show that P is perfect, let x ∈ P and U a nbhd of x in X . Then U isuncountable, so it contains some other condensation points of X .For uniqueness, let X = P1 ∪ C1 be another decomposition. Since P1 isperfect Polish, any open nbhd of x in P1 is perfect Polish as well, souncountable, and x is a condensation point of P1,

hence of X . ThusP1 ⊆ P. If x ∈ C1, since C1 is countable open, x ∈ C , so C1 ⊆ C .Consequently, P = P1,C = C1.





To show that P is perfect, let x ∈ P and U a nbhd of x in X . Then U isuncountable, so it contains some other condensation points of X .For uniqueness, let X = P1 ∪ C1 be another decomposition. Since P1 isperfect Polish, any open nbhd of x in P1 is perfect Polish as well, souncountable, and x is a condensation point of P1, hence of X .

ThusP1 ⊆ P. If x ∈ C1, since C1 is countable open, x ∈ C , so C1 ⊆ C .Consequently, P = P1,C = C1.





To show that P is perfect, let x ∈ P and U a nbhd of x in X . Then U isuncountable, so it contains some other condensation points of X .For uniqueness, let X = P1 ∪ C1 be another decomposition. Since P1 isperfect Polish, any open nbhd of x in P1 is perfect Polish as well, souncountable, and x is a condensation point of P1, hence of X . ThusP1 ⊆ P.

If x ∈ C1, since C1 is countable open, x ∈ C , so C1 ⊆ C .Consequently, P = P1,C = C1.





To show that P is perfect, let x ∈ P and U a nbhd of x in X . Then U isuncountable, so it contains some other condensation points of X .For uniqueness, let X = P1 ∪ C1 be another decomposition. Since P1 isperfect Polish, any open nbhd of x in P1 is perfect Polish as well, souncountable, and x is a condensation point of P1, hence of X . ThusP1 ⊆ P. If x ∈ C1, since C1 is countable open, x ∈ C ,

so C1 ⊆ C .Consequently, P = P1,C = C1.





To show that P is perfect, let x ∈ P and U a nbhd of x in X . Then U isuncountable, so it contains some other condensation points of X .For uniqueness, let X = P1 ∪ C1 be another decomposition. Since P1 isperfect Polish, any open nbhd of x in P1 is perfect Polish as well, souncountable, and x is a condensation point of P1, hence of X . ThusP1 ⊆ P. If x ∈ C1, since C1 is countable open, x ∈ C , so C1 ⊆ C .

Consequently, P = P1,C = C1.

CH for Polish spaces

CorollaryAny uncountable Polish space contains a subspace homeomorphic to 2N.In particular, any uncountable Polish space has cardinality 2ℵ0 .

CorollaryAny uncountable Gδ or Fσ subset of a Polish space contains a subspacehomeomorphic to 2N; in particular, it has cardinality 2ℵ0 .

Exercise. In the previous notation, show that P is the largest perfectsubset of X .

DefinitionIf X is a Polish space and X = P ∪ C with P perfect and C countable,then P is the perfect kernel of X .

Cantor-Bendixson derivatives and ranks

TheoremLet X be second countable and {Fα | α < ρ} be a strictly monotone(transfinite) sequence of closed sets.

The ρ is countable.So the same holds for strictly monotone sequences of open sets.

Proof.Let {Un}n be a basis for X . For F closed in X , letN(F ) = {n | F ∩ Un 6= ∅}. Since X \ F =

⋃n/∈N(F ) Un, the map

F 7→ N(F ) is injective and F ⊆ G ⇔ N(F ) ⊆ N(G ). So if {Fα | α < ρ}is strictly monotone, the same is true for {N(Fα) | α < ρ}, consequentlyρ is countable.


TheoremLet X be second countable and {Fα | α < ρ} be a strictly monotone(transfinite) sequence of closed sets. The ρ is countable.

So the same holds for strictly monotone sequences of open sets.





TheoremLet X be second countable and {Fα | α < ρ} be a strictly monotone(transfinite) sequence of closed sets. The ρ is countable.So the same holds for strictly monotone sequences of open sets.






Proof.Let {Un}n be a basis for X .

For F closed in X , letN(F ) = {n | F ∩ Un 6= ∅}. Since X \ F =





Proof.Let {Un}n be a basis for X . For F closed in X , letN(F ) = {n | F ∩ Un 6= ∅}.

Since X \ F =⋃

n/∈N(F ) Un, the map





⋃n/∈N(F ) Un,

the map






F 7→ N(F ) is injective

and F ⊆ G ⇔ N(F ) ⊆ N(G ). So if {Fα | α < ρ}is strictly monotone, the same is true for {N(Fα) | α < ρ}, consequentlyρ is countable.





F 7→ N(F ) is injective and F ⊆ G ⇔ N(F ) ⊆ N(G ).

So if {Fα | α < ρ}is strictly monotone, the same is true for {N(Fα) | α < ρ}, consequentlyρ is countable.





F 7→ N(F ) is injective and F ⊆ G ⇔ N(F ) ⊆ N(G ). So if {Fα | α < ρ}is strictly monotone, the same is true for {N(Fα) | α < ρ},

consequentlyρ is countable.


DefinitionGiven a topological space X , let the Cantor-Bendixson derivative of X bethe set X ′ = {x ∈ X | x is a limit point of X}.

So X ′ is closed. Moreover X is perfect iff X = X ′.

DefinitionBy transfinite recursion, define the iterated Cantor-Bendixson derivatives:

I X (0) = X

I X (α+1) = (X (α))′

I X (λ) =⋂α<λ X (α) for λ limit

So {X (α)}α∈On is a decreasing transfinite sequence of closed subsets ofX .

Exercise. Show that K (X )→ K (X ),K 7→ K ′ is continuous.


DefinitionGiven a topological space X , let the Cantor-Bendixson derivative of X bethe set X ′ = {x ∈ X | x is a limit point of X}.So X ′ is closed. Moreover X is perfect iff X = X ′.

DefinitionBy transfinite recursion, define the iterated Cantor-Bendixson derivatives:

I X (0) = X

I X (α+1) = (X (α))′

I X (λ) =⋂α<λ X (α) for λ limit

So {X (α)}α∈On is a decreasing transfinite sequence of closed subsets ofX .

Exercise. Show that K (X )→ K (X ),K 7→ K ′ is continuous.


TheoremLet X be Polish. There exists a countable ordinal α0 s.t.

I ∀α ≥ α0 X (α) = X (α0)

I X (α0) is the perfect kernel of X

Proof.By induction, if P is the perfect kernel of X , then ∀α P ⊆ X (α). Since{X (α)}α∈On is a decreasing sequence of closed sets, there exists α0

countable s.t. ∀α ≥ α0 X (α) = X (α0). Then (X (α0))′ = X (α0+1) = X (α0),so X (α0) is perfect and therefore X (α0) ⊆ P.

DefinitionThe least α0 such that ∀α ≥ α0 X (α) = X (α0) is the Cantor-Bendixsonrank of X .



I ∀α ≥ α0 X (α) = X (α0)


Proof.By induction, if P is the perfect kernel of X , then ∀α P ⊆ X (α).

Since{X (α)}α∈On is a decreasing sequence of closed sets, there exists α0





I ∀α ≥ α0 X (α) = X (α0)


Proof.By induction, if P is the perfect kernel of X , then ∀α P ⊆ X (α). Since{X (α)}α∈On is a decreasing sequence of closed sets,

there exists α0





I ∀α ≥ α0 X (α) = X (α0)



countable s.t. ∀α ≥ α0 X (α) = X (α0).

Then (X (α0))′ = X (α0+1) = X (α0),so X (α0) is perfect and therefore X (α0) ⊆ P.




I ∀α ≥ α0 X (α) = X (α0)



countable s.t. ∀α ≥ α0 X (α) = X (α0). Then (X (α0))′ = X (α0+1) = X (α0),

so X (α0) is perfect and therefore X (α0) ⊆ P.




I ∀α ≥ α0 X (α) = X (α0)



countable s.t. ∀α ≥ α0 X (α) = X (α0). Then (X (α0))′ = X (α0+1) = X (α0),so X (α0) is perfect

and therefore X (α0) ⊆ P.




I ∀α ≥ α0 X (α) = X (α0)





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