Infinite combinatorics
Almost disjointness
Definition
Sets x, y are disjoint if x ∩ y = ∅.
Remark. Let X be a set of cardinality κ (for example κ itself). Let A be
a family of pairwise disjoint subsets of X . How many members can A
contain?
Almost disjointness
Definition
Sets x, y are disjoint if x ∩ y = ∅.
Remark. Let X be a set of cardinality κ (for example κ itself). Let A be
a family of pairwise disjoint subsets of X . How many members can A
contain?
A family of disjoint subsets of a set of cardinality κ has cardinalty at
most κ.
Almost disjointness
Definition
Sets x, y are disjoint if x ∩ y = ∅.
Remark. Let X be a set of cardinality κ (for example κ itself). Let A be
a family of pairwise disjoint subsets of X . How many members can A
contain?
A family of disjoint subsets of a set of cardinality κ has cardinalty at
most κ.
In fact there exists a maximal disjoint family A of cardinality κ:
Almost disjointness
Definition
Sets x, y are disjoint if x ∩ y = ∅.
Remark. Let X be a set of cardinality κ (for example κ itself). Let A be
a family of pairwise disjoint subsets of X . How many members can A
contain?
A family of disjoint subsets of a set of cardinality κ has cardinalty at
most κ.
In fact there exists a maximal disjoint family A of cardinality κ: working
on X = κ × κ, let A = {Aα | α < κ} where
Aα = {(α, β) ∈ κ × κ | β < κ}.
Almost disjointness
Definition
Sets x, y are disjoint if x ∩ y = ∅.
Remark. Let X be a set of cardinality κ (for example κ itself). Let A be
a family of pairwise disjoint subsets of X . How many members can A
contain?
A family of disjoint subsets of a set of cardinality κ has cardinalty at
most κ.
In fact there exists a maximal disjoint family A of cardinality κ: working
on X = κ × κ, let A = {Aα | α < κ} where
S
Aα = {(α, β) ∈ κ × κ | β < κ}. Since A = κ, family A is maximal.
Almost disjointness
Definition
Sets x, y are disjoint if x ∩ y = ∅.
Remark. Let X be a set of cardinality κ (for example κ itself). Let A be
a family of pairwise disjoint subsets of X . How many members can A
contain?
A family of disjoint subsets of a set of cardinality κ has cardinalty at
most κ.
In fact there exists a maximal disjoint family A of cardinality κ: working
on X = κ × κ, let A = {Aα | α < κ} where
S
Aα = {(α, β) ∈ κ × κ | β < κ}. Since A = κ, family A is maximal.
What if we weaken the condition of disjointness?
Almost disjointness
Definition
Let κ be an infinite cardinal, x, y ⊆ κ. Then x, y are almost disjoint
(a.d.) if |x ∩ y | < κ.
Almost disjointness
Definition
Let κ be an infinite cardinal, x, y ⊆ κ. Then x, y are almost disjoint
(a.d.) if |x ∩ y | < κ.
An almost disjoint family is some A ⊆ P(κ) such that:
I
∀x ∈ A |x| = κ
I
∀x, y ∈ A (x 6= y ⇒ |x ∩ y | < κ)
Almost disjointness
Definition
Let κ be an infinite cardinal, x, y ⊆ κ. Then x, y are almost disjoint
(a.d.) if |x ∩ y | < κ.
An almost disjoint family is some A ⊆ P(κ) such that:
I
∀x ∈ A |x| = κ
I
∀x, y ∈ A (x 6= y ⇒ |x ∩ y | < κ)
A maximal almost disjoint (m.a.d.) family is an ad family A such that for
no ad family B one has A ⊂ B.
Almost disjointness
Theorem
Let κ be regular. Then:
1. If A ⊆ P(κ) is an ad family with |A| = κ, then A is not maximal
2. There is a mad family B ⊆ P(κ) of cardinality ≥ κ+
Almost disjointness
Theorem
Let κ be regular. Then:
1. If A ⊆ P(κ) is an ad family with |A| = κ, then A is not maximal
2. There is a mad family B ⊆ P(κ) of cardinality ≥ κ+
Proof.
(1) It is proved that there is some D ⊆ κ, D ∈
/ A such that D is a.d. with
all elements of A.
Almost disjointness
Theorem
Let κ be regular. Then:
1. If A ⊆ P(κ) is an ad family with |A| = κ, then A is not maximal
2. There is a mad family B ⊆ P(κ) of cardinality ≥ κ+
Proof.
(1) It is proved that there is some D ⊆ κ, D ∈
/ A such that D is a.d. with
all elements of A. Let A = {Aξ | ξ < κ}.
Almost disjointness
Theorem
Let κ be regular. Then:
1. If A ⊆ P(κ) is an ad family with |A| = κ, then A is not maximal
2. There is a mad family B ⊆ P(κ) of cardinality ≥ κ+
Proof.
(1) It is proved that there is some D ⊆ κ, D ∈
/ A such that D is a.d. with
{Aξ | ξ < κ}. Let
all elementsSof A. Let A =S
Bξ = Aξ \ η<ξ Aη = Aξ \ η<ξ (Aξ ∩ Aη ).
Almost disjointness
Theorem
Let κ be regular. Then:
1. If A ⊆ P(κ) is an ad family with |A| = κ, then A is not maximal
2. There is a mad family B ⊆ P(κ) of cardinality ≥ κ+
Proof.
(1) It is proved that there is some D ⊆ κ, D ∈
/ A such that D is a.d. with
{Aξ | ξ < κ}. Let
all elementsSof A. Let A =S
Bξ = Aξ \ η<ξ Aη = Aξ \ η<ξ (Aξ ∩ Aη ). So Bξ 6= ∅ by regularity of κ.
Almost disjointness
Theorem
Let κ be regular. Then:
1. If A ⊆ P(κ) is an ad family with |A| = κ, then A is not maximal
2. There is a mad family B ⊆ P(κ) of cardinality ≥ κ+
Proof.
(1) It is proved that there is some D ⊆ κ, D ∈
/ A such that D is a.d. with
{Aξ | ξ < κ}. Let
all elementsSof A. Let A =S
Bξ = Aξ \ η<ξ Aη = Aξ \ η<ξ (Aξ ∩ Aη ). So Bξ 6= ∅ by regularity of κ.
Let bξ ∈ Bξ .
Almost disjointness
Theorem
Let κ be regular. Then:
1. If A ⊆ P(κ) is an ad family with |A| = κ, then A is not maximal
2. There is a mad family B ⊆ P(κ) of cardinality ≥ κ+
Proof.
(1) It is proved that there is some D ⊆ κ, D ∈
/ A such that D is a.d. with
{Aξ | ξ < κ}. Let
all elementsSof A. Let A =S
Bξ = Aξ \ η<ξ Aη = Aξ \ η<ξ (Aξ ∩ Aη ). So Bξ 6= ∅ by regularity of κ.
Let bξ ∈ Bξ . Since the Bξ are pairwise disjoint, the bξ are all distinct.
Almost disjointness
Theorem
Let κ be regular. Then:
1. If A ⊆ P(κ) is an ad family with |A| = κ, then A is not maximal
2. There is a mad family B ⊆ P(κ) of cardinality ≥ κ+
Proof.
(1) It is proved that there is some D ⊆ κ, D ∈
/ A such that D is a.d. with
{Aξ | ξ < κ}. Let
all elementsSof A. Let A =S
Bξ = Aξ \ η<ξ Aη = Aξ \ η<ξ (Aξ ∩ Aη ). So Bξ 6= ∅ by regularity of κ.
Let bξ ∈ Bξ . Since the Bξ are pairwise disjoint, the bξ are all distinct. So
D = {bξ | ξ < κ} has cardinality κ.
Almost disjointness
Theorem
Let κ be regular. Then:
1. If A ⊆ P(κ) is an ad family with |A| = κ, then A is not maximal
2. There is a mad family B ⊆ P(κ) of cardinality ≥ κ+
Proof.
(1) It is proved that there is some D ⊆ κ, D ∈
/ A such that D is a.d. with
{Aξ | ξ < κ}. Let
all elementsSof A. Let A =S
Bξ = Aξ \ η<ξ Aη = Aξ \ η<ξ (Aξ ∩ Aη ). So Bξ 6= ∅ by regularity of κ.
Let bξ ∈ Bξ . Since the Bξ are pairwise disjoint, the bξ are all distinct. So
D = {bξ | ξ < κ} has cardinality κ. Finally, if bξ ∈ Aη then η ≥ ξ.
Almost disjointness
Theorem
Let κ be regular. Then:
1. If A ⊆ P(κ) is an ad family with |A| = κ, then A is not maximal
2. There is a mad family B ⊆ P(κ) of cardinality ≥ κ+
Proof.
(1) It is proved that there is some D ⊆ κ, D ∈
/ A such that D is a.d. with
{Aξ | ξ < κ}. Let
all elementsSof A. Let A =S
Bξ = Aξ \ η<ξ Aη = Aξ \ η<ξ (Aξ ∩ Aη ). So Bξ 6= ∅ by regularity of κ.
Let bξ ∈ Bξ . Since the Bξ are pairwise disjoint, the bξ are all distinct. So
D = {bξ | ξ < κ} has cardinality κ. Finally, if bξ ∈ Aη then η ≥ ξ. Thus
D ∩ Aη ⊆ {bξ | ξ ≤ η} has cardinality less than κ.
Almost disjointness
Theorem
Let κ be regular. Then:
1. If A ⊆ P(κ) is an ad family with |A| = κ, then A is not maximal
2. There is a mad family B ⊆ P(κ) of cardinality ≥ κ+
Proof.
(1) It is proved that there is some D ⊆ κ, D ∈
/ A such that D is a.d. with
{Aξ | ξ < κ}. Let
all elementsSof A. Let A =S
Bξ = Aξ \ η<ξ Aη = Aξ \ η<ξ (Aξ ∩ Aη ). So Bξ 6= ∅ by regularity of κ.
Let bξ ∈ Bξ . Since the Bξ are pairwise disjoint, the bξ are all distinct. So
D = {bξ | ξ < κ} has cardinality κ. Finally, if bξ ∈ Aη then η ≥ ξ. Thus
D ∩ Aη ⊆ {bξ | ξ ≤ η} has cardinality less than κ.
(2) Let A ⊆ P(κ) be an a.d. family of cardinality κ (for example a
disjoint family). By Zorn’s lemma, let B a m.a.d.f. with A ⊆ B. By part
(1), |B| > κ.
Almost disjointness
So under GCH any m.a.d. family in P(κ) has cardinality κ+ = 2κ .
Almost disjointness
So under GCH any m.a.d. family in P(κ) has cardinality κ+ = 2κ .
In general, the following holds:
Theorem
Let κ be an infinite cardinal. If 2<κ = κ,
Almost disjointness
So under GCH any m.a.d. family in P(κ) has cardinality κ+ = 2κ .
In general, the following holds:
Theorem
Let κ be an infinite cardinal. If 2<κ = κ, then there is an ad family
A ⊆ P(κ) of cardinality 2κ .
Almost disjointness
So under GCH any m.a.d. family in P(κ) has cardinality κ+ = 2κ .
In general, the following holds:
Theorem
Let κ be an infinite cardinal. If 2<κ = κ, then there is an ad family
A ⊆ P(κ) of cardinality 2κ .
This holds, for example, for κ = ω; or for κ = ω1 under CH.
Proof.
Let I = {x ⊆ κ | sup x < κ}. So, |I | = κ, as 2<κ = κ.
Almost disjointness
So under GCH any m.a.d. family in P(κ) has cardinality κ+ = 2κ .
In general, the following holds:
Theorem
Let κ be an infinite cardinal. If 2<κ = κ, then there is an ad family
A ⊆ P(κ) of cardinality 2κ .
This holds, for example, for κ = ω; or for κ = ω1 under CH.
Proof.
Let I = {x ⊆ κ | sup x < κ}. So, |I | = κ, as 2<κ = κ. For any X ⊆ κ
with |X | = κ, let AX = {X ∩ α | α < κ} be the family of initial cuts of
X.
Almost disjointness
So under GCH any m.a.d. family in P(κ) has cardinality κ+ = 2κ .
In general, the following holds:
Theorem
Let κ be an infinite cardinal. If 2<κ = κ, then there is an ad family
A ⊆ P(κ) of cardinality 2κ .
This holds, for example, for κ = ω; or for κ = ω1 under CH.
Proof.
Let I = {x ⊆ κ | sup x < κ}. So, |I | = κ, as 2<κ = κ. For any X ⊆ κ
with |X | = κ, let AX = {X ∩ α | α < κ} be the family of initial cuts of
X . So |AX | = κ.
Almost disjointness
So under GCH any m.a.d. family in P(κ) has cardinality κ+ = 2κ .
In general, the following holds:
Theorem
Let κ be an infinite cardinal. If 2<κ = κ, then there is an ad family
A ⊆ P(κ) of cardinality 2κ .
This holds, for example, for κ = ω; or for κ = ω1 under CH.
Proof.
Let I = {x ⊆ κ | sup x < κ}. So, |I | = κ, as 2<κ = κ. For any X ⊆ κ
with |X | = κ, let AX = {X ∩ α | α < κ} be the family of initial cuts of
X . So |AX | = κ. Moreover, suppose X 6= Y and let β be an element
belonging to exactly one of the two sets. Then
AX ∩ AY ⊆ {X ∩ α | α ≤ β}.
Almost disjointness
So under GCH any m.a.d. family in P(κ) has cardinality κ+ = 2κ .
In general, the following holds:
Theorem
Let κ be an infinite cardinal. If 2<κ = κ, then there is an ad family
A ⊆ P(κ) of cardinality 2κ .
This holds, for example, for κ = ω; or for κ = ω1 under CH.
Proof.
Let I = {x ⊆ κ | sup x < κ}. So, |I | = κ, as 2<κ = κ. For any X ⊆ κ
with |X | = κ, let AX = {X ∩ α | α < κ} be the family of initial cuts of
X . So |AX | = κ. Moreover, suppose X 6= Y and let β be an element
belonging to exactly one of the two sets. Then
AX ∩ AY ⊆ {X ∩ α | α ≤ β}. This means that
A = {AX | X ⊆ κ ∧ |X | = κ} is an a.d. family of subsets of I of
cardinality 2κ .
Almost disjointness
So under GCH any m.a.d. family in P(κ) has cardinality κ+ = 2κ .
In general, the following holds:
Theorem
Let κ be an infinite cardinal. If 2<κ = κ, then there is an ad family
A ⊆ P(κ) of cardinality 2κ .
This holds, for example, for κ = ω; or for κ = ω1 under CH.
Proof.
Let I = {x ⊆ κ | sup x < κ}. So, |I | = κ, as 2<κ = κ. For any X ⊆ κ
with |X | = κ, let AX = {X ∩ α | α < κ} be the family of initial cuts of
X . So |AX | = κ. Moreover, suppose X 6= Y and let β be an element
belonging to exactly one of the two sets. Then
AX ∩ AY ⊆ {X ∩ α | α ≤ β}. This means that
A = {AX | X ⊆ κ ∧ |X | = κ} is an a.d. family of subsets of I of
cardinality 2κ . Trasferring this to κ via a bijection I → κ, one gets the
assertion.
Almost disjointness: independent statements
In the theorem, the hypothesis 2<κ = κ, used to grant that |I | = κ,
which allowed to work on |I | instead of κ, cannot be dropped. For
example, the existence of an a.d. family in P(ω1 ) of cardinality 2ω1 is
independent of ZFC .
Almost disjointness: independent statements
In the theorem, the hypothesis 2<κ = κ, used to grant that |I | = κ,
which allowed to work on |I | instead of κ, cannot be dropped. For
example, the existence of an a.d. family in P(ω1 ) of cardinality 2ω1 is
independent of ZFC .
A somehow dual question:
Suppose 2κ > κ+ . Is there a m.a.d. family of cardinality κ+ ?
Almost disjointness: independent statements
In the theorem, the hypothesis 2<κ = κ, used to grant that |I | = κ,
which allowed to work on |I | instead of κ, cannot be dropped. For
example, the existence of an a.d. family in P(ω1 ) of cardinality 2ω1 is
independent of ZFC .
A somehow dual question:
Suppose 2κ > κ+ . Is there a m.a.d. family of cardinality κ+ ? This too is
independent of ZFC. This is related to Martin’s axiom.
Quasi-disjointness
Another weakening of disjointness is quasi-disjointness.
Definition
A family A of sets is quasi-disjoint, or a ∆-system, if there is a fixed set
r , called the root of the ∆-system, such that
∀a, b ∈ A (a 6= b ⇒ a ∩ b = r ).
The ∆-system lemma
Theorem
Let κ be an infinite cardinal.
The ∆-system lemma
Theorem
Let κ be an infinite cardinal. Let θ > κ be regular and such
The ∆-system lemma
Theorem
Let κ be an infinite cardinal. Let θ > κ be regular and such that
∀λ < θ λ<κ < θ.
The ∆-system lemma
Theorem
Let κ be an infinite cardinal. Let θ > κ be regular and such that
∀λ < θ λ<κ < θ. If |A| ≥ θ and ∀x ∈ A |x| < κ,
The ∆-system lemma
Theorem
Let κ be an infinite cardinal. Let θ > κ be regular and such that
∀λ < θ λ<κ < θ. If |A| ≥ θ and ∀x ∈ A |x| < κ, there there is B ⊆ A
such that |B| = θ and B is a ∆-system.
The ∆-system lemma
Theorem
Let κ be an infinite cardinal. Let θ > κ be regular and such that
∀λ < θ λ<κ < θ. If |A| ≥ θ and ∀x ∈ A |x| < κ, there there is B ⊆ A
such that |B| = θ and B is a ∆-system.
For instance: let κ = ω, θ = ω1 . Then:
The ∆-system lemma
Theorem
Let κ be an infinite cardinal. Let θ > κ be regular and such that
∀λ < θ λ<κ < θ. If |A| ≥ θ and ∀x ∈ A |x| < κ, there there is B ⊆ A
such that |B| = θ and B is a ∆-system.
For instance: let κ = ω, θ = ω1 . Then:
Corollary
If A is an uncountable family of finite sets, there is an uncountable
∆-system B ⊆ A.
The ∆-system lemma
Theorem
Let κ be an infinite cardinal. Let θ > κ be regular and such that
∀λ < θ λ<κ < θ. If |A| ≥ θ and ∀x ∈ A |x| < κ, there there is B ⊆ A
such that |B| = θ and B is a ∆-system.
For instance: let κ = ω, θ = ω1 . Then:
Corollary
If A is an uncountable family of finite sets, there is an uncountable
∆-system B ⊆ A.
Exercise. Give a direct proof of this corollary.
The ∆-system lemma
Proof.
By shrinking, one can assume |A| = θ,
The ∆-system lemma
Proof.
By shrinking, one can assume |A| = θ, so that |
S
A| ≤ θ,
The ∆-system lemma
Proof.
By shrinking, one S
can assume |A| = θ, so that |
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ).
S
A| ≤ θ, so that one
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ.
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ.
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ.
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. The subsets of α of cardinality less than κ are at most
|α|<κ < θ.
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ.
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ. For any x ∈ A1 , ξ < ρ, let x(ξ)
be the ξ-th element of x.
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ. For any x ∈ A1 , ξ < ρ, let x(ξ)
be the ξ-th element of x. Since θ is regular, there is a least ξ0 such that
{x(ξ0 ) | x ∈ A1 } is unbounded in θ.
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ. For any x ∈ A1 , ξ < ρ, let x(ξ)
be the ξ-th element of x. Since θ is regular, there is a least ξ0 such that
{x(ξ0 ) | x ∈ A1 } is unbounded in θ. Let
α0 = sup{x(η) + 1 | x ∈ A1 , η < ξ0 }.
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ. For any x ∈ A1 , ξ < ρ, let x(ξ)
be the ξ-th element of x. Since θ is regular, there is a least ξ0 such that
{x(ξ0 ) | x ∈ A1 } is unbounded in θ. Let
α0 = sup{x(η) + 1 | x ∈ A1 , η < ξ0 }. Then α0 < θ and
∀x ∈ A1 ∀η < ξ0 x(η) < α0 .
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ. For any x ∈ A1 , ξ < ρ, let x(ξ)
be the ξ-th element of x. Since θ is regular, there is a least ξ0 such that
{x(ξ0 ) | x ∈ A1 } is unbounded in θ. Let
α0 = sup{x(η) + 1 | x ∈ A1 , η < ξ0 }. Then α0 < θ and
∀x ∈ A1 ∀η < ξ0 x(η) < α0 .
Define inductively, for µ < θ, elements xµ ∈ A1 so that
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ. For any x ∈ A1 , ξ < ρ, let x(ξ)
be the ξ-th element of x. Since θ is regular, there is a least ξ0 such that
{x(ξ0 ) | x ∈ A1 } is unbounded in θ. Let
α0 = sup{x(η) + 1 | x ∈ A1 , η < ξ0 }. Then α0 < θ and
∀x ∈ A1 ∀η < ξ0 x(η) < α0 .
Define inductively, for µ < θ, elements xµ ∈ A1 so that xµ (ξ0 ) is bigger
than α0 and of all elements of all xν , for ν < µ.
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ. For any x ∈ A1 , ξ < ρ, let x(ξ)
be the ξ-th element of x. Since θ is regular, there is a least ξ0 such that
{x(ξ0 ) | x ∈ A1 } is unbounded in θ. Let
α0 = sup{x(η) + 1 | x ∈ A1 , η < ξ0 }. Then α0 < θ and
∀x ∈ A1 ∀η < ξ0 x(η) < α0 .
Define inductively, for µ < θ, elements xµ ∈ A1 so that xµ (ξ0 ) is bigger
than α0 and of all elements of all xν , for ν < µ. Let A2 be the collection
of such elements.
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ. For any x ∈ A1 , ξ < ρ, let x(ξ)
be the ξ-th element of x. Since θ is regular, there is a least ξ0 such that
{x(ξ0 ) | x ∈ A1 } is unbounded in θ. Let
α0 = sup{x(η) + 1 | x ∈ A1 , η < ξ0 }. Then α0 < θ and
∀x ∈ A1 ∀η < ξ0 x(η) < α0 .
Define inductively, for µ < θ, elements xµ ∈ A1 so that xµ (ξ0 ) is bigger
than α0 and of all elements of all xν , for ν < µ. Let A2 be the collection
of such elements. Then |A2 | = θ and ∀x, y ∈ A2 (x 6= y ⇒ x ∩ y ⊆ α0 ).
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ. For any x ∈ A1 , ξ < ρ, let x(ξ)
be the ξ-th element of x. Since θ is regular, there is a least ξ0 such that
{x(ξ0 ) | x ∈ A1 } is unbounded in θ. Let
α0 = sup{x(η) + 1 | x ∈ A1 , η < ξ0 }. Then α0 < θ and
∀x ∈ A1 ∀η < ξ0 x(η) < α0 .
Define inductively, for µ < θ, elements xµ ∈ A1 so that xµ (ξ0 ) is bigger
than α0 and of all elements of all xν , for ν < µ. Let A2 be the collection
of such elements. Then |A2 | = θ and ∀x, y ∈ A2 (x 6= y ⇒ x ∩ y ⊆ α0 ).
Each x ∩ α0 is a subset of α0 of cardinality less that κ.
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ. For any x ∈ A1 , ξ < ρ, let x(ξ)
be the ξ-th element of x. Since θ is regular, there is a least ξ0 such that
{x(ξ0 ) | x ∈ A1 } is unbounded in θ. Let
α0 = sup{x(η) + 1 | x ∈ A1 , η < ξ0 }. Then α0 < θ and
∀x ∈ A1 ∀η < ξ0 x(η) < α0 .
Define inductively, for µ < θ, elements xµ ∈ A1 so that xµ (ξ0 ) is bigger
than α0 and of all elements of all xν , for ν < µ. Let A2 be the collection
of such elements. Then |A2 | = θ and ∀x, y ∈ A2 (x 6= y ⇒ x ∩ y ⊆ α0 ).
Each x ∩ α0 is a subset of α0 of cardinality less that κ. Since there are at
most |α0 |<κ < θ of them,
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ. For any x ∈ A1 , ξ < ρ, let x(ξ)
be the ξ-th element of x. Since θ is regular, there is a least ξ0 such that
{x(ξ0 ) | x ∈ A1 } is unbounded in θ. Let
α0 = sup{x(η) + 1 | x ∈ A1 , η < ξ0 }. Then α0 < θ and
∀x ∈ A1 ∀η < ξ0 x(η) < α0 .
Define inductively, for µ < θ, elements xµ ∈ A1 so that xµ (ξ0 ) is bigger
than α0 and of all elements of all xν , for ν < µ. Let A2 be the collection
of such elements. Then |A2 | = θ and ∀x, y ∈ A2 (x 6= y ⇒ x ∩ y ⊆ α0 ).
Each x ∩ α0 is a subset of α0 of cardinality less that κ. Since there are at
most |α0 |<κ < θ of them, there are r ⊆ α0 , B ⊆ A2 such that |B| = θ
and ∀x ∈ B x ∩ α0 = r ,
The ∆-system lemma
Proof.
S
By shrinking, one S
can assume |A| = θ, so that | A| ≤ θ, so that one
can also assume | A| ⊆ θ, i.e. A ⊆ P(θ). Then each x ∈ A has some
order type < κ. Since θ is regular and θ > κ, there is ρ < κ such that
A1 = {x ∈ A | x has order type ρ} has cardinality θ. It is enough to
extract the ∆-system from A1 .
Fix α < θ. TheSsubsets of α of cardinality less than κ are at most
|α|<κ < θ. So A1 is unbounded in θ. For any x ∈ A1 , ξ < ρ, let x(ξ)
be the ξ-th element of x. Since θ is regular, there is a least ξ0 such that
{x(ξ0 ) | x ∈ A1 } is unbounded in θ. Let
α0 = sup{x(η) + 1 | x ∈ A1 , η < ξ0 }. Then α0 < θ and
∀x ∈ A1 ∀η < ξ0 x(η) < α0 .
Define inductively, for µ < θ, elements xµ ∈ A1 so that xµ (ξ0 ) is bigger
than α0 and of all elements of all xν , for ν < µ. Let A2 be the collection
of such elements. Then |A2 | = θ and ∀x, y ∈ A2 (x 6= y ⇒ x ∩ y ⊆ α0 ).
Each x ∩ α0 is a subset of α0 of cardinality less that κ. Since there are at
most |α0 |<κ < θ of them, there are r ⊆ α0 , B ⊆ A2 such that |B| = θ
and ∀x ∈ B x ∩ α0 = r , which implies ∀x, y ∈ B(x 6= y ⇒ x ∩ y = r ).
Some topology
Recall some basic definitions and facts.
I
A topological space is a (non-empty) set X together with a family
T ⊆ P(X ) such that
I
I
I
∅, X ∈ T
T is closed under arbitrary unions
T is closed under finite intersections
Elements of T are called open subset of X
Some topology
Recall some basic definitions and facts.
I
A topological space is a (non-empty) set X together with a family
T ⊆ P(X ) such that
I
I
I
∅, X ∈ T
T is closed under arbitrary unions
T is closed under finite intersections
Elements of T are called open subset of X
I
The interior of a set A ⊆ X is the biggest open set contained in A
Some topology
Recall some basic definitions and facts.
I
A topological space is a (non-empty) set X together with a family
T ⊆ P(X ) such that
I
I
I
∅, X ∈ T
T is closed under arbitrary unions
T is closed under finite intersections
Elements of T are called open subset of X
I
The interior of a set A ⊆ X is the biggest open set contained in A
I
A closed set is the complement of an open set
Some topology
Recall some basic definitions and facts.
I
A topological space is a (non-empty) set X together with a family
T ⊆ P(X ) such that
I
I
I
∅, X ∈ T
T is closed under arbitrary unions
T is closed under finite intersections
Elements of T are called open subset of X
I
The interior of a set A ⊆ X is the biggest open set contained in A
I
A closed set is the complement of an open set
The closure Ā of a set A ⊆ X is the smallest closed set containing A
I
Some topology
Recall some basic definitions and facts.
I
A topological space is a (non-empty) set X together with a family
T ⊆ P(X ) such that
I
I
I
∅, X ∈ T
T is closed under arbitrary unions
T is closed under finite intersections
Elements of T are called open subset of X
I
The interior of a set A ⊆ X is the biggest open set contained in A
I
A closed set is the complement of an open set
The closure Ā of a set A ⊆ X is the smallest closed set containing A
I
I
A subset D of space X is dense if D intersects all non-empty open
subsets of X
Some topology
Recall some basic definitions and facts.
I
A topological space is a (non-empty) set X together with a family
T ⊆ P(X ) such that
I
I
I
∅, X ∈ T
T is closed under arbitrary unions
T is closed under finite intersections
Elements of T are called open subset of X
I
The interior of a set A ⊆ X is the biggest open set contained in A
I
A closed set is the complement of an open set
The closure Ā of a set A ⊆ X is the smallest closed set containing A
I
I
A subset D of space X is dense if D intersects all non-empty open
subsets of X
I
Space X is separable if it has a countable dense subset
Some topology
Recall some basic definitions and facts.
I
A topological space is a (non-empty) set X together with a family
T ⊆ P(X ) such that
I
I
I
∅, X ∈ T
T is closed under arbitrary unions
T is closed under finite intersections
Elements of T are called open subset of X
I
The interior of a set A ⊆ X is the biggest open set contained in A
I
A closed set is the complement of an open set
The closure Ā of a set A ⊆ X is the smallest closed set containing A
I
I
A subset D of space X is dense if D intersects all non-empty open
subsets of X
I
Space X is separable if it has a countable dense subset
I
A set A ⊆ X is nowhere dense if its closure has empy interior;
Some topology
Recall some basic definitions and facts.
I
A topological space is a (non-empty) set X together with a family
T ⊆ P(X ) such that
I
I
I
∅, X ∈ T
T is closed under arbitrary unions
T is closed under finite intersections
Elements of T are called open subset of X
I
The interior of a set A ⊆ X is the biggest open set contained in A
I
A closed set is the complement of an open set
The closure Ā of a set A ⊆ X is the smallest closed set containing A
I
I
A subset D of space X is dense if D intersects all non-empty open
subsets of X
I
Space X is separable if it has a countable dense subset
I
A set A ⊆ X is nowhere dense if its closure has empy interior; it is
meagre if it is a countable union of nowhere dense sets
Some topology
I
If A ⊆ P(X ), there is a smallest topology T on X containing A.
Then A is a sub-basis for T .
Some topology
I
If A ⊆ P(X ), there is a smallest topology T on X containing A.
Then A is a sub-basis for T .
I
If A is closed under finite intersections, then A is a basis for T .
Every open set is a union of elements of A.
Some topology
I
If A ⊆ P(X ), there is a smallest topology T on X containing A.
Then A is a sub-basis for T .
I
If A is closed under finite intersections, then A is a basis for T .
Every open set is a union of elements of A.
I
If Xi , for i ∈ I are topological spaces, the product topology on
Y
Xi
i∈I
is
Some topology
I
If A ⊆ P(X ), there is a smallest topology T on X containing A.
Then A is a sub-basis for T .
I
If A is closed under finite intersections, then A is a basis for T .
Every open set is a union of elements of A.
I
If Xi , for i ∈ I are topological spaces, the product topology on
Y
Xi
i∈I
is the topology whose basis consists of sets of the form
Y
Y
( Uj ) ×
Xi
j∈r
i∈I \r
for r a finite subset of I and Uj open in Xj .
An application to topology
Definition
A topological space has the countable chain condition (c.c.c.) if there is
no uncountable family of pairwise disjoint open subsets.
An application to topology
Definition
A topological space has the countable chain condition (c.c.c.) if there is
no uncountable family of pairwise disjoint open subsets.
Example. A separable space is c.c.c.
An application to topology
Definition
A topological space has the countable chain condition (c.c.c.) if there is
no uncountable family of pairwise disjoint open subsets.
Example. A separable space is c.c.c.
An example of a c.c.c. non-separable space will be provided in a moment.
An application to topology
Definition
A topological space has the countable chain condition (c.c.c.) if there is
no uncountable family of pairwise disjoint open subsets.
Example. A separable space is c.c.c.
An example of a c.c.c. non-separable space will be provided in a moment.
Exercise The product of at most 2ℵ0 separable spaces is separable.
The product of more than 2ℵ0 separable spaces need not be separable.
Example: κ 2, where κ > 2ℵ0 and 2 has the discrete topology.
An application to topology
Definition
A topological space has the countable chain condition (c.c.c.) if there is
no uncountable family of pairwise disjoint open subsets.
Example. A separable space is c.c.c.
An example of a c.c.c. non-separable space will be provided in a moment.
Exercise The product of at most 2ℵ0 separable spaces is separable.
The product of more than 2ℵ0 separable spaces need not be separable.
Example: κ 2, where κ > 2ℵ0 and 2 has the discrete topology.
On the other hand, the fact that the product of two c.c.c. spaces is c.c.c.
is independent of ZFC :
An application to topology
Definition
A topological space has the countable chain condition (c.c.c.) if there is
no uncountable family of pairwise disjoint open subsets.
Example. A separable space is c.c.c.
An example of a c.c.c. non-separable space will be provided in a moment.
Exercise The product of at most 2ℵ0 separable spaces is separable.
The product of more than 2ℵ0 separable spaces need not be separable.
Example: κ 2, where κ > 2ℵ0 and 2 has the discrete topology.
On the other hand, the fact that the product of two c.c.c. spaces is c.c.c.
is independent of ZFC : in fact, it holds under MA+¬CH; it fails under
CH or the negation of Suslin’s Hypothesis.
An application to topology
However, the following holds
Theorem
Let Xi , for i Q
∈ I be topological spaces.
Q Suppose that for all finite r ⊆ I
the product i∈r Xi is c.c.c. Then i∈I Xi is c.c.c.
An application to topology
However, the following holds
Theorem
Let Xi , for i Q
∈ I be topological spaces.
Q Suppose that for all finite r ⊆ I
the product i∈r Xi is c.c.c. Then i∈I Xi is c.c.c.
Proof.
Let Uα , for α < ω1 be pairwise disjoint non-empty sets.
An application to topology
However, the following holds
Theorem
Let Xi , for i Q
∈ I be topological spaces.
Q Suppose that for all finite r ⊆ I
the product i∈r Xi is c.c.c. Then i∈I Xi is c.c.c.
Proof.
Let Uα , for α < ω1 be pairwise disjoint
non-empty
Q
Q sets. Each Uα
contains a basic open set Vα = ( i∈aα Viα ) × i ∈a
/ α Xi for some finite
set aα ⊆ I .
An application to topology
However, the following holds
Theorem
Let Xi , for i Q
∈ I be topological spaces.
Q Suppose that for all finite r ⊆ I
the product i∈r Xi is c.c.c. Then i∈I Xi is c.c.c.
Proof.
Let Uα , for α < ω1 be pairwise disjoint
non-empty
Q
Q sets. Each Uα
contains a basic open set Vα = ( i∈aα Viα ) × i ∈a
/ α Xi for some finite
set aα ⊆ I . So {aα | α < ω1 } is an uncountable family of finite sets,
which by the ∆-system lemma contains an uncountable ∆-system
{aαβ | β < ω1 }with some root r .
An application to topology
However, the following holds
Theorem
Let Xi , for i Q
∈ I be topological spaces.
Q Suppose that for all finite r ⊆ I
the product i∈r Xi is c.c.c. Then i∈I Xi is c.c.c.
Proof.
Let Uα , for α < ω1 be pairwise disjoint
non-empty
Q
Q sets. Each Uα
contains a basic open set Vα = ( i∈aα Viα ) × i ∈a
/ α Xi for some finite
set aα ⊆ I . So {aα | α < ω1 } is an uncountable family of finite sets,
which by the ∆-system lemma contains an uncountable ∆-system
{aαβ | β < ω1 }with some root r . Since aα ∩ aα0 = ∅ implies
Vα ∩ Vα0 6= ∅,
An application to topology
However, the following holds
Theorem
Let Xi , for i Q
∈ I be topological spaces.
Q Suppose that for all finite r ⊆ I
the product i∈r Xi is c.c.c. Then i∈I Xi is c.c.c.
Proof.
Let Uα , for α < ω1 be pairwise disjoint
non-empty
Q
Q sets. Each Uα
contains a basic open set Vα = ( i∈aα Viα ) × i ∈a
/ α Xi for some finite
set aα ⊆ I . So {aα | α < ω1 } is an uncountable family of finite sets,
which by the ∆-system lemma contains an uncountable ∆-system
{aαβ | β < ω1 }with some root r . Since aα ∩ aα0 = ∅ implies
Vα ∩ Vα0 6= ∅, it follows r 6= ∅.
An application to topology
However, the following holds
Theorem
Let Xi , for i Q
∈ I be topological spaces.
Q Suppose that for all finite r ⊆ I
the product i∈r Xi is c.c.c. Then i∈I Xi is c.c.c.
Proof.
Let Uα , for α < ω1 be pairwise disjoint
non-empty
Q
Q sets. Each Uα
contains a basic open set Vα = ( i∈aα Viα ) × i ∈a
/ α Xi for some finite
set aα ⊆ I . So {aα | α < ω1 } is an uncountable family of finite sets,
which by the ∆-system lemma contains an uncountable ∆-system
{aαβ | β < ω1 }with some root r . Since aα ∩ aα0 = ∅ implies
Q
αβ
Vα ∩ Vα0 6= ∅, it follows r 6= ∅. But then { i∈r
QVi | β < ω1 } would be
an uncountable disjoint family of open sets in i∈r Xi .
An application to topology
Some consequences:
I
If the product of any two c.c.c. spaces is c.c.c., then any product of
c.c.c. spaces is c.c.c.
An application to topology
Some consequences:
I
If the product of any two c.c.c. spaces is c.c.c., then any product of
c.c.c. spaces is c.c.c.
I
Whenever κ > 2ℵ0 , then κ 2 is an example of a c.c.c. non-separable
space
σ-algebras
Let X 6= ∅.
Definition
A σ-algebra on X is a family A ⊆ P(X ) such that:
I
∅, X ∈ A
I
if A ∈ A, then X \ A ∈ A
S
if ∀n ∈ ω An ∈ A then n∈ω An ∈ A.
I
σ-algebras
Let X 6= ∅.
Definition
A σ-algebra on X is a family A ⊆ P(X ) such that:
I
∅, X ∈ A
I
if A ∈ A, then X \ A ∈ A
S
if ∀n ∈ ω An ∈ A then n∈ω An ∈ A.
I
Given any B ⊆ P(X ) there is a smallest σ-algebra containing B.
σ-algebras
Let X 6= ∅.
Definition
A σ-algebra on X is a family A ⊆ P(X ) such that:
I
∅, X ∈ A
I
if A ∈ A, then X \ A ∈ A
S
if ∀n ∈ ω An ∈ A then n∈ω An ∈ A.
I
Given any B ⊆ P(X ) there is a smallest σ-algebra containing B.
Example. In a topological space, the σ-algebra generated by the open
sets is the σ-algebra of Borel sets.
Measures
Definition
A measure is a function µ : A → [0, +∞] defined a some σ-algebra of
subsets of some set X such that
I µ(∅) = 0
S
P∞
I µ(
n=0 µ(An ) if the An are pairwise disjoint
n∈ω An ) =
Measures
Definition
A measure is a function µ : A → [0, +∞] defined a some σ-algebra of
subsets of some set X such that
I µ(∅) = 0
S
P∞
I µ(
n=0 µ(An ) if the An are pairwise disjoint
n∈ω An ) =
If B ⊆ A ∈ A, µ(A) = 0, then B is null.
Measures
Definition
A measure is a function µ : A → [0, +∞] defined a some σ-algebra of
subsets of some set X such that
I µ(∅) = 0
S
P∞
I µ(
n=0 µ(An ) if the An are pairwise disjoint
n∈ω An ) =
If B ⊆ A ∈ A, µ(A) = 0, then B is null. If NULL is the collection of null
sets, then the σ-algebra M generated by A ∪ NULL consists of sets of
the form A ∪ N, where A ∈ A, N ∈ NULL.
Measures
Definition
A measure is a function µ : A → [0, +∞] defined a some σ-algebra of
subsets of some set X such that
I µ(∅) = 0
S
P∞
I µ(
n=0 µ(An ) if the An are pairwise disjoint
n∈ω An ) =
If B ⊆ A ∈ A, µ(A) = 0, then B is null. If NULL is the collection of null
sets, then the σ-algebra M generated by A ∪ NULL consists of sets of
the form A ∪ N, where A ∈ A, N ∈ NULL. Defining µ̃ : M → [0, +∞] as
µ̃(A ∪ N) = µ(A).
Measures
Definition
A measure is a function µ : A → [0, +∞] defined a some σ-algebra of
subsets of some set X such that
I µ(∅) = 0
S
P∞
I µ(
n=0 µ(An ) if the An are pairwise disjoint
n∈ω An ) =
If B ⊆ A ∈ A, µ(A) = 0, then B is null. If NULL is the collection of null
sets, then the σ-algebra M generated by A ∪ NULL consists of sets of
the form A ∪ N, where A ∈ A, N ∈ NULL. Defining µ̃ : M → [0, +∞] as
µ̃(A ∪ N) = µ(A).
M is the σ-algebra of µ-measurable sets.
Measures
Definition
A measure is a function µ : A → [0, +∞] defined a some σ-algebra of
subsets of some set X such that
I µ(∅) = 0
S
P∞
I µ(
n=0 µ(An ) if the An are pairwise disjoint
n∈ω An ) =
If B ⊆ A ∈ A, µ(A) = 0, then B is null. If NULL is the collection of null
sets, then the σ-algebra M generated by A ∪ NULL consists of sets of
the form A ∪ N, where A ∈ A, N ∈ NULL. Defining µ̃ : M → [0, +∞] as
µ̃(A ∪ N) = µ(A).
M is the σ-algebra of µ-measurable sets.
µ̃ is the completion of µ. It is also denoted simply by µ.
Measures
Definition
A measure is a function µ : A → [0, +∞] defined a some σ-algebra of
subsets of some set X such that
I µ(∅) = 0
S
P∞
I µ(
n=0 µ(An ) if the An are pairwise disjoint
n∈ω An ) =
If B ⊆ A ∈ A, µ(A) = 0, then B is null. If NULL is the collection of null
sets, then the σ-algebra M generated by A ∪ NULL consists of sets of
the form A ∪ N, where A ∈ A, N ∈ NULL. Defining µ̃ : M → [0, +∞] as
µ̃(A ∪ N) = µ(A).
M is the σ-algebra of µ-measurable sets.
µ̃ is the completion of µ. It is also denoted simply by µ.
Example. Lebesgue measure µ is the measure on the measurable subsets
of R such that µ([a, b]) = b − a.
Some questions
Let ℵ0 ≤ κ < 2ℵ0 .
Questions.
Some questions
Let ℵ0 ≤ κ < 2ℵ0 .
Questions.
I
Is 2κ = 2ℵ0 ?
Some questions
Let ℵ0 ≤ κ < 2ℵ0 .
Questions.
I
Is 2κ = 2ℵ0 ?
I
Does an a.d. family A ⊆ P(ω) of cardinality κ fail to be maximal?
Some questions
Let ℵ0 ≤ κ < 2ℵ0 .
Questions.
I
Is 2κ = 2ℵ0 ?
I
Does an a.d. family A ⊆ P(ω) of cardinality κ fail to be maximal?
I
Is the union of κ null subsets of R null?
Some questions
Let ℵ0 ≤ κ < 2ℵ0 .
Questions.
I
Is 2κ = 2ℵ0 ?
I
Does an a.d. family A ⊆ P(ω) of cardinality κ fail to be maximal?
I
Is the union of κ null subsets of R null?
I
Is the union of κ meagre subsets of R meagre?
Some questions
Let ℵ0 ≤ κ < 2ℵ0 .
Questions.
I
Is 2κ = 2ℵ0 ?
I
Does an a.d. family A ⊆ P(ω) of cardinality κ fail to be maximal?
I
Is the union of κ null subsets of R null?
I
Is the union of κ meagre subsets of R meagre?
All answer are positive for κ = ω. So they make sense under ¬CH.
Some questions
Let ℵ0 ≤ κ < 2ℵ0 .
Questions.
I
Is 2κ = 2ℵ0 ?
I
Does an a.d. family A ⊆ P(ω) of cardinality κ fail to be maximal?
I
Is the union of κ null subsets of R null?
I
Is the union of κ meagre subsets of R meagre?
All answer are positive for κ = ω. So they make sense under ¬CH. They
are all independent from ZFC + ¬CH.
Martin’s axiom
Martin’s axiom (MA) ia a principle consistent with ¬CH and implying a
positive answer to all these questions. It has a number of consequences,
mainly in combinatorics and set-theoretic topology, but also in algebra
and analysis.
Martin’s axiom (topological form). No c.c.c. compact space is the
union of less than continuum many closed nowhere dense sets.
This formulation of MA is neat, but it is hard to apply. There is a
combinatorial formulation of MA, which requires some preliminaries and
terminology (which is sometimes in conflict with the terminology used
elsewhere).
Partial orders
Definition
I
A partial order (or pre-order, or quasi-order), abbreviated p.o.,
(P, ≤) is a reflexive and transitive relation on a non-empty set:
Partial orders
Definition
I
A partial order (or pre-order, or quasi-order), abbreviated p.o.,
(P, ≤) is a reflexive and transitive relation on a non-empty set:
I
∀p ∈ P p ≤ p
Partial orders
Definition
I
A partial order (or pre-order, or quasi-order), abbreviated p.o.,
(P, ≤) is a reflexive and transitive relation on a non-empty set:
I
I
∀p ∈ P p ≤ p
∀p, q, r ∈ P (p ≤ q ∧ q ≤ r ⇒ p ≤ r )
Partial orders
Definition
I
A partial order (or pre-order, or quasi-order), abbreviated p.o.,
(P, ≤) is a reflexive and transitive relation on a non-empty set:
I
I
∀p ∈ P p ≤ p
∀p, q, r ∈ P (p ≤ q ∧ q ≤ r ⇒ p ≤ r )
Elements of P are called conditions. When p ≤ q one says that p
extends q.
Partial orders
Definition
I
A partial order (or pre-order, or quasi-order), abbreviated p.o.,
(P, ≤) is a reflexive and transitive relation on a non-empty set:
I
I
I
∀p ∈ P p ≤ p
∀p, q, r ∈ P (p ≤ q ∧ q ≤ r ⇒ p ≤ r )
Elements of P are called conditions. When p ≤ q one says that p
extends q.
If moreover relation ≤ is anti-symmetric, i.e.
∀p, q ∈ P (p ≤ q ∧ q ≤ p ⇒ p = q),
then it is sometimes called a partial order in the strict sense; some
other times, just a partial order.
Partial orders
Definition
I
A partial order (or pre-order, or quasi-order), abbreviated p.o.,
(P, ≤) is a reflexive and transitive relation on a non-empty set:
I
I
I
∀p ∈ P p ≤ p
∀p, q, r ∈ P (p ≤ q ∧ q ≤ r ⇒ p ≤ r )
Elements of P are called conditions. When p ≤ q one says that p
extends q.
If moreover relation ≤ is anti-symmetric, i.e.
∀p, q ∈ P (p ≤ q ∧ q ≤ p ⇒ p = q),
then it is sometimes called a partial order in the strict sense; some
other times, just a partial order.
In this case, let p < q ⇔ p ≤ q ∧ p 6= q.
Partial orders
Definition
I
A chain in a p.o. P is a totally (pre-)ordered subset, i.e. a subset
C ⊆ P such that ∀p, q ∈ C (p ≤ q ∨ q ≤ p).
Partial orders
Definition
I
A chain in a p.o. P is a totally (pre-)ordered subset, i.e. a subset
C ⊆ P such that ∀p, q ∈ C (p ≤ q ∨ q ≤ p).
I
Elements p, q ∈ P are compatible if they have a common extension,
i.e. ∃r ∈ P (r ≤ p ∧ r ≤ q).
Partial orders
Definition
I
A chain in a p.o. P is a totally (pre-)ordered subset, i.e. a subset
C ⊆ P such that ∀p, q ∈ C (p ≤ q ∨ q ≤ p).
I
Elements p, q ∈ P are compatible if they have a common extension,
i.e. ∃r ∈ P (r ≤ p ∧ r ≤ q). Otherwise they are incompatible: p⊥q.
Partial orders
Definition
I
A chain in a p.o. P is a totally (pre-)ordered subset, i.e. a subset
C ⊆ P such that ∀p, q ∈ C (p ≤ q ∨ q ≤ p).
I
Elements p, q ∈ P are compatible if they have a common extension,
i.e. ∃r ∈ P (r ≤ p ∧ r ≤ q). Otherwise they are incompatible: p⊥q.
I
An antichain A in P is a set of pairwise incompatible elements, i.e.
∀p, q ∈ A (p 6= q ⇒ p⊥q).
Partial orders
Definition
I
A chain in a p.o. P is a totally (pre-)ordered subset, i.e. a subset
C ⊆ P such that ∀p, q ∈ C (p ≤ q ∨ q ≤ p).
I
Elements p, q ∈ P are compatible if they have a common extension,
i.e. ∃r ∈ P (r ≤ p ∧ r ≤ q). Otherwise they are incompatible: p⊥q.
I
An antichain A in P is a set of pairwise incompatible elements, i.e.
∀p, q ∈ A (p 6= q ⇒ p⊥q).
I
P has the countable chain condition (c.c.c.) if every antichain is
countable.
Partial orders
Examples.
1. The discrete p.o.: ≤= P × P, i.e. ∀p, q p ≤ q.
Partial orders
Examples.
1. The discrete p.o.: ≤= P × P, i.e. ∀p, q p ≤ q.
2. A linear order. Then every subset of P is a chain and antichains
have at most 1 element. So P has the c.c.c.
Partial orders
Examples.
1. The discrete p.o.: ≤= P × P, i.e. ∀p, q p ≤ q.
2. A linear order. Then every subset of P is a chain and antichains
have at most 1 element. So P has the c.c.c.
3. Let X be a non-empty set and P = P(X ) \ {∅}, with
p ≤ q ⇔ p ⊆ q.
Partial orders
Examples.
1. The discrete p.o.: ≤= P × P, i.e. ∀p, q p ≤ q.
2. A linear order. Then every subset of P is a chain and antichains
have at most 1 element. So P has the c.c.c.
3. Let X be a non-empty set and P = P(X ) \ {∅}, with
p ≤ q ⇔ p ⊆ q. Then p⊥q ⇔ p ∩ q = ∅.
Partial orders
Examples.
1. The discrete p.o.: ≤= P × P, i.e. ∀p, q p ≤ q.
2. A linear order. Then every subset of P is a chain and antichains
have at most 1 element. So P has the c.c.c.
3. Let X be a non-empty set and P = P(X ) \ {∅}, with
p ≤ q ⇔ p ⊆ q. Then p⊥q ⇔ p ∩ q = ∅. So A ⊆ P is an antichain
iff the elements of A are pairwise disjoint, consequently P is c.c.c.
iff |X | ≤ ℵ0 .
Partial orders
Examples.
1. The discrete p.o.: ≤= P × P, i.e. ∀p, q p ≤ q.
2. A linear order. Then every subset of P is a chain and antichains
have at most 1 element. So P has the c.c.c.
3. Let X be a non-empty set and P = P(X ) \ {∅}, with
p ≤ q ⇔ p ⊆ q. Then p⊥q ⇔ p ∩ q = ∅. So A ⊆ P is an antichain
iff the elements of A are pairwise disjoint, consequently P is c.c.c.
iff |X | ≤ ℵ0 .
4. Let X be a topological space, P = {p ⊆ X | p is open ∧ p 6= ∅}, and
p ≤ q ⇔ p ⊆ q.
Partial orders
Examples.
1. The discrete p.o.: ≤= P × P, i.e. ∀p, q p ≤ q.
2. A linear order. Then every subset of P is a chain and antichains
have at most 1 element. So P has the c.c.c.
3. Let X be a non-empty set and P = P(X ) \ {∅}, with
p ≤ q ⇔ p ⊆ q. Then p⊥q ⇔ p ∩ q = ∅. So A ⊆ P is an antichain
iff the elements of A are pairwise disjoint, consequently P is c.c.c.
iff |X | ≤ ℵ0 .
4. Let X be a topological space, P = {p ⊆ X | p is open ∧ p 6= ∅}, and
p ≤ q ⇔ p ⊆ q. Again, p⊥q ⇔ p ∩ q = ∅, so P is c.c.c. if X is
c.c.c. as a topological space.
5. Let B be a Boolean algebra, P = B \ {0B } with the same order as in
B.
Partial orders
Examples.
1. The discrete p.o.: ≤= P × P, i.e. ∀p, q p ≤ q.
2. A linear order. Then every subset of P is a chain and antichains
have at most 1 element. So P has the c.c.c.
3. Let X be a non-empty set and P = P(X ) \ {∅}, with
p ≤ q ⇔ p ⊆ q. Then p⊥q ⇔ p ∩ q = ∅. So A ⊆ P is an antichain
iff the elements of A are pairwise disjoint, consequently P is c.c.c.
iff |X | ≤ ℵ0 .
4. Let X be a topological space, P = {p ⊆ X | p is open ∧ p 6= ∅}, and
p ≤ q ⇔ p ⊆ q. Again, p⊥q ⇔ p ∩ q = ∅, so P is c.c.c. if X is
c.c.c. as a topological space.
5. Let B be a Boolean algebra, P = B \ {0B } with the same order as in
B. Then p⊥q iff p ∧ q = 0.
Partial orders
Exercises.
1. Show that a topological space X is c.c.c. if and only if there is no
sequence (Uα )α<ω1 of open sets such that α < β ⇒ Uα ⊂ Uβ .
2. A Boolean algebra is complete if every non-empty subset has a
supremum. Show that a complete Boolean algebra B is c.c.c. if it
contains no sequence (bα )α<ω1 such that α < β ⇒ bα < bβ .
Martin’s axiom: a combinatorial formulation
Definition
Let (P, ≤) be a p.o.
Martin’s axiom: a combinatorial formulation
Definition
Let (P, ≤) be a p.o.
I
A subset D ⊆ P is dense in P if ∀p ∈ P ∃q ≤ p q ∈ D
Martin’s axiom: a combinatorial formulation
Definition
Let (P, ≤) be a p.o.
I
I
A subset D ⊆ P is dense in P if ∀p ∈ P ∃q ≤ p q ∈ D
A subset G ⊆ P is a filter in P if:
Martin’s axiom: a combinatorial formulation
Definition
Let (P, ≤) be a p.o.
I
I
A subset D ⊆ P is dense in P if ∀p ∈ P ∃q ≤ p q ∈ D
A subset G ⊆ P is a filter in P if:
1. ∀p, q ∈ G ∃r ∈ G (r ≤ p ∧ r ≤ q)
Martin’s axiom: a combinatorial formulation
Definition
Let (P, ≤) be a p.o.
I
I
A subset D ⊆ P is dense in P if ∀p ∈ P ∃q ≤ p q ∈ D
A subset G ⊆ P is a filter in P if:
1. ∀p, q ∈ G ∃r ∈ G (r ≤ p ∧ r ≤ q) (in particular, any two elements of
a filter are compatible)
Martin’s axiom: a combinatorial formulation
Definition
Let (P, ≤) be a p.o.
I
I
A subset D ⊆ P is dense in P if ∀p ∈ P ∃q ≤ p q ∈ D
A subset G ⊆ P is a filter in P if:
1. ∀p, q ∈ G ∃r ∈ G (r ≤ p ∧ r ≤ q) (in particular, any two elements of
a filter are compatible)
2. ∀p ∈ G ∀q ∈ P (p ≤ q ⇒ q ∈ G )
Martin’s axiom: a combinatorial formulation
Definition
Let (P, ≤) be a p.o.
I
I
A subset D ⊆ P is dense in P if ∀p ∈ P ∃q ≤ p q ∈ D
A subset G ⊆ P is a filter in P if:
1. ∀p, q ∈ G ∃r ∈ G (r ≤ p ∧ r ≤ q) (in particular, any two elements of
a filter are compatible)
2. ∀p ∈ G ∀q ∈ P (p ≤ q ⇒ q ∈ G )
Definition
Let κ be a cardinal. Then MA(κ) is the statement:
Martin’s axiom: a combinatorial formulation
Definition
Let (P, ≤) be a p.o.
I
I
A subset D ⊆ P is dense in P if ∀p ∈ P ∃q ≤ p q ∈ D
A subset G ⊆ P is a filter in P if:
1. ∀p, q ∈ G ∃r ∈ G (r ≤ p ∧ r ≤ q) (in particular, any two elements of
a filter are compatible)
2. ∀p ∈ G ∀q ∈ P (p ≤ q ⇒ q ∈ G )
Definition
Let κ be a cardinal. Then MA(κ) is the statement:
• If (P, ≤) is a c.c.c. p.o.
Martin’s axiom: a combinatorial formulation
Definition
Let (P, ≤) be a p.o.
I
I
A subset D ⊆ P is dense in P if ∀p ∈ P ∃q ≤ p q ∈ D
A subset G ⊆ P is a filter in P if:
1. ∀p, q ∈ G ∃r ∈ G (r ≤ p ∧ r ≤ q) (in particular, any two elements of
a filter are compatible)
2. ∀p ∈ G ∀q ∈ P (p ≤ q ⇒ q ∈ G )
Definition
Let κ be a cardinal. Then MA(κ) is the statement:
• If (P, ≤) is a c.c.c. p.o. and D is a family of dense subsets of P with
|D| ≤ κ,
Martin’s axiom: a combinatorial formulation
Definition
Let (P, ≤) be a p.o.
I
I
A subset D ⊆ P is dense in P if ∀p ∈ P ∃q ≤ p q ∈ D
A subset G ⊆ P is a filter in P if:
1. ∀p, q ∈ G ∃r ∈ G (r ≤ p ∧ r ≤ q) (in particular, any two elements of
a filter are compatible)
2. ∀p ∈ G ∀q ∈ P (p ≤ q ⇒ q ∈ G )
Definition
Let κ be a cardinal. Then MA(κ) is the statement:
• If (P, ≤) is a c.c.c. p.o. and D is a family of dense subsets of P with
|D| ≤ κ, then there is a filter G in P such that ∀D ∈ D G ∩ D 6= ∅
Martin’s axiom: a combinatorial formulation
Definition
Let (P, ≤) be a p.o.
I
I
A subset D ⊆ P is dense in P if ∀p ∈ P ∃q ≤ p q ∈ D
A subset G ⊆ P is a filter in P if:
1. ∀p, q ∈ G ∃r ∈ G (r ≤ p ∧ r ≤ q) (in particular, any two elements of
a filter are compatible)
2. ∀p ∈ G ∀q ∈ P (p ≤ q ⇒ q ∈ G )
Definition
Let κ be a cardinal. Then MA(κ) is the statement:
• If (P, ≤) is a c.c.c. p.o. and D is a family of dense subsets of P with
|D| ≤ κ, then there is a filter G in P such that ∀D ∈ D G ∩ D 6= ∅
Martin’s axiom MA is the statement:
• ∀κ < 2ℵ0 MA(κ)
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}.
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}. Let
p≤q⇔q⊆p
i.e., p is stronger than q iff p extends q as a function.
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}. Let
p≤q⇔q⊆p
i.e., p is stronger than q iff p extends q as a function.
Conditions p, q are compatible iff they agree on domp ∩ domq. In this
case p ∪ q is a common extension.
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}. Let
p≤q⇔q⊆p
i.e., p is stronger than q iff p extends q as a function.
Conditions p, q are compatible iff they agree on domp ∩ domq. In this
case p ∪ q is a common extension.
Since |P| = ℵ0 , the p.o. is c.c.c.
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}. Let
p≤q⇔q⊆p
i.e., p is stronger than q iff p extends q as a function.
Conditions p, q are compatible iff they agree on domp ∩ domq. In this
case p ∪ q is a common extension.
Since |P| = ℵ0 , the p.o. is c.c.c.
S
Given a filter G , since any two of its elements are compatible, fG = G
is a function: fG : domfG ⊆ ω → 2.
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}. Let
p≤q⇔q⊆p
i.e., p is stronger than q iff p extends q as a function.
Conditions p, q are compatible iff they agree on domp ∩ domq. In this
case p ∪ q is a common extension.
Since |P| = ℵ0 , the p.o. is c.c.c.
S
Given a filter G , since any two of its elements are compatible, fG = G
is a function: fG : domfG ⊆ ω → 2. Any p ∈ G fixes the values of fG on
domp.
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}. Let
p≤q⇔q⊆p
i.e., p is stronger than q iff p extends q as a function.
Conditions p, q are compatible iff they agree on domp ∩ domq. In this
case p ∪ q is a common extension.
Since |P| = ℵ0 , the p.o. is c.c.c.
S
Given a filter G , since any two of its elements are compatible, fG = G
is a function: fG : domfG ⊆ ω → 2. Any p ∈ G fixes the values of fG on
domp.
Let Dn = {p ∈ P | n ∈ domp}.
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}. Let
p≤q⇔q⊆p
i.e., p is stronger than q iff p extends q as a function.
Conditions p, q are compatible iff they agree on domp ∩ domq. In this
case p ∪ q is a common extension.
Since |P| = ℵ0 , the p.o. is c.c.c.
S
Given a filter G , since any two of its elements are compatible, fG = G
is a function: fG : domfG ⊆ ω → 2. Any p ∈ G fixes the values of fG on
domp.
Let Dn = {p ∈ P | n ∈ domp}. Then Dn is dense.
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}. Let
p≤q⇔q⊆p
i.e., p is stronger than q iff p extends q as a function.
Conditions p, q are compatible iff they agree on domp ∩ domq. In this
case p ∪ q is a common extension.
Since |P| = ℵ0 , the p.o. is c.c.c.
S
Given a filter G , since any two of its elements are compatible, fG = G
is a function: fG : domfG ⊆ ω → 2. Any p ∈ G fixes the values of fG on
domp.
Let Dn = {p ∈ P | n ∈ domp}. Then Dn is dense. So if
∀n ∈ ω G ∩ Dn 6= ∅, then domfG = ω.
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}. Let
p≤q⇔q⊆p
i.e., p is stronger than q iff p extends q as a function.
Conditions p, q are compatible iff they agree on domp ∩ domq. In this
case p ∪ q is a common extension.
Since |P| = ℵ0 , the p.o. is c.c.c.
S
Given a filter G , since any two of its elements are compatible, fG = G
is a function: fG : domfG ⊆ ω → 2. Any p ∈ G fixes the values of fG on
domp.
Let Dn = {p ∈ P | n ∈ domp}. Then Dn is dense. So if
∀n ∈ ω G ∩ Dn 6= ∅, then domfG = ω. The existence of such a G is a
consequence of MA(ℵ0 ).
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}. Let
p≤q⇔q⊆p
i.e., p is stronger than q iff p extends q as a function.
Conditions p, q are compatible iff they agree on domp ∩ domq. In this
case p ∪ q is a common extension.
Since |P| = ℵ0 , the p.o. is c.c.c.
S
Given a filter G , since any two of its elements are compatible, fG = G
is a function: fG : domfG ⊆ ω → 2. Any p ∈ G fixes the values of fG on
domp.
Let Dn = {p ∈ P | n ∈ domp}. Then Dn is dense. So if
∀n ∈ ω G ∩ Dn 6= ∅, then domfG = ω. The existence of such a G is a
consequence of MA(ℵ0 ).
Similarly one can use Martin’s axiom to produce filters G whose
associated function fG has some prescribed property.
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}. Let
p≤q⇔q⊆p
i.e., p is stronger than q iff p extends q as a function.
Conditions p, q are compatible iff they agree on domp ∩ domq. In this
case p ∪ q is a common extension.
Since |P| = ℵ0 , the p.o. is c.c.c.
S
Given a filter G , since any two of its elements are compatible, fG = G
is a function: fG : domfG ⊆ ω → 2. Any p ∈ G fixes the values of fG on
domp.
Let Dn = {p ∈ P | n ∈ domp}. Then Dn is dense. So if
∀n ∈ ω G ∩ Dn 6= ∅, then domfG = ω. The existence of such a G is a
consequence of MA(ℵ0 ).
Similarly one can use Martin’s axiom to produce filters G whose
associated function fG has some prescribed property. For example, if
Ei = {p ∈ P | ∃n ∈ domp p(n) = i}, then E0 , E1 are dense in P.
A critical example
Let
P = {p ⊆ ω × 2 | p is a function ∧ |p| < ℵ0 }
be the set of finite partial functions from ω to {0, 1}. Let
p≤q⇔q⊆p
i.e., p is stronger than q iff p extends q as a function.
Conditions p, q are compatible iff they agree on domp ∩ domq. In this
case p ∪ q is a common extension.
Since |P| = ℵ0 , the p.o. is c.c.c.
S
Given a filter G , since any two of its elements are compatible, fG = G
is a function: fG : domfG ⊆ ω → 2. Any p ∈ G fixes the values of fG on
domp.
Let Dn = {p ∈ P | n ∈ domp}. Then Dn is dense. So if
∀n ∈ ω G ∩ Dn 6= ∅, then domfG = ω. The existence of such a G is a
consequence of MA(ℵ0 ).
Similarly one can use Martin’s axiom to produce filters G whose
associated function fG has some prescribed property. For example, if
Ei = {p ∈ P | ∃n ∈ domp p(n) = i}, then E0 , E1 are dense in P. If a filter
G intersects all Dn and also E0 , E1 , then fG is everywhere defined and
non-constant.
Martin’s axiom: basics
Theorem
1. κ < κ0 ∧ MA(κ0 ) ⇒ MA(κ)
2. MA(ℵ0 )
3. ¬MA(2ℵ0 )
Martin’s axiom: basics
Theorem
1. κ < κ0 ∧ MA(κ0 ) ⇒ MA(κ)
2. MA(ℵ0 )
3. ¬MA(2ℵ0 )
Proof.
(2) Let Dn , for n ∈ ω, be dense sets.
Martin’s axiom: basics
Theorem
1. κ < κ0 ∧ MA(κ0 ) ⇒ MA(κ)
2. MA(ℵ0 )
3. ¬MA(2ℵ0 )
Proof.
(2) Let Dn , for n ∈ ω, be dense sets. Define inductively a sequence pn :
Martin’s axiom: basics
Theorem
1. κ < κ0 ∧ MA(κ0 ) ⇒ MA(κ)
2. MA(ℵ0 )
3. ¬MA(2ℵ0 )
Proof.
(2) Let Dn , for n ∈ ω, be dense sets. Define inductively a sequence pn :
I
Let p0 be any element of P.
Martin’s axiom: basics
Theorem
1. κ < κ0 ∧ MA(κ0 ) ⇒ MA(κ)
2. MA(ℵ0 )
3. ¬MA(2ℵ0 )
Proof.
(2) Let Dn , for n ∈ ω, be dense sets. Define inductively a sequence pn :
I
Let p0 be any element of P.
I
Having defined pn , let pn+1 ∈ Dn be any extension of pn .
Martin’s axiom: basics
Theorem
1. κ < κ0 ∧ MA(κ0 ) ⇒ MA(κ)
2. MA(ℵ0 )
3. ¬MA(2ℵ0 )
Proof.
(2) Let Dn , for n ∈ ω, be dense sets. Define inductively a sequence pn :
I
Let p0 be any element of P.
I
Having defined pn , let pn+1 ∈ Dn be any extension of pn .
So p0 ≥ p1 ≥ p2 ≥ . . .
Martin’s axiom: basics
Theorem
1. κ < κ0 ∧ MA(κ0 ) ⇒ MA(κ)
2. MA(ℵ0 )
3. ¬MA(2ℵ0 )
Proof.
(2) Let Dn , for n ∈ ω, be dense sets. Define inductively a sequence pn :
I
Let p0 be any element of P.
I
Having defined pn , let pn+1 ∈ Dn be any extension of pn .
So p0 ≥ p1 ≥ p2 ≥ . . .
Let G = {p ∈ P | ∃n pn ≤ p}. Then G is a filter that intersects every Dn .
(Notice that the hypothesis of P being c.c.c. has not been used.)
Proof (cont’d)
(3: ¬MA(2ℵ0 )) Take the previous example of finite partial functions from
ω to 2.
Proof (cont’d)
(3: ¬MA(2ℵ0 )) Take the previous example of finite partial functions from
ω to 2. For each h : ω → 2 define
Eh = {p ∈ P | ∃n ∈ domp p(n) 6= h(n)}.
Proof (cont’d)
(3: ¬MA(2ℵ0 )) Take the previous example of finite partial functions from
ω to 2. For each h : ω → 2 define
Eh = {p ∈ P | ∃n ∈ domp p(n) 6= h(n)}. Then Eh is dense and if G is a
filter intersecting every Dn and Eh , then fG : ω → 2, fG 6= h.
Proof (cont’d)
(3: ¬MA(2ℵ0 )) Take the previous example of finite partial functions from
ω to 2. For each h : ω → 2 define
Eh = {p ∈ P | ∃n ∈ domp p(n) 6= h(n)}. Then Eh is dense and if G is a
filter intersecting
every
S
S Dn and Eh , then fG : ω → 2, fG 6= h.
Let D = n∈ω Dn ∪ h∈ω 2 Eh . Then |D| = 2ℵ0 . If G is a filter meeting
all elements of D, then fG : ω → 2 is a function different from all
functions ω → 2, which is impossible.
Martin’s axiom: basics
So, under CH, Martin’s axiom MA holds since it is equivalent to MA(ℵ0 ).
Martin’s axiom: basics
So, under CH, Martin’s axiom MA holds since it is equivalent to MA(ℵ0 ).
The proof of MA(ℵ0 ) did not require that the p.o. be c.c.c. However, for
κ > ℵ0 , the strengthening of MA(κ) to any p.o. is false:
Martin’s axiom: basics
So, under CH, Martin’s axiom MA holds since it is equivalent to MA(ℵ0 ).
The proof of MA(ℵ0 ) did not require that the p.o. be c.c.c. However, for
κ > ℵ0 , the strengthening of MA(κ) to any p.o. is false:
Example. P = {p ⊆ ω × ω1 | p is a function ∧ |p| < ω}, with
p ≤ q ⇔ q ⊆ p.
Martin’s axiom: basics
So, under CH, Martin’s axiom MA holds since it is equivalent to MA(ℵ0 ).
The proof of MA(ℵ0 ) did not require that the p.o. be c.c.c. However, for
κ > ℵ0 , the strengthening of MA(κ) to any p.o. is false:
Example. P = {p ⊆ ω × ω1 | p is a function ∧ |p| < ω}, with
p ≤ q ⇔ q ⊆ p.
P is not c.c.c.: functions pα : {0} → ω1 , pα (0) = α form an antichain.
Martin’s axiom: basics
So, under CH, Martin’s axiom MA holds since it is equivalent to MA(ℵ0 ).
The proof of MA(ℵ0 ) did not require that the p.o. be c.c.c. However, for
κ > ℵ0 , the strengthening of MA(κ) to any p.o. is false:
Example. P = {p ⊆ ω × ω1 | p is a function ∧ |p| < ω}, with
p ≤ q ⇔ q ⊆ p.
P is not c.c.c.: functions pα : {0}
S → ω1 , pα (0) = α form an antichain.
If G is a filter in P, then fG = G is a function from a subset of ω to ω1 .
Martin’s axiom: basics
So, under CH, Martin’s axiom MA holds since it is equivalent to MA(ℵ0 ).
The proof of MA(ℵ0 ) did not require that the p.o. be c.c.c. However, for
κ > ℵ0 , the strengthening of MA(κ) to any p.o. is false:
Example. P = {p ⊆ ω × ω1 | p is a function ∧ |p| < ω}, with
p ≤ q ⇔ q ⊆ p.
P is not c.c.c.: functions pα : {0}
S → ω1 , pα (0) = α form an antichain.
If G is a filter in P, then fG = G is a function from a subset of ω to ω1 .
For α < ω1 , let Dα = {p ∈ P | α ∈ imp}. These are dense.
Martin’s axiom: basics
So, under CH, Martin’s axiom MA holds since it is equivalent to MA(ℵ0 ).
The proof of MA(ℵ0 ) did not require that the p.o. be c.c.c. However, for
κ > ℵ0 , the strengthening of MA(κ) to any p.o. is false:
Example. P = {p ⊆ ω × ω1 | p is a function ∧ |p| < ω}, with
p ≤ q ⇔ q ⊆ p.
P is not c.c.c.: functions pα : {0}
S → ω1 , pα (0) = α form an antichain.
If G is a filter in P, then fG = G is a function from a subset of ω to ω1 .
For α < ω1 , let Dα = {p ∈ P | α ∈ imp}. These are dense. If a filter G
meets all of them, then imfG = ω1 , which is impossible.
Martin’s axiom: applications
Back to the questions. Let ℵ0 ≤ κ < 2ℵ0 .
1. 2κ = 2ℵ0 ?
2. Does any a.d. family A ⊆ P(ω) of cardinality κ fail to be maximal?
3. Is the union of κ null subsets of R null?
4. Is the union of κ meagre subsets of R meagre?
Focus on question 2: given A, try and build a d ⊆ ω a.d. from all
elements of A.
Martin’s axiom: applications
Back to the questions. Let ℵ0 ≤ κ < 2ℵ0 .
1. 2κ = 2ℵ0 ?
2. Does any a.d. family A ⊆ P(ω) of cardinality κ fail to be maximal?
3. Is the union of κ null subsets of R null?
4. Is the union of κ meagre subsets of R meagre?
Focus on question 2: given A, try and build a d ⊆ ω a.d. from all
elements of A.
Definition
Given A ⊆ P(ω), let
PA = {(s, F ) | s ⊆ ω, s finite ∧ F ⊆ A, F finite}
with
Martin’s axiom: applications
Back to the questions. Let ℵ0 ≤ κ < 2ℵ0 .
1. 2κ = 2ℵ0 ?
2. Does any a.d. family A ⊆ P(ω) of cardinality κ fail to be maximal?
3. Is the union of κ null subsets of R null?
4. Is the union of κ meagre subsets of R meagre?
Focus on question 2: given A, try and build a d ⊆ ω a.d. from all
elements of A.
Definition
Given A ⊆ P(ω), let
PA = {(s, F ) | s ⊆ ω, s finite ∧ F ⊆ A, F finite}
with
(s 0 , F 0 ) ≤ (s, F ) ⇔ s ⊆ s 0 ∧ F ⊆ F 0 ∧
Martin’s axiom: applications
Back to the questions. Let ℵ0 ≤ κ < 2ℵ0 .
1. 2κ = 2ℵ0 ?
2. Does any a.d. family A ⊆ P(ω) of cardinality κ fail to be maximal?
3. Is the union of κ null subsets of R null?
4. Is the union of κ meagre subsets of R meagre?
Focus on question 2: given A, try and build a d ⊆ ω a.d. from all
elements of A.
Definition
Given A ⊆ P(ω), let
PA = {(s, F ) | s ⊆ ω, s finite ∧ F ⊆ A, F finite}
with
(s 0 , F 0 ) ≤ (s, F ) ⇔ s ⊆ s 0 ∧ F ⊆ F 0 ∧ ∀x ∈ F x ∩ s 0 ⊆ s
Martin’s axiom: applications
Lemma
(s0 , F0 ), (s1 , F1 ) are compatible iff
∀x ∈ F0 x ∩ s1 ⊆ s0 ∧ ∀x ∈ F1 x ∩ s0 ⊆ s1
In this case, (s0 ∪ s1 , F0 ∪ F1 ) is a common extension.
Martin’s axiom: applications
Lemma
(s0 , F0 ), (s1 , F1 ) are compatible iff
∀x ∈ F0 x ∩ s1 ⊆ s0 ∧ ∀x ∈ F1 x ∩ s0 ⊆ s1
In this case, (s0 ∪ s1 , F0 ∪ F1 ) is a common extension.
Proof.
(Forward implication) If (s0 , F0 ), (s1 , F1 ) are compatible, let (t, H) be a
common extension.
Martin’s axiom: applications
Lemma
(s0 , F0 ), (s1 , F1 ) are compatible iff
∀x ∈ F0 x ∩ s1 ⊆ s0 ∧ ∀x ∈ F1 x ∩ s0 ⊆ s1
In this case, (s0 ∪ s1 , F0 ∪ F1 ) is a common extension.
Proof.
(Forward implication) If (s0 , F0 ), (s1 , F1 ) are compatible, let (t, H) be a
common extension. Then s0 ∪ s1 ⊆ t, F0 ∪ F1 ⊆ H.
Martin’s axiom: applications
Lemma
(s0 , F0 ), (s1 , F1 ) are compatible iff
∀x ∈ F0 x ∩ s1 ⊆ s0 ∧ ∀x ∈ F1 x ∩ s0 ⊆ s1
In this case, (s0 ∪ s1 , F0 ∪ F1 ) is a common extension.
Proof.
(Forward implication) If (s0 , F0 ), (s1 , F1 ) are compatible, let (t, H) be a
common extension. Then s0 ∪ s1 ⊆ t, F0 ∪ F1 ⊆ H. Moreover, if (t, H)
satisfies the condition of extending both (s0 , F0 ) and (s1 , F1 ) then any
(t 0 , H 0 ), with s0 ∪ s1 ⊆ t 0 ⊆ t, F0 ∪ F1 ⊆ H 0 ⊆ H, does. So
(s0 ∪ s1 , F0 ∪ F1 ) too is a common extension of (s0 , F0 ), (s1 , F1 ).
Martin’s axiom: applications
Lemma
(s0 , F0 ), (s1 , F1 ) are compatible iff
∀x ∈ F0 x ∩ s1 ⊆ s0 ∧ ∀x ∈ F1 x ∩ s0 ⊆ s1
In this case, (s0 ∪ s1 , F0 ∪ F1 ) is a common extension.
Proof.
(Forward implication) If (s0 , F0 ), (s1 , F1 ) are compatible, let (t, H) be a
common extension. Then s0 ∪ s1 ⊆ t, F0 ∪ F1 ⊆ H. Moreover, if (t, H)
satisfies the condition of extending both (s0 , F0 ) and (s1 , F1 ) then any
(t 0 , H 0 ), with s0 ∪ s1 ⊆ t 0 ⊆ t, F0 ∪ F1 ⊆ H 0 ⊆ H, does. So
(s0 ∪ s1 , F0 ∪ F1 ) too is a common extension of (s0 , F0 ), (s1 , F1 ). This
entails ∀x ∈ Fi x ∩ (s0 ∪ s1 ) ⊆ si , i.e. ∀x ∈ Fi x ∩ s1−i ⊆ si .
Martin’s axiom: applications
Corollary
PA is c.c.c.
Martin’s axiom: applications
Corollary
PA is c.c.c.
Proof.
If (sξ , Fξ ) are pairwise incompatible, in particular the sξ are pairwise
distinct, but there are only ℵ0 finite subsets of ω.
Martin’s axiom: applications
Corollary
PA is c.c.c.
Proof.
If (sξ , Fξ ) are pairwise incompatible, in particular the sξ are pairwise
distinct, but there are only ℵ0 finite subsets of ω.
S
For G a filter in PA , let dG = {s | ∃F (s, F ) ∈ G }.
Martin’s axiom: applications
Corollary
PA is c.c.c.
Proof.
If (sξ , Fξ ) are pairwise incompatible, in particular the sξ are pairwise
distinct, but there are only ℵ0 finite subsets of ω.
S
For G a filter in PA , let dG = {s | ∃F (s, F ) ∈ G }.
Lemma
If G is a filter in PA and (s, F ) ∈ G , then ∀x ∈ F x ∩ dG ⊆ s.
Martin’s axiom: applications
Corollary
PA is c.c.c.
Proof.
If (sξ , Fξ ) are pairwise incompatible, in particular the sξ are pairwise
distinct, but there are only ℵ0 finite subsets of ω.
S
For G a filter in PA , let dG = {s | ∃F (s, F ) ∈ G }.
Lemma
If G is a filter in PA and (s, F ) ∈ G , then ∀x ∈ F x ∩ dG ⊆ s.
Proof.
If n ∈ x ∩ dG , there is (s 0 , F 0 ) ∈ G with n ∈ s 0 .
Martin’s axiom: applications
Corollary
PA is c.c.c.
Proof.
If (sξ , Fξ ) are pairwise incompatible, in particular the sξ are pairwise
distinct, but there are only ℵ0 finite subsets of ω.
S
For G a filter in PA , let dG = {s | ∃F (s, F ) ∈ G }.
Lemma
If G is a filter in PA and (s, F ) ∈ G , then ∀x ∈ F x ∩ dG ⊆ s.
Proof.
If n ∈ x ∩ dG , there is (s 0 , F 0 ) ∈ G with n ∈ s 0 . Since (s, F ), (s 0 , F 0 ) are
compatible,
Martin’s axiom: applications
Corollary
PA is c.c.c.
Proof.
If (sξ , Fξ ) are pairwise incompatible, in particular the sξ are pairwise
distinct, but there are only ℵ0 finite subsets of ω.
S
For G a filter in PA , let dG = {s | ∃F (s, F ) ∈ G }.
Lemma
If G is a filter in PA and (s, F ) ∈ G , then ∀x ∈ F x ∩ dG ⊆ s.
Proof.
If n ∈ x ∩ dG , there is (s 0 , F 0 ) ∈ G with n ∈ s 0 . Since (s, F ), (s 0 , F 0 ) are
compatible, x ∩ s 0 ⊆ s; consequently, n ∈ s.
Martin’s axiom: applications
Corollary
PA is c.c.c.
Proof.
If (sξ , Fξ ) are pairwise incompatible, in particular the sξ are pairwise
distinct, but there are only ℵ0 finite subsets of ω.
S
For G a filter in PA , let dG = {s | ∃F (s, F ) ∈ G }.
Lemma
If G is a filter in PA and (s, F ) ∈ G , then ∀x ∈ F x ∩ dG ⊆ s.
Proof.
If n ∈ x ∩ dG , there is (s 0 , F 0 ) ∈ G with n ∈ s 0 . Since (s, F ), (s 0 , F 0 ) are
compatible, x ∩ s 0 ⊆ s; consequently, n ∈ s.
For any x ∈ A, let Dx = {(s, F ) ∈ PA | x ∈ F }.
Martin’s axiom: applications
Corollary
PA is c.c.c.
Proof.
If (sξ , Fξ ) are pairwise incompatible, in particular the sξ are pairwise
distinct, but there are only ℵ0 finite subsets of ω.
S
For G a filter in PA , let dG = {s | ∃F (s, F ) ∈ G }.
Lemma
If G is a filter in PA and (s, F ) ∈ G , then ∀x ∈ F x ∩ dG ⊆ s.
Proof.
If n ∈ x ∩ dG , there is (s 0 , F 0 ) ∈ G with n ∈ s 0 . Since (s, F ), (s 0 , F 0 ) are
compatible, x ∩ s 0 ⊆ s; consequently, n ∈ s.
For any x ∈ A, let Dx = {(s, F ) ∈ PA | x ∈ F }. This is dense in PA .
Martin’s axiom: applications
Corollary
PA is c.c.c.
Proof.
If (sξ , Fξ ) are pairwise incompatible, in particular the sξ are pairwise
distinct, but there are only ℵ0 finite subsets of ω.
S
For G a filter in PA , let dG = {s | ∃F (s, F ) ∈ G }.
Lemma
If G is a filter in PA and (s, F ) ∈ G , then ∀x ∈ F x ∩ dG ⊆ s.
Proof.
If n ∈ x ∩ dG , there is (s 0 , F 0 ) ∈ G with n ∈ s 0 . Since (s, F ), (s 0 , F 0 ) are
compatible, x ∩ s 0 ⊆ s; consequently, n ∈ s.
For any x ∈ A, let Dx = {(s, F ) ∈ PA | x ∈ F }. This is dense in PA .
If G is a filter in PA such that G ∩ Dx 6= ∅,
Martin’s axiom: applications
Corollary
PA is c.c.c.
Proof.
If (sξ , Fξ ) are pairwise incompatible, in particular the sξ are pairwise
distinct, but there are only ℵ0 finite subsets of ω.
S
For G a filter in PA , let dG = {s | ∃F (s, F ) ∈ G }.
Lemma
If G is a filter in PA and (s, F ) ∈ G , then ∀x ∈ F x ∩ dG ⊆ s.
Proof.
If n ∈ x ∩ dG , there is (s 0 , F 0 ) ∈ G with n ∈ s 0 . Since (s, F ), (s 0 , F 0 ) are
compatible, x ∩ s 0 ⊆ s; consequently, n ∈ s.
For any x ∈ A, let Dx = {(s, F ) ∈ PA | x ∈ F }. This is dense in PA .
If G is a filter in PA such that G ∩ Dx 6= ∅, then |x ∩ dG | < ℵ0 .
Martin’s axiom: applications
Corollary
PA is c.c.c.
Proof.
If (sξ , Fξ ) are pairwise incompatible, in particular the sξ are pairwise
distinct, but there are only ℵ0 finite subsets of ω.
S
For G a filter in PA , let dG = {s | ∃F (s, F ) ∈ G }.
Lemma
If G is a filter in PA and (s, F ) ∈ G , then ∀x ∈ F x ∩ dG ⊆ s.
Proof.
If n ∈ x ∩ dG , there is (s 0 , F 0 ) ∈ G with n ∈ s 0 . Since (s, F ), (s 0 , F 0 ) are
compatible, x ∩ s 0 ⊆ s; consequently, n ∈ s.
For any x ∈ A, let Dx = {(s, F ) ∈ PA | x ∈ F }. This is dense in PA .
If G is a filter in PA such that G ∩ Dx 6= ∅, then |x ∩ dG | < ℵ0 . So such
a dG would do,
Martin’s axiom: applications
Corollary
PA is c.c.c.
Proof.
If (sξ , Fξ ) are pairwise incompatible, in particular the sξ are pairwise
distinct, but there are only ℵ0 finite subsets of ω.
S
For G a filter in PA , let dG = {s | ∃F (s, F ) ∈ G }.
Lemma
If G is a filter in PA and (s, F ) ∈ G , then ∀x ∈ F x ∩ dG ⊆ s.
Proof.
If n ∈ x ∩ dG , there is (s 0 , F 0 ) ∈ G with n ∈ s 0 . Since (s, F ), (s 0 , F 0 ) are
compatible, x ∩ s 0 ⊆ s; consequently, n ∈ s.
For any x ∈ A, let Dx = {(s, F ) ∈ PA | x ∈ F }. This is dense in PA .
If G is a filter in PA such that G ∩ Dx 6= ∅, then |x ∩ dG | < ℵ0 . So such
a dG would do, provided it is infinite.
Martin’s axiom: applications
Theorem (Solovay)
Assume MA(κ). Let C, A ⊆ P(ω) with |C| ≤ κ, |A| ≤ κ.
Martin’s axiom: applications
Theorem (Solovay)
Assume MA(κ). Let C, A ⊆ P(ω) with |C| ≤ κ, |A| S
≤ κ. Suppose that
for all y ∈ C and all finite F ⊆ A, one has that y \ F is infinite.
Martin’s axiom: applications
Theorem (Solovay)
Assume MA(κ). Let C, A ⊆ P(ω) with |C| ≤ κ, |A| S
≤ κ. Suppose that
for all y ∈ C and all finite F ⊆ A, one has that y \ F is infinite. Then
∃d ⊆ ω (∀y ∈ C |d ∩ y | is infinite ∧ ∀x ∈ A |d ∩ x| is finite)
Martin’s axiom: applications
Theorem (Solovay)
Assume MA(κ). Let C, A ⊆ P(ω) with |C| ≤ κ, |A| S
≤ κ. Suppose that
for all y ∈ C and all finite F ⊆ A, one has that y \ F is infinite. Then
∃d ⊆ ω (∀y ∈ C |d ∩ y | is infinite ∧ ∀x ∈ A |d ∩ x| is finite)
Proof.
For y ∈ C, n ∈ ω, set Eny = {(s, F ) ∈ PA | s ∩ y * n}.
Martin’s axiom: applications
Theorem (Solovay)
Assume MA(κ). Let C, A ⊆ P(ω) with |C| ≤ κ, |A| S
≤ κ. Suppose that
for all y ∈ C and all finite F ⊆ A, one has that y \ F is infinite. Then
∃d ⊆ ω (∀y ∈ C |d ∩ y | is infinite ∧ ∀x ∈ A |d ∩ x| is finite)
Proof.
For y ∈ C, n ∈ ω, set Eny = {(s, F ) ∈ PA | s ∩ y * n}. This is dense in
PA :
Martin’s axiom: applications
Theorem (Solovay)
Assume MA(κ). Let C, A ⊆ P(ω) with |C| ≤ κ, |A| S
≤ κ. Suppose that
for all y ∈ C and all finite F ⊆ A, one has that y \ F is infinite. Then
∃d ⊆ ω (∀y ∈ C |d ∩ y | is infinite ∧ ∀x ∈ A |d ∩ x| is finite)
Proof.
For y ∈ C, n ∈ ω, set Eny =S{(s, F ) ∈ PA | s ∩ y * n}. ThisSis dense in
PA : if (s, F ) ∈ PA , as y \ F is infinite, there is m ∈ y \ F with
m > n;
Martin’s axiom: applications
Theorem (Solovay)
Assume MA(κ). Let C, A ⊆ P(ω) with |C| ≤ κ, |A| S
≤ κ. Suppose that
for all y ∈ C and all finite F ⊆ A, one has that y \ F is infinite. Then
∃d ⊆ ω (∀y ∈ C |d ∩ y | is infinite ∧ ∀x ∈ A |d ∩ x| is finite)
Proof.
For y ∈ C, n ∈ ω, set Eny =S{(s, F ) ∈ PA | s ∩ y * n}. ThisSis dense in
PA : if (s, F ) ∈ PA , as y \ F is infinite, there is m ∈ y \ F with
m > n; then (s ∪ {m}, F ) ∈ Eny extends (s, F ).
Martin’s axiom: applications
Theorem (Solovay)
Assume MA(κ). Let C, A ⊆ P(ω) with |C| ≤ κ, |A| S
≤ κ. Suppose that
for all y ∈ C and all finite F ⊆ A, one has that y \ F is infinite. Then
∃d ⊆ ω (∀y ∈ C |d ∩ y | is infinite ∧ ∀x ∈ A |d ∩ x| is finite)
Proof.
For y ∈ C, n ∈ ω, set Eny =S{(s, F ) ∈ PA | s ∩ y * n}. ThisSis dense in
PA : if (s, F ) ∈ PA , as y \ F is infinite, there is m ∈ y \ F with
m > n; then (s ∪ {m}, F ) ∈ Eny extends (s, F ).
By MA(κ) let G be a filter meeting all members of
{Dx | x ∈ A} ∪ {Eny | y ∈ C, n ∈ ω}
Martin’s axiom: applications
Theorem (Solovay)
Assume MA(κ). Let C, A ⊆ P(ω) with |C| ≤ κ, |A| S
≤ κ. Suppose that
for all y ∈ C and all finite F ⊆ A, one has that y \ F is infinite. Then
∃d ⊆ ω (∀y ∈ C |d ∩ y | is infinite ∧ ∀x ∈ A |d ∩ x| is finite)
Proof.
For y ∈ C, n ∈ ω, set Eny =S{(s, F ) ∈ PA | s ∩ y * n}. ThisSis dense in
PA : if (s, F ) ∈ PA , as y \ F is infinite, there is m ∈ y \ F with
m > n; then (s ∪ {m}, F ) ∈ Eny extends (s, F ).
By MA(κ) let G be a filter meeting all members of
{Dx | x ∈ A} ∪ {Eny | y ∈ C, n ∈ ω}
Then dG is almost disjoint from every element of A.
Martin’s axiom: applications
Theorem (Solovay)
Assume MA(κ). Let C, A ⊆ P(ω) with |C| ≤ κ, |A| S
≤ κ. Suppose that
for all y ∈ C and all finite F ⊆ A, one has that y \ F is infinite. Then
∃d ⊆ ω (∀y ∈ C |d ∩ y | is infinite ∧ ∀x ∈ A |d ∩ x| is finite)
Proof.
For y ∈ C, n ∈ ω, set Eny =S{(s, F ) ∈ PA | s ∩ y * n}. ThisSis dense in
PA : if (s, F ) ∈ PA , as y \ F is infinite, there is m ∈ y \ F with
m > n; then (s ∪ {m}, F ) ∈ Eny extends (s, F ).
By MA(κ) let G be a filter meeting all members of
{Dx | x ∈ A} ∪ {Eny | y ∈ C, n ∈ ω}
Then dG is almost disjoint from every element of A. If y ∈ C, then for
every n there is (s, F ) ∈ G such that s ∩ y * n.
Martin’s axiom: applications
Theorem (Solovay)
Assume MA(κ). Let C, A ⊆ P(ω) with |C| ≤ κ, |A| S
≤ κ. Suppose that
for all y ∈ C and all finite F ⊆ A, one has that y \ F is infinite. Then
∃d ⊆ ω (∀y ∈ C |d ∩ y | is infinite ∧ ∀x ∈ A |d ∩ x| is finite)
Proof.
For y ∈ C, n ∈ ω, set Eny =S{(s, F ) ∈ PA | s ∩ y * n}. ThisSis dense in
PA : if (s, F ) ∈ PA , as y \ F is infinite, there is m ∈ y \ F with
m > n; then (s ∪ {m}, F ) ∈ Eny extends (s, F ).
By MA(κ) let G be a filter meeting all members of
{Dx | x ∈ A} ∪ {Eny | y ∈ C, n ∈ ω}
Then dG is almost disjoint from every element of A. If y ∈ C, then for
every n there is (s, F ) ∈ G such that s ∩ y * n. This implies dG ∩ y * n,
so dG ∩ y is infinite.
Martin’s axiom: applications
Corollary (Solovay)
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Then no a.d. A ⊆ P(ω) of
cardinality κ can be m.a.d.
Martin’s axiom: applications
Corollary (Solovay)
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Then no a.d. A ⊆ P(ω) of
cardinality κ can be m.a.d.
Proof.
Since A is a.d. and infinite, for any finite F ⊆ A one has that ω \
infinite.
S
F is
Martin’s axiom: applications
Corollary (Solovay)
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Then no a.d. A ⊆ P(ω) of
cardinality κ can be m.a.d.
Proof.
S
Since A is a.d. and infinite, for any finite F ⊆ A one has that ω \ F is
infinite. So apply the theorem to C = {ω} to get an infinite dG almost
disjoint from all elements of A.
Martin’s axiom: applications
Now to the question: ℵ0 ≤ κ < 2ℵ0 ⇒ 2κ = 2ℵ0 ?
Martin’s axiom: applications
Now to the question: ℵ0 ≤ κ < 2ℵ0 ⇒ 2κ = 2ℵ0 ?
Lemma (Solovay)
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ).
Martin’s axiom: applications
Now to the question: ℵ0 ≤ κ < 2ℵ0 ⇒ 2κ = 2ℵ0 ?
Lemma (Solovay)
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Let B ⊆ P(ω) be an a.d. family
of cardinality κ
Martin’s axiom: applications
Now to the question: ℵ0 ≤ κ < 2ℵ0 ⇒ 2κ = 2ℵ0 ?
Lemma (Solovay)
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Let B ⊆ P(ω) be an a.d. family
of cardinality κ and A ⊆ B.
Martin’s axiom: applications
Now to the question: ℵ0 ≤ κ < 2ℵ0 ⇒ 2κ = 2ℵ0 ?
Lemma (Solovay)
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Let B ⊆ P(ω) be an a.d. family
of cardinality κ and A ⊆ B. Then there is d ⊆ ω such that
∀x ∈ A d ∩ x is finite ∧ ∀x ∈ B \ A d ∩ x is infinite.
Martin’s axiom: applications
Now to the question: ℵ0 ≤ κ < 2ℵ0 ⇒ 2κ = 2ℵ0 ?
Lemma (Solovay)
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Let B ⊆ P(ω) be an a.d. family
of cardinality κ and A ⊆ B. Then there is d ⊆ ω such that
∀x ∈ A d ∩ x is finite ∧ ∀x ∈ B \ A d ∩ x is infinite.
Proof.
Let C = B \ A in Solovay’s theorem.
Martin’s axiom: applications
Now to the question: ℵ0 ≤ κ < 2ℵ0 ⇒ 2κ = 2ℵ0 ?
Lemma (Solovay)
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Let B ⊆ P(ω) be an a.d. family
of cardinality κ and A ⊆ B. Then there is d ⊆ ω such that
∀x ∈ A d ∩ x is finite ∧ ∀x ∈ B \ A d ∩ x is infinite.
Proof.
Let C = B \ A in Solovay’s theorem. Since B is almost disjoint, the
hypothesis
[
∀y ∈ C ∀F ⊆ A (|F | < ℵ0 ⇒ |y \
F | = ℵ0 )
is satisfied.
Martin’s axiom: applications
Theorem
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Then 2κ = 2ℵ0 .
Martin’s axiom: applications
Theorem
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Then 2κ = 2ℵ0 .
Proof.
Since there is an a.d. subfamily of P(ω) of cardinality 2ℵ0 , let B ⊆ P(ω)
be any a.d. family of cardinality κ.
Martin’s axiom: applications
Theorem
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Then 2κ = 2ℵ0 .
Proof.
Since there is an a.d. subfamily of P(ω) of cardinality 2ℵ0 , let B ⊆ P(ω)
be any a.d. family of cardinality κ. Define Φ : P(ω) → P(B) by
Martin’s axiom: applications
Theorem
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Then 2κ = 2ℵ0 .
Proof.
Since there is an a.d. subfamily of P(ω) of cardinality 2ℵ0 , let B ⊆ P(ω)
be any a.d. family of cardinality κ. Define Φ : P(ω) → P(B) by
Φ(d) = {x ∈ B | d ∩ x is finite}.
Martin’s axiom: applications
Theorem
Let ℵ0 ≤ κ < 2ℵ0 and assume MA(κ). Then 2κ = 2ℵ0 .
Proof.
Since there is an a.d. subfamily of P(ω) of cardinality 2ℵ0 , let B ⊆ P(ω)
be any a.d. family of cardinality κ. Define Φ : P(ω) → P(B) by
Φ(d) = {x ∈ B | d ∩ x is finite}. By previous lemma, Φ is surjective.
Regularity of the continuum
Corollary
MA ⇒ 2ℵ0 is regular.
Regularity of the continuum
Corollary
MA ⇒ 2ℵ0 is regular.
For the proof, one needs first to establish König’ lemma:
Lemma
If λ is an infinite cardinal and cof (λ) ≤ κ, then λ < λκ .
Regularity of the continuum
Corollary
MA ⇒ 2ℵ0 is regular.
For the proof, one needs first to establish König’ lemma:
Lemma
If λ is an infinite cardinal and cof (λ) ≤ κ, then λ < λκ .
Proof.
It is shown that no G : λ → κ λ is surjective.
Regularity of the continuum
Corollary
MA ⇒ 2ℵ0 is regular.
For the proof, one needs first to establish König’ lemma:
Lemma
If λ is an infinite cardinal and cof (λ) ≤ κ, then λ < λκ .
Proof.
It is shown that no G : λ → κ λ is surjective. Let f : κ → λ be cofinal.
Regularity of the continuum
Corollary
MA ⇒ 2ℵ0 is regular.
For the proof, one needs first to establish König’ lemma:
Lemma
If λ is an infinite cardinal and cof (λ) ≤ κ, then λ < λκ .
Proof.
It is shown that no G : λ → κ λ is surjective. Let f : κ → λ be cofinal.
Define h : κ → λ
Regularity of the continuum
Corollary
MA ⇒ 2ℵ0 is regular.
For the proof, one needs first to establish König’ lemma:
Lemma
If λ is an infinite cardinal and cof (λ) ≤ κ, then λ < λκ .
Proof.
It is shown that no G : λ → κ λ is surjective. Let f : κ → λ be cofinal.
Define h : κ → λ by setting
h(α) = min(λ \ {G (µ)(α) | µ < f (α)})
This exists by cardinality reasons.
Regularity of the continuum
Corollary
MA ⇒ 2ℵ0 is regular.
For the proof, one needs first to establish König’ lemma:
Lemma
If λ is an infinite cardinal and cof (λ) ≤ κ, then λ < λκ .
Proof.
It is shown that no G : λ → κ λ is surjective. Let f : κ → λ be cofinal.
Define h : κ → λ by setting
h(α) = min(λ \ {G (µ)(α) | µ < f (α)})
This exists by cardinality reasons. Moreover h is different from any G (µ):
Regularity of the continuum
Corollary
MA ⇒ 2ℵ0 is regular.
For the proof, one needs first to establish König’ lemma:
Lemma
If λ is an infinite cardinal and cof (λ) ≤ κ, then λ < λκ .
Proof.
It is shown that no G : λ → κ λ is surjective. Let f : κ → λ be cofinal.
Define h : κ → λ by setting
h(α) = min(λ \ {G (µ)(α) | µ < f (α)})
This exists by cardinality reasons. Moreover h is different from any G (µ):
choose α such that f (α) > µ.
Regularity of the continuum
Corollary
If κ ≥ ℵ0 , then κ < cof (2κ ).
Regularity of the continuum
Corollary
If κ ≥ ℵ0 , then κ < cof (2κ ).
Proof.
2
(2κ )κ = 2κ = 2κ .
Regularity of the continuum
Corollary
If κ ≥ ℵ0 , then κ < cof (2κ ).
Proof.
2
(2κ )κ = 2κ = 2κ . Apply König’ lemma with λ = 2κ .
Regularity of the continuum
Corollary
If κ ≥ ℵ0 , then κ < cof (2κ ).
Proof.
2
(2κ )κ = 2κ = 2κ . Apply König’ lemma with λ = 2κ .
To conclude the proof of the regularity of 2ω under MA, consider any
cardinal κ with ℵ0 ≤ κ < 2ℵ0 .
Regularity of the continuum
Corollary
If κ ≥ ℵ0 , then κ < cof (2κ ).
Proof.
2
(2κ )κ = 2κ = 2κ . Apply König’ lemma with λ = 2κ .
To conclude the proof of the regularity of 2ω under MA, consider any
cardinal κ with ℵ0 ≤ κ < 2ℵ0 . Then 2κ = 2ℵ0 ,
Regularity of the continuum
Corollary
If κ ≥ ℵ0 , then κ < cof (2κ ).
Proof.
2
(2κ )κ = 2κ = 2κ . Apply König’ lemma with λ = 2κ .
To conclude the proof of the regularity of 2ω under MA, consider any
cardinal κ with ℵ0 ≤ κ < 2ℵ0 . Then 2κ = 2ℵ0 , so by the corollary
κ < cof (2κ ) = cof (2ℵ0 ).
Regularity of the continuum
Corollary
If κ ≥ ℵ0 , then κ < cof (2κ ).
Proof.
2
(2κ )κ = 2κ = 2κ . Apply König’ lemma with λ = 2κ .
To conclude the proof of the regularity of 2ω under MA, consider any
cardinal κ with ℵ0 ≤ κ < 2ℵ0 . Then 2κ = 2ℵ0 , so by the corollary
κ < cof (2κ ) = cof (2ℵ0 ).
On the other hand it is consistent with ZFC that 2ℵ0 is a singular
cardinal.
κ-unions of meagre sets
Theorem
Let ℵ0 ≤ κ < 2ℵ0 and
S assume MA(κ). Let Mα , for α < κ, be meagre
subsets of R. Then α<κ Mα is meagre.
κ-unions of meagre sets
Theorem
Let ℵ0 ≤ κ < 2ℵ0 and
S assume MA(κ). Let Mα , for α < κ, be meagre
subsets of R. Then α<κ Mα is meagre.
Proof.
A set M is meagre if M ⊆
S
n∈ω
Kn , with Kn closed nowhere dense.
κ-unions of meagre sets
Theorem
Let ℵ0 ≤ κ < 2ℵ0 and
S assume MA(κ). Let Mα , for α < κ, be meagre
subsets of R. Then α<κ Mα is meagre.
Proof.
S
A set M is meagre if M ⊆ n∈ω Kn , with Kn closed nowhere dense. So it
is enough to prove that if Kα , for α < κ, are closed nowhere dense, then
there are closed
S
S nowhere dense Hn , for n < ω, such that
α<κ Kα ⊆
n<ω Hn .
κ-unions of meagre sets
Theorem
Let ℵ0 ≤ κ < 2ℵ0 and
S assume MA(κ). Let Mα , for α < κ, be meagre
subsets of R. Then α<κ Mα is meagre.
Proof.
S
A set M is meagre if M ⊆ n∈ω Kn , with Kn closed nowhere dense. So it
is enough to prove that if Kα , for α < κ, are closed nowhere dense, then
there are closed
S
S nowhere dense Hn , for n < ω, such that
α<κ Kα ⊆
n<ω Hn . By passing to the complements, it is enough to
show that given open dense
for α < κ, there are open dense sets
T sets Uα , T
Vn , for n < ω, such that n<ω Vn ⊆ α<κ Uα .
κ-unions of meagre sets
Theorem
Let ℵ0 ≤ κ < 2ℵ0 and
S assume MA(κ). Let Mα , for α < κ, be meagre
subsets of R. Then α<κ Mα is meagre.
Proof.
S
A set M is meagre if M ⊆ n∈ω Kn , with Kn closed nowhere dense. So it
is enough to prove that if Kα , for α < κ, are closed nowhere dense, then
there are closed
S
S nowhere dense Hn , for n < ω, such that
α<κ Kα ⊆
n<ω Hn . By passing to the complements, it is enough to
show that given open dense
for α < κ, there are open dense sets
T sets Uα , T
Vn , for n < ω, such that n<ω Vn ⊆ α<κ Uα .
Let (Bi )i<ω enumerate all open intervals with rational endpoints.
κ-unions of meagre sets
Theorem
Let ℵ0 ≤ κ < 2ℵ0 and
S assume MA(κ). Let Mα , for α < κ, be meagre
subsets of R. Then α<κ Mα is meagre.
Proof.
S
A set M is meagre if M ⊆ n∈ω Kn , with Kn closed nowhere dense. So it
is enough to prove that if Kα , for α < κ, are closed nowhere dense, then
there are closed
S
S nowhere dense Hn , for n < ω, such that
α<κ Kα ⊆
n<ω Hn . By passing to the complements, it is enough to
show that given open dense
for α < κ, there are open dense sets
T sets Uα , T
Vn , for n < ω, such that n<ω Vn ⊆ α<κ Uα .
Let (Bi )i<ω enumerate all open intervals with rational endpoints.
Solovay’s
S theorem will provide a suitable d ⊆ ω that will allow to define
Vn = i∈d,i>n Bi .
Proof (cont’d)
Let cj = {i ∈ ω | Bi ⊆ Bj }.
Proof (cont’d)
Let cj = {i ∈ ω | Bi ⊆ Bj }.
If d ∩ cj is infinite, then ∀n∃i > n (i ∈ d ∧ Bi ⊆ Bj ), so Vn ∩ Bj 6= ∅.
Proof (cont’d)
Let cj = {i ∈ ω | Bi ⊆ Bj }.
If d ∩ cj is infinite, then ∀n∃i > n (i ∈ d ∧ Bi ⊆ Bj ), so Vn ∩ Bj 6= ∅. If
d ∩ cj is infinite for all j, then ∀n∀j Vn ∩ Bj 6= ∅, so each Vn is dense.
Proof (cont’d)
Let cj = {i ∈ ω | Bi ⊆ Bj }.
If d ∩ cj is infinite, then ∀n∃i > n (i ∈ d ∧ Bi ⊆ Bj ), so Vn ∩ Bj 6= ∅. If
d ∩ cj is infinite for all j, then ∀n∀j Vn ∩ Bj 6= ∅, so each Vn is dense.
For α < κ, let aα = {i ∈ ω | Bi * Uα }.
Proof (cont’d)
Let cj = {i ∈ ω | Bi ⊆ Bj }.
If d ∩ cj is infinite, then ∀n∃i > n (i ∈ d ∧ Bi ⊆ Bj ), so Vn ∩ Bj 6= ∅. If
d ∩ cj is infinite for all j, then ∀n∀j Vn ∩ Bj 6= ∅, so each Vn is dense.
For α < κ, let aα = {i ∈ ω | Bi * Uα }.
If d ∩ aα is finite, then ∃n d ∩ aα ⊆ n + 1, i.e.
∀i ∈ d (i ≥ n + 1 ⇒ Bi ⊆ Uα ), whence Vn ⊆ Uα .
Proof (cont’d)
Let cj = {i ∈ ω | Bi ⊆ Bj }.
If d ∩ cj is infinite, then ∀n∃i > n (i ∈ d ∧ Bi ⊆ Bj ), so Vn ∩ Bj 6= ∅. If
d ∩ cj is infinite for all j, then ∀n∀j Vn ∩ Bj 6= ∅, so each Vn is dense.
For α < κ, let aα = {i ∈ ω | Bi * Uα }.
If d ∩ aα is finite, then ∃n d ∩ aα ⊆ n + 1, i.e.
∀i ∈ d (i ≥ n +T1 ⇒ Bi ⊆ U
Tα ), whence Vn ⊆ Uα . Thus if d ∩ aα is finite
for all α, then n<ω Vn ⊆ α<κ Uα .
Proof (cont’d)
Let cj = {i ∈ ω | Bi ⊆ Bj }.
If d ∩ cj is infinite, then ∀n∃i > n (i ∈ d ∧ Bi ⊆ Bj ), so Vn ∩ Bj 6= ∅. If
d ∩ cj is infinite for all j, then ∀n∀j Vn ∩ Bj 6= ∅, so each Vn is dense.
For α < κ, let aα = {i ∈ ω | Bi * Uα }.
If d ∩ aα is finite, then ∃n d ∩ aα ⊆ n + 1, i.e.
∀i ∈ d (i ≥ n +T1 ⇒ Bi ⊆ U
Tα ), whence Vn ⊆ Uα . Thus if d ∩ aα is finite
for all α, then n<ω Vn ⊆ α<κ Uα .
Set A = {aα | α < κ}, C = {cj | j < ω}.
Proof (cont’d)
Let cj = {i ∈ ω | Bi ⊆ Bj }.
If d ∩ cj is infinite, then ∀n∃i > n (i ∈ d ∧ Bi ⊆ Bj ), so Vn ∩ Bj 6= ∅. If
d ∩ cj is infinite for all j, then ∀n∀j Vn ∩ Bj 6= ∅, so each Vn is dense.
For α < κ, let aα = {i ∈ ω | Bi * Uα }.
If d ∩ aα is finite, then ∃n d ∩ aα ⊆ n + 1, i.e.
∀i ∈ d (i ≥ n +T1 ⇒ Bi ⊆ U
Tα ), whence Vn ⊆ Uα . Thus if d ∩ aα is finite
for all α, then n<ω Vn ⊆ α<κ Uα .
Set A = {aα | α < κ}, C = {cj | j < ω}. The existence of d as above
follows from Solovay’s
S theorem after checking that for all finite F ⊆ κ
and for all j, cj \ α∈F aα is infinite.
Proof (cont’d)
Let cj = {i ∈ ω | Bi ⊆ Bj }.
If d ∩ cj is infinite, then ∀n∃i > n (i ∈ d ∧ Bi ⊆ Bj ), so Vn ∩ Bj 6= ∅. If
d ∩ cj is infinite for all j, then ∀n∀j Vn ∩ Bj 6= ∅, so each Vn is dense.
For α < κ, let aα = {i ∈ ω | Bi * Uα }.
If d ∩ aα is finite, then ∃n d ∩ aα ⊆ n + 1, i.e.
∀i ∈ d (i ≥ n +T1 ⇒ Bi ⊆ U
Tα ), whence Vn ⊆ Uα . Thus if d ∩ aα is finite
for all α, then n<ω Vn ⊆ α<κ Uα .
Set A = {aα | α < κ}, C = {cj | j < ω}. The existence of d as above
follows from Solovay’s
S theorem after checking that for all finite F ⊆ κ
and for all j, cj \ α∈F aα is infinite. Indeed,
cj \
[
α∈F
aα = {i | Bi ⊆ Bj ∩
\
α∈F
Uα }
Proof (cont’d)
Let cj = {i ∈ ω | Bi ⊆ Bj }.
If d ∩ cj is infinite, then ∀n∃i > n (i ∈ d ∧ Bi ⊆ Bj ), so Vn ∩ Bj 6= ∅. If
d ∩ cj is infinite for all j, then ∀n∀j Vn ∩ Bj 6= ∅, so each Vn is dense.
For α < κ, let aα = {i ∈ ω | Bi * Uα }.
If d ∩ aα is finite, then ∃n d ∩ aα ⊆ n + 1, i.e.
∀i ∈ d (i ≥ n +T1 ⇒ Bi ⊆ U
Tα ), whence Vn ⊆ Uα . Thus if d ∩ aα is finite
for all α, then n<ω Vn ⊆ α<κ Uα .
Set A = {aα | α < κ}, C = {cj | j < ω}. The existence of d as above
follows from Solovay’s
S theorem after checking that for all finite F ⊆ κ
and for all j, cj \ α∈F aα is infinite. Indeed,
cj \
[
α∈F
is infinite as Bj ∩
T
α∈F Uα .
T
α∈F
aα = {i | Bi ⊆ Bj ∩
\
Uα }
α∈F
Uα is open non-empty by the density of
κ-unions of null sets
To answer the remaining motivating question, a different p.o. is used.
Theorem
Let ℵ0 ≤ κ < 2ℵ0 and assume
MA(κ). Let Mα , for α < κ, be subsets of
S
R of measure 0. Then α<κ Mα has measure 0.
Proof.
Set M has measure 0 iff for all ε > 0 there is open U such that
M ⊆ U, µ(U)
S ≤ ε. Fix ε > 0, in order to find open U with
µ(U) ≤ ε, α<κ Mα . Let
P = {p ⊆ R (p open ∧ µ(p) < ε)}
with p ≤ q ⇔ q ⊆ p. Then p, q are compatible iff µ(p ∪ q) ≤ ε, in which
case p ∪ q is a common extension.
Proof (cont’d)
S
If G is a filter on P, then UG = G is open. To check µ(UG ) ≤ ε, notice
that if p, q ∈ G and r ∈ G is a common extension, then r ≤ p ∪ q, so
p ∪ q ∈ G . So, ∀p1 , . . . , pn ∈ G p1S∪ . . . ∪ pn ∈ G and
µ(p1 ∪ . . . ∪ pn ) S
< ε. Since UG = A for some countable A ⊆ G , one
has µ(UG ) = µ( A) ≤ ε.
To show that P is c.c.c. use the following property: let W be the
collection of all finite unions of open intervals with rational endpoints;
then for every measurable M and any δ > 0 there is W ∈ W with
µ(M4W ) < δ. Notice that W is countable.
Proof (cont’d)
So suppose {pα | α < ω1 } is an antichain in P. There is δ > 0 such that
the set X = {α ∈ ω1 | µ(pα ) ≤ ε − δ} is uncountable.
If α, β ∈ X , α 6= β, since µ(pα ∪ pβ ) ≥ ε, it follows µ(pα 4pβ ) ≥ δ.
If for each α, Wα ∈ W is such that µ(pα 4Wα ) < δ2 , then
X → W, α 7→ Wα would be injective, which is impossible.
For each α < κ, let Dα = {p ∈ P | Mα ⊆ p}. These are dense, since the
Mα are null.
By MA(κ),
let G be a filter meeting every Dα , so Mα ⊆ UG for each α,
S
i.e. α<κ Mα ⊆ UG . Moreover, µ(UG ) ≤ ε.
Compact spaces
Definition
A topological space X is compact if
1. For all distinct x, y ∈ X there are disjoint open sets U, V such that
x ∈ U, y ∈ V
2. Every open cover {Uα | α ∈ A} admits a finite subcover
{Uαi | i ∈ I }
Compact spaces
Theorem
1. A closed (non-empty) subset C of a compact space X is compact
under the induced topology {A ∩ C | A open in X }
2. A compact space X is regular: if C is closed in X and x ∈
/ C then
there are disjoint open sets U, V such that x ∈ U, C ⊆ V
(equivalently, for all point x and open set W with x ∈ W , there is
open U such that x ∈ U, Ū ⊆ W )
Proof.
(1) If {Aα | α ∈ J} is an open cover of C , extract a finite sub-cover of X
from {Aα ∪ (X \ C ) | α ∈ J}. Let it be {Aα ∪ (X \ C ) | α ∈ I }. Then
{Aα | α ∈ I } is a finite open cover of C .
(2) For each y ∈ C let Uy , Vy be disjoint, open sets such that
x ∈ Uy , y ∈ Vy . So {Vy ∩ C | y ∈ C } is an open cover of C . Let
{Vyi ∩
TC | i ∈ I } beSa finite sub-cover. Set
U = i∈I Uyi , V = i∈I Vyi .
Other topological applications
Theorem
Assume MA(κ). Let X be aTcompact c.c.c. space and let Uα , for α < κ,
be dense open in X . Then α<κ Uα 6= ∅.
Proof.
Let
P = {p ⊆ X | p open, p 6= ∅}
with p ≤ q ⇔ p ⊆ q. So p⊥q ⇔ p ∩ q = ∅ and P is c.c.c.
If G is a filter in P, then the intersection
of finitely many elements of G is
T
non-empty.
By
compactness,
{p̄
|
p
∈
G
}=
6 ∅. Indeed, if
T
{p̄ | p ∈ G } = ∅, then {X \ p̄ | p ∈ G } would
be an open cover of X . If
T
{X \ pi | i ∈ I } is a finite sub-cover, then i∈I pi = ∅, which is
impossible.
For each α < κ, let Dα = {p ∈ P | p̄ ⊆ Uα }. Then Dα is dense in P,
since Uα is dense
T in X and X isTregular. If G is a filter intersecting all
Dα , then ∅ =
6
{p̄ | p ∈ G } ⊆ α<κ Uα .
Other topological applications
The statement is actually equivalent to MA(κ). This was the easy
implication.
For κ = ω, this is Baire category theorem and it also without the
hypothesis of X being c.c.c.
Without the hypothesis of c.c.c., the theorem is false in ZFC :
Exercise. Show that if ω1 + 1 is given the order topology then ω (ω1 + 1)
is a counterexample, being the union of ω1 closed nowhere dense sets.
Remark. Passing to complements X \ Uα in last theorem shows that X
is not the union of κ closed nowhere dense sets. In contrast as what
happened for R, it cannot be concluded that such union is meagre in
general.
Exercise. Show that [0, 1]ω1 is a counterexample.
Products of c.c.c. spaces
Lemma
Assume MA(ω1 ). Let X be a c.c.c. topological space and {Uα | α < ω1 }
be non-empty
T open subsets of X . Then there is an uncountable A ⊆ ω1
such that α∈B Uα 6= ∅ for all finite B ⊆ A.
Proof.
S
Let Vα = γ>α Uα , so α < β ⇒ Vβ ⊆ Vα .
∃α∀β > α Vβ = Vα : otherwise there would be an increasing sequence
αξ , for ξ < ω1 , such that ∀ξ Vαξ 6= Vαξ+1 , whence Vαξ \ Vαξ+1 6= ∅.
These sets are open and pairwise disjoint, contradicting c.c.c.
Fix such an α and let P = {p ⊆ Vα | p open, p 6= ∅} ordered by ⊆. P is
c.c.c. as X is.
Proof (cont’d)
If G is a filter in P, finite intersections of elements of G are non-empty.
So if A = {α < ω1 | ∃p ∈ G p ⊆ Uγ } then finite intersections of
members of {Uγ | γ ∈ A} are non-empty. It remains to show that A is
uncountable.
For each β let Dβ = {p ∈ P | ∃γ > β p ⊆ Uγ }. To see that it is dense,
notice that Vα ⊆ Vβ . So if p ∈ P, then p ∩ Vβ 6= ∅, so
∃γ > β p ∩ Uγ 6= ∅; thus p ∩ Uγ ∈ Dβ extends p.
Finally, if G ∩ Dβ 6= ∅, then set A contains some γ > β. If
∀β < ω1 G ∩ Dβ 6= ∅, set A is unbounded in ω1 .
Products of c.c.c. spaces
Theorem
Assume MA(ℵ1 ). Then any product of c.c.c. spaces is c.c.c.
Proof.
It is enough to prove that if X , Y are c.c.c. then so is X × Y . Suppose
not. Let {Wα α < ω1 } be a family of pairwise disjoint, non-empty open
subsets of X × Y . For any α there are non-empty Uα , Vα open in X , Y ,
respectively such that Uα × Vα ⊆ Wα .
By the preceding lemma there is an uncountable family {Uα | α ∈ A}
whose finite intersections are non-empty, in particular Uα ∩ Uβ 6= ∅ for
α 6= β. Since (Uα × Vα ) ∩ (Uβ × Vβ ) = ∅, this implies Vα ∩ Vβ = ∅.
This contadicts the fact that Y is c.c.c.
Equivalents of MA
Theorem
MA(κ) is equivalent to MA(κ) restricted to p.o. of cardinality ≤ κ. In
particular, MA is equivalent to MA restricted to p.o. of cardinality < 2ℵ0 .
Proof.
Assume the restrict form of MA(κ). Let Q be any p.o. and D be a family
of dense subsets of Q, with |D| ≤ κ.
There is a p.o. P ⊆ Q such that
1. |P| ≤ κ
2. D ∩ P is dense in P for every D ∈ D
3. if p, q are compatible in Q then they are compatible in P
Proof (cont’d)
For this, take a substructure of Q closed under operations g , fD , for
D ∈ D, where
I
if p, q ∈ Q are compatible, then g (p, q) ≤ p, g (p, q) ≤ q
I
∀p ∈ Q (fD (p) ∈ D ∧ fD (p) ≤ p)
Property (3) implies that P is c.c.c.
Let G be a filter in P such that ∀D ∈ D G ∩ (D ∩ P) 6= ∅.
Let G be the filter in Q generated by G : H = {q ∈ Q | ∃p ∈ G p ≤ q}.
This meets every D ∈ D.
MA and Boolean algebras
Definition
Let B = (B, ∨, ∧,0 , 0, 1, ≤) be a Boolean algebra.
I
An antichain A in B is an antichain in the p.o. (B \ {0}, ≤):
∀a, b ∈ A (a 6= b ⇒ a ∧ b = 0)
I
B is c.c.c. iff B has no uncountable antichain iff the p.o. B \ {0} is
c.c.c.
I
A (proper) filter on B is a filter in the p.o. B \ {0}
MA(κ) (or MA) restricted to complete Boolean algebras is the restriction
of MA(κ) (or MA) to p.o. of the form B \ {0} for some complete c.c.c.
Boolean algebra B.
The regular open algebra
Definition
If X is a topological space, the regular open algebra of X is
ro(X ) = {b ⊆ X | b is open, b = Int(b̄)}
with the operations
I
b∧c =b∩c
I
b ∨ c = Int(b ∪ c)
I
b 0 = Int(X \ b)
ro(X ) is complete: given ∅ =
6 S ⊆ ro(X ),
[
sup S = Int( S)
\
inf S = Int( S)
P.o.’s and Boolean algebras
Lemma
Let P be a p.o. There is a complete Boolean algebra B and
i : P → B \ {0} such that
1. i(P) is dense in B \ {0}
2. ∀p, q ∈ P (p ≤ q ⇒ i(p) ≤ i(q))
3. ∀p, q ∈ P (p⊥q ⇔ i(p) ∧ i(q) = 0)
Moreover, P is c.c.c. iff B is c.c.c.
Proof.
For p ∈ P let Np = {q ∈ P | q ≤ p}. As q ∈ Np ⇒ Nq ⊆ Np , the family
{Np | p ∈ P} is a basis for a topology on P. Np is the smallest open set
containing p.
Let B = ro(P), i(p) = Int(Np ). So (2) follows directly from the definition.
(1) Given b regular, open, non-empty, let p ∈ b. Then Np ⊆ b, whence
i(p) = Int(Np ) ⊆ Int(b̄) = b.
Proof (cont’d)
(3) If p, q ∈ P are compatible, let r ≤ p, r ≤ q. By (2),
i(r ) ≤ i(p) ∧ i(q) 6= ∅. Conversely, if p⊥q, then Np ∩ Nq = ∅. Since Nq
is open, Np ∩ Nq = ∅, so i(p) ∩ Nq = Int(Np ) ∩ Nq = ∅. As i(p) is open,
i(p) ∧ i(q) = i(p) ∩ i(q) = i(p) ∩ Int(Nq ) = ∅.
For the final statement notice that:
I
If {pα | α < ω1 } is an antichain in P, then the i(pα ) are all distinct
and they form an antichain in B.
I
If {bα | α < ω1 } is an antichain in B, by density for every α let
pα ∈ P with i(pα ) ≤ bα . Then {i(pα ) | α < ω1 } is an antichain in
B, and {pα | α < ω1 } is an antichain in P.
P.o.’s and Boolean algebras
Such a Boolean algebra si called a completion of P.
Exercise. Show that the completion of a p.o. P is unique: if B1 , B2 are
completions of P, witnessed by maps ij : P → Bj , then there is an
isomorphism of Boolean algebras h : B1 → B2 such that hi1 = i2 .
Remark. Function i : P → B needs not be surjective. For instance, if all
elements of P are compatible, then B = {0, 1} and ∀p ∈ P i(p) = 1.
Exercise. P.o. P is separative if ≤ is antisymmetric and
∀p, q (p q ⇒ ∃r (r ≤ p, r ⊥q)). Show that P is separative iff function i
is injective and ∀p, q ∈ P (p ≤ q ⇔ i(p) ≤ i(q)).
MA and Boolean algebras
Theorem
For any κ ≥ ℵ0 , MA(κ) is equivalent to MA(κ) restricted to complete
Boolean algebras.
Proof.
Assume MA(κ) restricted to complete Boolean algebras. Let P be a
c.c.c. p.o., which can be assumed of cardinality ≤ κ, and D a family of
dense subsets of D, with |D| ≤ κ. So i(D) is dense in B.
Let G be a filter in B such that ∀D ∈ D G ∩ i(D) 6= ∅. Let H = i −1 (G ).
Then ∀D ∈ D H ∩ D 6= ∅. Moreover ∀p ∈ H ∀q ∈ P (p ≤ q ⇒ g ∈ H).
However, this construction is not enough, since it does not grant
∀p, q ∈ H ∃r ∈ H (r ≤ p, r ≤ q). This difficulty is overpassed by
enlarging — not too much — D and consequently G and H.
Proof (cont’d)
For each p, q ∈ P, let Dpq = {r ∈ P | (r ≤ p, r ≤ q) or r ⊥p or r ⊥q}.
Then Dpq is dense in P. Indeed, fix r0 ∈ P. If r0 has an extension
incompatible with either p or q, then such an extension is in Dpq .
Otherwise, in particular r0 is compatible with p, so let r1 ∈ P with
r1 ≤ r0 , r1 ≤ p. Then, r1 , being an extension of r0 , is compatible with q:
let r2 ∈ P such that r2 ≤ r1 , r2 ≤ q. Since r2 ≤ r1 ≤ p, it follows
r2 ∈ Dpq , r2 ≤ r0 .
Since |{Dpq | p, q ∈ P}| ≤ κ, by enlarging it can be assumed that all
Dpq ∈ D. Now, since H meets all elements of D, let p, q ∈ H and fix
some r ∈ H ∩ Dpq . As i(r ), i(p) are compatible, being elements of the
filter G , and similarly for i(r ), i(q), the same holds for r , p and r , q in P.
So the only possibility is r ≤ p, r ≤ q.
The Stone space
Definition
A proper filter F in a Boolean algebra B is an ultrafilter if it is maximal
(with respect to inclusion).
Example. If a is an atom in B, i.e the only element less than a is 0, then
{b ∈ B | a ≤ b} is the principal ultrafilter generated by a.
If B is finite, these are the only ultrafilter of B.
Definition
The Stone space of the Boolean algebra B is the set X of ultrafilters of B
endowed with the topology generated by the basis {Nb | b ∈ B}, where
Nb = {G ∈ X | b ∈ G }.
X turns out to be a compact space. Moreover Nb ∩ Nc = ∅ ⇔ b ∧ c = 0,
so X is c.c.c. iff B is c.c.c.
Back to the topological formulation
Theorem
Let κ ≥ ℵ0 . Assume that whenever {Uα | α ≤ T
κ} are dense open subsets
of a compact c.c.c. space X their intersection α<κ Uα is non-empty.
Then MA(κ).
Proof.
It is enough to establish MA(κ) restricted to complete Boolean algebras.
In fact, if B is a c.c.c. Boolean algebra (non necessarily complete) and D
a family of ≤ κ dense subsets of the order B \ {0}, a filter G will be
produced intersecting all members of D.
S
Let X be the Stone space of B. For each D ∈ D, let WD = b∈D Nb .
This is an open subset of X . Moreover WD is a (topologically) dense
subset of X . Suppose indeed WD ∩ Nc = ∅ for some c ∈ B. Then
∀b ∈ D c ∧ b = 0, but this is impossible as D is dense in the order
B \ {0}, so it must contain T
an extension of c.
By the hypothesis, let G ∈ D∈D WD . Then G is an ultrafilter and for
all D ∈ D, one has G ∈ WD , so ∃b ∈ G G ∈ Nb , which means
∃b ∈ D b ∈ G , whence G ∩ D 6= ∅.
The order topology and Suslin problem
Definition
Let (X , ≤) be a total (or linear) ordering. The order topology on X is the
one generated by open intervals: ]a, b[, ] ←, a[, or ]a, → [.
Recall that every separable topological spaces is c.c.c., but there are
c.c.c. spaces that are not separable (ex.: κ 2 when κ > 2ℵ0 ). This
motivates the following:
Definition
A Suslin line is a total ordering whose order topology is c.c.c. but not
separable.
Suslin’s hypothesis (SH) is the statement: there are no Suslin lines.
In other words, SH states that for order topologies c.c.c. is equivalent to
being separable.
The order topology and Suslin problem
SH turns out to be independent of ZFC :
I
It follows from MA + ¬CH
I
Its negation follows from ♦
Theorem
If X is a Suslin line, then X 2 is not c.c.c.
From this is follows:
Corollary
MA(ω1 ) ⇒ SH
Proof.
From MA(ω1 ) it follows that the product of c.c.c. spaces is c.c.c. So a
Suslin line cannot exist.
The order topology and Suslin problem
Proof of theorem.
By induction on α < ω1 , define aα , bα , cα ∈ X such that:
1. aα < bα < cα
2. ]aα , bα [6= ∅ =
6 ]bα , cα [
3. ]aα , cα [∩{bξ | ξ < α} = ∅
To do this, let W be the set of isolated points in the order topology.
Since X is c.c.c., W is countable.
Assume that for some α < ω1 , aξ , bξ , cξ have been chosen for all ξ < α.
Since X is not separable, the open set X \ W ∪ {bξ | ξ < α} in
non-empty, so it contains an open interval ]aα , cα [. Since ]aα , cα [ does
not contain isolated points, it is infinite and one can choose bα ∈]aα , cα [
so that the intervals ]aα , bα [, ]bα , cα [ are non-empty.
Let Uα =]aα , bα [×]bα , cα [. So Uα 6= ∅.
If ξ < α, then Uξ ∩ Uα = ∅. Indeed, either bξ ≤ aα , in which case
]aξ , bξ [∩]aα , bα [= ∅; or cα ≤ bξ and then ]bξ , cξ [∩]bα , cα [= ∅.
So {Uα | α < ω1 } witnesses the failure of c.c.c. for X 2 .
Condensations
Definition
An equivalence relation on a (non-empty) set A is a reflexive, transitive
and symmetric relation E :
1. ∀x ∈ A xEx (reflexivity)
2. ∀x, y , x ∈ A (xEyEz ⇒ xEz) (transitivity)
3. ∀x, y ∈ A (xEy ⇒ yEx) (symmetry)
Given x ∈ I , the set [x]E = {y ∈ A | xEy } is the E -equivalence class of x.
The equivalence classes form a partition A/E = {[x]E | x ∈ A} of A,
called the quotient set.
Definition
If (L, ≤) is a total order, E an equivalence relation on L, call E a
condensation if the equivalence classes are convex, i.e.
∀x, y , z ∈ E (xEz, x ≤ y ≤ z ⇒ xEy ).
In this case, the quotient set can be ordered by letting
[x]E < [y ]E ⇔ x < y
Condensations
Here [x, y ] is the set of elements between x, y , regardless of their mutual
order. Tipical examples of condensations are defined by requiring some
smallness conditions on closed intervals. The critical point is then to
establish transitivity.
Examples. The following are condensations:
I
x ∼κ ⇔ |[x, y ]| < κ
I
x ∼κ ↔ [x, y ] is scattered (i.e., the rational do not embed in [x, y ])
I
x ∼c y ⇔ [x, y ] is complete (i.e., all upperly bounded sets have a
supremum)
I
x ∼sep y ⇔ [x, y ] is separable
Nicer Suslin lines
Lemma
If there is a Suslin line Y , there exists a Suslin line X such that
I
X is dense in itself: ∀a, b ∈ X (a < b ⇒ ∃c ∈ X a < c < b)
I
no non-empty open subset of X is separable
Proof.
Let X = Y / ∼sep . Then each I ∈ X is separable. Indeed, let M be a
maximal disjoint collection of non-empty open intervals ]x, y [, with
x, y ∈ I . M is countable, by c.c.c. of Y , so M = {]xn , yn [| n ∈ ω}.
SinceSxn ∼sep yn , let Dn beSa countable dense subset of ]xn , yn [. Then
D = n∈ω Dn is dense in n∈<ω ]xn , yn [. If z in not an endpoint of I ,
then given any ]x, y [⊆ I with z ∈]x, y [, there is n such that
]x, y [∩]xn , yn [6= ∅, by maximality of M. Thus D, possible together with
the endpoints of I , is a countable dense subsets of I .
X is dense in itself: suppose I < J were consecutive elements of X ;
picking any x ∈ I , y ∈ J, the set ]x, y [ is included in I ∪ J and would be
separable, so x ∼sep y , a contradiction.
Proof (cont’d)
To verify that no non-empty open subset of X is separable, it is enough
to check that no ]I , J[ is separable. Suppose ]I , J[ is separable. As Y is
c.c.c., only countably many elements of ]I , J[ can have more then two
points; moreover no two elements of ]I , J[ with at most two points can
be consecutive. Set K0 = I , K1 = J and let {Kn | 2 ≤ n < ω} be a dense
set in ]I , J[. For each n < ω, let Dn be a countable dense set of Kn . Take
also a countable dense set Ei in anySIi ∈]I , J[ which
is countableSbut does
S
not belong to {Kn | n ≥ 2}. Then n<ω Dn ∪ i Ei is dense in [I , J].
So points of I are ∼epi equivalent to points in J, a contradiction.
Finally, to see that X is c.c.c., suppose ]Iα , Jα [, for α < ω1 are disjoint
open intervals in X . Pick xα ∈ Iα , yα ∈ Jα , with the condition that if
Iα = Jβ then xα = yβ . Then ]xα , yα [ are pairwise disjoint intervals of Y ,
which is impossible.
Exercise. Show that if there is a Suslin line Y , then there is a Suslin line
X such that
I
X has no first or last element
I
X is connected in the order topology
The order topology and Suslin problem
Exercise. Let (X , ≤) be a total ordering such that:
1. X has no first or last element
2. X is connected in the order topology
3. X is separable in the order topology
Show that the order (X , ≤) is isomorphic to (R, ≤).
Suslin asked whether (3) may be replaced by
3’ X is c.c.c. in the order topology
So, under SH, conditions (3) and (3’) are equivalent.
By the exercise in the previous slide, SH is equivalent to the statement
that (1), (2) and (3’) characterise R.
Trees
Trees turn out to be an important combinatorial tool.
Definition
A tree (sometimes called a forest) (T , ≤) is an anti-symmetric p.o. such
that, for each x ∈ T , {y ∈ T | y < x} is well ordered.
Definition
Let T be a tree:
I
For x ∈ T , the height of x in T , denoted ht(x, T ), is the order type
of {y ∈ T | y < x}
I
For each ordinal α, the α-th level of T , denoted Levα (T ) is
{x ∈ T | ht(x, T ) = α}
I
The height of T , denoted ht(T ) is the least α such that
Levα (T ) = 0
I
A subtree of T is a subset T 0 ⊆ T such that
∀x ∈ T 0 ∀y ∈ T (y < x ⇒ y ∈ T 0 )
Trees
Easy facts.
I
ht(T ) = sup{ht(x, T ) + 1 | x ∈ T }
I
If T 0 is a subtree of T , then ∀x ∈ T 0 ht(x, T ) = ht(x, T 0 )
Trivial examples
I
Any non-empty set T with x ≤ y iff x = y . Then
∀x ∈ T ht(x, T ) = 0 and ht(T ) = 1.
I
Any ordinal δ. Then ht(α, δ) = α and ht(δ) = δ.
A better example. I <δ : the set of I -sequences of length some ordinal
α < δ, ordered by inclusion.