Nonseparable growth of the integers supporting a measure

0.13 MB·English

by  Piotr Drygier

NONSEPARABLE GROWTH OF THE INTEGERS
SUPPORTING A MEASURE
PIOTR DRYGIER AND GRZEGORZ PLEBANEK
5 Abstract. Assuming b = c (or some weaker statement), we construct a compactifi-
1
cation γω of ω such that its remainder γω \ω is nonseparable and carries a strictly
0
2 positive measure.

n
a
J
AMS subject classification: 54D35,54D65
8
2
1. Introduction
]
N
Weconsider here compactifications γω of thediscrete space ω. Given a compact space
G
. K, we say that K is a growth of ω if there is a compactification γω with the remainder
h
t γω\ω homeomorphic to the space K. It is well-known that every separable compactum
a
m is a growth of ω.
[ By a well-known theorem due to Parovi˘cenko [10], under the continuum hypothesis

1 every compact space K oftopological weight ≤ c is ancontinuous image ofthe remainder
v
ˇ
2 βω \ω of the Cech-Stone compactification of ω and, consequently, K is homeomorphic
7
to the remainder γω \ω of some compactification γω.
9
6 Let S be the Stone space of the measure algebra which may be seen as the quotient
0
. of Bor(2ω) modulo the ideal of null sets. Then S is a nonseparable compact space
1
0 that carries a strictly positive (regular probability Borel) measure, i.e. a measure that
5
1 is positive on every nonempty open subset of S. Since the topological weight of S is c,
:
v CH implies that S is a growth of ω. In fact, this may be done in such a way that the
i
X canonical measure on S is defined by the asymptotic density defined for subsets of ω,
r see Frankiewicz and Gutek [5].
a
Dow and Hart [3] proved that the space S is not a growth of ω if one assumes the
OpenColoring Axiom (OCA). Therefore it seems tobe interesting to investigate whether
one can always, in the usual set theory, construct a compactification γω such that the
remainder γω \ω is nonseparable but carries a strictly positive measure.
If a compact space carries a strictly positive measure then it satisfies ccc; the converse
does not hold which was already demonstrated by Gaifman [6]. Later Bell [2], van Mill
[9], Todorˇcevi´c [11] constructed several interesting examples of compactifications of ω
G. Plebanek was partially supported by NCN grant 2013/11/B/ST1/03596(2014-2017).
1
having nonseparable ccc remainders; cf. Todorˇcevi´c [12]. It seems that the structure of
all those examples exclude the possibility that ccc could be strenghten to saying that
the remainder in question supports a measure, see e.g. Lemma 3.2 in Dˇzamonja and
Plebanek [4].
Before we state our main result we need to fix some notation and terminology con-
cerning the set-theoretic assumption that we use. We denote by λ the usual product
measure on the Cantor set 2ω. Let E be an ideal of subsets A of 2ω for which λ(A) = 0.
Recall that the covering number cov(E) is the least cardinality of a covering of 2ω by
sets from E; cf. Bartoszyn´ski and Shelah [1] for cardinal invariants of the ideal E. We
shall sometimes write κ = cov(E) for simplicity. As usual, [κ ]≤ω denotes the family of
0 0
all countable subsets of κ . Recall that cof[κ ]≤ω, the cofinality of this partially ordered
0 0
set, is the least size of a family J ⊆ [κ ]≤ω such that every countable subset of κ is
0 0
contained in some J ∈ J. Our set-theoretic assumption (*) involves also b, the familiar
bounding number and reads as follows.
(∗) writing κ = cov(E) we have cof[κ ]≤ω ≤ b.
0 0
Theorem 1.1. Assuming (*) there is a compactification γω of the set of natural numbers
such that its remainder γω \ ω is not separable but carries a strictly positive regular
probability Borel measure.
Note that (∗) holds whenever b = c or κ = ω . We do not know whether Theorem
0 1
1.1 can be proved in the usual set theory. In connection with the result of Dow and Hart
mentioned above it is worth remarking that under OCA, b = ω and we can further
2
assume ω = c (see Moore [8]) so the compactification we construct here may exist even
2
when the Stone space of the measure algebra is not a growth of ω.
WeremarkthatinourproofofTheorem 1.1we construct γω such thatγω\ω supports
a measure µ of countable Maharam type (meaning that L (µ) is separable). We might,
1
however, modify the construction so that the resulting µ will be of type κ.
The paper is organized as follows. In section 2 we formulate Theorem 1.1 in terms of
subalgebras of P(ω) and finitely additive measures defined on them, see Theorem 2.2.
Section 3 describes an inductive construction using (*) that leads to 2.2. The key lemma
showing that the inductions works is postponed to the final section 4.
2. Compactifications and Boolean algebras
We denote by fin a family of finite subsets of ω. In the sequel, we shall consider
Boolean algebras (of sets) A such that fin ⊆ A ⊆ P(ω). Every such an algebra A
determines a compactification K of ω, where K may be seen as the Stone space of
A A
all ultrafilters on A. Then the algebra Clop(K∗) of the clopen subsets of the remainder
A
2
K∗ = K \ω is isomorphic to the quotient algebra A/fin. Hence K∗ is not separable if
A A A
and only if A/fin is not σ-centred.
Given an algebra A such that fin ⊆ A ⊆ P(ω), we shall consider finitely additive
probability measures µ on A that vanish on finite sets. Such a measure µ defines, via the
Stone isomorphism, a finitely additive measure µ on Clop(K ). Then µ extends uniquely
A
to a regular probability Borel measure µ on KbA such that µ(KA∗) =b1. Note that the
resulting Borel measure will be strictly positive whenever µ has the property mentioned
in the following definition.
Definition 2.1. If fin ⊆ A ⊆ P(ω) and µ is finitely additive measure on A then we shall
say that µ is almost strictly positive if for every A ∈ A, µ(A) = 0 if and only if A ∈ fin.
We can summarise our preliminary remarks and conclude that Theorem 1.1 is an
immediate consequence of the following result.
Theorem 2.2. Assume (∗). There exists a Boolean algebra A such that fin ⊆ A ⊆ P(ω)
and a finitely additive probability measure µ on A such that
(a) A/fin is not σ-centred;
(b) µ is almost strictly positive.
We shall prove Theorem 2.2 by inductive construction, gradually enlarging Boolean
algebras and extending measures. If X ⊆ ω then A[X] stands for an algebra generated
by A∪{X}. It is easy to check that A[X] = {(A∩X)∪(A′ \X): A,A′ ∈ A}.
If µ is finitely additive on A and Z ⊆ ω then we write
µ (Z) = sup{µ(A): A ∈ A,A ⊆ Z}, µ∗(Z) = inf{µ(A): A ∈ A,A ⊇ Z},
∗
for the corresponding inner- and outer-measure. Recall the following well-known fact
about extension of measures, see L o´s and Marczewski [7].
Proposition 2.3. The formula
µ (A∩X)∪(A′ \X) = µ (A∩X)+µ∗(A′ \X),
∗
(cid:0) (cid:1)
defines ean extension of µ to a finitely additive measure µ on A[X].
e
3. A construction
In this section we prove Theorem 2.2. We start by fixing a countable dense subset
D of the Cantor space 2ω playing the role of ω (and fin stands for finite subsets of D).
Let A be the subalgebra of P(D) generated by {C ∩D : C ∈ Clop(2ω)} and all finite
0
subsets of D. Further let µ be a measure on A defined by
0 0
µ (C △F) = λ(C) for C ∈ Clop(2ω) and F ∈ fin.
0
3
Clearly, µ is well-defined and almost strictly positive on A . Note that µ is nonatomic,
0 0 0
in the sense that every element of A may be written as a finite union of sets of arbitrary
0
small measure.
Our construction requires a certain bookkeeping of families of sequences in A . It will
0
be convenient to use the following definition.
Definition 3.1. A countable family S ⊆ (A )ω will be called an s−family if for every
0
S = (S(k)) ∈ S
k
(i) S(0) ⊇ S(1) ⊇ ... and S(k) = ∅;
Tk
(ii) lim λ (S(k)) = 0.
k 0
Let S be a countable infinite s−family together with some fixed enumeration S =
{S : n ∈ ω}. Then for any ϕ ∈ ωω we write
n
X (S) = S ϕ(n) .
ϕ [ n(cid:0) (cid:1)
n∈ω
We shall write below κ = cov(E) and κ = cof[κ ]≤ω for simplicity.
0 0
Choose a family {Z : α < κ } of closed subsets of 2ω such that λ(Z ) = 0 for every
α 0 α
α < κ and Z = 2ω. To every Z we associate S ∈ Aω as follows. Fix a bijection
0 Sα<κ0 α α α 0
d: ω → D. Write Z as an intersection of a decreasing sequence of C ∈ Clop(2ω) and
α α,k
set
S (k) = C ∩D \{d(0),...,d(k −1)}.
α α,k
Let J be a cofinal family in [κ ]≤ω of cardinality κ. Given J ∈ J, write SJ = {S :
0 α
α ∈ J}. Then SJ is an s−family in our terminology. For any ϕ ∈ ωJ we put
X (SJ) = S (ϕ(α)).
ϕ [ α
α∈J
Lemma 3.2. Assume (∗). There are a family (A ) of Boolean subalgebras of P(D)
ξ ξ<κ
and a family (µ ) , where every µ is a finitely additive measure defined on A , such
ξ ξ<κ ξ ξ
that, writing A = A , we have
Sξ<κ ξ
(i) |A | < κ for every ξ < κ;
ξ
(ii) µ is almost strictly positive on A ;
ξ ξ
(iii) µ |A = µ whenever ξ < η < κ;
η ξ ξ
(iv) for every J ∈ J there is ϕ ∈ ωJ such that X (SJ) ∈ A and ω\X (SJ) is infinite.
ϕ ϕ
Proof. Enumerate J as (J ) . We start from A and µ defined at the beginning of
ξ 1≤ξ<κ 0 0
this section. Fix ξ < κ. Given A and µ for β < ξ we apply Lemma 4.3 from section
β β
4 to the algebra B = ∪ A , the measure ν on B which extends all µ , β < ξ and an
β<ξ β β
4
s−family S = SJξ. Then Lemma 4.3 gives us a suitable X = X (SJξ) and we let A be
ϕ ξ
B[X] and µ be an extension of ν to an almost strictly positive measure on A . (cid:3)
ξ ξ
Now we check that the algebra A satisfying (i)-(iv) of Lemma 3.2 is the one we are
looking for.
Lemma 3.3. If A is an algebra is in 3.2 then A carries an almost strictly positive finitely
additive probability measure and A/fin is not σ-centred.
Proof. It is clear that if we let µ be the unique measure on A extending all µ ’s then µ
ξ
is as required.
Checking that A/fin is not σ-centred amounts to verifying that if {p : k ∈ ω} is a
k
family of nonprincipial ultrafilters on A then there is an infinite A ∈ A such that A ∈/ p
k
for every k.
Clearly every p defines the unique t ∈ 2ω which is in the intersection
k k
{C ∈ Clop(2ω) : C ∩D ∈ p }.
\ k
Then t ∈ Z for some α < κ . An s−family {S : k ∈ ω} is contained in SJ for some
k αk k 0 αk
J ∈ J. Take X (SJ) as in 3.2(iv). Then A = D\X (SJ) is an infinite set lying outside
ϕ ϕ
every p , as required. (cid:3)
k
4. Key lemma
We shall now prove an auxiliary result, stated below as Lemma 4.3, showing that the
inductive construction of Lemma 3.2 can be carried out. We follow here the notation
introduced in section 3; in particular, the notion of an s-family was introduced in 3.1.
Recall also that, given an s-family S = {S : i ∈ I}, any J ⊆ I and a function ϕ ∈ ωJ,
i
we write X (S) = S (ϕ(i)). We sometimes write X rather than X (S) if S is clear
ϕ Si∈J i ϕ ϕ
from the context.
Lemma 4.1. Let S = {S ,...,S } be a fixed finite s-family. For a given infinite set
0 n−1
A ⊆ ω we put
ΦA = {ϕ ∈ ωn: |A\X (S)| = ω}.
n ϕ
If we consider ΦA with the natural partial order then it has finitely many minimal ele-
n
ments.
Proof. It is pretty obvious that for every ϕ ∈ ΦA there is minimal ϕ′ ∈ ΦA such that
n n
ϕ′ ≤ ϕ.
Fix an infinite set A ⊆ ω. We prove the lemma by induction on n. For n = 1 it is
trivial since ΦA ⊆ ω is well-ordered. Assume that for any k < n and any infinite B ⊆ ω
1
the family ΦB has finitely many minimal elements.
k
5
Fix some minimal ϕ ∈ ΦA. Consider any non-empty set I ⊆ n of size less than n and
0 n
any function ψ ∈ ωI such that ψ ≤ ϕ ↾I and A\X is infinite (it is important that there
0 ψ
is only finitely many such ψ and sets I ⊆ n). By inductive assumption applied to A\X
ψ
and an s−family indexed by n\I, there exist finitely many minimal elements in ΦA\Xψ
n\I
(call them ϕj). Observe that any minimal element ϕ ∈ ΦA is of the form ϕ = ψ∪ϕj for
n
some j, so the proof is complete. (cid:3)
Lemma 4.2. Let B ⊆ P(D) be a Boolean algebra of size less than b and let S = {S :
n
n ∈ ω} be a fixed s−family in Bω.
There exists g ∈ ωω such that for any g ∈ ωω with g ≥ g , whenever A ∈ B and
0 0
A ⊆ X then A ⊆ S (g(j)) for some N ∈ ω.
g Sj≤N j
Proof. Fix A ∈ B. It follows from Lemma 4.1 that for any n ∈ ω there exists a finite
set I ⊆ ω such that if ϕ ∈ ωn and |A\X | = ω then (A\X )∩I 6= ∅. We inductively
n ϕ ϕ n
define a function h ∈ ωω such that
A
S h (n) ∩ I = ∅,
n(cid:0) A (cid:1) [ j
j≤n
which can be done since (S (k)) is a decreasing sequence with empty intersection.
n k
Now the family of functions {h : A ∈ B} is of size less than b so there exists g ∈ ωω
A 0
such that h ≤∗ g for any A ∈ B. We shall check that g is as required.
A 0 0
Take any A ∈ B and g ≥ g , and suppose that A ⊆ X . Let N ∈ ω be such that
0 g
h (n) ≤ g(n) for any n ≥ N. Write ϕ for the restriction of g to N. Note that the
A
set A \ X must be finite; indeed, otherwise ϕ ∈ ΦA so J = (A \ X ) ∩ I 6= ∅. But
ϕ N ϕ N
I ∩ S (g(n)) = ∅ for every n ≥ N, hence J ⊆ A \ X which means that A is not
N n g
contained in X , contrary to our assumption.
g
As A\X is finite, the lemma follows. (cid:3)
ϕ
We are now ready for the key lemma; recall that the measure µ on A was introduced
0 0
in section 3.
Lemma 4.3. Let B ⊆ P(D) be a Boolean algebra of size < b containing A and let ν
0
be a finitely additive almost strictly positive probability measure on B extending µ . Let
0
S = {S : n ∈ ω} be a fixed s−family contained in Aω.
n 0
There exists g ∈ ωω such that for X = X (S), ω\X is infinite and ν can be extended
g g g
to an almost strictly positive measure on B[X ].
g
Proof. For any X ⊆ P(ω) we may consider an extension ν of ν to a finitely additive
measure on B[X] given by the formula as in Propositione2.3. The plan is to find a
function g ∈ ωω such that the measure ν defined in this way is almost strictly positive
e
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on B[X ]. Observe that the extended measure ν will be almost strictly positive if for
g
every A ∈ B e
4.3(i) ν (A∩X ) > 0 whenever A∩X infinite,
∗ g g
4.3(ii) ν∗(A\X ) > 0 whenever A\X infinite,
ϕ ϕ
so we need to find g for which (i) and (ii) are satisfied..
Fix infinite A ∈ B and n ∈ ω. Consider ΦA from Lemma 4.1; as this set contains
n
finitely many minimal elements and ν is almost strictly positive, we can choose εA > 0
n
such that ν(A\X ) ≥ εA for every ϕ ∈ ΦA. Note that the sequence of εA is decreasing.
ϕ n n n
We inductively define h ∈ ωω so that
A
εA
ν S h (n) ≤ n .
(cid:16) n(cid:0) A (cid:1)(cid:17) 2n+2
This can be done since S (k) ∈ A , ν|A = µ and lim µ S (k) = 0 by our
n 0 0 0 k→∞ 0 n
(cid:0) (cid:1)
definition of an s−family.
Let g ∈ ωω be a function as in Lemma 4.2 Since |B| < b, there exists a function
0
g ∈ ωω such that g (n) ≥ g (n) for every n and h ≤∗ g for every A ∈ B; say that
1 1 0 A 1
h (n) ≤ g (n) for every n ≥ N .
a 1 A
Claim. For every g ≥ g , if A ∈ B and A\X is infinite then ν∗(A\X ) ≥ 1εA .
1 g g 2 NA
Let ϕ be the restriction of g to N ; write Y = ∞ S (g(n)). With this notation
A Sn=NA n
we have A\X = (A\X )\Y, where A\X ∈ B.
g ϕ ϕ
Note first that ν∗(A\X ) ≥ ν(A\X )−ν (Y); indeed if B ∈ B, B ⊇ A\X then
g ϕ ∗ g
(A\X )\B ⊆ Y so
ϕ
ν(A\X )−ν(B) ≤ ν((A\X )\B) ≤ ν (Y).
ϕ ϕ ∗
Further note that ν(A \ X ) ≥ εA so to verify Claim it is sufficient to check that
ϕ NA
ν (Y) ≤ εA /2. Since g ≥ g , it follows from Lemma 4.2 that whenever B ∈ B and
∗ NA 1 0
B ⊆ Y then B is contained in S (g(n)) for some k. Therefore, using finite
SNA≤n≤k n
additivity of ν, we get
k k
1 1 1
ν(B) ≤ ν S g(n) ≤ εA ≤ εA ≤ εA .
X (cid:16) n(cid:0) (cid:1)(cid:17) X 2n+2 n 2NA+1 NA 2 NA
n=NA n=NA
This shows that indeed ν (Y) ≤ εA /2, and the proof of Claim is complete.
∗ NA
Note that by an analogous argument we check that if g ≥ g then ν (X ) ≤ 1/2 so
1 ∗ g
ω \X must be infinite.
g
We have proved that every g ≥ g guarantees 4.3(ii) so to complete the proof we need
1
to find g ≥ g for which 4.3(i) holds.
1
7
We again apply a diagonal argument using |B| < b. Given A ∈ B, let f (n) = g (n)
A 1
if |A ∩ S (k)| = ω for every k. Otherwise, if A ∩ S (k) is finite for some k we can
n n
take f (n) ≥ f (n) such that A ∩ S (f (n)) = ∅ (recall that S (k) = ∅). Finally,
A 1 n A Tk n
let g ∈ ωω be a function that eventually dominates every f , A ∈ B. If A ∈ B and
A
ν (A∩X ) = 0 then A∩S (g(n)) is finite for every n. Taking N such that g(n) ≥ f (n)
∗ g n A
for every n ≥ N, and writing ϕ for the restriction of g to N we get A ⊆ X so A is finite
ϕ
itself. Now g is so that both 4.3(i)-(ii) are satisfied, and the proof is complete. (cid:3)
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Instytut Matematyczny, Uniwersytet Wrocl awski
E-mail address: [email protected]
Instytut Matematyczny, Uniwersytet Wrocl awski
E-mail address: [email protected]
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