Forcing: How to prove unprovability The Continuum Hypothesis Regula Krapf

Forcing: How to prove unprovability

The Continuum Hypothesis

Regula Krapf

University of Koblenz-Landau, Campus Koblenz

June 26, 2018

Content

1 The technique of forcing

2 Cohen forcing

3 Lebesgue measure

4 Forcing¬CH

Motivation

Theorem (Cantor)

There is no surjection from Nonto P(N).

This lead to the study ofinfinite cardinals.

Definition

ℵ₀ denotes the cardinality ofω =N. ℵ₁ denotes the least uncountable cardinal.

ℵ₂ denotes the second-least uncountable cardinal.

c=2^ℵ0 denotes the cardinality of P(ω).

The Continuum Hypothesis

In 1878, Georg Cantor conjectured the Continuum Hypothesis:

Conjecture (The Continuum Hypothesis (CH))

There is no cardinal between ℵ₀ and2^ℵ0, i.e. 2^ℵ0 =ℵ₁.

The Continuum Hypothesis was the first of Hilbert’s famous list of 23 open Problems.

In 1940, Kurt Gödel constructed a model of ZFC (denoted L) in which the Continuum Hypothesis holds.

In 1963, Paul Cohen proved that CH is actually independent of the axioms of ZFC, i.e. both ZFC+CH and ZFC+¬CH are consistent.

The method he used (and developed for this result) isforcing.

Forcing notions

Definition

A forcing (notion) is a non-atomic partial order P= (P,≤_P,1P) with a maximal element 1_P, i.e.

≤_P is reflexive, transitive and antisymmetric, for every p ∈P,p≤_P 1P,

for every p ∈Pthere are p₀,p₁≤_P p which are incompatible, i.e.

there is nor ∈P withr ≤_P p0,p1.

Forcing notions

Example (Cohen forcing)

Let P=Fn(ω,2,ℵ₀) be the set of partial functionsp:dom(p)→2 with dom(p)⊆ω finite, ordered by reverse inclusion, i.e.

p ≤_P q ⇐⇒dom(p)⊇dom(q) andpdom(q) =q. Furthermore, let 1_P=∅, the empty function.

Generic filters

Definition

LetM be a countable transitive model of ZFC andP∈M be a forcing notion.

1 A subsetD⊆Pis said to bedense, if for everyp∈P there is someq≤_P p withq∈D.

2 A subsetG ⊆Pis said to be aP-generic filter, if it has the following properties:

Ifp≤_Pqandp∈G, thenq∈G.

Ifp,q∈G then there isr ∈G such thatr ≤_Pp,q.

IfD⊆Pis a dense set which is inM, thenG∩D6=∅.

Example

Consider Cohen forcingP=Fn(ω,2,ℵ0). Then for eachn∈ω, the set Dn={p∈P|n∈dom(p)}is dense.

Names

Definition

Let Pbe a forcing notion. A P-name is a set whose elements are of the form (σ,p), where σ is a P-name andp∈P.

This definition requires transfinite recursion.

Example

LetM |=ZFC and P∈M a forcing notion.

∅ is aP-name.

Letx ∈M. Then there is acanonicalP-name forx given by

ˇx={(ˇy,1P)|y∈x}.

The setG˙ ={(ˇp,p)|p∈P}is aP-name, the canonicalP-name for a P-generic filter.

Evaluations of names

Let Pbe a forcing notion and G aP-generic filter. We can evaluate a P-nameσ as follows:

σ^G ={τ^G | ∃p ∈G : (τ,p)∈σ}.

Example

∅^G =∅.

Letx ∈M. Then ˇ

x^G ={ˇy^G |y ∈x}={y |y ∈x}=x by transfinite induction.

G˙^G ={ˇp^G |p∈G}=G.

Generic extensions

Let M |=ZFC,P∈M be a non-trivial forcing notion andG aP-generic filter. Then we define

M[G] ={σ^G |σ is a P-name}.

Fact

1 M∪ {G} ⊆M[G]

2 G ∈/ M.

3 M[G]|=ZFC.

Proof.

For (2) suppose that G ∈M. Then the set D =P\G is inM. Moreover, D is dense: Letp∈P. Since Pis non-atomic, there arep₀,p₁≤_P p such that p₀ andp₁ are incompatible. But then at least one of p₀,p₁ is not in G.

Where do generic filters live?

Suppose that V|=ZFC and V contains countable, transitive modelsM ∈V of ZFC. Let P∈M be a forcing notion. We extend M to M[G], whereG is a P-generic filter contained in V.

Fact

For every p ∈Pthere is a P-generic filter G in Vwith p∈G .

Proof.

SinceM is countable, M contains only countably many dense subsets ofP. Let (D_n|n∈ω) enumerate them (in V). Now we inductively construct a sequence of conditions (p_n|n∈ω)by

p0 =p

Givenp_n, letp_n+1 ≤_Pp_n withp_n+1 ∈D_n.

Then G ={q∈P| ∃n ∈ω:p_n≤_Pq} is a P-generic filter overM.

The idea behind names

Every element of M[G]has aP-nameσ∈M butM does not know howσ will be evaluated. This is similar to the case of extensions of fields:

Consider Qand an algebraic closureQ¯ of Q.

InQ, the polynomialX²−2 names the root √ 2∈Q¯. If we extendQto Q[√

2], then√

2 is the evaluation ofX²−2.

Cohen forcing

LetP denoteFn(ω,2,ℵ0)and letG beP-generic overM. InM[G], consider the function

c= [

p∈G

p. (c is called aCohen real)

Fact

1 c is a functionω→2.

2 If f ∈M is a function f :ω→2then f 6=c.

Proof.

(1)

Letn∈ω. If there are p,q∈G such thatn∈dom(p)∩dom(q)and p(n)6=q(n), thenp andq are incompatible.

To see that dom(c) =ω, note thatDn={p∈P|n∈dom(p)}is dense.

Letp∈G∩Dn. Thenn∈dom(c)andc(n) =p(n).

Cohen forcing

LetP denoteFn(ω,2,ℵ0)and letG beP-generic overM. InM[G], consider the function

c= [

p∈G

p. (c is called aCohen real)

Fact

1 c is a functionω→2.

2 If f ∈M is a function f :ω→2then f 6=c.

Proof.

(2) ConsiderDf ={p∈P| ∃n∈dom(p) :p(n)6=f(n)}. We claim thatDf is a dense subset of P. Letp∈P. Since dom(p)is finite, there isn∈ω\dom(p).

Then q=p∪ {(n,1−f(n))} ≤_Pp andq∈Df.

Takep0∈G∩Df and letn0∈dom(p0)such thatp0(n0)6=f(n0). Then c(n0) =p0(n0)and soc6=f.

A nice application: Cantor’s Theorem

Theorem (Cantor)

There is no surjection from ω onto P(ω).

Proof.

Suppose the contrary. Observe that P=Fn(ω,2,ℵ₀)is countable. Then so is P(P), so we can enumerate (in M) all dense subsets of P. But then we can construct in M aP-generic filter over M.

The Cantor space

We identify Rwith ^ω2, the space of functions ω→2.

Observation

If P is a forcing notion andG isP-generic overM, thenR^M andR^M[G] do not have to be the same.

Intervals in the Cantor space

We define intervals in the following way: For

s ∈^<ω2={t :dom(t)→2|dom(t)∈ω} we define I_s ={x ∈R|s ⊆x} ⊆R and we define µ(I_s) =2^−dom(s).

Definition

We say that a set X ⊆Rhas measure zero, if for every ε >0 there exists a sequence (I_n|n∈ω) of intervals in Rsuch that X ⊆S

n∈ωI_n and P

n∈ωµ(I_s)< ε.

Theorem

Let M[G]be a generic extension of M by Cohen forcing, and let c ∈M[G] be the canonical Cohen real. Then in M[G],R^M has measure zero.

Proof.

Let ε >0, let k∈ω such that ε >2^−k. For n∈ω define s_n:k+n+1→2,s_n(l) =c(n+l).

Now define I_n=I_sn. Then X

n∈ω

µ(I_n) =X

n∈ω

2^−(k+n+1) =2^−k < ε.

Proof, continued.

sn:k+n+1→2,sn(l) =c(n+l),In=Isn. It remains to check: R^M ⊆S

n∈ωI_n.Let x ∈R^M. Then

D_x ={p∈P|∃n∈ω∀l <k+n+1:n+l ∈dom(p)

∧p(n+l) =x(l)} ∈M

is a dense subset of Cohen forcing P. Takep ∈G ∩D_x andn∈ω which witnesses thatp ∈D_x. Then

∀l <k+n+1:s_n(l) =c(n+l) =p(n+l) =x(l) and so sn⊆x andx ∈In.

Forcing ¬CH

We have used Cohen forcing to add one real c which is not an element of the ground modelM. What happens if we addℵ₂-many Cohen reals in this way?

Definition

Let Fn(ℵ^M₂ ×ω,2,ℵ₀)denote the set of finite partial functions from ℵ^M₂ ×ω→2, ordered by reverse inclusion.

Forcing ¬CH

Let Pdenote Fn(ℵ^M₂ ×ω,2,ℵ₀)and let G beP-generic over M.

Lemma

The following statements hold:

1 F =S

p∈Gp is a function F :ℵ^M₂ ×ω →2.

2 For each α <ℵ^M₂ , let c_α(n) =F(α,n). Then for all α < β <ℵ^M₂ , c_α 6=c_β.

Proof.

(1) Follows from the density of the sets

D_α,n={p ∈P|(α,n)∈dom(p)}

for all α <ℵ^M₂ andn ∈ω.

Forcing ¬CH

Let Pdenote Fn(ℵ^M₂ ×ω,2,ℵ₀)and let G beP-generic over M.

Lemma

The following statements hold:

1 F =S

p∈Gp is a function F :ℵ^M₂ ×ω →2.

2 For α <ℵ^M₂ , let c_α(n) =F(α,n). Then for all α < β <ℵ^M₂ , c_α6=c_β. Proof.

(2) For all α < β <ℵ^M₂ consider

D_αβ ={p∈P| ∃n ∈ω : (α,n),(β,n)∈dom(p)andp(α,n)6=p(β,n)}.

Then D_αβ is dense in P. By genericity, takep∈G ∩D_αβ andn∈ω such that (α,n),(β,n)∈dom(p)with p(α,n)6=p(β,n). Then

c_α(n) =F(α,n) =p(α,n)6=p(β,n) =F(β,n) =c_β(n).

Forcing ¬CH

The previous lemma shows that there are at leastℵ^M₂ many reals inM[G], so if we can show thatℵ^M₂ =ℵ^M[G₂ ^] then inM[G]

(2^ℵ0)^M[G]=|R^M[G]| ≥ ℵ^M₂ =ℵ^M[G₂ ^].

Definition

Let Pbe a forcing notion.

1 An antichain inP is a subsetA⊆Pwith the property that all

elements of Aare incompatible, i.e. for allp,q ∈A, there is nor ∈P with r≤_P p,q.

2 P satisfies thecountable chain condition (ccc), if every antichain inP is at most countable.

Forcing ¬CH

Theorem

If Psatisfies the ccc then Ppreserves all cardinals, i.e. |κ|^M[G]=κin every P-generic extension M[G]. In particular,ℵ^M[G₂ ^]=ℵ^M₂ .

How can it happen thatℵ^M₂ is collapsed?

Forcing ¬CH

Theorem

If Psatisfies the ccc then Ppreserves all cardinals, i.e. |κ|^M[G]=κin every P-generic extension M[G]. In particular,ℵ^M[G₂ ^]=ℵ^M₂ .

Example

Consider the forcing notion

Col(ω,ℵ^M₂ ) ={p:dom(p)→ ℵ^M₂ |dom(p)⊆ω finite}

ordered by reverse inclusion. If G is Col(ω,ℵ^M₂ )-generic, thenF =S

p∈G p is a function ω→ ℵ^M₂ . ThenF is surjective because for eachα <ℵ^M₂ the set

D_α={p ∈Col(ω,ℵ^M₂ )| ∃n∈dom(p) :p(n) =α}

is dense. So in M[G]we have ℵ^M₂ =ℵ^M[G₀ ^].

Forcing ¬CH

Lemma

P=Fn(ℵ^M₂ ×ω,2,ℵ₀) has the ccc.

Proof.

Suppose that (p_i |i <ℵ^M₁ )enumerates an uncountable antichain. Then there is a setX ⊆ ℵ^M₁ of size ℵ^M₁ andn∈ω such that

∀i ∈X : |dom(p_i)|=n.

Now letm≤n be maximal such that there is a set b with |b|=m and Y ⊆X of sizeℵ^M₁ such that for alli ∈Y,b⊆dom(p_i).

We claim that for all x ∈/ b there isi(x)<ℵ^M₁ such that for all i >i(x)in Y,x ∈/ dom(p_i). This holds because otherwiseb∪ {x}would contradict the maximality ofm.

Forcing ¬CH

Proof, continued.

Y ⊆ ℵ₁ of sizeℵ^M₁ withb⊆dom(p_i) for alli ∈Y.

for all x ∈/ b,i(x)<ℵ^M₁ with x ∈/ dom(p_i) for alli >i(x) in Y. We construct an increasing sequence (i_ξ|ξ <ℵ₁) withi_ξ∈Y by recursion:

Suppose that (i_ξ |ξ < ζ) has already been defined. Then leti_ζ be the minimal ordinal i ∈Y such that

i >sup{i(x)| ∃ξ < ζ :x ∈dom(p_ξ)}.

Now letξ < ζ <ℵ^M₁ . We claim that dom(p_iξ)∩dom(p_iζ)⊆b. Let x ∈dom(p_iξ)∩dom(p_iζ) andx ∈/ b. Theni_ζ>i(x) and so x ∈/ dom(p_iζ).

But there are only countably many possibilities for p_iξ b, so there must be some ξ 6=ζ such that p_iξ andp_iζ are compatible.

Forcing ¬CH

Thus we have proved Theorem (Gödel,Cohen)

CH is independent of the axioms ofZFC.

Thank you for your attention!