[PPT] - Suslins Problem and Martin Axiom Forcing Direct Limit Construction PowerPoint Presentation

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Suslin’s Problem and Martin Axiom Suslin’s Problem Martin Axiom Iterated Forcing Direct Limit Construction

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Suslin’s Problem and Martin Axiom

23 July 2014

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Suslin’s Problem

Is there a linearly ordered set which satisfies the countable chain condition (ccc) and is not separable?

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Suslin’s Problem

Is there a linearly ordered set which satisfies the countable chain condition (ccc) and is not separable? Such a set is called a Suslin line. The existence of a Suslin line is equivalent to the existence of a normal Suslin tree.

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Tree

A tree is a poset (P, <) such that ∀x ∈ T {y : y < x} is well

rdered by <.

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Tree

A tree is a poset (P, <) such that ∀x ∈ T {y : y < x} is well

rdered by <.

Suslin Tree

A tree is called a Suslin tree if:

1 height(T) = ω1 2 every branch in T is at most countable 3 every antichain in T is at most countable

A Suslin tree is called normal if:

1 T has a unique least point 2 each level of T is at most countable 3 x not maximal has infinitely many immediate successors 4 ∀x ∈ T there is some z > x at each greater level 5 if o(x) = o(y) = β with β limit and

{z : z < x} = {z : z < y} then x = y

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MAk

If a poset (P, <) satisfies ccc and D is a collection of at most k dense subsets of P, then there exists a D-generic filter on P.

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MAk

If a poset (P, <) satisfies ccc and D is a collection of at most k dense subsets of P, then there exists a D-generic filter on P.

Lemma

If MAℵ1 holds then there is no Suslin tree.

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MAk

If a poset (P, <) satisfies ccc and D is a collection of at most k dense subsets of P, then there exists a D-generic filter on P.

Lemma

If MAℵ1 holds then there is no Suslin tree.

Solovay-Tennenbaum

There is a model M of ZFC such that M | = MA + 2ℵ0 > ℵ1.

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Suslin’s Problem and Martin Axiom Suslin’s Problem Martin Axiom Iterated Forcing Direct Limit Construction

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Let P be a forcing notion in M and G1 ⊆ P a M-generic filter.

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Let P be a forcing notion in M and G1 ⊆ P a M-generic filter. Let Q be a poset in M[G1] and G2 ⊆ Q a M[G1]-generic filter.

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Let P be a forcing notion in M and G1 ⊆ P a M-generic filter. Let Q be a poset in M[G1] and G2 ⊆ Q a M[G1]-generic filter. I want to show that there exists a G M-generic filter on R such that: M[G1][G2] = M[G] We will define this filter using Boolean algebras.

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Suslin’s Problem and Martin Axiom Suslin’s Problem Martin Axiom Iterated Forcing Direct Limit Construction

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Let B be a complete Boolean algebra in M.

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Let B be a complete Boolean algebra in M. Let C ∈ MB such that ||C is a complete Boolean algebra || = 1.

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Let B be a complete Boolean algebra in M. Let C ∈ MB such that ||C is a complete Boolean algebra || = 1. D is a maximal subset in MB such that:

1 ||c ∈ C|| = 1 ∀c ∈ D 2 c1, c2 ∈ D, c1 = c2 ⇒ ||c1 = c2|| < 1

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Suslin’s Problem and Martin Axiom Suslin’s Problem Martin Axiom Iterated Forcing Direct Limit Construction

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Let B be a complete Boolean algebra in M. Let C ∈ MB such that ||C is a complete Boolean algebra || = 1. D is a maximal subset in MB such that:

1 ||c ∈ C|| = 1 ∀c ∈ D 2 c1, c2 ∈ D, c1 = c2 ⇒ ||c1 = c2|| < 1

I define +D: ∀c1, c2 ∈ D ∃c ∈ D such that ||c = c1 +C c2|| = 1 this c is unique and I define c = c1 +D c2. The operations ·D and −D are defined similarly.

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Let B be a complete Boolean algebra in M. Let C ∈ MB such that ||C is a complete Boolean algebra || = 1. D is a maximal subset in MB such that:

1 ||c ∈ C|| = 1 ∀c ∈ D 2 c1, c2 ∈ D, c1 = c2 ⇒ ||c1 = c2|| < 1

I define +D: ∀c1, c2 ∈ D ∃c ∈ D such that ||c = c1 +C c2|| = 1 this c is unique and I define c = c1 +D c2. The operations ·D and −D are defined similarly. With this operations D is a complete Boolean algebra ( in M ). I define B ∗ C = D.

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Theorem

Let B be a complete Boolean algebra in M, let C ∈ MB be such that ||C is a complete Boolean algebra|| = 1 and let D = B ∗ C such that B is a complete subalgebra of D. Then

1 If G1 is an M-generic ultrafilter on B, C = iG1(C) and G2

is an M[G1]-generic ultrafilter on C then there is an M-generic ultrafilter G on B ∗ C such that: M[G1][G2] = M[G]

2 If G is an M-generic ultrafilter on B ∗ C. G1 = G ∩ B and

C = iG1(C) then there is an M[G1]-generic ultrafilter G2

n C such that:

M[G1][G2] = M[G]

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Lemma

B satisfies ccc and ||C satisfies ccc || = 1 iff B ∗ C satisfies ccc.

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Let α be a limit ordinal. Let {Bi}i<α a sequence such that

Bi is a complete Boolean algebra
if i < j Bi is a complete subalgebra of Bj

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Let α be a limit ordinal. Let {Bi}i<α a sequence such that

Bi is a complete Boolean algebra
if i < j Bi is a complete subalgebra of Bj

Direct limit

The completion B of

i<α Bi is called direct limit of {Bi}.

B = limdiri≤αBi.

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Let α be a limit ordinal. Let {Bi}i<α a sequence such that

Bi is a complete Boolean algebra
if i < j Bi is a complete subalgebra of Bj

Direct limit

The completion B of

i<α Bi is called direct limit of {Bi}.

B = limdiri≤αBi.

Lemma

Then if each Bi is k-saturated then B is k-saturated. In particular if each Bi satisfies ccc then B satisfies ccc.

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Let M be a transitive model of ZFC + GCH. We will construct a complete Boolean algebra B such that if G is an M-generic filter on B then M[G] | = MA + 2ℵ0 ≤ ℵ2

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Bα

Let {Bα} be a sequence such that:

1 α < β ⇒ Bα is a complete subalgebra of Bβ 2 γ limit ⇒ Bγ = limdiri≤γBi 3 each Bα satisfies ccc 4 |Bα| ≤ ℵ2

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Bα

Let {Bα} be a sequence such that:

1 α < β ⇒ Bα is a complete subalgebra of Bβ 2 γ limit ⇒ Bγ = limdiri≤γBi 3 each Bα satisfies ccc 4 |Bα| ≤ ℵ2

I define B = limdiri<ω2Bi.

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Bα

Let {Bα} be a sequence such that:

1 α < β ⇒ Bα is a complete subalgebra of Bβ 2 γ limit ⇒ Bγ = limdiri≤γBi 3 each Bα satisfies ccc 4 |Bα| ≤ ℵ2

I define B = limdiri<ω2Bi. So we have:

1 B satisfies ccc 2 |B| = ℵ2

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Bα

Let {Bα} be a sequence such that:

1 α < β ⇒ Bα is a complete subalgebra of Bβ 2 γ limit ⇒ Bγ = limdiri≤γBi 3 each Bα satisfies ccc 4 |Bα| ≤ ℵ2

I define B = limdiri<ω2Bi. So we have:

1 B satisfies ccc 2 |B| = ℵ2

M[G] perserves cardinals and M[G] | = 2ℵ0 ≤ ℵ2 ( Jech, lemma 19.4 ).

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Let α → (βα, γα) canonical mapping of ω2 onto ω2 × ω2.

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Let α → (βα, γα) canonical mapping of ω2 onto ω2 × ω2. B0 = {0, 1} and Bγ = limdiri<γBi for γ limit.

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Let α → (βα, γα) canonical mapping of ω2 onto ω2 × ω2. B0 = {0, 1} and Bγ = limdiri<γBi for γ limit. I construct Bα+1.

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Let α → (βα, γα) canonical mapping of ω2 onto ω2 × ω2. B0 = {0, 1} and Bγ = limdiri<γBi for γ limit. I construct Bα+1. D = Bβα and R = RD

γα γα-th relationship on ˇ

ω1, R ∈ MBα.

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Suslin’s Problem and Martin Axiom Suslin’s Problem Martin Axiom Iterated Forcing Direct Limit Construction

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Let α → (βα, γα) canonical mapping of ω2 onto ω2 × ω2. B0 = {0, 1} and Bγ = limdiri<γBi for γ limit. I construct Bα+1. D = Bβα and R = RD

γα γα-th relationship on ˇ

ω1, R ∈ MBα. b = ||R is a partial ordering of ˇ ω1 and (ˇ ω1, R) satisfies ccc||.

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Suslin’s Problem and Martin Axiom Suslin’s Problem Martin Axiom Iterated Forcing Direct Limit Construction

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Let α → (βα, γα) canonical mapping of ω2 onto ω2 × ω2. B0 = {0, 1} and Bγ = limdiri<γBi for γ limit. I construct Bα+1. D = Bβα and R = RD

γα γα-th relationship on ˇ

ω1, R ∈ MBα. b = ||R is a partial ordering of ˇ ω1 and (ˇ ω1, R) satisfies ccc||. Let C ∈ MBα be the complete Boolean algebra such that:

||C is the trivial algebra || = −b
||C = r.o.(ˇ

ω1, R)|| = b

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Suslin’s Problem and Martin Axiom Suslin’s Problem Martin Axiom Iterated Forcing Direct Limit Construction

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Let α → (βα, γα) canonical mapping of ω2 onto ω2 × ω2. B0 = {0, 1} and Bγ = limdiri<γBi for γ limit. I construct Bα+1. D = Bβα and R = RD

γα γα-th relationship on ˇ

ω1, R ∈ MBα. b = ||R is a partial ordering of ˇ ω1 and (ˇ ω1, R) satisfies ccc||. Let C ∈ MBα be the complete Boolean algebra such that:

||C is the trivial algebra || = −b
||C = r.o.(ˇ

ω1, R)|| = b I define Bα+1 = Bα ∗ C.

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Let G be a generic ultrafilter on B and Gα = G ∩ Bα.

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Let G be a generic ultrafilter on B and Gα = G ∩ Bα. Let (P, <) be a poset in M[G] that satisfies ccc, we assume (P, <) = (ω1, R).

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Let G be a generic ultrafilter on B and Gα = G ∩ Bα. Let (P, <) be a poset in M[G] that satisfies ccc, we assume (P, <) = (ω1, R). Let D ∈ M[G] be a collection of at most ℵ1 dense subsets of ω1.

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Let G be a generic ultrafilter on B and Gα = G ∩ Bα. Let (P, <) be a poset in M[G] that satisfies ccc, we assume (P, <) = (ω1, R). Let D ∈ M[G] be a collection of at most ℵ1 dense subsets of ω1.

Lemma

If X ∈ M[G] is a subset of ω1 then exists α < ω2 such that X ∈ M[Gα].

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Let G be a generic ultrafilter on B and Gα = G ∩ Bα. Let (P, <) be a poset in M[G] that satisfies ccc, we assume (P, <) = (ω1, R). Let D ∈ M[G] be a collection of at most ℵ1 dense subsets of ω1.

Lemma

If X ∈ M[G] is a subset of ω1 then exists α < ω2 such that X ∈ M[Gα]. Let β < ω2 such that D, R ∈ M[Gβ].

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Suslin’s Problem and Martin Axiom Suslin’s Problem Martin Axiom Iterated Forcing Direct Limit Construction

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Let G be a generic ultrafilter on B and Gα = G ∩ Bα. Let (P, <) be a poset in M[G] that satisfies ccc, we assume (P, <) = (ω1, R). Let D ∈ M[G] be a collection of at most ℵ1 dense subsets of ω1.

Lemma

If X ∈ M[G] is a subset of ω1 then exists α < ω2 such that X ∈ M[Gα]. Let β < ω2 such that D, R ∈ M[Gβ]. Let RBβ

γ

∈ MBβ be a name for R.

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Suslin’s Problem and Martin Axiom Suslin’s Problem Martin Axiom Iterated Forcing Direct Limit Construction

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Let G be a generic ultrafilter on B and Gα = G ∩ Bα. Let (P, <) be a poset in M[G] that satisfies ccc, we assume (P, <) = (ω1, R). Let D ∈ M[G] be a collection of at most ℵ1 dense subsets of ω1.

Lemma

If X ∈ M[G] is a subset of ω1 then exists α < ω2 such that X ∈ M[Gα]. Let β < ω2 such that D, R ∈ M[Gβ]. Let RBβ

γ

∈ MBβ be a name for R. Let α < ω2 be such that α → (βα, γα) = (β, γ).

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Now, since M[Gα] is a submodel of M[G], we have M[G] | = (ω1, R) satisfies ccc ⇒ M[Gα] | = (ω1, R) satisfies ccc

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Now, since M[Gα] is a submodel of M[G], we have M[G] | = (ω1, R) satisfies ccc ⇒ M[Gα] | = (ω1, R) satisfies ccc So we have b = ||(ˇ ω1, R) satisfies ccc || ∈ Gα.

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Now, since M[Gα] is a submodel of M[G], we have M[G] | = (ω1, R) satisfies ccc ⇒ M[Gα] | = (ω1, R) satisfies ccc So we have b = ||(ˇ ω1, R) satisfies ccc || ∈ Gα. By construction Bα+1 = Bα ∗ C and ||C = r.o.(ˇ ω1, R)|| = b

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Now, since M[Gα] is a submodel of M[G], we have M[G] | = (ω1, R) satisfies ccc ⇒ M[Gα] | = (ω1, R) satisfies ccc So we have b = ||(ˇ ω1, R) satisfies ccc || ∈ Gα. By construction Bα+1 = Bα ∗ C and ||C = r.o.(ˇ ω1, R)|| = b Using a previous Theorem exists H M[Gα]-generic filter on (ω1, R) such that M[Gα+1] = M[Gα][H]

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Now, since M[Gα] is a submodel of M[G], we have M[G] | = (ω1, R) satisfies ccc ⇒ M[Gα] | = (ω1, R) satisfies ccc So we have b = ||(ˇ ω1, R) satisfies ccc || ∈ Gα. By construction Bα+1 = Bα ∗ C and ||C = r.o.(ˇ ω1, R)|| = b Using a previous Theorem exists H M[Gα]-generic filter on (ω1, R) such that M[Gα+1] = M[Gα][H] So H is D-generic on (ω1, R) and we conclude.

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