set theory of the continuum - its.caltech.edupanagio/116c/w2d1.pdf · set theory of the continuum...

Set theory of the continuum

Caltech

Week 2, day 1

(Caltech) Set theory of the continuum Week 2, day 1 1 / 18

Table of Contents

1 “Probabilistic method” for Polish spaces

2 The Baire space

Comeager sets are “large”

Let X be a Polish space. A subset M is meager if it is a countable union

M =⋃n

where every set Nn is nowhere dense.

A set N ⊆ X is nowhere dense, if for every open non-empty U ⊆ X,there is an open non-empty V ⊂ U , so that V ∩N = ∅

A subset C ⊆ X is comeager, if C is meager.

C is comeager ⇐⇒ C contains a dense Gδ set.

Consequence of: nowhere dense sets have “closed envelopes”:A is nowhere dense ⇐⇒ A is nowhere dense ⇐⇒ A has empty interior.

M =⋃n

Showing that a property P is generic

Recall that we say “the generic element of X has property P” if:

{x ∈ X | x satisfies P} is comeager subset of X

How does one show that a property P is generic?

In practice there are three ways to do that (of increasing difficulty):

1 show that P is dense and Gδ;

2 find some property Q with Q ⊆ P which is dense Gδ;

3 show that Player II has winning strategy in the Banach-Mazur game.

We will see some examples of (1) and (2) today and leave (3) for later.

Generic properties (via method 1)

Theorem

The generic countable linear ordering is isomorphic to (Q,≤).

Proof. Recall that the space LO(N) is a closed subset of {0, 1}N×N.Recall that the set IsoQ of all elements of LO(N) which are isomorphic to(Q,≤) is the intersection of the following sets

DENSE ⊆ LO(N), NoMin ⊆ LO(N), NoMax ⊆ LO(N)

Which are the intersection of the following sets (m,n range over N):

NoMax(n) := {(N,≤A) ∈ LO(N) | n is not the ≤A -max}NoMin(n) := {(N,≤A) ∈ LO(N) | n is not the ≤A -min}

DENSE(n,m) := {(N,≤A) ∈ LO(N) | m is not the ≤A -successor of n}Each of the latter sets is dense Gδ. For example,

NoMax(n) :=⋃k

{(N,≤A) ∈ LO(N) | n < k}

Where the Red Set is easily seen to be dense open.

Theorem

Proof. Recall that the space LO(N) is a closed subset of {0, 1}N×N.

Recall that the set IsoQ of all elements of LO(N) which are isomorphic to(Q,≤) is the intersection of the following sets

NoMax(n) :=⋃k

{(N,≤A) ∈ LO(N) | n < k}

Theorem

NoMax(n) :=⋃k

{(N,≤A) ∈ LO(N) | n < k}

Theorem

DENSE(n,m) := {(N,≤A) ∈ LO(N) | m is not the ≤A -successor of n}

Each of the latter sets is dense Gδ. For example,

NoMax(n) :=⋃k

{(N,≤A) ∈ LO(N) | n < k}

Theorem

DENSE(n,m) := {(N,≤A) ∈ LO(N) | m is not the ≤A -successor of n}Each of the latter sets is dense Gδ.

For example,

NoMax(n) :=⋃k

{(N,≤A) ∈ LO(N) | n < k}

Theorem

NoMax(n) :=⋃k

{(N,≤A) ∈ LO(N) | n < k}

Theorem

NoMax(n) :=⋃k

{(N,≤A) ∈ LO(N) | n < k}

Where the Red Set is easily seen to be dense open.(Caltech) Set theory of the continuum Week 2, day 1 5 / 18

Theorem

NoMax(n) :=⋃k

{(N,≤A) ∈ LO(N) | n < k}

Where the Red Set is easily seen to be dense open.(Caltech) Set theory of the continuum Week 2, day 1 5 / 18

Theorem

The generic continuous function f : [0, 1]→ R is nowhere differentiable.

For the proof of this we will need to use a very important fact about

projections of closed sets

Recall Lebesgue’s mistake that projections “increase complexity” i.e., ifB ⊆ R2 is Borel then projR(B) is not always Borel.

In fact there is a closed C ⊂ NN × NN so that projNN(C) is not Borel!That being said, we have the following very important lemma:

Lemma (Tube Lemma)

Let X,K be Polish, with K being compact. Let also C ⊆ X ×K beclosed. Then the projection projX(C) of C on X is closed.

Exercise If C ⊆ R2 is closed then projR(C) is Borel but not always closed.

Theorem

Lemma (Tube Lemma)

Theorem

Lemma (Tube Lemma)

Theorem

In fact there is a closed C ⊂ NN × NN so that projNN(C) is not Borel!

That being said, we have the following very important lemma:

Lemma (Tube Lemma)

Theorem

Lemma (Tube Lemma)

Theorem

Lemma (Tube Lemma)

Theorem

Lemma (Tube Lemma)

Exercise If C ⊆ R2 is closed then projR(C) is Borel but not always closed.(Caltech) Set theory of the continuum Week 2, day 1 6 / 18

Tube Lemma

Lemma (Tube Lemma)

Proof. Let x 6∈ projX(C). We will find U ⊆ X open with x ∈ U .

Consider the fiber {x} ×K ⊆ X ×K.

Since {x} ×K ∩ C = ∅, for each (x, k) ∈ {x} ×K there are open setsUk ⊂ X, Wk ⊆ K, so that (x, k) ∈ Uk ×Wk ⊆ Cc.By compactness we need only finitely many k1, . . . , kn so that the union ofthe Wk covers K.

Set U := Uk1 ∩ Uk2 ∩ · · · ∩ Ukn .

Tube Lemma

Lemma (Tube Lemma)

Set U := Uk1 ∩ Uk2 ∩ · · · ∩ Ukn .

Tube Lemma

Lemma (Tube Lemma)

Set U := Uk1 ∩ Uk2 ∩ · · · ∩ Ukn .

Tube Lemma

Lemma (Tube Lemma)

Since {x} ×K ∩ C = ∅, for each (x, k) ∈ {x} ×K there are open setsUk ⊂ X, Wk ⊆ K, so that (x, k) ∈ Uk ×Wk ⊆ Cc.

By compactness we need only finitely many k1, . . . , kn so that the union ofthe Wk covers K.

Set U := Uk1 ∩ Uk2 ∩ · · · ∩ Ukn .

Tube Lemma

Lemma (Tube Lemma)

Set U := Uk1 ∩ Uk2 ∩ · · · ∩ Ukn .

Tube Lemma

Lemma (Tube Lemma)

Set U := Uk1 ∩ Uk2 ∩ · · · ∩ Ukn .

Theorem

Proof. For every n ∈ N consider the set:

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

∣∣ ≤ n)}Notice that if f is somewhere differentiable then f ∈ Nn for some large n.

So it suffices to show that⋃nNn is meager!

We will show that Nn is closed and nowhere dense.

Question: If we show Nn is closed, will it help with Nn is nowhere dense?

YES! Enough to show that Nn is closed and int(Nn) = ∅.

Theorem

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

∣∣ ≤ n)}

Notice that if f is somewhere differentiable then f ∈ Nn for some large n.

Theorem

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

Theorem

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

Theorem

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

Theorem

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

Theorem

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

Theorem

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

Generic properties (via method 2)Claim 1. For every n ∈ N the set Nn is closed, where:

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

∣∣ ≤ n)}

Fix n ∈ N and fix δ ∈ (0, 1/n] and consider the set:

An,δ = {(x, f) ∈ [0, 1− 1

n]× C([0, 1])

∣∣ (∣∣f(x+ δ)− f(x)δ

∣∣ ≤ n)}An,δ is closed since if (xn, fn) ∈ An,δ and (xn, fn)→ (x, f), then:

fn(xn)→ f(x) and fn(xn + δ)→ f(x+ δ)

It follows that (x, f) ∈ An,δ, and therefore the latter is closed. But:

Nn = projC([0,1])

( ⋂δ∈(0, 1

An,δ)

By Tube Lemma we have that Nn is closed.

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

∣∣ ≤ n)}Fix n ∈ N and fix δ ∈ (0, 1/n] and consider the set:

An,δ = {(x, f) ∈ [0, 1− 1

n]× C([0, 1])

∣∣ (∣∣f(x+ δ)− f(x)δ

Nn = projC([0,1])

( ⋂δ∈(0, 1

An,δ)

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

An,δ = {(x, f) ∈ [0, 1− 1

n]× C([0, 1])

∣∣ (∣∣f(x+ δ)− f(x)δ

∣∣ ≤ n)}

An,δ is closed since if (xn, fn) ∈ An,δ and (xn, fn)→ (x, f), then:

Nn = projC([0,1])

( ⋂δ∈(0, 1

An,δ)

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

An,δ = {(x, f) ∈ [0, 1− 1

n]× C([0, 1])

∣∣ (∣∣f(x+ δ)− f(x)δ

Nn = projC([0,1])

( ⋂δ∈(0, 1

An,δ)

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

An,δ = {(x, f) ∈ [0, 1− 1

n]× C([0, 1])

∣∣ (∣∣f(x+ δ)− f(x)δ

Nn = projC([0,1])

( ⋂δ∈(0, 1

An,δ)

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

An,δ = {(x, f) ∈ [0, 1− 1

n]× C([0, 1])

∣∣ (∣∣f(x+ δ)− f(x)δ

It follows that (x, f) ∈ An,δ, and therefore the latter is closed.

Nn = projC([0,1])

( ⋂δ∈(0, 1

An,δ)

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

An,δ = {(x, f) ∈ [0, 1− 1

n]× C([0, 1])

∣∣ (∣∣f(x+ δ)− f(x)δ

Nn = projC([0,1])

( ⋂δ∈(0, 1

An,δ)

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

An,δ = {(x, f) ∈ [0, 1− 1

n]× C([0, 1])

∣∣ (∣∣f(x+ δ)− f(x)δ

Nn = projC([0,1])

( ⋂δ∈(0, 1

An,δ)

Generic properties (via method 2)Claim 2. For every n ∈ N we have that int(Nn) = ∅, where:

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

∣∣ ≤ n)}

We will show that if U is a basic open set then there is some g ∈ U \Nn.

By Weierstrass: may assume that there is polynomial p and ε > 0 so that

U = Up,ε := {f ∈ C([0, 1]) | ∀x ∈ [0, 1] |f(x)− p(x)| < ε}.Let M be an upper bound on the derivative of p.

Let l(x) be a PL-function with slope ±(M + n+ 1) and 0 ≤ |l(x)| < ε

Set g(x) = p(x) + l(x). Clearly g ∈ Up,ε.For any fixed x, if δ is small enough so that l is linear on [x, x+ δ] then:∣∣g(x+ δ)− g(x)

∣∣ ≥ ∣∣ l(x+ δ)− l(x)δ

∣∣− ∣∣p(x+ δ)− p(x)δ

∣∣ ≥≥ (M + n+ 1)−M > n. Hence, g 6∈ Nn.

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

∣∣ ≤ n)}We will show that if U is a basic open set then there is some g ∈ U \Nn.

∣∣ ≥ ∣∣ l(x+ δ)− l(x)δ

∣∣− ∣∣p(x+ δ)− p(x)δ

∣∣ ≥≥ (M + n+ 1)−M > n. Hence, g 6∈ Nn.

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

U = Up,ε := {f ∈ C([0, 1]) | ∀x ∈ [0, 1] |f(x)− p(x)| < ε}.

Let M be an upper bound on the derivative of p.

∣∣ ≥ ∣∣ l(x+ δ)− l(x)δ

∣∣− ∣∣p(x+ δ)− p(x)δ

∣∣ ≥≥ (M + n+ 1)−M > n. Hence, g 6∈ Nn.

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

∣∣ ≥ ∣∣ l(x+ δ)− l(x)δ

∣∣− ∣∣p(x+ δ)− p(x)δ

∣∣ ≥≥ (M + n+ 1)−M > n. Hence, g 6∈ Nn.

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

∣∣ ≥ ∣∣ l(x+ δ)− l(x)δ

∣∣− ∣∣p(x+ δ)− p(x)δ

∣∣ ≥≥ (M + n+ 1)−M > n. Hence, g 6∈ Nn.

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

Set g(x) = p(x) + l(x). Clearly g ∈ Up,ε.

For any fixed x, if δ is small enough so that l is linear on [x, x+ δ] then:∣∣g(x+ δ)− g(x)δ

∣∣ ≥ ∣∣ l(x+ δ)− l(x)δ

∣∣− ∣∣p(x+ δ)− p(x)δ

∣∣ ≥≥ (M + n+ 1)−M > n. Hence, g 6∈ Nn.

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

Set g(x) = p(x) + l(x). Clearly g ∈ Up,ε.For any fixed x, if δ is small enough so that l is linear on [x, x+ δ] then:

∣∣g(x+ δ)− g(x)δ

∣∣ ≥ ∣∣ l(x+ δ)− l(x)δ

∣∣− ∣∣p(x+ δ)− p(x)δ

∣∣ ≥≥ (M + n+ 1)−M > n. Hence, g 6∈ Nn.

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

∣∣ ≥ ∣∣ l(x+ δ)− l(x)δ

∣∣− ∣∣p(x+ δ)− p(x)δ

∣∣

≥ (M + n+ 1)−M > n. Hence, g 6∈ Nn.

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

∣∣ ≥ ∣∣ l(x+ δ)− l(x)δ

∣∣− ∣∣p(x+ δ)− p(x)δ

∣∣ ≥≥ (M + n+ 1)−M > n.

Hence, g 6∈ Nn.

Nn = {f ∈ C([0, 1]) |(∃x ∈ [0, 1− 1

n])(∀δ ∈ (0,

n])(∣∣f(x+ δ)− f(x)

∣∣ ≥ ∣∣ l(x+ δ)− l(x)δ

∣∣− ∣∣p(x+ δ)− p(x)δ

∣∣ ≥≥ (M + n+ 1)−M > n. Hence, g 6∈ Nn.

More generic properties (Exercise)

1 The generic “countable graph” is connected.

2 The generic “countable graph” is isomorphic to the Random graph.3 The generic “countable locally-finite group” is isomorphic to Hall’s

universal group. This is the unique up to isomorphism countablelocally-finite group G so that:

1 every finite group A embeds into G;2 if i : A→ B is an embedding between finite groups and f : A→ G is

an emebedding then there is embedding f̃ : B → G so that f = (f̃ ◦ i).4 There is a generic permutation of N up to “change of coordinates,”

i.e., there is some g ∈ Perm(N) whose conjugacy class is comeager.

2 The generic “countable graph” is isomorphic to the Random graph.

3 The generic “countable locally-finite group” is isomorphic to Hall’suniversal group. This is the unique up to isomorphism countablelocally-finite group G so that:

universal group.

This is the unique up to isomorphism countablelocally-finite group G so that:

an emebedding then there is embedding f̃ : B → G so that f = (f̃ ◦ i).

4 There is a generic permutation of N up to “change of coordinates,”i.e., there is some g ∈ Perm(N) whose conjugacy class is comeager.

Table of Contents

1 “Probabilistic method” for Polish spaces

2 The Baire space

Let A be a non-empty set, the alphabet. Consider the set

A<N :=⋃n∈N

of all finite sequences from A, where A0 := {∅}. Let

s = (s0, . . . , sn−1), t = (t0, . . . , tm−1) be elements of A<N.

The length |s| of s is n.

We write s ⊆ t, if t extends s, i.e., |s| ≤ |t| and si = ti for all i < |s|.

The concatenation s_t is the sequence (s0, . . . sn−1, t0, . . . , tm−1).

We also write s_a instead of s_(a), when a ∈ A.

A tree on A is any subset T ⊆ A<N that is downward closed:

t ∈ T and s ⊆ t then s ∈ T

Notice that ∅ belongs to every non-empty tree.

TreesLet A be a non-empty set, the alphabet. Consider the set

A<N :=⋃n∈N

of all finite sequences from A, where A0 := {∅}.

A<N :=⋃n∈N

Notice that ∅ belongs to every non-empty tree.(Caltech) Set theory of the continuum Week 2, day 1 13 / 18

Infinite branches

Let now AN be the set of all infinite sequences from A:

α = (α0, α1, α2, . . .)

We view AN as a topological space: we endow A with the discretetopology and AN inherits the product topology.Every s ∈ A<ω defines a basic open subset of AN, the set:

Ns := {α ∈ AN | s ⊆ α}. It is clopen!

Moreover, the collection of all Ns, forms a basis for the topology.

Definition

We denote by N the Baire space NN.We denote by C the Cantor space 2N.

Notice that both N , C are Polish and C is compact.

Infinite branchesLet now AN be the set of all infinite sequences from A:

α = (α0, α1, α2, . . .)

Definition

α = (α0, α1, α2, . . .)

We view AN as a topological space: we endow A with the discretetopology and AN inherits the product topology.

Every s ∈ A<ω defines a basic open subset of AN, the set:

Definition

α = (α0, α1, α2, . . .)

Ns := {α ∈ AN | s ⊆ α}.

It is clopen!

Definition

α = (α0, α1, α2, . . .)

Definition

α = (α0, α1, α2, . . .)

Definition

α = (α0, α1, α2, . . .)

Definition

α = (α0, α1, α2, . . .)

Definition

Notice that both N , C are Polish and C is compact.(Caltech) Set theory of the continuum Week 2, day 1 14 / 18

N is far from compact

A subset Z of a Polish space is Kσ if Z =⋃nKn with Kn compact.

Theorem

The Baire space N is not Kσ.

Proof.Assume that Z =

⋃nKn with Kn compact.

For every l ∈ N the collection {Ns | |s| = l} forms an open cover of N.

There is some n0 ∈ N so that N(n0) ∩K0 = ∅.

· · ·

Assume Ns avoids K0, . . . ,Kl−1. There is nl so that Ns_(nl) ∩Kl = ∅.

The element (n0, . . . , nl, . . .) avoids Z.

Theorem

· · ·

Theorem

· · ·

Theorem

· · ·

Theorem

· · ·

Theorem

· · ·

Theorem

· · ·

Theorem

· · ·

Assume Ns avoids K0, . . . ,Kl−1.

There is nl so that Ns_(nl) ∩Kl = ∅.

Theorem

· · ·

Theorem

· · ·

A combinatorial characterization of closed sets

The body [T ] of T is the set of all α ∈ AN which are approximable by T :

[T ] := {α ∈ AN | ∀n ∈ N(α|n ∈ T

Notice. [T ] is a closed subset of AN.Indeed, if α 6∈ [T ] then there is n ∈ N so that s := α|n 6∈ T .Hence Ns is an open neighborhood of α avoiding [T ].

Going now the other direction, if X ⊆ AN then we have the tree

TX := {s ∈ A<N | s ⊆ x for some x ∈ X}.

Notice. TX is a pruned tree.T is pruned if for every s ∈ T there is a ∈ A so that s_a ∈ T .

A combinatorial characterization of closed setsA tree on A is any subset T ⊆ A<N that is downward closed:

[T ] := {α ∈ AN | ∀n ∈ N(α|n ∈ T

Notice. [T ] is a closed subset of AN.

Indeed, if α 6∈ [T ] then there is n ∈ N so that s := α|n 6∈ T .Hence Ns is an open neighborhood of α avoiding [T ].

[T ] := {α ∈ AN | ∀n ∈ N(α|n ∈ T

Notice. [T ] is a closed subset of AN.Indeed, if α 6∈ [T ] then there is n ∈ N so that s := α|n 6∈ T .

Hence Ns is an open neighborhood of α avoiding [T ].

[T ] := {α ∈ AN | ∀n ∈ N(α|n ∈ T

Going now the other direction,

if X ⊆ AN then we have the tree

[T ] := {α ∈ AN | ∀n ∈ N(α|n ∈ T

Notice. TX is a pruned tree.

T is pruned if for every s ∈ T there is a ∈ A so that s_a ∈ T .

[T ] := {α ∈ AN | ∀n ∈ N(α|n ∈ T

A combinatorial characterization of closed sets

There is a one to one correspondence:

closed subsets of AN

⇐⇒pruned subtrees of A<N

Retracts

Theorem

Every closed subset F of N is a retract.

That is, there is a continuousmap f : N → F with f�F = idF .

Retracts

Theorem

Every closed subset F of N is a retract. That is, there is a continuousmap f : N → F with f�F = idF .

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