Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
Rado’s selection Lemma implies Hahn-Banach
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
Marianne Morillon
University of La Réunion (France)
Kielce, 25-30 july 2010
1/11
Rado’ selection Lemma
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
Notation
For every set I , we denote by fin∗ (I ) the set of non-empty finite
subsets of I .
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
2/11
Rado’ selection Lemma
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
Notation
For every set I , we denote by fin∗ (I ) the set of non-empty finite
subsets of I .
Rado (1949) Axiomatic treatment of rank in infinite sets
RL: Given a family (Xi )i∈I of finite sets and a family
Q
∗
(σF )F ∈fin∗ (I ) such
Qthat for every F ∈ fin (I ), σF ∈ i∈F Xi ,
there exists f ∈ i∈I Xi which “respects” (σF )F ∈fin∗ (I ) :
∀F ∈ fin∗ (I ) ∃G ∈ fin∗ (I ) F ⊆ G and fG = σG
2/11
Rado’ selection Lemma
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
Notation
For every set I , we denote by fin∗ (I ) the set of non-empty finite
subsets of I .
Rado (1949) Axiomatic treatment of rank in infinite sets
RL: Given a family (Xi )i∈I of finite sets and a family
Q
∗
(σF )F ∈fin∗ (I ) such
Qthat for every F ∈ fin (I ), σF ∈ i∈F Xi ,
there exists f ∈ i∈I Xi which “respects” (σF )F ∈fin∗ (I ) :
∀F ∈ fin∗ (I ) ∃G ∈ fin∗ (I ) F ⊆ G and fG = σG
Remark
“f respects (σF )F ∈fin∗ (I ) ” means that the set
{G ∈ fin∗ (I ) : fG = σG } is cofinal in the poset (fin∗ (I ), ⊆).
2/11
Tychonov implies RL
Rado’s
selection
Lemma
implies
Hahn-Banach
We work in set-theory ZF (without the Axiom of Choice AC).
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
3/11
Tychonov implies RL
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
We work in set-theory ZF (without the Axiom of Choice AC).
The classical proof of RL relies on Tychonov’s axiom for
families of (finite) compact Hausdorff spaces:
T2 : For every infinite family (XiQ
)i∈I of compact Hausdorff
spaces, the topological product i∈I Xi is compact. Thus:
AC ⇒ T2 ⇒ RL
RL ⇒ HB:
idea
RL ⇒ HB:
proof
3/11
Tychonov implies RL
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
We work in set-theory ZF (without the Axiom of Choice AC).
The classical proof of RL relies on Tychonov’s axiom for
families of (finite) compact Hausdorff spaces:
T2 : For every infinite family (XiQ
)i∈I of compact Hausdorff
spaces, the topological product i∈I Xi is compact. Thus:
AC ⇒ T2 ⇒ RL
Consider the following consequence of T2 :
ACfin : “Every infinite family of finite non-empty sets has a
non-empty product.”
Blass noticed that:
T2 ⇔ (RL+ACfin )
3/11
RL does not hold for infinite sets
Rado’s
selection
Lemma
implies
Hahn-Banach
Let R be the following bi-partite graph (“play-boy graph”):
R = {(i + 1, i) : i ∈ N} ∪ {(0, i) : i ∈ N}
M. Morillon
RL
For every F ∈ fin∗ (N), consider a R-marriage σF : F → N.
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
4/11
RL does not hold for infinite sets
Rado’s
selection
Lemma
implies
Hahn-Banach
Let R be the following bi-partite graph (“play-boy graph”):
R = {(i + 1, i) : i ∈ N} ∪ {(0, i) : i ∈ N}
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
For every F ∈ fin∗ (N), consider a R-marriage σF : F → N.
Consider the family (Xi )i∈N defined by X0Q= N and Xi+1 = {i}
for every i ∈ N. There is no element f ∈ i∈N Xi which
respects the family (σF )F ∈fin∗ (N) .
Remark
RL ⇒ HB:
proof
4/11
RL does not hold for infinite sets
Rado’s
selection
Lemma
implies
Hahn-Banach
Let R be the following bi-partite graph (“play-boy graph”):
R = {(i + 1, i) : i ∈ N} ∪ {(0, i) : i ∈ N}
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
For every F ∈ fin∗ (N), consider a R-marriage σF : F → N.
Consider the family (Xi )i∈N defined by X0Q= N and Xi+1 = {i}
for every i ∈ N. There is no element f ∈ i∈N Xi which
respects the family (σF )F ∈fin∗ (N) .
Remark
1
T2 implies Hall’s infinite marriage axiom H.
4/11
RL does not hold for infinite sets
Rado’s
selection
Lemma
implies
Hahn-Banach
Let R be the following bi-partite graph (“play-boy graph”):
R = {(i + 1, i) : i ∈ N} ∪ {(0, i) : i ∈ N}
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
For every F ∈ fin∗ (N), consider a R-marriage σF : F → N.
Consider the family (Xi )i∈N defined by X0Q= N and Xi+1 = {i}
for every i ∈ N. There is no element f ∈ i∈N Xi which
respects the family (σF )F ∈fin∗ (N) .
Remark
1
T2 implies Hall’s infinite marriage axiom H.
2
In turn, H implies that in a vector space (or more
generally in a finitary matroid), all bases are equipotent
(one of the aims of Rado’s paper [3]).
4/11
Does RL imply T2 in ZF?
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
Let ZFA be the set-theory without AC and with atoms: thus
ZF is (ZFA+“There are no atoms”), and ZFA is weaker (i.e.
has less axioms) than ZF.
Theorem (P. Howard (1984), [2])
There is a model of ZFA where RL holds and T2 does not hold.
RL ⇒ HB:
idea
RL ⇒ HB:
proof
5/11
Does RL imply T2 in ZF?
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
Let ZFA be the set-theory without AC and with atoms: thus
ZF is (ZFA+“There are no atoms”), and ZFA is weaker (i.e.
has less axioms) than ZF.
Theorem (P. Howard (1984), [2])
There is a model of ZFA where RL holds and T2 does not hold.
RL ⇒ HB:
idea
RL ⇒ HB:
proof
5/11
Does RL imply T2 in ZF?
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
Let ZFA be the set-theory without AC and with atoms: thus
ZF is (ZFA+“There are no atoms”), and ZFA is weaker (i.e.
has less axioms) than ZF.
Theorem (P. Howard (1984), [2])
There is a model of ZFA where RL holds and T2 does not hold.
RL ⇒ HB:
idea
Remark
RL ⇒ HB:
proof
We do not know whether RL implies T2 in ZF.
5/11
Does RL imply T2 in ZF?
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
Let ZFA be the set-theory without AC and with atoms: thus
ZF is (ZFA+“There are no atoms”), and ZFA is weaker (i.e.
has less axioms) than ZF.
Theorem (P. Howard (1984), [2])
There is a model of ZFA where RL holds and T2 does not hold.
RL ⇒ HB:
idea
Remark
RL ⇒ HB:
proof
We do not know whether RL implies T2 in ZF.
5/11
Does RL imply T2 in ZF?
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
Let ZFA be the set-theory without AC and with atoms: thus
ZF is (ZFA+“There are no atoms”), and ZFA is weaker (i.e.
has less axioms) than ZF.
Theorem (P. Howard (1984), [2])
There is a model of ZFA where RL holds and T2 does not hold.
RL ⇒ HB:
idea
Remark
RL ⇒ HB:
proof
We do not know whether RL implies T2 in ZF.
We shall prove in ZF (and even in ZFA) that RL implies the
Hahn-Banach axiom HB, a consequence of T2 .
5/11
The Hahn-Banach axiom HB
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
Given a boolean algebra B, a measure on B is a mapping
m : B → [0, 1] such that for every x, y ∈ B:
x ∧ y = 0B ⇒ m(x ∨ y ) = m(x) + m(y )
If moreover m(1B ) = 1, the measure m is said to be unitary.
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
6/11
The Hahn-Banach axiom HB
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
Given a boolean algebra B, a measure on B is a mapping
m : B → [0, 1] such that for every x, y ∈ B:
x ∧ y = 0B ⇒ m(x ∨ y ) = m(x) + m(y )
RL
If moreover m(1B ) = 1, the measure m is said to be unitary.
T2 ⇒ RL
Uniform probability on a finite non-trivial bool. algebra B
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
We denote by PB the unitary measure on B giving the same
measure to all atoms of B.
RL ⇒ HB:
proof
6/11
The Hahn-Banach axiom HB
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
Given a boolean algebra B, a measure on B is a mapping
m : B → [0, 1] such that for every x, y ∈ B:
x ∧ y = 0B ⇒ m(x ∨ y ) = m(x) + m(y )
RL
If moreover m(1B ) = 1, the measure m is said to be unitary.
T2 ⇒ RL
Uniform probability on a finite non-trivial bool. algebra B
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
We denote by PB the unitary measure on B giving the same
measure to all atoms of B.
HB (Hahn-Banach in the “boolean” setting)
Given an infinite boolean algebra B, there exists a unitary
measure µ : B → [0, 1].
6/11
The Hahn-Banach axiom HB
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
Given a boolean algebra B, a measure on B is a mapping
m : B → [0, 1] such that for every x, y ∈ B:
x ∧ y = 0B ⇒ m(x ∨ y ) = m(x) + m(y )
RL
If moreover m(1B ) = 1, the measure m is said to be unitary.
T2 ⇒ RL
Uniform probability on a finite non-trivial bool. algebra B
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
We denote by PB the unitary measure on B giving the same
measure to all atoms of B.
HB (Hahn-Banach in the “boolean” setting)
Given an infinite boolean algebra B, there exists a unitary
measure µ : B → [0, 1].
6/11
The Hahn-Banach axiom HB
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
Given a boolean algebra B, a measure on B is a mapping
m : B → [0, 1] such that for every x, y ∈ B:
x ∧ y = 0B ⇒ m(x ∨ y ) = m(x) + m(y )
RL
If moreover m(1B ) = 1, the measure m is said to be unitary.
T2 ⇒ RL
Uniform probability on a finite non-trivial bool. algebra B
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
We denote by PB the unitary measure on B giving the same
measure to all atoms of B.
HB (Hahn-Banach in the “boolean” setting)
Given an infinite boolean algebra B, there exists a unitary
measure µ : B → [0, 1].
Luxembourg (see [1], 1969) proved the HB is equivalent (in
ZF) to the classical forms of the Hahn-Banach property
(analytic form).
6/11
The Hahn-Banach axiom HB
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
Given a boolean algebra B, a measure on B is a mapping
m : B → [0, 1] such that for every x, y ∈ B:
x ∧ y = 0B ⇒ m(x ∨ y ) = m(x) + m(y )
RL
If moreover m(1B ) = 1, the measure m is said to be unitary.
T2 ⇒ RL
Uniform probability on a finite non-trivial bool. algebra B
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
We denote by PB the unitary measure on B giving the same
measure to all atoms of B.
HB (Hahn-Banach in the “boolean” setting)
Given an infinite boolean algebra B, there exists a unitary
measure µ : B → [0, 1].
Luxembourg (see [1], 1969) proved the HB is equivalent (in
ZF) to the classical forms of the Hahn-Banach property
(analytic form). Notice that T2 ⇒ HB.
6/11
RL implies HB: a first tentative
Rado’s
selection
Lemma
implies
Hahn-Banach
RL implies HB: a first idea
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
7/11
RL implies HB: a first tentative
Rado’s
selection
Lemma
implies
Hahn-Banach
RL implies HB: a first idea
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
7/11
RL implies HB: a first tentative
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
RL implies HB: a first idea
Given an infinite boolean algebra B, for every F ∈ fin∗ (B),
denote by bool(F ) the boolean sub-algebra of B which is
generated by F , and consider the mapping σF : F → [0, 1]
which is the restriction of Pbool(F ) .
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
7/11
RL implies HB: a first tentative
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL implies HB: a first idea
Given an infinite boolean algebra B, for every F ∈ fin∗ (B),
denote by bool(F ) the boolean sub-algebra of B which is
generated by F , and consider the mapping σF : F → [0, 1]
which is the restriction of Pbool(F ) . If we could apply RL to the
product of infinite sets [0, 1]B , we would obtain a mapping
m : B → [0, 1] respecting the family (σF )F ∈fin∗ (B) .
RL ⇒ HB:
proof
7/11
RL implies HB: a first tentative
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
RL implies HB: a first idea
Given an infinite boolean algebra B, for every F ∈ fin∗ (B),
denote by bool(F ) the boolean sub-algebra of B which is
generated by F , and consider the mapping σF : F → [0, 1]
which is the restriction of Pbool(F ) . If we could apply RL to the
product of infinite sets [0, 1]B , we would obtain a mapping
m : B → [0, 1] respecting the family (σF )F ∈fin∗ (B) . Such a
mapping would be a unitary measure on B: indeed, given
x, y ∈ B such that x ∧ y = 0B , consider some G ∈ fin(B)
containing {x, y , x ∨ y , 1B } such that mG = σG ;
7/11
RL implies HB: a first tentative
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
RL implies HB: a first idea
Given an infinite boolean algebra B, for every F ∈ fin∗ (B),
denote by bool(F ) the boolean sub-algebra of B which is
generated by F , and consider the mapping σF : F → [0, 1]
which is the restriction of Pbool(F ) . If we could apply RL to the
product of infinite sets [0, 1]B , we would obtain a mapping
m : B → [0, 1] respecting the family (σF )F ∈fin∗ (B) . Such a
mapping would be a unitary measure on B: indeed, given
x, y ∈ B such that x ∧ y = 0B , consider some G ∈ fin(B)
containing {x, y , x ∨ y , 1B } such that mG = σG ; then
m(1B ) = σG (1B ) = 1;
7/11
RL implies HB: a first tentative
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
RL implies HB: a first idea
Given an infinite boolean algebra B, for every F ∈ fin∗ (B),
denote by bool(F ) the boolean sub-algebra of B which is
generated by F , and consider the mapping σF : F → [0, 1]
which is the restriction of Pbool(F ) . If we could apply RL to the
product of infinite sets [0, 1]B , we would obtain a mapping
m : B → [0, 1] respecting the family (σF )F ∈fin∗ (B) . Such a
mapping would be a unitary measure on B: indeed, given
x, y ∈ B such that x ∧ y = 0B , consider some G ∈ fin(B)
containing {x, y , x ∨ y , 1B } such that mG = σG ; then
m(1B ) = σG (1B ) = 1; moreover,
m(x ∨ y ) = σG (x ∨ y )
7/11
RL implies HB: a first tentative
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
RL implies HB: a first idea
Given an infinite boolean algebra B, for every F ∈ fin∗ (B),
denote by bool(F ) the boolean sub-algebra of B which is
generated by F , and consider the mapping σF : F → [0, 1]
which is the restriction of Pbool(F ) . If we could apply RL to the
product of infinite sets [0, 1]B , we would obtain a mapping
m : B → [0, 1] respecting the family (σF )F ∈fin∗ (B) . Such a
mapping would be a unitary measure on B: indeed, given
x, y ∈ B such that x ∧ y = 0B , consider some G ∈ fin(B)
containing {x, y , x ∨ y , 1B } such that mG = σG ; then
m(1B ) = σG (1B ) = 1; moreover,
m(x ∨ y ) = σG (x ∨ y ) = σG (x) + σG (y )
7/11
RL implies HB: a first tentative
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
RL implies HB: a first idea
Given an infinite boolean algebra B, for every F ∈ fin∗ (B),
denote by bool(F ) the boolean sub-algebra of B which is
generated by F , and consider the mapping σF : F → [0, 1]
which is the restriction of Pbool(F ) . If we could apply RL to the
product of infinite sets [0, 1]B , we would obtain a mapping
m : B → [0, 1] respecting the family (σF )F ∈fin∗ (B) . Such a
mapping would be a unitary measure on B: indeed, given
x, y ∈ B such that x ∧ y = 0B , consider some G ∈ fin(B)
containing {x, y , x ∨ y , 1B } such that mG = σG ; then
m(1B ) = σG (1B ) = 1; moreover,
m(x ∨ y ) = σG (x ∨ y ) = σG (x) + σG (y ) = m(x) + m(y ).
7/11
The finite sets Dn
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
Notation
k
For every n ∈ N, let Dn := { n+1
: k ∈ N and 0 ≤ k ≤ n + 1}.
Notice that ∪n∈N Dn is countable and dense in [0, 1].
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
8/11
The finite sets Dn
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
Notation
k
For every n ∈ N, let Dn := { n+1
: k ∈ N and 0 ≤ k ≤ n + 1}.
Notice that ∪n∈N Dn is countable and dense in [0, 1].
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
8/11
The finite sets Dn
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
Notation
k
For every n ∈ N, let Dn := { n+1
: k ∈ N and 0 ≤ k ≤ n + 1}.
Notice that ∪n∈N Dn is countable and dense in [0, 1].
¿RL ⇒ T2 ?
n-approximation
HB
For every x ∈ [0, 1], there is a unique k ∈ {0, . . . , n} such that
k
k+1
k
n ≤ x < n ; call the number n the n-approximation of x (in
Dn ).
RL ⇒ HB:
idea
RL ⇒ HB:
proof
8/11
The finite sets Dn
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
Notation
k
For every n ∈ N, let Dn := { n+1
: k ∈ N and 0 ≤ k ≤ n + 1}.
Notice that ∪n∈N Dn is countable and dense in [0, 1].
¿RL ⇒ T2 ?
n-approximation
HB
For every x ∈ [0, 1], there is a unique k ∈ {0, . . . , n} such that
k
k+1
k
n ≤ x < n ; call the number n the n-approximation of x (in
Dn ).
RL ⇒ HB:
idea
RL ⇒ HB:
proof
8/11
The finite sets Dn
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
Notation
k
For every n ∈ N, let Dn := { n+1
: k ∈ N and 0 ≤ k ≤ n + 1}.
Notice that ∪n∈N Dn is countable and dense in [0, 1].
¿RL ⇒ T2 ?
n-approximation
HB
For every x ∈ [0, 1], there is a unique k ∈ {0, . . . , n} such that
k
k+1
k
n ≤ x < n ; call the number n the n-approximation of x (in
Dn ).
RL ⇒ HB:
idea
RL ⇒ HB:
proof
Let B be an (infinite) boolean algebra. For every n ∈ N, let
Bn := B × {n}. Let Bω := ∪n∈N Bn Q
. Thus Bω is the union of ω
copies of B. We shall apply RL to n∈N DnBn .
8/11
Definition of σF for F ∈ fin∗ (Bω )
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
For every non-empty finite subset F of Bω , we define σF as
follows. Since F is of the form ∪0≤i≤n (Fi × {i}) where n ∈ N
and Fn is non-empty, consider the uniform probability P on
boolB (∪0≤i≤n Fi ), and, for every (x, i) ∈ Fi × {i}, let σF ((x, i))
be the i-approximation of P(x) (in Di ).
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
9/11
Definition of σF for F ∈ fin∗ (Bω )
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
For every non-empty finite subset F of Bω , we define σF as
follows. Since F is of the form ∪0≤i≤n (Fi × {i}) where n ∈ N
and Fn is non-empty, consider the uniform probability P on
boolB (∪0≤i≤n Fi ), and, for every (x, i) ∈ Fi × {i}, let σF ((x, i))
be the i-approximation of P(x) (in Di ).
For every F ∈ fin∗ (Bω ), for every (x, i), (x, j), (y , k) ∈ F :
1
(1B , i) ∈ F ⇒ σF ((1B , i)) = 1
RL ⇒ HB:
proof
9/11
Definition of σF for F ∈ fin∗ (Bω )
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
For every non-empty finite subset F of Bω , we define σF as
follows. Since F is of the form ∪0≤i≤n (Fi × {i}) where n ∈ N
and Fn is non-empty, consider the uniform probability P on
boolB (∪0≤i≤n Fi ), and, for every (x, i) ∈ Fi × {i}, let σF ((x, i))
be the i-approximation of P(x) (in Di ).
For every F ∈ fin∗ (Bω ), for every (x, i), (x, j), (y , k) ∈ F :
RL ⇒ HB:
idea
1
(1B , i) ∈ F ⇒ σF ((1B , i)) = 1
RL ⇒ HB:
proof
2
(0B , i) ∈ F ⇒ σF ((0B , i)) = 0
9/11
Definition of σF for F ∈ fin∗ (Bω )
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
For every non-empty finite subset F of Bω , we define σF as
follows. Since F is of the form ∪0≤i≤n (Fi × {i}) where n ∈ N
and Fn is non-empty, consider the uniform probability P on
boolB (∪0≤i≤n Fi ), and, for every (x, i) ∈ Fi × {i}, let σF ((x, i))
be the i-approximation of P(x) (in Di ).
For every F ∈ fin∗ (Bω ), for every (x, i), (x, j), (y , k) ∈ F :
RL ⇒ HB:
idea
1
(1B , i) ∈ F ⇒ σF ((1B , i)) = 1
RL ⇒ HB:
proof
2
(0B , i) ∈ F ⇒ σF ((0B , i)) = 0
3
|σF ((x, i)) − σF ((x, j))| ≤
1
i
+
1
j
9/11
Definition of σF for F ∈ fin∗ (Bω )
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
For every non-empty finite subset F of Bω , we define σF as
follows. Since F is of the form ∪0≤i≤n (Fi × {i}) where n ∈ N
and Fn is non-empty, consider the uniform probability P on
boolB (∪0≤i≤n Fi ), and, for every (x, i) ∈ Fi × {i}, let σF ((x, i))
be the i-approximation of P(x) (in Di ).
For every F ∈ fin∗ (Bω ), for every (x, i), (x, j), (y , k) ∈ F :
RL ⇒ HB:
idea
1
(1B , i) ∈ F ⇒ σF ((1B , i)) = 1
RL ⇒ HB:
proof
2
(0B , i) ∈ F ⇒ σF ((0B , i)) = 0
3
|σF ((x, i)) − σF ((x, j))| ≤
4
If x ∧ y = 0B and (x ∨ y , l) ∈ F then
|σF ((x ∨ y , l)) − σF ((x, i)) − σF ((y , k))| ≤
1
i
+
1
j
1
l
+
1
i
+ k1 .
9/11
RL implies HB
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
Q
Using RL, let f ∈ n∈N DnBn be a mapping respecting the
family (σF )F ∈fin∗ (Bω ) . For every n ∈ N, let fn : B → Dn be the
mapping x 7→ f (x, n).
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
10/11
RL implies HB
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
Q
Using RL, let f ∈ n∈N DnBn be a mapping respecting the
family (σF )F ∈fin∗ (Bω ) . For every n ∈ N, let fn : B → Dn be the
mapping x 7→ f (x, n).
Theorem
For every x ∈ B, the sequence (fn (x))n∈N is Cauchy so the
sequence (fn (x))n∈N converges to a real number m(x) ∈ [0, 1].
The mapping m : B → [0, 1] is a unitary measure on B.
RL ⇒ HB:
proof
10/11
RL implies HB
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
Q
Using RL, let f ∈ n∈N DnBn be a mapping respecting the
family (σF )F ∈fin∗ (Bω ) . For every n ∈ N, let fn : B → Dn be the
mapping x 7→ f (x, n).
Theorem
For every x ∈ B, the sequence (fn (x))n∈N is Cauchy so the
sequence (fn (x))n∈N converges to a real number m(x) ∈ [0, 1].
The mapping m : B → [0, 1] is a unitary measure on B.
Proof.
Given x ∈ B, Condition (3) implies that the sequence
(fn (x))n∈N is Cauchy. Conditions (1) and (2) imply that
m(1B ) = 1 and m(0B ) = 0. Condition (4) implies that m is a
measure.
10/11
References
Rado’s
selection
Lemma
implies
Hahn-Banach
M. Morillon
RL
T2 ⇒ RL
¿RL ⇒ T2 ?
HB
RL ⇒ HB:
idea
RL ⇒ HB:
proof
Luxemburg, W.A.J.
Reduced powers of the real number system and equivalents
of the Hahn-Banach extension theorem. Holt, Rinehart and
Winston, New York (1969).
Howard, Paul E.
Rado’s selection lemma does not imply the Boolean prime
ideal theorem. Z. Math. Logik Grundlag. Math., (30), no 2,
p. 129-132, (1984).
Rado, R.
Axiomatic treatment of rank in infinite sets, Canadian J.
Math., (1) p. 337-343 (1949).
11/11