A tale of three Boolean algebras - Analysis Group …hart/37/publications/talks/...2008/11/13 ·...

The three algebrasNo difference

Large differencesTopology

Other relationsEmbedding into P(N)/fin

A tale of three Boolean algebrasNon impeditus ab ulla scientia

K. P. Hart

Faculty EEMCSTU Delft

Delft, 13 November, 2008: 16:00–17:00

K. P. Hart A tale of three Boolean algebras

Outline

1 The three algebras

2 No difference

3 Large differences

4 Topology

5 Other relations

6 Embedding into P(N)/fin

Outline

2 No difference

3 Large differences

4 Topology

5 Other relations

Borel sets and ideals

Our starting point is B, the σ-algebra of Borel sets of [0, 1].

It has two natural ideals:

M, the sets of first category

N , the sets of measure zero

Ideal: closed under finite unions and taking subsets.M and N are even σ-ideals: closed under countable unions.

It has two natural ideals

Ideal: closed under finite unions and taking subsets.

M and N are even σ-ideals: closed under countable unions.

Category algebra

The Category algebra, C, is the quotient algebra B/M.

That is, A ∼ B iff A M B ∈M

It is isomorphic to the regular open algebra of [0, 1].This follows from

{A : A is equivalent to an open set} is a σ-algebra, hence

every Borel set is equivalent to an open set

if O is open then it is equivalent to int clO

a regular open set — O = int cl O — is only equivalent toitself

Operations: U ∨ V = int cl(U ∪ V ), U ∧ V = U ∩ V andU ′ = [0, 1] \ cl U.

Category algebra

The Category algebra, C, is the quotient algebra B/M.That is, A ∼ B iff A M B ∈M

Category algebra

It is isomorphic to the regular open algebra of [0, 1].

This follows from

Category algebra

{A : A is equivalent to an open set} is a σ-algebra

, hence

Category algebra

Measure algebra

The Measure algebra, M, is the quotient algebra B/N .

That is, A ∼ B iff A M B ∈ N

It is not isomorphic to any easliy recognizable algebra.

Unless you know M very well, in which case . . .

Measure algebra

The Measure algebra, M, is the quotient algebra B/N .That is, A ∼ B iff A M B ∈ N

Measure algebra

P(N)/fin

This is clearly a quotient: the power set of N modulo the ideal, fin,of finite sets.

Some common notations in this case

A =∗ B for A M B ∈ fin

A ⊆∗ B for A \ B ∈ fin

etcetera

Also P(N)/fin is not isomorphic to an easily recognizable algebra,unless of course . . .

P(N)/fin

etcetera

P(N)/fin

etcetera

P(N)/fin

etcetera

P(N)/fin

etcetera

P(N)/fin

etcetera

Also P(N)/fin is not isomorphic to an easily recognizable algebra

,unless of course . . .

P(N)/fin

etcetera

They are familiar, though

The algebras correspond to three ‘familiar’ objects from FucntionalAnalysis.

C goes with the Dedekind completion of C([0, 1]

)M goes with L∞[0, 1]

P(N)/fin goes with `∞/c0

We’ll see how that works later.

M goes with L∞[0, 1]

Outline

2 No difference

3 Large differences

4 Topology

5 Other relations

Tarski’s theorem

All three algebras are atomless.

This means that they are elementarily equivalent.That is, they satisfy the exact same Boolean algebraic sentences.

This was proved by Tarski in the 1930’s.

Tarski’s theorem

This means that they are elementarily equivalent.

That is, they satisfy the exact same Boolean algebraic sentences.

Tarski’s theorem

Outline

2 No difference

3 Large differences

4 Topology

5 Other relations

Separability, etc

Theorem

The algebra C is the union of countably many filters

For each rational q the family

Fq = {U : q ∈ U}

is a filter andC =

⋃{Fq : q ∈ Q ∩ [0, 1]

}We say C is σ-centered

Separability, etc

Theorem

The algebra C is the union of countably many filters:

Fq = {U : q ∈ U}

is a filter

andC =

⋃{Fq : q ∈ Q ∩ [0, 1]

Separability, etc

Theorem

Fq = {U : q ∈ U}

is a filter andC =

⋃{Fq : q ∈ Q ∩ [0, 1]

We say C is σ-centered

Separability, etc

Theorem

Fq = {U : q ∈ U}

is a filter andC =

⋃{Fq : q ∈ Q ∩ [0, 1]

Separability, etc

Theorem

The algebra M is not σ-centered.

Let {Fn : n ∈ N} be a family of filters in M; we assume they areultrafilters.

For each n pick Bn in Fn of measure less that 3−n.(Think of the intervals [i3−n, (i + 1)3−n) for i < 3n.)

The union⋃

n Bn has measure at most 12 ; its complement belongs

to no Fn.

Separability, etc

Theorem

Let {Fn : n ∈ N} be a family of filters in M

; we assume they areultrafilters.

The union⋃

to no Fn.

Separability, etc

Theorem

The union⋃

to no Fn.

Separability, etc

Theorem

The union⋃

to no Fn.

Separability, etc

Theorem

The union⋃

n Bn has measure at most 12

; its complement belongsto no Fn.

Separability, etc

Theorem

The union⋃

to no Fn.

Separability, etc

Theorem

The algebra M is σ-linked.

For each rational interval [p, q) put

Lp,q ={B : µ

(B ∩ [p, q)

2(q − p)

}if B1,B2 ∈ Lp,q then µ(B1 ∩ B2) > 0, so Lp.q is linked.

M is the union of the countably many Lp,q.

Separability, etc

Theorem

Lp,q ={B : µ

(B ∩ [p, q)

2(q − p)

if B1,B2 ∈ Lp,q then µ(B1 ∩ B2) > 0, so Lp.q is linked.

Separability, etc

Theorem

Lp,q ={B : µ

(B ∩ [p, q)

2(q − p)

Separability, etc

Theorem

Lp,q ={B : µ

(B ∩ [p, q)

2(q − p)

Separability, etc

Theorem

The algebra P(N)/fin has a pairwise disjont subset of size |R|.

Sierpinski: if x is irrational and larger than 1 put

Sx ={1

nbnxc : n ∈ N

}Sx is (the range of) a sequence of rational numbers thatconverges to x .

if x 6= y then Sx ∩ Sy =∗ ∅.so {Sx : x > 0 and x irrational} gives the set we seek.

Separability, etc

Theorem

Sx ={1

nbnxc : n ∈ N

Sx is (the range of) a sequence of rational numbers thatconverges to x .

Separability, etc

Theorem

Sx ={1

nbnxc : n ∈ N

Separability, etc

Theorem

Sx ={1

nbnxc : n ∈ N

if x 6= y then Sx ∩ Sy =∗ ∅.

so {Sx : x > 0 and x irrational} gives the set we seek.

Separability, etc

Theorem

Sx ={1

nbnxc : n ∈ N

Outline

2 No difference

3 Large differences

4 Topology

5 Other relations

Stone space

If B is a Boolean algebra then we let St(B) denote the set ofultrafilters on B.

For a ∈ B put a = {u ∈ St(B) : a ∈ u}

{a : a ∈ B} is a base for a topology on St(B) that iszero-dimensional compact Hausdorff and has {a : a ∈ B} as itsfamily of clopen sets.

Stone space

Three spaces

St(C) is separable

St(M) is not separable but it satisfies the countable chaincondition

St(P(N)/fin) does not satisfy the countable chain condition

Three spaces

St(C) is separable

St(M) is not separable

but it satisfies the countable chaincondition

Three spaces

St(C) is separable

Three spaces

St(C) is separable

Three Banach algebras

As promised:

C(St(C)

)is the Dedekind completion of C

([0, 1]

(St(M)

)is L∞[0, 1]

C(St(P(N)/fin)

)is `∞/c0

As promised:

C(St(C)

([0, 1]

C(St(M)

)is L∞[0, 1]

C(St(P(N)/fin)

)is `∞/c0

As promised:

C(St(C)

([0, 1]

(St(M)

)is L∞[0, 1]

C(St(P(N)/fin)

)is `∞/c0

As promised:

C(St(C)

([0, 1]

(St(M)

)is L∞[0, 1]

C(St(P(N)/fin)

)is `∞/c0

Outline

2 No difference

3 Large differences

4 Topology

5 Other relations

Embeddability

We investigate mutual embeddability of the algebras.

Useful observation:

Theorem

B is isomorphic to a subalgebra of A iff St(B) is a continuousimage of St(A)

So P(N)/fin cannot be embedded into M and C,and M cannot be embedded into C.

Embeddability

We investigate mutual embeddability of the algebras.Useful observation:

Theorem

Embeddability

Theorem

Embeddability

Theorem

So P(N)/fin cannot be embedded into M and C,

and M cannot be embedded into C.

Embeddability

Theorem

C into M?

A false proof would be to restrict the qutient homomrphismq : B → M to the regular open sets.Why?We can have disjoint regular open sets U and V such thatµ(U ∪ V ) < 1 yet U ∪ V is dense in [0, 1].Then U ∨ V = 1 in C, but q(U) ∨ q(V ) < 1 in M.

C into M?

Yes.A false proof would be to restrict the qutient homomrphismq : B → M to the regular open sets.

Why?We can have disjoint regular open sets U and V such thatµ(U ∪ V ) < 1 yet U ∪ V is dense in [0, 1].Then U ∨ V = 1 in C, but q(U) ∨ q(V ) < 1 in M.

C into M?

Yes.A false proof would be to restrict the qutient homomrphismq : B → M to the regular open sets.Why?

We can have disjoint regular open sets U and V such thatµ(U ∪ V ) < 1 yet U ∪ V is dense in [0, 1].Then U ∨ V = 1 in C, but q(U) ∨ q(V ) < 1 in M.

C into M?

Yes.A false proof would be to restrict the qutient homomrphismq : B → M to the regular open sets.Why?We can have disjoint regular open sets U and V such thatµ(U ∪ V ) < 1 yet U ∪ V is dense in [0, 1].

Then U ∨ V = 1 in C, but q(U) ∨ q(V ) < 1 in M.

C into M?

Yes.A false proof would be to restrict the qutient homomrphismq : B → M to the regular open sets.Why?We can have disjoint regular open sets U and V such thatµ(U ∪ V ) < 1 yet U ∪ V is dense in [0, 1].Then U ∨ V = 1 in C, but q(U) ∨ q(V ) < 1 in M.

C into M?

A correct proof is by transfinite recursion, using the completenessof M to find new values at each step.

Start with the natural inclusion of the algebra generated by therational intervals.

Sikorski’s extension criterion applies at each step.

C into M?

A correct proof is by transfinite recursion, using the completenessof M to find new values at each step.Start with the natural inclusion of the algebra generated by therational intervals.

C into M?

A correct proof is by transfinite recursion, using the completenessof M to find new values at each step.Start with the natural inclusion of the algebra generated by therational intervals.

Outline

2 No difference

3 Large differences

4 Topology

5 Other relations

Parovicenko’s theorem

Theorem

Every Boolean algebra of cardinality ℵ1 (or less) embeds intoP(N)/fin.

Hence the Continuum Hypothesis implies that C and M can beembedded into P(N)/fin.

Do we need the Continuum Hypothesis?

Theorem

Every Boolean algebra of cardinality ℵ1 (or less) embeds intoP(N)/fin.Hence the Continuum Hypothesis implies that C and M can beembedded into P(N)/fin.

Theorem

Every Boolean algebra of cardinality ℵ1 (or less) embeds intoP(N)/fin.Hence the Continuum Hypothesis implies that C and M can beembedded into P(N)/fin.

C into P(N)/fin

For C we don’t need CH; let Q = Q ∩ [0, 1]

Let {Vq : q ∈ Q} partition N into infinite sets.

Define Φ(U) =⋃{Vq : q ∈ Q ∩ U} (U regular open).

φ : U 7→ Φ(U)/fin is an embedding.

C into P(N)/fin

M into P(N)/fin

Assume φ : M → P(N)/fin is an embedding and take a lifting

Φ : B → P(N)

That is: φ([B]N

[Φ(B)

for all B.

Also Φ(A ∩ B) =∗ Φ(A) ∩ Φ(B), etc.We can ensure equality on the (countable) algebra C generated bythe rational intervals.

M into P(N)/fin

Φ : B → P(N)

That is: φ([B]N

[Φ(B)

for all B.

M into P(N)/fin

Φ : B → P(N)

That is: φ([B]N

[Φ(B)

for all B.

Also Φ(A ∩ B) =∗ Φ(A) ∩ Φ(B), etc.

We can ensure equality on the (countable) algebra C generated bythe rational intervals.

M into P(N)/fin

Φ : B → P(N)

That is: φ([B]N

[Φ(B)

for all B.

M into P(N)/fin

For convenience replace [0, 1] by R.

Enumerate Φ([n, n + 1)

)as 〈k(n, i) : i ∈ N〉

Choose m(n, i) < 3i such that

k(n, i) ∈ Φ(n + [m(n, i)3−i , (m(n, i) + 1)3−i )

)and put

Un = n +∞⋃i=1

[m(n, i)3−i , (m(n, i) + 1)3−i )

Observe µ(Un) ≤∑∞

i=1 3−i = 12

M into P(N)/fin

For convenience replace [0, 1] by R.Enumerate Φ

([n, n + 1)

)as 〈k(n, i) : i ∈ N〉

k(n, i) ∈ Φ(n + [m(n, i)3−i , (m(n, i) + 1)3−i )

)and put

Un = n +∞⋃i=1

[m(n, i)3−i , (m(n, i) + 1)3−i )

i=1 3−i = 12

M into P(N)/fin

([n, n + 1)

)as 〈k(n, i) : i ∈ N〉

k(n, i) ∈ Φ(n + [m(n, i)3−i , (m(n, i) + 1)3−i )

and put

Un = n +∞⋃i=1

[m(n, i)3−i , (m(n, i) + 1)3−i )

i=1 3−i = 12

M into P(N)/fin

([n, n + 1)

)as 〈k(n, i) : i ∈ N〉

k(n, i) ∈ Φ(n + [m(n, i)3−i , (m(n, i) + 1)3−i )

)and put

Un = n +∞⋃i=1

[m(n, i)3−i , (m(n, i) + 1)3−i )

i=1 3−i = 12

M into P(N)/fin

([n, n + 1)

)as 〈k(n, i) : i ∈ N〉

k(n, i) ∈ Φ(n + [m(n, i)3−i , (m(n, i) + 1)3−i )

)and put

Un = n +∞⋃i=1

[m(n, i)3−i , (m(n, i) + 1)3−i )

i=1 3−i = 12

M into P(N)/fin

Let U =⋃

n Un and F = R \ U.

Then µ(F ∩ [n, n + 1)) ≥ 12 for all n

So Φ(F ) ∩ Φ([n, n + 1)) is infinite.

Take the first in with k(n, in) ∈ Φ(F ) and put

In = n + [m(n, in)3−in , (m(n, in) + 1)3−in)

M into P(N)/fin

Let U =⋃

n Un and F = R \ U.

M into P(N)/fin

Let U =⋃

n Un and F = R \ U.

M into P(N)/fin

Let U =⋃

n Un and F = R \ U.

M into P(N)/fin

Let U =⋃

n Un and F = R \ U.

M into P(N)/fin

On the one hand⋃

n In ∩ F = ∅, hence Φ(⋃

n In) ∩Φ(F ) =∗ ∅.

On the other hand:⋃

n Φ(In) ∩ Φ(F ) 6=∗ ∅(it contains {k(n, in) : n ∈ N}).

So, if Φ is an embedding of C into P(N) then there will besequence like that of the In above.

That is, with Φ(⋃

n In) 6=∗ ⋃n Φ(In).

M into P(N)/fin

On the one hand⋃

n In) ∩Φ(F ) =∗ ∅.On the other hand:

⋃n Φ(In) ∩ Φ(F ) 6=∗ ∅

(it contains {k(n, in) : n ∈ N}).

n In) 6=∗ ⋃n Φ(In).

M into P(N)/fin

On the one hand⋃

⋃n Φ(In) ∩ Φ(F ) 6=∗ ∅

n In) 6=∗ ⋃n Φ(In).

M into P(N)/fin

On the one hand⋃

⋃n Φ(In) ∩ Φ(F ) 6=∗ ∅

n In) 6=∗ ⋃n Φ(In).

M into P(N)/fin

On the one hand⋃

⋃n Φ(In) ∩ Φ(F ) 6=∗ ∅

n In) 6=∗ ⋃n Φ(In).

Open Colouring Axiom

If X is separable and metrizable and X 2 = K0 ∪ K1 with K0 openand symmetric then

either there is an uncountable Y such that Y 2 ⊆ K0

or X =⋃

n Xn with X 2n ⊆ K1 for all n

This contradicts CH but it is consistent with the usual axioms ofSet Theory.

or X =⋃

The Open Colouring Axiom implies:

If there is an embedding of M into P(N)/finthen there are an embedding ψ and a lifting Ψ such that

Ψ embeds C into P(N)⋃n Ψ(In) =∗ Ψ(

⋃n In) for all sequences of intervals as above.

Which means that OCA implies there are no embeddings at all.

If there is an embedding of M into P(N)/fin

then there are an embedding ψ and a lifting Ψ such that

Ψ embeds C into P(N)

⋃n Ψ(In) =∗ Ψ(

A tale of three Boolean algebras - Analysis Group …hart/37/publications/talks/...2008/11/13 ·...

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