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Iterated Forcing and Forcing Axioms Mirna Dˇ zamonja, Tutorial 2 Recall and MA PFA How about ω 2 , or something larger Finite conditions Work with Gregor Dolinar (APAL 2013) Neeman’s revolution Iterated Forcing and Forcing Axioms Mirna Dˇ zamonja, Tutorial 2 31 January, 2020

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Page 1: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated Forcing and Forcing Axioms

Mirna Dzamonja, Tutorial 2

31 January, 2020

Page 2: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing and iterated forcing

Forcing is a technique to extend a universe V of settheory=ZFC (or ZFC∗) to another one, V[G], so that V[G]

has the same ordinals(most often) has the same cardinalssatisfies a desired formula φ.

For example, φ could be the failure of CH, or somethingmore involved such as there are no Suslin trees or “everyccc Boolean algebra of size < c supports a measure”.For such more involved statements ¬∃ or ∀∃ we need touse iterated forcing.This can get a little complicated because of two issues:preservation of the axioms and preservation of cardinals.

Page 3: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing and iterated forcing

Forcing is a technique to extend a universe V of settheory=ZFC (or ZFC∗) to another one, V[G], so that V[G]

has the same ordinals

(most often) has the same cardinalssatisfies a desired formula φ.

For example, φ could be the failure of CH, or somethingmore involved such as there are no Suslin trees or “everyccc Boolean algebra of size < c supports a measure”.For such more involved statements ¬∃ or ∀∃ we need touse iterated forcing.This can get a little complicated because of two issues:preservation of the axioms and preservation of cardinals.

Page 4: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing and iterated forcing

Forcing is a technique to extend a universe V of settheory=ZFC (or ZFC∗) to another one, V[G], so that V[G]

has the same ordinals(most often) has the same cardinals

satisfies a desired formula φ.For example, φ could be the failure of CH, or somethingmore involved such as there are no Suslin trees or “everyccc Boolean algebra of size < c supports a measure”.For such more involved statements ¬∃ or ∀∃ we need touse iterated forcing.This can get a little complicated because of two issues:preservation of the axioms and preservation of cardinals.

Page 5: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing and iterated forcing

Forcing is a technique to extend a universe V of settheory=ZFC (or ZFC∗) to another one, V[G], so that V[G]

has the same ordinals(most often) has the same cardinalssatisfies a desired formula φ.

For example, φ could be the failure of CH, or somethingmore involved such as there are no Suslin trees or “everyccc Boolean algebra of size < c supports a measure”.For such more involved statements ¬∃ or ∀∃ we need touse iterated forcing.This can get a little complicated because of two issues:preservation of the axioms and preservation of cardinals.

Page 6: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing and iterated forcing

Forcing is a technique to extend a universe V of settheory=ZFC (or ZFC∗) to another one, V[G], so that V[G]

has the same ordinals(most often) has the same cardinalssatisfies a desired formula φ.

For example, φ could be the failure of CH, or somethingmore involved such as

there are no Suslin trees or “everyccc Boolean algebra of size < c supports a measure”.For such more involved statements ¬∃ or ∀∃ we need touse iterated forcing.This can get a little complicated because of two issues:preservation of the axioms and preservation of cardinals.

Page 7: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing and iterated forcing

Forcing is a technique to extend a universe V of settheory=ZFC (or ZFC∗) to another one, V[G], so that V[G]

has the same ordinals(most often) has the same cardinalssatisfies a desired formula φ.

For example, φ could be the failure of CH, or somethingmore involved such as there are no Suslin trees or

“everyccc Boolean algebra of size < c supports a measure”.For such more involved statements ¬∃ or ∀∃ we need touse iterated forcing.This can get a little complicated because of two issues:preservation of the axioms and preservation of cardinals.

Page 8: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing and iterated forcing

Forcing is a technique to extend a universe V of settheory=ZFC (or ZFC∗) to another one, V[G], so that V[G]

has the same ordinals(most often) has the same cardinalssatisfies a desired formula φ.

For example, φ could be the failure of CH, or somethingmore involved such as there are no Suslin trees or “everyccc Boolean algebra of size < c supports a measure”.

For such more involved statements ¬∃ or ∀∃ we need touse iterated forcing.This can get a little complicated because of two issues:preservation of the axioms and preservation of cardinals.

Page 9: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing and iterated forcing

Forcing is a technique to extend a universe V of settheory=ZFC (or ZFC∗) to another one, V[G], so that V[G]

has the same ordinals(most often) has the same cardinalssatisfies a desired formula φ.

For example, φ could be the failure of CH, or somethingmore involved such as there are no Suslin trees or “everyccc Boolean algebra of size < c supports a measure”.For such more involved statements ¬∃ or ∀∃ we need touse iterated forcing.

This can get a little complicated because of two issues:preservation of the axioms and preservation of cardinals.

Page 10: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing and iterated forcing

Forcing is a technique to extend a universe V of settheory=ZFC (or ZFC∗) to another one, V[G], so that V[G]

has the same ordinals(most often) has the same cardinalssatisfies a desired formula φ.

For example, φ could be the failure of CH, or somethingmore involved such as there are no Suslin trees or “everyccc Boolean algebra of size < c supports a measure”.For such more involved statements ¬∃ or ∀∃ we need touse iterated forcing.This can get a little complicated because of two issues:preservation of the axioms and preservation of cardinals.

Page 11: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated forcing

Iteration of forcing is NOT constructing a sequence ofextensions V0 ⊆ V1 = V0[G0] ⊆ V1[G1] . . . ⊆ Vα[Gα] . . .such that each Gα is Pα-generic for some forcing Pα ∈ Vαchosen independently of the previous ones.

This approach runs into problems already at the stage ω.For example, if we take Vω =

⋃Vn[Gn], in general this will

not even contain 〈Gn : n < ω〉. See Kunen’s book 1stedition.

Instead we need to deal with a forcing which is entirely inV0 and consists of names.

Page 12: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated forcing

Iteration of forcing is NOT constructing a sequence ofextensions V0 ⊆ V1 = V0[G0] ⊆ V1[G1] . . . ⊆ Vα[Gα] . . .such that each Gα is Pα-generic for some forcing Pα ∈ Vαchosen independently of the previous ones.

This approach runs into problems already at the stage ω.

For example, if we take Vω =⋃

Vn[Gn], in general this willnot even contain 〈Gn : n < ω〉. See Kunen’s book 1stedition.

Instead we need to deal with a forcing which is entirely inV0 and consists of names.

Page 13: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated forcing

Iteration of forcing is NOT constructing a sequence ofextensions V0 ⊆ V1 = V0[G0] ⊆ V1[G1] . . . ⊆ Vα[Gα] . . .such that each Gα is Pα-generic for some forcing Pα ∈ Vαchosen independently of the previous ones.

This approach runs into problems already at the stage ω.For example, if we take Vω =

⋃Vn[Gn], in general this will

not even contain 〈Gn : n < ω〉.

See Kunen’s book 1stedition.

Instead we need to deal with a forcing which is entirely inV0 and consists of names.

Page 14: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated forcing

Iteration of forcing is NOT constructing a sequence ofextensions V0 ⊆ V1 = V0[G0] ⊆ V1[G1] . . . ⊆ Vα[Gα] . . .such that each Gα is Pα-generic for some forcing Pα ∈ Vαchosen independently of the previous ones.

This approach runs into problems already at the stage ω.For example, if we take Vω =

⋃Vn[Gn], in general this will

not even contain 〈Gn : n < ω〉. See Kunen’s book 1stedition.

Instead we need to deal with a forcing which is entirely inV0 and consists of names.

Page 15: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated forcing

Iteration of forcing is NOT constructing a sequence ofextensions V0 ⊆ V1 = V0[G0] ⊆ V1[G1] . . . ⊆ Vα[Gα] . . .such that each Gα is Pα-generic for some forcing Pα ∈ Vαchosen independently of the previous ones.

This approach runs into problems already at the stage ω.For example, if we take Vω =

⋃Vn[Gn], in general this will

not even contain 〈Gn : n < ω〉. See Kunen’s book 1stedition.

Instead we need to deal with a forcing which is entirely inV0 and consists of names.

Page 16: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Names

DefinitionNames are defined recursively. A P-name is a set of pairs(p, σ

˜) where p ∈ P and σ

˜is a P-name.

Names, say τ˜

, are in V and are going to be calculatedinto objects, τ

˜G in V [G]. In fact, V [G] consists entirely of

such calculated objects.

Definition(1) The values of τ

˜G are defined recursively. If G is a filter

on P and τ˜

a P-name, then

τ˜

G = {σ˜

G : (∃p ∈ G)(p, σ˜) ∈ τ

˜}.

(2) V [G] = {τ˜

G : τ˜∈ V}.

Page 17: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Names

DefinitionNames are defined recursively. A P-name is a set of pairs(p, σ

˜) where p ∈ P and σ

˜is a P-name.

Names, say τ˜

, are in V and are going to be calculatedinto objects, τ

˜G in V [G]. In fact, V [G] consists entirely of

such calculated objects.

Definition(1) The values of τ

˜G are defined recursively. If G is a filter

on P and τ˜

a P-name, then

τ˜

G = {σ˜

G : (∃p ∈ G)(p, σ˜) ∈ τ

˜}.

(2) V [G] = {τ˜

G : τ˜∈ V}.

Page 18: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Names

DefinitionNames are defined recursively. A P-name is a set of pairs(p, σ

˜) where p ∈ P and σ

˜is a P-name.

Names, say τ˜

, are in V and are going to be calculatedinto objects, τ

˜G in V [G]. In fact, V [G] consists entirely of

such calculated objects.

Definition(1) The values of τ

˜G are defined recursively. If G is a filter

on P and τ˜

a P-name, then

τ˜

G = {σ˜

G : (∃p ∈ G)(p, σ˜) ∈ τ

˜}.

(2) V [G] = {τ˜

G : τ˜∈ V}.

Page 19: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Names

DefinitionNames are defined recursively. A P-name is a set of pairs(p, σ

˜) where p ∈ P and σ

˜is a P-name.

Names, say τ˜

, are in V and are going to be calculatedinto objects, τ

˜G in V [G]. In fact, V [G] consists entirely of

such calculated objects.

Definition(1) The values of τ

˜G are defined recursively. If G is a filter

on P and τ˜

a P-name, then

τ˜

G = {σ˜

G : (∃p ∈ G)(p, σ˜) ∈ τ

˜}.

(2) V [G] = {τ˜

G : τ˜∈ V}.

Page 20: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

The forcing relation

DefinitionSuppose that ϕ(x0, x1, . . . xn−1) is a formula in thelanguage of set theory L = {∈} and τ

˜0, τ˜

1, . . . τ˜

n−1 areP-names, while p ∈ P.

We say that p ϕ(τ˜

0, τ˜

1, . . . τ˜

n−1)iff for every relevant V , for every P-generic G over V withp ∈ G, we have V [G] |= ϕ(τ

˜0G, τ˜

1G, . . . τ˜

n−1G ).

A good definition but seems too hard to satisfy. That iswhere the magic comes:

Forcing Theorem, part 2 (1) For any ϕ andτ˜

0, τ˜

1, . . . τ˜

n−1, V as above, V [G] |= ϕ(τ˜

0G, τ˜

1G, . . . τ˜

n−1G ) iff

for some p ∈ G we have p ϕ(τ˜

0, τ˜

1, . . . τ˜

n − 1).(2) There is a relation ∗ definable in the ground modelsuch that p ϕ iff p ∗ ϕ.

Page 21: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

The forcing relation

DefinitionSuppose that ϕ(x0, x1, . . . xn−1) is a formula in thelanguage of set theory L = {∈} and τ

˜0, τ˜

1, . . . τ˜

n−1 areP-names, while p ∈ P. We say that p ϕ(τ

˜0, τ˜

1, . . . τ˜

n−1)iff

for every relevant V , for every P-generic G over V withp ∈ G, we have V [G] |= ϕ(τ

˜0G, τ˜

1G, . . . τ˜

n−1G ).

A good definition but seems too hard to satisfy. That iswhere the magic comes:

Forcing Theorem, part 2 (1) For any ϕ andτ˜

0, τ˜

1, . . . τ˜

n−1, V as above, V [G] |= ϕ(τ˜

0G, τ˜

1G, . . . τ˜

n−1G ) iff

for some p ∈ G we have p ϕ(τ˜

0, τ˜

1, . . . τ˜

n − 1).(2) There is a relation ∗ definable in the ground modelsuch that p ϕ iff p ∗ ϕ.

Page 22: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

The forcing relation

DefinitionSuppose that ϕ(x0, x1, . . . xn−1) is a formula in thelanguage of set theory L = {∈} and τ

˜0, τ˜

1, . . . τ˜

n−1 areP-names, while p ∈ P. We say that p ϕ(τ

˜0, τ˜

1, . . . τ˜

n−1)iff for every relevant V , for every P-generic G over V withp ∈ G, we have V [G] |= ϕ(τ

˜0G, τ˜

1G, . . . τ˜

n−1G ).

A good definition but seems too hard to satisfy. That iswhere the magic comes:

Forcing Theorem, part 2 (1) For any ϕ andτ˜

0, τ˜

1, . . . τ˜

n−1, V as above, V [G] |= ϕ(τ˜

0G, τ˜

1G, . . . τ˜

n−1G ) iff

for some p ∈ G we have p ϕ(τ˜

0, τ˜

1, . . . τ˜

n − 1).(2) There is a relation ∗ definable in the ground modelsuch that p ϕ iff p ∗ ϕ.

Page 23: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

The forcing relation

DefinitionSuppose that ϕ(x0, x1, . . . xn−1) is a formula in thelanguage of set theory L = {∈} and τ

˜0, τ˜

1, . . . τ˜

n−1 areP-names, while p ∈ P. We say that p ϕ(τ

˜0, τ˜

1, . . . τ˜

n−1)iff for every relevant V , for every P-generic G over V withp ∈ G, we have V [G] |= ϕ(τ

˜0G, τ˜

1G, . . . τ˜

n−1G ).

A good definition but seems too hard to satisfy.

That iswhere the magic comes:

Forcing Theorem, part 2 (1) For any ϕ andτ˜

0, τ˜

1, . . . τ˜

n−1, V as above, V [G] |= ϕ(τ˜

0G, τ˜

1G, . . . τ˜

n−1G ) iff

for some p ∈ G we have p ϕ(τ˜

0, τ˜

1, . . . τ˜

n − 1).(2) There is a relation ∗ definable in the ground modelsuch that p ϕ iff p ∗ ϕ.

Page 24: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

The forcing relation

DefinitionSuppose that ϕ(x0, x1, . . . xn−1) is a formula in thelanguage of set theory L = {∈} and τ

˜0, τ˜

1, . . . τ˜

n−1 areP-names, while p ∈ P. We say that p ϕ(τ

˜0, τ˜

1, . . . τ˜

n−1)iff for every relevant V , for every P-generic G over V withp ∈ G, we have V [G] |= ϕ(τ

˜0G, τ˜

1G, . . . τ˜

n−1G ).

A good definition but seems too hard to satisfy. That iswhere the magic comes:

Forcing Theorem, part 2 (1) For any ϕ andτ˜

0, τ˜

1, . . . τ˜

n−1, V as above, V [G] |= ϕ(τ˜

0G, τ˜

1G, . . . τ˜

n−1G ) iff

for some p ∈ G we have p ϕ(τ˜

0, τ˜

1, . . . τ˜

n − 1).(2) There is a relation ∗ definable in the ground modelsuch that p ϕ iff p ∗ ϕ.

Page 25: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

The forcing relation

DefinitionSuppose that ϕ(x0, x1, . . . xn−1) is a formula in thelanguage of set theory L = {∈} and τ

˜0, τ˜

1, . . . τ˜

n−1 areP-names, while p ∈ P. We say that p ϕ(τ

˜0, τ˜

1, . . . τ˜

n−1)iff for every relevant V , for every P-generic G over V withp ∈ G, we have V [G] |= ϕ(τ

˜0G, τ˜

1G, . . . τ˜

n−1G ).

A good definition but seems too hard to satisfy. That iswhere the magic comes:

Forcing Theorem, part 2 (1) For any ϕ andτ˜

0, τ˜

1, . . . τ˜

n−1, V as above, V [G] |= ϕ(τ˜

0G, τ˜

1G, . . . τ˜

n−1G ) iff

for some p ∈ G we have p ϕ(τ˜

0, τ˜

1, . . . τ˜

n − 1).

(2) There is a relation ∗ definable in the ground modelsuch that p ϕ iff p ∗ ϕ.

Page 26: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

The forcing relation

DefinitionSuppose that ϕ(x0, x1, . . . xn−1) is a formula in thelanguage of set theory L = {∈} and τ

˜0, τ˜

1, . . . τ˜

n−1 areP-names, while p ∈ P. We say that p ϕ(τ

˜0, τ˜

1, . . . τ˜

n−1)iff for every relevant V , for every P-generic G over V withp ∈ G, we have V [G] |= ϕ(τ

˜0G, τ˜

1G, . . . τ˜

n−1G ).

A good definition but seems too hard to satisfy. That iswhere the magic comes:

Forcing Theorem, part 2 (1) For any ϕ andτ˜

0, τ˜

1, . . . τ˜

n−1, V as above, V [G] |= ϕ(τ˜

0G, τ˜

1G, . . . τ˜

n−1G ) iff

for some p ∈ G we have p ϕ(τ˜

0, τ˜

1, . . . τ˜

n − 1).(2) There is a relation ∗ definable in the ground modelsuch that p ϕ iff p ∗ ϕ.

Page 27: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated ForcingTwo Step Iteration Suppose that P is a forcing notionand Q

˜is a P-name for a forcing notion (i.e. ∅P Q

˜is a

forcing notion.)

We define P ∗Q˜

to consist of all pairs(p,q

˜) such that p ∈ P and p P q

˜∈ Q

˜, ordered by

(p0,q˜

0) ≤ (p1,q1˜) iff p0 ≤ p1 and p1 q0

˜≤ q1

˜.

Finite Support Iteration A finite support iteration offorcing 〈Pα,Q

˜β : α ≤ α∗, β < α∗〉 is defined by induction.

The Inductive Hypotheses are that each Pα ∈ V and thatQ˜α is a Pα-name for a forcing notion.

P0 = {∅}. Given Pα we have that Pα+1 = Pα ∗Q˜α. If

α > 0 is a limit ordinal then,

Pα = all functions p such that :

dom(p) = α,∀γ < α, p � γ p(γ) ∈ Q

˜γ ,

{γ < α : ¬(p � γ p(γ) = ∅˜γ)} is finite.

Page 28: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated ForcingTwo Step Iteration Suppose that P is a forcing notionand Q

˜is a P-name for a forcing notion (i.e. ∅P Q

˜is a

forcing notion.) We define P ∗Q˜

to consist of all pairs(p,q

˜) such that p ∈ P and p P q

˜∈ Q

˜, ordered by

(p0,q˜

0) ≤ (p1,q1˜) iff p0 ≤ p1 and p1 q0

˜≤ q1

˜.

Finite Support Iteration A finite support iteration offorcing 〈Pα,Q

˜β : α ≤ α∗, β < α∗〉 is defined by induction.

The Inductive Hypotheses are that each Pα ∈ V and thatQ˜α is a Pα-name for a forcing notion.

P0 = {∅}. Given Pα we have that Pα+1 = Pα ∗Q˜α. If

α > 0 is a limit ordinal then,

Pα = all functions p such that :

dom(p) = α,∀γ < α, p � γ p(γ) ∈ Q

˜γ ,

{γ < α : ¬(p � γ p(γ) = ∅˜γ)} is finite.

Page 29: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated ForcingTwo Step Iteration Suppose that P is a forcing notionand Q

˜is a P-name for a forcing notion (i.e. ∅P Q

˜is a

forcing notion.) We define P ∗Q˜

to consist of all pairs(p,q

˜) such that p ∈ P and p P q

˜∈ Q

˜, ordered by

(p0,q˜

0) ≤ (p1,q1˜) iff p0 ≤ p1 and p1 q0

˜≤ q1

˜.

Finite Support Iteration A finite support iteration offorcing 〈Pα,Q

˜β : α ≤ α∗, β < α∗〉 is defined by induction.

The Inductive Hypotheses are that each Pα ∈ V and thatQ˜α is a Pα-name for a forcing notion.

P0 = {∅}. Given Pα we have that Pα+1 = Pα ∗Q˜α. If

α > 0 is a limit ordinal then,

Pα = all functions p such that :

dom(p) = α,∀γ < α, p � γ p(γ) ∈ Q

˜γ ,

{γ < α : ¬(p � γ p(γ) = ∅˜γ)} is finite.

Page 30: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated ForcingTwo Step Iteration Suppose that P is a forcing notionand Q

˜is a P-name for a forcing notion (i.e. ∅P Q

˜is a

forcing notion.) We define P ∗Q˜

to consist of all pairs(p,q

˜) such that p ∈ P and p P q

˜∈ Q

˜, ordered by

(p0,q˜

0) ≤ (p1,q1˜) iff p0 ≤ p1 and p1 q0

˜≤ q1

˜.

Finite Support Iteration A finite support iteration offorcing 〈Pα,Q

˜β : α ≤ α∗, β < α∗〉 is defined by induction.

The Inductive Hypotheses are that each Pα ∈ V and thatQ˜α is a Pα-name for a forcing notion.

P0 = {∅}. Given Pα we have that Pα+1 = Pα ∗Q˜α. If

α > 0 is a limit ordinal then,

Pα = all functions p such that :

dom(p) = α,∀γ < α, p � γ p(γ) ∈ Q

˜γ ,

{γ < α : ¬(p � γ p(γ) = ∅˜γ)} is finite.

Page 31: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated ForcingTwo Step Iteration Suppose that P is a forcing notionand Q

˜is a P-name for a forcing notion (i.e. ∅P Q

˜is a

forcing notion.) We define P ∗Q˜

to consist of all pairs(p,q

˜) such that p ∈ P and p P q

˜∈ Q

˜, ordered by

(p0,q˜

0) ≤ (p1,q1˜) iff p0 ≤ p1 and p1 q0

˜≤ q1

˜.

Finite Support Iteration A finite support iteration offorcing 〈Pα,Q

˜β : α ≤ α∗, β < α∗〉 is defined by induction.

The Inductive Hypotheses are that each Pα ∈ V and thatQ˜α is a Pα-name for a forcing notion.

P0 = {∅}. Given Pα we have that Pα+1 = Pα ∗Q˜α. If

α > 0 is a limit ordinal then,

Pα = all functions p such that :

dom(p) = α,∀γ < α, p � γ p(γ) ∈ Q

˜γ ,

{γ < α : ¬(p � γ p(γ) = ∅˜γ)} is finite.

Page 32: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated ForcingTwo Step Iteration Suppose that P is a forcing notionand Q

˜is a P-name for a forcing notion (i.e. ∅P Q

˜is a

forcing notion.) We define P ∗Q˜

to consist of all pairs(p,q

˜) such that p ∈ P and p P q

˜∈ Q

˜, ordered by

(p0,q˜

0) ≤ (p1,q1˜) iff p0 ≤ p1 and p1 q0

˜≤ q1

˜.

Finite Support Iteration A finite support iteration offorcing 〈Pα,Q

˜β : α ≤ α∗, β < α∗〉 is defined by induction.

The Inductive Hypotheses are that each Pα ∈ V and thatQ˜α is a Pα-name for a forcing notion.

P0 = {∅}.

Given Pα we have that Pα+1 = Pα ∗Q˜α. If

α > 0 is a limit ordinal then,

Pα = all functions p such that :

dom(p) = α,∀γ < α, p � γ p(γ) ∈ Q

˜γ ,

{γ < α : ¬(p � γ p(γ) = ∅˜γ)} is finite.

Page 33: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated ForcingTwo Step Iteration Suppose that P is a forcing notionand Q

˜is a P-name for a forcing notion (i.e. ∅P Q

˜is a

forcing notion.) We define P ∗Q˜

to consist of all pairs(p,q

˜) such that p ∈ P and p P q

˜∈ Q

˜, ordered by

(p0,q˜

0) ≤ (p1,q1˜) iff p0 ≤ p1 and p1 q0

˜≤ q1

˜.

Finite Support Iteration A finite support iteration offorcing 〈Pα,Q

˜β : α ≤ α∗, β < α∗〉 is defined by induction.

The Inductive Hypotheses are that each Pα ∈ V and thatQ˜α is a Pα-name for a forcing notion.

P0 = {∅}. Given Pα we have that Pα+1 = Pα ∗Q˜α.

Ifα > 0 is a limit ordinal then,

Pα = all functions p such that :

dom(p) = α,∀γ < α, p � γ p(γ) ∈ Q

˜γ ,

{γ < α : ¬(p � γ p(γ) = ∅˜γ)} is finite.

Page 34: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated ForcingTwo Step Iteration Suppose that P is a forcing notionand Q

˜is a P-name for a forcing notion (i.e. ∅P Q

˜is a

forcing notion.) We define P ∗Q˜

to consist of all pairs(p,q

˜) such that p ∈ P and p P q

˜∈ Q

˜, ordered by

(p0,q˜

0) ≤ (p1,q1˜) iff p0 ≤ p1 and p1 q0

˜≤ q1

˜.

Finite Support Iteration A finite support iteration offorcing 〈Pα,Q

˜β : α ≤ α∗, β < α∗〉 is defined by induction.

The Inductive Hypotheses are that each Pα ∈ V and thatQ˜α is a Pα-name for a forcing notion.

P0 = {∅}. Given Pα we have that Pα+1 = Pα ∗Q˜α. If

α > 0 is a limit ordinal then,

Pα = all functions p such that :

dom(p) = α,∀γ < α, p � γ p(γ) ∈ Q

˜γ ,

{γ < α : ¬(p � γ p(γ) = ∅˜γ)} is finite.

Page 35: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated ForcingTwo Step Iteration Suppose that P is a forcing notionand Q

˜is a P-name for a forcing notion (i.e. ∅P Q

˜is a

forcing notion.) We define P ∗Q˜

to consist of all pairs(p,q

˜) such that p ∈ P and p P q

˜∈ Q

˜, ordered by

(p0,q˜

0) ≤ (p1,q1˜) iff p0 ≤ p1 and p1 q0

˜≤ q1

˜.

Finite Support Iteration A finite support iteration offorcing 〈Pα,Q

˜β : α ≤ α∗, β < α∗〉 is defined by induction.

The Inductive Hypotheses are that each Pα ∈ V and thatQ˜α is a Pα-name for a forcing notion.

P0 = {∅}. Given Pα we have that Pα+1 = Pα ∗Q˜α. If

α > 0 is a limit ordinal then,

Pα = all functions p such that :

dom(p) = α,

∀γ < α, p � γ p(γ) ∈ Q˜γ ,

{γ < α : ¬(p � γ p(γ) = ∅˜γ)} is finite.

Page 36: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated ForcingTwo Step Iteration Suppose that P is a forcing notionand Q

˜is a P-name for a forcing notion (i.e. ∅P Q

˜is a

forcing notion.) We define P ∗Q˜

to consist of all pairs(p,q

˜) such that p ∈ P and p P q

˜∈ Q

˜, ordered by

(p0,q˜

0) ≤ (p1,q1˜) iff p0 ≤ p1 and p1 q0

˜≤ q1

˜.

Finite Support Iteration A finite support iteration offorcing 〈Pα,Q

˜β : α ≤ α∗, β < α∗〉 is defined by induction.

The Inductive Hypotheses are that each Pα ∈ V and thatQ˜α is a Pα-name for a forcing notion.

P0 = {∅}. Given Pα we have that Pα+1 = Pα ∗Q˜α. If

α > 0 is a limit ordinal then,

Pα = all functions p such that :

dom(p) = α,∀γ < α, p � γ p(γ) ∈ Q

˜γ ,

{γ < α : ¬(p � γ p(γ) = ∅˜γ)} is finite.

Page 37: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Iterated ForcingTwo Step Iteration Suppose that P is a forcing notionand Q

˜is a P-name for a forcing notion (i.e. ∅P Q

˜is a

forcing notion.) We define P ∗Q˜

to consist of all pairs(p,q

˜) such that p ∈ P and p P q

˜∈ Q

˜, ordered by

(p0,q˜

0) ≤ (p1,q1˜) iff p0 ≤ p1 and p1 q0

˜≤ q1

˜.

Finite Support Iteration A finite support iteration offorcing 〈Pα,Q

˜β : α ≤ α∗, β < α∗〉 is defined by induction.

The Inductive Hypotheses are that each Pα ∈ V and thatQ˜α is a Pα-name for a forcing notion.

P0 = {∅}. Given Pα we have that Pα+1 = Pα ∗Q˜α. If

α > 0 is a limit ordinal then,

Pα = all functions p such that :

dom(p) = α,∀γ < α, p � γ p(γ) ∈ Q

˜γ ,

{γ < α : ¬(p � γ p(γ) = ∅˜γ)} is finite.

Page 38: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Killing Souslin trees

Solovay and Tennenbaum 1970:

It is (relatively)consistent that there are no Souslin trees.

Start with say V = L and do an iteration of length ω2 inwhich each individual step Q

˜α comes equipped with a

Pα-name R˜α for a Souslin tree and it achieves that R

˜α is

no longer a Souslin tree in Pα+1.

It is possible to arrange the “bookkeeping” so that〈R˜α : α < ω2〉 goes over all relevant names, i.e. any

Souslin tree in the extension by Pω2 is named at somestage as R

˜α.

For all said so far can use as reference ””Proper andImproper Forcing” by Shelah and for iterated forcing alsothe article ”Iterated Forcing” by Jim Baumgartner.

Page 39: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Killing Souslin trees

Solovay and Tennenbaum 1970: It is (relatively)consistent that there are no Souslin trees.

Start with say V = L and do an iteration of length ω2 inwhich each individual step Q

˜α comes equipped with a

Pα-name R˜α for a Souslin tree and it achieves that R

˜α is

no longer a Souslin tree in Pα+1.

It is possible to arrange the “bookkeeping” so that〈R˜α : α < ω2〉 goes over all relevant names, i.e. any

Souslin tree in the extension by Pω2 is named at somestage as R

˜α.

For all said so far can use as reference ””Proper andImproper Forcing” by Shelah and for iterated forcing alsothe article ”Iterated Forcing” by Jim Baumgartner.

Page 40: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Killing Souslin trees

Solovay and Tennenbaum 1970: It is (relatively)consistent that there are no Souslin trees.

Start with say V = L and do an iteration of length ω2 inwhich each individual step Q

˜α comes equipped with a

Pα-name R˜α for a Souslin tree and it achieves that R

˜α is

no longer a Souslin tree in Pα+1.

It is possible to arrange the “bookkeeping” so that〈R˜α : α < ω2〉 goes over all relevant names, i.e. any

Souslin tree in the extension by Pω2 is named at somestage as R

˜α.

For all said so far can use as reference ””Proper andImproper Forcing” by Shelah and for iterated forcing alsothe article ”Iterated Forcing” by Jim Baumgartner.

Page 41: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Killing Souslin trees

Solovay and Tennenbaum 1970: It is (relatively)consistent that there are no Souslin trees.

Start with say V = L and do an iteration of length ω2 inwhich each individual step Q

˜α comes equipped with a

Pα-name R˜α for a Souslin tree and it achieves that R

˜α is

no longer a Souslin tree in Pα+1.

It is possible to arrange the “bookkeeping” so that〈R˜α : α < ω2〉 goes over all relevant names, i.e. any

Souslin tree in the extension by Pω2 is named at somestage as R

˜α.

For all said so far can use as reference ””Proper andImproper Forcing” by Shelah and for iterated forcing alsothe article ”Iterated Forcing” by Jim Baumgartner.

Page 42: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Killing Souslin trees

Solovay and Tennenbaum 1970: It is (relatively)consistent that there are no Souslin trees.

Start with say V = L and do an iteration of length ω2 inwhich each individual step Q

˜α comes equipped with a

Pα-name R˜α for a Souslin tree and it achieves that R

˜α is

no longer a Souslin tree in Pα+1.

It is possible to arrange the “bookkeeping” so that〈R˜α : α < ω2〉 goes over all relevant names, i.e. any

Souslin tree in the extension by Pω2 is named at somestage as R

˜α.

For all said so far can use as reference ””Proper andImproper Forcing” by Shelah and for iterated forcing alsothe article ”Iterated Forcing” by Jim Baumgartner.

Page 43: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

But how to preserve cardinals?

ccc is a property of forcing that guarantees that theforcing preserves all cardinals.

Theorem(Martin, Solovay 1970) An iteration of ccc forcing withfinite supports is ccc.******************* Do you find iterated forcing a bit hairy?

We have got a solution for you ....

Page 44: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

But how to preserve cardinals?

ccc is a property of forcing that guarantees that theforcing preserves all cardinals.

Theorem(Martin, Solovay 1970) An iteration of ccc forcing withfinite supports is ccc.

******************* Do you find iterated forcing a bit hairy?

We have got a solution for you ....

Page 45: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

But how to preserve cardinals?

ccc is a property of forcing that guarantees that theforcing preserves all cardinals.

Theorem(Martin, Solovay 1970) An iteration of ccc forcing withfinite supports is ccc.*******************

Do you find iterated forcing a bit hairy?

We have got a solution for you ....

Page 46: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

But how to preserve cardinals?

ccc is a property of forcing that guarantees that theforcing preserves all cardinals.

Theorem(Martin, Solovay 1970) An iteration of ccc forcing withfinite supports is ccc.******************* Do you find iterated forcing a bit hairy?

We have got a solution for you ....

Page 47: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

But how to preserve cardinals?

ccc is a property of forcing that guarantees that theforcing preserves all cardinals.

Theorem(Martin, Solovay 1970) An iteration of ccc forcing withfinite supports is ccc.******************* Do you find iterated forcing a bit hairy?

We have got a solution for you ....

Page 48: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing Axioms or forcing without forcing

Martin’s Axiom (MA) : For every ccc forcing notion P

andevery family F of < c many dense sets in P, there is a filterin P which intersects all elements of F.

Under CH, MA is true. It is consistent to have MA + ¬CH,as one can prove using an iteration of ccc forcing.

Using MA + ¬CH set theorists and non-set theorists haveproved a variety of consistency results, mostly about ω1.

Page 49: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing Axioms or forcing without forcing

Martin’s Axiom (MA) : For every ccc forcing notion P andevery family F of < c many dense sets in P,

there is a filterin P which intersects all elements of F.

Under CH, MA is true. It is consistent to have MA + ¬CH,as one can prove using an iteration of ccc forcing.

Using MA + ¬CH set theorists and non-set theorists haveproved a variety of consistency results, mostly about ω1.

Page 50: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing Axioms or forcing without forcing

Martin’s Axiom (MA) : For every ccc forcing notion P andevery family F of < c many dense sets in P, there is a filterin P which intersects all elements of F.

Under CH, MA is true. It is consistent to have MA + ¬CH,as one can prove using an iteration of ccc forcing.

Using MA + ¬CH set theorists and non-set theorists haveproved a variety of consistency results, mostly about ω1.

Page 51: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing Axioms or forcing without forcing

Martin’s Axiom (MA) : For every ccc forcing notion P andevery family F of < c many dense sets in P, there is a filterin P which intersects all elements of F.

Under CH, MA is true.

It is consistent to have MA + ¬CH,as one can prove using an iteration of ccc forcing.

Using MA + ¬CH set theorists and non-set theorists haveproved a variety of consistency results, mostly about ω1.

Page 52: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing Axioms or forcing without forcing

Martin’s Axiom (MA) : For every ccc forcing notion P andevery family F of < c many dense sets in P, there is a filterin P which intersects all elements of F.

Under CH, MA is true. It is consistent to have MA + ¬CH,

as one can prove using an iteration of ccc forcing.

Using MA + ¬CH set theorists and non-set theorists haveproved a variety of consistency results, mostly about ω1.

Page 53: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing Axioms or forcing without forcing

Martin’s Axiom (MA) : For every ccc forcing notion P andevery family F of < c many dense sets in P, there is a filterin P which intersects all elements of F.

Under CH, MA is true. It is consistent to have MA + ¬CH,as one can prove using an iteration of ccc forcing.

Using MA + ¬CH set theorists and non-set theorists haveproved a variety of consistency results, mostly about ω1.

Page 54: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Forcing Axioms or forcing without forcing

Martin’s Axiom (MA) : For every ccc forcing notion P andevery family F of < c many dense sets in P, there is a filterin P which intersects all elements of F.

Under CH, MA is true. It is consistent to have MA + ¬CH,as one can prove using an iteration of ccc forcing.

Using MA + ¬CH set theorists and non-set theorists haveproved a variety of consistency results, mostly about ω1.

Page 55: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

05/06/12 12:4131yk4LxV7rL._BO2,204,203,200_PIsitb-sticker-arrow-click,TopRight,35,-76_AA300_SH20_OU01_.jpg 300 × 300 pixels

Page 1 sur 2http://ecx.images-amazon.com/images/I/31yk4LxV7rL._BO2,204,203,200_PIsitb-sticker-arrow-click,TopRight,35,-76_AA300_SH20_OU01_.jpg

Page 56: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some new results under MA

Fifty years after its discovery now, it is rather rare to seenew applications of Martin’s Axiom, but there are stillsome.

In our recent work with Shelah (preprint available uponrequest)where we answered an old question by Meklerand Vaananen (1993).

We showed: under MA(ω1) + ¬CH there is no tree of sizeand height ω1 and with no uncountable branches thatweakly embeds all such trees.

Page 57: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some new results under MA

Fifty years after its discovery now, it is rather rare to seenew applications of Martin’s Axiom, but there are stillsome.

In our recent work with Shelah (preprint available uponrequest)

where we answered an old question by Meklerand Vaananen (1993).

We showed: under MA(ω1) + ¬CH there is no tree of sizeand height ω1 and with no uncountable branches thatweakly embeds all such trees.

Page 58: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some new results under MA

Fifty years after its discovery now, it is rather rare to seenew applications of Martin’s Axiom, but there are stillsome.

In our recent work with Shelah (preprint available uponrequest)where we answered an old question by Meklerand Vaananen (1993).

We showed: under MA(ω1) + ¬CH there is no tree of sizeand height ω1 and with no uncountable branches thatweakly embeds all such trees.

Page 59: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some new results under MA

Fifty years after its discovery now, it is rather rare to seenew applications of Martin’s Axiom, but there are stillsome.

In our recent work with Shelah (preprint available uponrequest)where we answered an old question by Meklerand Vaananen (1993).

We showed:

under MA(ω1) + ¬CH there is no tree of sizeand height ω1 and with no uncountable branches thatweakly embeds all such trees.

Page 60: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some new results under MA

Fifty years after its discovery now, it is rather rare to seenew applications of Martin’s Axiom, but there are stillsome.

In our recent work with Shelah (preprint available uponrequest)where we answered an old question by Meklerand Vaananen (1993).

We showed: under MA(ω1) + ¬CH there is no tree of sizeand height ω1 and with no uncountable branches thatweakly embeds all such trees.

Page 61: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In the absence of cccThere are nice forcings that preserve cardinals,

yet theyare not ccc. For example, adding a Sacks real.To iteratethose we need a more involved notion.

Properness is a property that guarantees that ω1 ispreserved. The definition of properness is a strike ofgenius, by Shelah.

Theorem(Shelah 1980) Properness is preserved under countablesupport iterations.

PFA The same as MA but with “ccc” replaced by properand “< c” with ω1 dense sets.[Why ω1 ? Todorcevic and Velickovic proved that PFAimplies c = ω2.]

Theorem(Baumgartner 1984) Modulo a supercompact cardinal,PFA is consistent.

Page 62: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In the absence of cccThere are nice forcings that preserve cardinals, yet theyare not ccc.

For example, adding a Sacks real.To iteratethose we need a more involved notion.

Properness is a property that guarantees that ω1 ispreserved. The definition of properness is a strike ofgenius, by Shelah.

Theorem(Shelah 1980) Properness is preserved under countablesupport iterations.

PFA The same as MA but with “ccc” replaced by properand “< c” with ω1 dense sets.[Why ω1 ? Todorcevic and Velickovic proved that PFAimplies c = ω2.]

Theorem(Baumgartner 1984) Modulo a supercompact cardinal,PFA is consistent.

Page 63: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In the absence of cccThere are nice forcings that preserve cardinals, yet theyare not ccc. For example, adding a Sacks real.

To iteratethose we need a more involved notion.

Properness is a property that guarantees that ω1 ispreserved. The definition of properness is a strike ofgenius, by Shelah.

Theorem(Shelah 1980) Properness is preserved under countablesupport iterations.

PFA The same as MA but with “ccc” replaced by properand “< c” with ω1 dense sets.[Why ω1 ? Todorcevic and Velickovic proved that PFAimplies c = ω2.]

Theorem(Baumgartner 1984) Modulo a supercompact cardinal,PFA is consistent.

Page 64: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In the absence of cccThere are nice forcings that preserve cardinals, yet theyare not ccc. For example, adding a Sacks real.To iteratethose we need a more involved notion.

Properness is a property that guarantees that ω1 ispreserved. The definition of properness is a strike ofgenius, by Shelah.

Theorem(Shelah 1980) Properness is preserved under countablesupport iterations.

PFA The same as MA but with “ccc” replaced by properand “< c” with ω1 dense sets.[Why ω1 ? Todorcevic and Velickovic proved that PFAimplies c = ω2.]

Theorem(Baumgartner 1984) Modulo a supercompact cardinal,PFA is consistent.

Page 65: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In the absence of cccThere are nice forcings that preserve cardinals, yet theyare not ccc. For example, adding a Sacks real.To iteratethose we need a more involved notion.

Properness is a property that guarantees that ω1 ispreserved.

The definition of properness is a strike ofgenius, by Shelah.

Theorem(Shelah 1980) Properness is preserved under countablesupport iterations.

PFA The same as MA but with “ccc” replaced by properand “< c” with ω1 dense sets.[Why ω1 ? Todorcevic and Velickovic proved that PFAimplies c = ω2.]

Theorem(Baumgartner 1984) Modulo a supercompact cardinal,PFA is consistent.

Page 66: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In the absence of cccThere are nice forcings that preserve cardinals, yet theyare not ccc. For example, adding a Sacks real.To iteratethose we need a more involved notion.

Properness is a property that guarantees that ω1 ispreserved. The definition of properness is a strike ofgenius, by Shelah.

Theorem(Shelah 1980) Properness is preserved under countablesupport iterations.

PFA The same as MA but with “ccc” replaced by properand “< c” with ω1 dense sets.[Why ω1 ? Todorcevic and Velickovic proved that PFAimplies c = ω2.]

Theorem(Baumgartner 1984) Modulo a supercompact cardinal,PFA is consistent.

Page 67: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In the absence of cccThere are nice forcings that preserve cardinals, yet theyare not ccc. For example, adding a Sacks real.To iteratethose we need a more involved notion.

Properness is a property that guarantees that ω1 ispreserved. The definition of properness is a strike ofgenius, by Shelah.

Theorem(Shelah 1980) Properness is preserved under countablesupport iterations.

PFA The same as MA but with “ccc” replaced by properand “< c” with ω1 dense sets.[Why ω1 ? Todorcevic and Velickovic proved that PFAimplies c = ω2.]

Theorem(Baumgartner 1984) Modulo a supercompact cardinal,PFA is consistent.

Page 68: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In the absence of cccThere are nice forcings that preserve cardinals, yet theyare not ccc. For example, adding a Sacks real.To iteratethose we need a more involved notion.

Properness is a property that guarantees that ω1 ispreserved. The definition of properness is a strike ofgenius, by Shelah.

Theorem(Shelah 1980) Properness is preserved under countablesupport iterations.

PFA The same as MA but with “ccc” replaced by properand “< c” with ω1 dense sets.

[Why ω1 ? Todorcevic and Velickovic proved that PFAimplies c = ω2.]

Theorem(Baumgartner 1984) Modulo a supercompact cardinal,PFA is consistent.

Page 69: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In the absence of cccThere are nice forcings that preserve cardinals, yet theyare not ccc. For example, adding a Sacks real.To iteratethose we need a more involved notion.

Properness is a property that guarantees that ω1 ispreserved. The definition of properness is a strike ofgenius, by Shelah.

Theorem(Shelah 1980) Properness is preserved under countablesupport iterations.

PFA The same as MA but with “ccc” replaced by properand “< c” with ω1 dense sets.[Why ω1 ? Todorcevic and Velickovic proved that PFAimplies c = ω2.]

Theorem(Baumgartner 1984) Modulo a supercompact cardinal,PFA is consistent.

Page 70: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In the absence of cccThere are nice forcings that preserve cardinals, yet theyare not ccc. For example, adding a Sacks real.To iteratethose we need a more involved notion.

Properness is a property that guarantees that ω1 ispreserved. The definition of properness is a strike ofgenius, by Shelah.

Theorem(Shelah 1980) Properness is preserved under countablesupport iterations.

PFA The same as MA but with “ccc” replaced by properand “< c” with ω1 dense sets.[Why ω1 ? Todorcevic and Velickovic proved that PFAimplies c = ω2.]

Theorem(Baumgartner 1984) Modulo a supercompact cardinal,PFA is consistent.

Page 71: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Properness and generic conditions

DefinitionSuppose that M ≺ (H(χ),∈) with P ∈ M.

A condition q is(P,M)-generic if for every maximal antichain A ∈ M, theantichain A ∩M is maximal above q (i.e. for conditions rwith q ≤ r ).

P is proper if for every countable M ≺ (H(χ),∈) withP ∈ M, for every p ∈ P ∩M, there is q ≥ p which is(P,M)-generic.

A posteriori, this is a lot like master conditions in largecardinal forcing.Note ccc implies proper.

Page 72: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Properness and generic conditions

DefinitionSuppose that M ≺ (H(χ),∈) with P ∈ M. A condition q is(P,M)-generic if for every maximal antichain A ∈ M, theantichain A ∩M is maximal above q (i.e. for conditions rwith q ≤ r ).

P is proper if for every countable M ≺ (H(χ),∈) withP ∈ M, for every p ∈ P ∩M, there is q ≥ p which is(P,M)-generic.

A posteriori, this is a lot like master conditions in largecardinal forcing.Note ccc implies proper.

Page 73: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Properness and generic conditions

DefinitionSuppose that M ≺ (H(χ),∈) with P ∈ M. A condition q is(P,M)-generic if for every maximal antichain A ∈ M, theantichain A ∩M is maximal above q (i.e. for conditions rwith q ≤ r ).

P is proper if for every countable M ≺ (H(χ),∈) withP ∈ M, for every p ∈ P ∩M, there is q ≥ p which is(P,M)-generic.

A posteriori, this is a lot like master conditions in largecardinal forcing.Note ccc implies proper.

Page 74: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Properness and generic conditions

DefinitionSuppose that M ≺ (H(χ),∈) with P ∈ M. A condition q is(P,M)-generic if for every maximal antichain A ∈ M, theantichain A ∩M is maximal above q (i.e. for conditions rwith q ≤ r ).

P is proper if for every countable M ≺ (H(χ),∈) withP ∈ M, for every p ∈ P ∩M, there is q ≥ p which is(P,M)-generic.

A posteriori, this is a lot like master conditions in largecardinal forcing.Note ccc implies proper.

Page 75: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

05/06/12 13:29416NGXCBCNL._SL500_AA300_.jpg 300 × 300 pixels

Page 1 sur 2http://ecx.images-amazon.com/images/I/416NGXCBCNL._SL500_AA300_.jpg

Page 76: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some facts about proper forcing

Proper forcing cannot be iterated with finite supports.

The iteration theorem for countable supports of properforcing is much more involved than the one for finitesupports of ccc forcing.

Proper forcing of size ℵ1 or with strong ℵ2-cc propertiespreserves cardinals, cofinalities and stationary subsets ofω1.

The natural applications of proper forcing are therefore onω1, with countable conditions as everybody knows :-)

We shall see.

Page 77: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some facts about proper forcing

Proper forcing cannot be iterated with finite supports.

The iteration theorem for countable supports of properforcing is much more involved than the one for finitesupports of ccc forcing.

Proper forcing of size ℵ1 or with strong ℵ2-cc propertiespreserves cardinals, cofinalities and stationary subsets ofω1.

The natural applications of proper forcing are therefore onω1, with countable conditions as everybody knows :-)

We shall see.

Page 78: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some facts about proper forcing

Proper forcing cannot be iterated with finite supports.

The iteration theorem for countable supports of properforcing is much more involved than the one for finitesupports of ccc forcing.

Proper forcing of size ℵ1 or with strong ℵ2-cc properties

preserves cardinals, cofinalities and stationary subsets ofω1.

The natural applications of proper forcing are therefore onω1, with countable conditions as everybody knows :-)

We shall see.

Page 79: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some facts about proper forcing

Proper forcing cannot be iterated with finite supports.

The iteration theorem for countable supports of properforcing is much more involved than the one for finitesupports of ccc forcing.

Proper forcing of size ℵ1 or with strong ℵ2-cc propertiespreserves cardinals, cofinalities and stationary subsets ofω1.

The natural applications of proper forcing are therefore onω1, with countable conditions as everybody knows :-)

We shall see.

Page 80: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some facts about proper forcing

Proper forcing cannot be iterated with finite supports.

The iteration theorem for countable supports of properforcing is much more involved than the one for finitesupports of ccc forcing.

Proper forcing of size ℵ1 or with strong ℵ2-cc propertiespreserves cardinals, cofinalities and stationary subsets ofω1.

The natural applications of proper forcing are therefore onω1,

with countable conditions as everybody knows :-)

We shall see.

Page 81: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some facts about proper forcing

Proper forcing cannot be iterated with finite supports.

The iteration theorem for countable supports of properforcing is much more involved than the one for finitesupports of ccc forcing.

Proper forcing of size ℵ1 or with strong ℵ2-cc propertiespreserves cardinals, cofinalities and stationary subsets ofω1.

The natural applications of proper forcing are therefore onω1, with countable conditions

as everybody knows :-)

We shall see.

Page 82: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some facts about proper forcing

Proper forcing cannot be iterated with finite supports.

The iteration theorem for countable supports of properforcing is much more involved than the one for finitesupports of ccc forcing.

Proper forcing of size ℵ1 or with strong ℵ2-cc propertiespreserves cardinals, cofinalities and stationary subsets ofω1.

The natural applications of proper forcing are therefore onω1, with countable conditions as everybody knows :-)

We shall see.

Page 83: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Some facts about proper forcing

Proper forcing cannot be iterated with finite supports.

The iteration theorem for countable supports of properforcing is much more involved than the one for finitesupports of ccc forcing.

Proper forcing of size ℵ1 or with strong ℵ2-cc propertiespreserves cardinals, cofinalities and stationary subsets ofω1.

The natural applications of proper forcing are therefore onω1, with countable conditions as everybody knows :-)

We shall see.

Page 84: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Above ω1

It turns out that naive analogues of MA *do not* work withω2.

For example, the iteration of κ+-cc < κ-closed (everyincreasing sequence of length < κ has an upper bound)forcing does not have to be κ+-cc (various examples, aknown one by Mitchell, involving Souslin trees).

To generalize MA to κ+ with κ<κ = κ we need to assumea strong form of κ+ − cc (Baumgartner, Shelah 1984)< κ-directed completeness or similar and some sort of“well met property” : every two compatible conditionshave a lub.

There is no, at least no popular, analogue of propernessfor ω2.

Solution, for adding an object *once* (no iteration) issometimes to use finite conditions. (Baumgartner-Shelah,Koszmider, Mitchell, Todorcevic, Velickovic-Venturi).Models as side conditions.

Page 85: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Above ω1

It turns out that naive analogues of MA *do not* work withω2. For example, the iteration of κ+-cc < κ-closed (everyincreasing sequence of length < κ has an upper bound)forcing does not have to be κ+-cc (various examples, aknown one by Mitchell, involving Souslin trees).

To generalize MA to κ+ with κ<κ = κ we need to assumea strong form of κ+ − cc (Baumgartner, Shelah 1984)< κ-directed completeness or similar and some sort of“well met property” : every two compatible conditionshave a lub.

There is no, at least no popular, analogue of propernessfor ω2.

Solution, for adding an object *once* (no iteration) issometimes to use finite conditions. (Baumgartner-Shelah,Koszmider, Mitchell, Todorcevic, Velickovic-Venturi).Models as side conditions.

Page 86: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Above ω1

It turns out that naive analogues of MA *do not* work withω2. For example, the iteration of κ+-cc < κ-closed (everyincreasing sequence of length < κ has an upper bound)forcing does not have to be κ+-cc (various examples, aknown one by Mitchell, involving Souslin trees).

To generalize MA to κ+ with κ<κ = κ

we need to assumea strong form of κ+ − cc (Baumgartner, Shelah 1984)< κ-directed completeness or similar and some sort of“well met property” : every two compatible conditionshave a lub.

There is no, at least no popular, analogue of propernessfor ω2.

Solution, for adding an object *once* (no iteration) issometimes to use finite conditions. (Baumgartner-Shelah,Koszmider, Mitchell, Todorcevic, Velickovic-Venturi).Models as side conditions.

Page 87: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Above ω1

It turns out that naive analogues of MA *do not* work withω2. For example, the iteration of κ+-cc < κ-closed (everyincreasing sequence of length < κ has an upper bound)forcing does not have to be κ+-cc (various examples, aknown one by Mitchell, involving Souslin trees).

To generalize MA to κ+ with κ<κ = κ we need to assumea strong form of κ+ − cc (Baumgartner, Shelah 1984)

< κ-directed completeness or similar and some sort of“well met property” : every two compatible conditionshave a lub.

There is no, at least no popular, analogue of propernessfor ω2.

Solution, for adding an object *once* (no iteration) issometimes to use finite conditions. (Baumgartner-Shelah,Koszmider, Mitchell, Todorcevic, Velickovic-Venturi).Models as side conditions.

Page 88: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Above ω1

It turns out that naive analogues of MA *do not* work withω2. For example, the iteration of κ+-cc < κ-closed (everyincreasing sequence of length < κ has an upper bound)forcing does not have to be κ+-cc (various examples, aknown one by Mitchell, involving Souslin trees).

To generalize MA to κ+ with κ<κ = κ we need to assumea strong form of κ+ − cc (Baumgartner, Shelah 1984)< κ-directed completeness or similar and

some sort of“well met property” : every two compatible conditionshave a lub.

There is no, at least no popular, analogue of propernessfor ω2.

Solution, for adding an object *once* (no iteration) issometimes to use finite conditions. (Baumgartner-Shelah,Koszmider, Mitchell, Todorcevic, Velickovic-Venturi).Models as side conditions.

Page 89: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Above ω1

It turns out that naive analogues of MA *do not* work withω2. For example, the iteration of κ+-cc < κ-closed (everyincreasing sequence of length < κ has an upper bound)forcing does not have to be κ+-cc (various examples, aknown one by Mitchell, involving Souslin trees).

To generalize MA to κ+ with κ<κ = κ we need to assumea strong form of κ+ − cc (Baumgartner, Shelah 1984)< κ-directed completeness or similar and some sort of“well met property” :

every two compatible conditionshave a lub.

There is no, at least no popular, analogue of propernessfor ω2.

Solution, for adding an object *once* (no iteration) issometimes to use finite conditions. (Baumgartner-Shelah,Koszmider, Mitchell, Todorcevic, Velickovic-Venturi).Models as side conditions.

Page 90: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Above ω1

It turns out that naive analogues of MA *do not* work withω2. For example, the iteration of κ+-cc < κ-closed (everyincreasing sequence of length < κ has an upper bound)forcing does not have to be κ+-cc (various examples, aknown one by Mitchell, involving Souslin trees).

To generalize MA to κ+ with κ<κ = κ we need to assumea strong form of κ+ − cc (Baumgartner, Shelah 1984)< κ-directed completeness or similar and some sort of“well met property” : every two compatible conditionshave a lub.

There is no, at least no popular, analogue of propernessfor ω2.

Solution, for adding an object *once* (no iteration) issometimes to use finite conditions. (Baumgartner-Shelah,Koszmider, Mitchell, Todorcevic, Velickovic-Venturi).Models as side conditions.

Page 91: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Above ω1

It turns out that naive analogues of MA *do not* work withω2. For example, the iteration of κ+-cc < κ-closed (everyincreasing sequence of length < κ has an upper bound)forcing does not have to be κ+-cc (various examples, aknown one by Mitchell, involving Souslin trees).

To generalize MA to κ+ with κ<κ = κ we need to assumea strong form of κ+ − cc (Baumgartner, Shelah 1984)< κ-directed completeness or similar and some sort of“well met property” : every two compatible conditionshave a lub.

There is no, at least no popular, analogue of propernessfor ω2.

Solution, for adding an object *once* (no iteration) issometimes to use finite conditions. (Baumgartner-Shelah,Koszmider, Mitchell, Todorcevic, Velickovic-Venturi).Models as side conditions.

Page 92: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Above ω1

It turns out that naive analogues of MA *do not* work withω2. For example, the iteration of κ+-cc < κ-closed (everyincreasing sequence of length < κ has an upper bound)forcing does not have to be κ+-cc (various examples, aknown one by Mitchell, involving Souslin trees).

To generalize MA to κ+ with κ<κ = κ we need to assumea strong form of κ+ − cc (Baumgartner, Shelah 1984)< κ-directed completeness or similar and some sort of“well met property” : every two compatible conditionshave a lub.

There is no, at least no popular, analogue of propernessfor ω2.

Solution, for adding an object *once* (no iteration) issometimes to use finite conditions. (Baumgartner-Shelah,Koszmider, Mitchell, Todorcevic, Velickovic-Venturi).

Models as side conditions.

Page 93: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Above ω1

It turns out that naive analogues of MA *do not* work withω2. For example, the iteration of κ+-cc < κ-closed (everyincreasing sequence of length < κ has an upper bound)forcing does not have to be κ+-cc (various examples, aknown one by Mitchell, involving Souslin trees).

To generalize MA to κ+ with κ<κ = κ we need to assumea strong form of κ+ − cc (Baumgartner, Shelah 1984)< κ-directed completeness or similar and some sort of“well met property” : every two compatible conditionshave a lub.

There is no, at least no popular, analogue of propernessfor ω2.

Solution, for adding an object *once* (no iteration) issometimes to use finite conditions. (Baumgartner-Shelah,Koszmider, Mitchell, Todorcevic, Velickovic-Venturi).Models as side conditions.

Page 94: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us try to add a club to ω1 using *finite* conditions,

sofinite subsets of the club. No reason that the genericshould be closed.

Baumgartner (1984) (alternative Abraham 1983) solvedthis: each condition gives finitely many ordinals that willbe in and finitely many intervals of the form (α, α′] fromwhich we have to keep out.

If we try this with ω2, then ω1 will get collapsed.

Friedman (2006) and Mitchell (2009) independently founda way how to add a club to ω2 using finite conditions andproper forcing! (in fact there is also earlier work ofRoslanowski and Shelah (2001)). The trick is that eachcondition in addition to the ordinals and the intervals hasa finite sequence of countable elementary submodels ofHθ that are used to prove that the forcing is actuallyproper.

Introduced the idea of strongly proper forcing.

Page 95: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us try to add a club to ω1 using *finite* conditions, sofinite subsets of the club.

No reason that the genericshould be closed.

Baumgartner (1984) (alternative Abraham 1983) solvedthis: each condition gives finitely many ordinals that willbe in and finitely many intervals of the form (α, α′] fromwhich we have to keep out.

If we try this with ω2, then ω1 will get collapsed.

Friedman (2006) and Mitchell (2009) independently founda way how to add a club to ω2 using finite conditions andproper forcing! (in fact there is also earlier work ofRoslanowski and Shelah (2001)). The trick is that eachcondition in addition to the ordinals and the intervals hasa finite sequence of countable elementary submodels ofHθ that are used to prove that the forcing is actuallyproper.

Introduced the idea of strongly proper forcing.

Page 96: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us try to add a club to ω1 using *finite* conditions, sofinite subsets of the club. No reason that the genericshould be closed.

Baumgartner (1984) (alternative Abraham 1983) solvedthis: each condition gives finitely many ordinals that willbe in and finitely many intervals of the form (α, α′] fromwhich we have to keep out.

If we try this with ω2, then ω1 will get collapsed.

Friedman (2006) and Mitchell (2009) independently founda way how to add a club to ω2 using finite conditions andproper forcing! (in fact there is also earlier work ofRoslanowski and Shelah (2001)). The trick is that eachcondition in addition to the ordinals and the intervals hasa finite sequence of countable elementary submodels ofHθ that are used to prove that the forcing is actuallyproper.

Introduced the idea of strongly proper forcing.

Page 97: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us try to add a club to ω1 using *finite* conditions, sofinite subsets of the club. No reason that the genericshould be closed.

Baumgartner (1984) (alternative Abraham 1983) solvedthis:

each condition gives finitely many ordinals that willbe in and finitely many intervals of the form (α, α′] fromwhich we have to keep out.

If we try this with ω2, then ω1 will get collapsed.

Friedman (2006) and Mitchell (2009) independently founda way how to add a club to ω2 using finite conditions andproper forcing! (in fact there is also earlier work ofRoslanowski and Shelah (2001)). The trick is that eachcondition in addition to the ordinals and the intervals hasa finite sequence of countable elementary submodels ofHθ that are used to prove that the forcing is actuallyproper.

Introduced the idea of strongly proper forcing.

Page 98: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us try to add a club to ω1 using *finite* conditions, sofinite subsets of the club. No reason that the genericshould be closed.

Baumgartner (1984) (alternative Abraham 1983) solvedthis: each condition gives finitely many ordinals that willbe in

and finitely many intervals of the form (α, α′] fromwhich we have to keep out.

If we try this with ω2, then ω1 will get collapsed.

Friedman (2006) and Mitchell (2009) independently founda way how to add a club to ω2 using finite conditions andproper forcing! (in fact there is also earlier work ofRoslanowski and Shelah (2001)). The trick is that eachcondition in addition to the ordinals and the intervals hasa finite sequence of countable elementary submodels ofHθ that are used to prove that the forcing is actuallyproper.

Introduced the idea of strongly proper forcing.

Page 99: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us try to add a club to ω1 using *finite* conditions, sofinite subsets of the club. No reason that the genericshould be closed.

Baumgartner (1984) (alternative Abraham 1983) solvedthis: each condition gives finitely many ordinals that willbe in and finitely many intervals of the form (α, α′] fromwhich we have to keep out.

If we try this with ω2, then ω1 will get collapsed.

Friedman (2006) and Mitchell (2009) independently founda way how to add a club to ω2 using finite conditions andproper forcing! (in fact there is also earlier work ofRoslanowski and Shelah (2001)). The trick is that eachcondition in addition to the ordinals and the intervals hasa finite sequence of countable elementary submodels ofHθ that are used to prove that the forcing is actuallyproper.

Introduced the idea of strongly proper forcing.

Page 100: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us try to add a club to ω1 using *finite* conditions, sofinite subsets of the club. No reason that the genericshould be closed.

Baumgartner (1984) (alternative Abraham 1983) solvedthis: each condition gives finitely many ordinals that willbe in and finitely many intervals of the form (α, α′] fromwhich we have to keep out.

If we try this with ω2, then ω1 will get collapsed.

Friedman (2006) and Mitchell (2009) independently founda way how to add a club to ω2 using finite conditions andproper forcing! (in fact there is also earlier work ofRoslanowski and Shelah (2001)). The trick is that eachcondition in addition to the ordinals and the intervals hasa finite sequence of countable elementary submodels ofHθ that are used to prove that the forcing is actuallyproper.

Introduced the idea of strongly proper forcing.

Page 101: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us try to add a club to ω1 using *finite* conditions, sofinite subsets of the club. No reason that the genericshould be closed.

Baumgartner (1984) (alternative Abraham 1983) solvedthis: each condition gives finitely many ordinals that willbe in and finitely many intervals of the form (α, α′] fromwhich we have to keep out.

If we try this with ω2, then ω1 will get collapsed.

Friedman (2006) and Mitchell (2009) independently founda way how to add

a club to ω2 using finite conditions andproper forcing! (in fact there is also earlier work ofRoslanowski and Shelah (2001)). The trick is that eachcondition in addition to the ordinals and the intervals hasa finite sequence of countable elementary submodels ofHθ that are used to prove that the forcing is actuallyproper.

Introduced the idea of strongly proper forcing.

Page 102: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us try to add a club to ω1 using *finite* conditions, sofinite subsets of the club. No reason that the genericshould be closed.

Baumgartner (1984) (alternative Abraham 1983) solvedthis: each condition gives finitely many ordinals that willbe in and finitely many intervals of the form (α, α′] fromwhich we have to keep out.

If we try this with ω2, then ω1 will get collapsed.

Friedman (2006) and Mitchell (2009) independently founda way how to add a club to ω2 using finite conditions andproper forcing! (in fact there is also earlier work ofRoslanowski and Shelah (2001)).

The trick is that eachcondition in addition to the ordinals and the intervals hasa finite sequence of countable elementary submodels ofHθ that are used to prove that the forcing is actuallyproper.

Introduced the idea of strongly proper forcing.

Page 103: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us try to add a club to ω1 using *finite* conditions, sofinite subsets of the club. No reason that the genericshould be closed.

Baumgartner (1984) (alternative Abraham 1983) solvedthis: each condition gives finitely many ordinals that willbe in and finitely many intervals of the form (α, α′] fromwhich we have to keep out.

If we try this with ω2, then ω1 will get collapsed.

Friedman (2006) and Mitchell (2009) independently founda way how to add a club to ω2 using finite conditions andproper forcing! (in fact there is also earlier work ofRoslanowski and Shelah (2001)). The trick is that eachcondition in addition to the ordinals and the intervals hasa finite sequence of countable elementary submodels ofHθ

that are used to prove that the forcing is actuallyproper.

Introduced the idea of strongly proper forcing.

Page 104: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us try to add a club to ω1 using *finite* conditions, sofinite subsets of the club. No reason that the genericshould be closed.

Baumgartner (1984) (alternative Abraham 1983) solvedthis: each condition gives finitely many ordinals that willbe in and finitely many intervals of the form (α, α′] fromwhich we have to keep out.

If we try this with ω2, then ω1 will get collapsed.

Friedman (2006) and Mitchell (2009) independently founda way how to add a club to ω2 using finite conditions andproper forcing! (in fact there is also earlier work ofRoslanowski and Shelah (2001)). The trick is that eachcondition in addition to the ordinals and the intervals hasa finite sequence of countable elementary submodels ofHθ that are used to prove that the forcing is actuallyproper.

Introduced the idea of strongly proper forcing.

Page 105: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us try to add a club to ω1 using *finite* conditions, sofinite subsets of the club. No reason that the genericshould be closed.

Baumgartner (1984) (alternative Abraham 1983) solvedthis: each condition gives finitely many ordinals that willbe in and finitely many intervals of the form (α, α′] fromwhich we have to keep out.

If we try this with ω2, then ω1 will get collapsed.

Friedman (2006) and Mitchell (2009) independently founda way how to add a club to ω2 using finite conditions andproper forcing! (in fact there is also earlier work ofRoslanowski and Shelah (2001)). The trick is that eachcondition in addition to the ordinals and the intervals hasa finite sequence of countable elementary submodels ofHθ that are used to prove that the forcing is actuallyproper.

Introduced the idea of strongly proper forcing.

Page 106: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

�ω1 using finite conditions and proper forcing

Recall Jensen’s definition:

�ω1 is the statement:

There is a sequence 〈Cα : α limit < ω2〉 such that:Cα is a club in αif α is a limit point of Cβ then Cα = Cβ ∩ αif cf(α) = ω then Cα is countable.

This holds in L. Adding a square is a classical forcingquestion, natural after adding a club.

Note: to destroy a square ”thread a club”.

Page 107: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

�ω1 using finite conditions and proper forcing

Recall Jensen’s definition:

�ω1 is the statement:

There is a sequence 〈Cα : α limit < ω2〉 such that:Cα is a club in αif α is a limit point of Cβ then Cα = Cβ ∩ αif cf(α) = ω then Cα is countable.

This holds in L. Adding a square is a classical forcingquestion, natural after adding a club.

Note: to destroy a square ”thread a club”.

Page 108: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

�ω1 using finite conditions and proper forcing

Recall Jensen’s definition:

�ω1 is the statement:

There is a sequence 〈Cα : α limit < ω2〉 such that:Cα is a club in α

if α is a limit point of Cβ then Cα = Cβ ∩ αif cf(α) = ω then Cα is countable.

This holds in L. Adding a square is a classical forcingquestion, natural after adding a club.

Note: to destroy a square ”thread a club”.

Page 109: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

�ω1 using finite conditions and proper forcing

Recall Jensen’s definition:

�ω1 is the statement:

There is a sequence 〈Cα : α limit < ω2〉 such that:Cα is a club in αif α is a limit point of Cβ then Cα = Cβ ∩ α

if cf(α) = ω then Cα is countable.

This holds in L. Adding a square is a classical forcingquestion, natural after adding a club.

Note: to destroy a square ”thread a club”.

Page 110: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

�ω1 using finite conditions and proper forcing

Recall Jensen’s definition:

�ω1 is the statement:

There is a sequence 〈Cα : α limit < ω2〉 such that:Cα is a club in αif α is a limit point of Cβ then Cα = Cβ ∩ αif cf(α) = ω then Cα is countable.

This holds in L. Adding a square is a classical forcingquestion, natural after adding a club.

Note: to destroy a square ”thread a club”.

Page 111: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

�ω1 using finite conditions and proper forcing

Recall Jensen’s definition:

�ω1 is the statement:

There is a sequence 〈Cα : α limit < ω2〉 such that:Cα is a club in αif α is a limit point of Cβ then Cα = Cβ ∩ αif cf(α) = ω then Cα is countable.

This holds in L.

Adding a square is a classical forcingquestion, natural after adding a club.

Note: to destroy a square ”thread a club”.

Page 112: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

�ω1 using finite conditions and proper forcing

Recall Jensen’s definition:

�ω1 is the statement:

There is a sequence 〈Cα : α limit < ω2〉 such that:Cα is a club in αif α is a limit point of Cβ then Cα = Cβ ∩ αif cf(α) = ω then Cα is countable.

This holds in L. Adding a square is a classical forcingquestion,

natural after adding a club.

Note: to destroy a square ”thread a club”.

Page 113: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

�ω1 using finite conditions and proper forcing

Recall Jensen’s definition:

�ω1 is the statement:

There is a sequence 〈Cα : α limit < ω2〉 such that:Cα is a club in αif α is a limit point of Cβ then Cα = Cβ ∩ αif cf(α) = ω then Cα is countable.

This holds in L. Adding a square is a classical forcingquestion, natural after adding a club.

Note: to destroy a square ”thread a club”.

Page 114: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

�ω1 using finite conditions and proper forcing

Recall Jensen’s definition:

�ω1 is the statement:

There is a sequence 〈Cα : α limit < ω2〉 such that:Cα is a club in αif α is a limit point of Cβ then Cα = Cβ ∩ αif cf(α) = ω then Cα is countable.

This holds in L. Adding a square is a classical forcingquestion, natural after adding a club.

Note: to destroy a square ”thread a club”.

Page 115: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club, aMitchell-Friedman club. It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp, for each α ∈ Fpwe choose a club, a finite set of intervals Op that keepthings out, and a finite set of modelsMp and”safeguards” Sp. Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2). Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 116: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club,

aMitchell-Friedman club. It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp, for each α ∈ Fpwe choose a club, a finite set of intervals Op that keepthings out, and a finite set of modelsMp and”safeguards” Sp. Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2). Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 117: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club, aMitchell-Friedman club.

It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp, for each α ∈ Fpwe choose a club, a finite set of intervals Op that keepthings out, and a finite set of modelsMp and”safeguards” Sp. Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2). Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 118: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club, aMitchell-Friedman club. It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp, for each α ∈ Fpwe choose a club, a finite set of intervals Op that keepthings out, and a finite set of modelsMp and”safeguards” Sp. Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2). Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 119: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club, aMitchell-Friedman club. It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp,

for each α ∈ Fpwe choose a club, a finite set of intervals Op that keepthings out, and a finite set of modelsMp and”safeguards” Sp. Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2). Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 120: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club, aMitchell-Friedman club. It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp, for each α ∈ Fpwe choose a club,

a finite set of intervals Op that keepthings out, and a finite set of modelsMp and”safeguards” Sp. Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2). Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 121: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club, aMitchell-Friedman club. It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp, for each α ∈ Fpwe choose a club, a finite set of intervals Op that keepthings out, and

a finite set of modelsMp and”safeguards” Sp. Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2). Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 122: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club, aMitchell-Friedman club. It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp, for each α ∈ Fpwe choose a club, a finite set of intervals Op that keepthings out, and a finite set of modelsMp

and”safeguards” Sp. Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2). Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 123: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club, aMitchell-Friedman club. It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp, for each α ∈ Fpwe choose a club, a finite set of intervals Op that keepthings out, and a finite set of modelsMp and”safeguards” Sp.

Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2). Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 124: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club, aMitchell-Friedman club. It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp, for each α ∈ Fpwe choose a club, a finite set of intervals Op that keepthings out, and a finite set of modelsMp and”safeguards” Sp. Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2). Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 125: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club, aMitchell-Friedman club. It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp, for each α ∈ Fpwe choose a club, a finite set of intervals Op that keepthings out, and a finite set of modelsMp and”safeguards” Sp. Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2). Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 126: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club, aMitchell-Friedman club. It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp, for each α ∈ Fpwe choose a club, a finite set of intervals Op that keepthings out, and a finite set of modelsMp and”safeguards” Sp. Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2).

Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 127: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

In a joint paper with Dolinar (APAL 2013), we add asquare with finite conditions.

Our forcing actually adds a square on a club, aMitchell-Friedman club. It can be shown that if there is asquare on a club, then there is a square.

Each condition has a finite domain Fp, for each α ∈ Fpwe choose a club, a finite set of intervals Op that keepthings out, and a finite set of modelsMp and”safeguards” Sp. Safeguards control the interactionbetween the models and the club.

For α < ω2 with cf(α) = ω1, let Eα denote some fixed clubin α of order type ω1.

Suppose thatM1,M2 ≺ Hω2 are countable and letδ := sup(M1 ∩M2). Then the set{min(M1 \ λ) | λ ∈ M2, δ < λ < sup(M1)} ∪ {min(M1 \ δ)}is called the set of M1-fences for M2;

Page 128: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

DefinitionThe forcing notion P is the set of conditions of the formp := (Fp, Sp, Op, Mp), where(1) Fp : Lim(ω2)→ P(ω2), |Fp| < ω and for allα ∈ Dp := dom(Fp), Fp(α) is a club Cα ⊂ α whose ordertype is < ω1 if cf(α) = ω and which satisfiesCα ∈ {Eα \ β | β ∈ Dp ∩ α} if cf(α) = ω1;(2) Sp ⊂ Dp and α ∈ Sp for every α ∈ Dp with cf(α) = ω1;(3)Mp is a finite set of sets M ∩ ω2 for some ocountableM[M] ≺ Hθ, and sup(M) ∈ Sp for every M ∈Mp;(4) for every α 6= β ∈ Dp, if µ ∈ Lim(Cα) ∩ Lim(Cβ) thenCα ∩ µ = Cβ ∩ µ;(5) if α ∈ Dp and σ ∈ Sp ∩ α, then Cα ∩ σ is a finite set;(6) for all α ∈ Dp and M ∈Mp:(a) if α ∈ M then Cα ∈M[M],(b) if α 6∈ M is such that α < sup(M), or if α ∈ M is suchthat sup(M ∩ α) < α, then min(M \ α) ∈ Sp andsup(M ∩ α) ∈ Dp

1,1Note that if α ∈ M then sup(M ∩ α) < α iff cf(α) = ω1.

Page 129: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Definition(c) if α 6∈ M, sup(M ∩ α) < α < sup(M) and there is noβ ∈ Dp \ (α+ 1), such that α ∈ Lim(Cβ), thenCα ∩ sup(M ∩ α) is a finite set,(d) if α 6∈ M, sup(M ∩ α) = α and there is noβ ∈ Dp \ (α+ 1), such that α ∈ Lim(Cβ), then Cα is somecofinal sequence in α of length ω;(7) Op is a finite set of half open nonempty intervals(β′, β] ⊂ ω2 such that Dp ∩

⋃Op = ∅;

(8) if (β′, β] ∈ Op and M ∈Mp then either (β′, β] ∈M or(β′, β] ∩M = ∅;(9) if M1,M2 ∈Mp then they are compatible, and theM1-fence for M2 is a subset of Sp.

For p,q ∈ P definep ≤ q def⇐⇒ Fp ⊂ Fq, Sp ⊂ Sq, Op ⊂ Oq,Mp ⊂Mq.

Page 130: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Neeman’s new way of iterating

Inspired by various developments in the forcing withmodels as side conditions,

Neeman (presentation 2010,preprint 2011, paper 2014) gave a completely new way toiterate: PFA with finite supports.

How is this possible?

The conditions in the iteration are a mixture of elementarymodels and conditions in a proper forcing.

Page 131: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Neeman’s new way of iterating

Inspired by various developments in the forcing withmodels as side conditions, Neeman (presentation 2010,preprint 2011, paper 2014) gave a completely new way toiterate:

PFA with finite supports.

How is this possible?

The conditions in the iteration are a mixture of elementarymodels and conditions in a proper forcing.

Page 132: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Neeman’s new way of iterating

Inspired by various developments in the forcing withmodels as side conditions, Neeman (presentation 2010,preprint 2011, paper 2014) gave a completely new way toiterate: PFA with finite supports.

How is this possible?

The conditions in the iteration are a mixture of elementarymodels and conditions in a proper forcing.

Page 133: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Neeman’s new way of iterating

Inspired by various developments in the forcing withmodels as side conditions, Neeman (presentation 2010,preprint 2011, paper 2014) gave a completely new way toiterate: PFA with finite supports.

How is this possible?

The conditions in the iteration are a mixture of elementarymodels and conditions in a proper forcing.

Page 134: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Neeman’s new way of iterating

Inspired by various developments in the forcing withmodels as side conditions, Neeman (presentation 2010,preprint 2011, paper 2014) gave a completely new way toiterate: PFA with finite supports.

How is this possible?

The conditions in the iteration are a mixture of elementarymodels and conditions in a proper forcing.

Page 135: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us fix a large regular χ and consider elementarysubmodels of Hχ.

Let θ be a supercompact cardinal andF a Laver function on θ.

Definition(Neeman) The forcing notion A consists of pairs (s,p)such that

s is a finite sequence of models which are eithercountable or of the form Hα THIS IS NEWs is ∈-increasing and closed under intersectionsp is a finite partial function from θ such that if p(α) isdefined, then A ∩ H(α) forces F (α) to be a properforcing and p(α) to be in F (α)

for such α, if M is a countable model in s and α ∈ Mthen p(α) is an M-generic condition for F (α).

Page 136: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us fix a large regular χ and consider elementarysubmodels of Hχ. Let θ be a supercompact cardinal andF a Laver function on θ.

Definition(Neeman) The forcing notion A consists of pairs (s,p)such that

s is a finite sequence of models which are eithercountable or of the form Hα THIS IS NEWs is ∈-increasing and closed under intersectionsp is a finite partial function from θ such that if p(α) isdefined, then A ∩ H(α) forces F (α) to be a properforcing and p(α) to be in F (α)

for such α, if M is a countable model in s and α ∈ Mthen p(α) is an M-generic condition for F (α).

Page 137: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us fix a large regular χ and consider elementarysubmodels of Hχ. Let θ be a supercompact cardinal andF a Laver function on θ.

Definition(Neeman) The forcing notion A consists of pairs (s,p)such that

s is a finite sequence of models which are eithercountable or of the form Hα THIS IS NEWs is ∈-increasing and closed under intersectionsp is a finite partial function from θ such that if p(α) isdefined, then A ∩ H(α) forces F (α) to be a properforcing and p(α) to be in F (α)

for such α, if M is a countable model in s and α ∈ Mthen p(α) is an M-generic condition for F (α).

Page 138: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us fix a large regular χ and consider elementarysubmodels of Hχ. Let θ be a supercompact cardinal andF a Laver function on θ.

Definition(Neeman) The forcing notion A consists of pairs (s,p)such that

s is a finite sequence of models which are eithercountable or of the form Hα THIS IS NEW

s is ∈-increasing and closed under intersectionsp is a finite partial function from θ such that if p(α) isdefined, then A ∩ H(α) forces F (α) to be a properforcing and p(α) to be in F (α)

for such α, if M is a countable model in s and α ∈ Mthen p(α) is an M-generic condition for F (α).

Page 139: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us fix a large regular χ and consider elementarysubmodels of Hχ. Let θ be a supercompact cardinal andF a Laver function on θ.

Definition(Neeman) The forcing notion A consists of pairs (s,p)such that

s is a finite sequence of models which are eithercountable or of the form Hα THIS IS NEWs is ∈-increasing and closed under intersections

p is a finite partial function from θ such that if p(α) isdefined, then A ∩ H(α) forces F (α) to be a properforcing and p(α) to be in F (α)

for such α, if M is a countable model in s and α ∈ Mthen p(α) is an M-generic condition for F (α).

Page 140: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us fix a large regular χ and consider elementarysubmodels of Hχ. Let θ be a supercompact cardinal andF a Laver function on θ.

Definition(Neeman) The forcing notion A consists of pairs (s,p)such that

s is a finite sequence of models which are eithercountable or of the form Hα THIS IS NEWs is ∈-increasing and closed under intersectionsp is a finite partial function from θ such that if p(α) isdefined, then A ∩ H(α) forces F (α) to be a properforcing

and p(α) to be in F (α)

for such α, if M is a countable model in s and α ∈ Mthen p(α) is an M-generic condition for F (α).

Page 141: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us fix a large regular χ and consider elementarysubmodels of Hχ. Let θ be a supercompact cardinal andF a Laver function on θ.

Definition(Neeman) The forcing notion A consists of pairs (s,p)such that

s is a finite sequence of models which are eithercountable or of the form Hα THIS IS NEWs is ∈-increasing and closed under intersectionsp is a finite partial function from θ such that if p(α) isdefined, then A ∩ H(α) forces F (α) to be a properforcing and p(α) to be in F (α)

for such α, if M is a countable model in s and α ∈ Mthen p(α) is an M-generic condition for F (α).

Page 142: Iterated Forcing and Forcing Axioms - ULisboagmferreira/Dzamonja2Lisboa2020.pdfIterated Forcing and Forcing Axioms Mirna Dzamonja,ˇ Tutorial 2 Recall and MA PFA How about !2, or something

Iterated Forcingand Forcing

Axioms

Mirna Dzamonja,Tutorial 2

Recall and MA

PFA

How about ω2, orsomething larger

Finite conditions

Work with GregorDolinar (APAL2013)

Neeman’srevolution

Let us fix a large regular χ and consider elementarysubmodels of Hχ. Let θ be a supercompact cardinal andF a Laver function on θ.

Definition(Neeman) The forcing notion A consists of pairs (s,p)such that

s is a finite sequence of models which are eithercountable or of the form Hα THIS IS NEWs is ∈-increasing and closed under intersectionsp is a finite partial function from θ such that if p(α) isdefined, then A ∩ H(α) forces F (α) to be a properforcing and p(α) to be in F (α)

for such α, if M is a countable model in s and α ∈ Mthen p(α) is an M-generic condition for F (α).