Weakly infinite dimensional subsets of R^N

0.22 MB·English

by  Liljana Babinkostova

WEAKLY INFINITE DIMENSIONAL SUBSETS OF RN
LILJANABABINKOSTOVAANDMARIONSCHEEPERS
9
0
0
2

n Abstract. TheContinuumHypothesisimpliesanErdo¨s-Sierpin´skilikedual-
a itybetweentheidealoffirstcategorysubsetsofRN,andtheidealofcountable
J dimensional subsets of RN. The algebraic sum of a Hurewicz subset - a di-

2 mensiontheoretic analogueofSierpinskisetsandLusinsets -ofRN withany
2 compactly countabledimensionalsubsetofRN hasfirstcategory.

]
N 1. Introduction.
G The main purpose of this paper is threefold: We point out strong analogies
. betweenthe well-knownclassofcountabledimensionalsubsets ofthe Hilbertcube,
h
and the classes of Lebesgue measure zero subsets of the real line and the class of
t
a
first category subsets of complete separable metric spaces. A classical theorem of
m
Erdo¨s and Sierpin´ski gives, under the Continuum Hypothesis, some explanation of
[ thisanalogy. WeshowalsothatthedimensiontheoreticanalogueofLusinsetsand

2 SierpinskisetshassomeofthedeeperpropertiessharedbyLusinsetsandSierpin´ski
v sets. And we give some information about a class of weakly infinite dimensional
1 spaces which is emerging as an important class in dimension theory. In particular
6
weshow,usingtheContinuumHypothesis,thatthemostrestrictiveoftheseclasses
6
contains metric spaces which are not countable dimensional.
3
. RN denotes the Tychonoff product of countably many copies of the real line, R.
1 The symbol [0, 1] denotes the unit interval and the subspace [0,1]N of RN denotes
1
the Hilbert cube. The symbol P denotes the set of irrationalnumbers, viewed as a
8
0 subspace of the real line.
:
v 2. The countable dimensional subsets of RN
i
X
Hurewicz [9] defined a subset of a topologicalspace to be countable dimensional
r
a ifitisaunionofcountablymanyzero-dimensionalsubsetsofthespace,andproved
that the Hilbert cube, thus also RN is not countable dimensional. Thus, the collec-
tion CD of countable dimensional subsets of RN is a σ-ideal. Since each metrizable
space of cardinality less than 2ℵ0 is zerodimensional, the cofinality of CD, denoted
cof(CD), is 2ℵ0. Smirnov and independently Nagami also showed that every sepa-
rable metric space is a union of ℵ zero-dimensionalsubsets, and thus the covering
1
number of CD, denoted cov(CD), is ℵ . This is unlike the σ-ideal M of first cate-
1
gorysubsets ofR, andthe σ-idealN ofLebesgue measurezerosubsetsofR, where
the cofinality and the covering number are not decided by ZFC.
If one assumes the Continuum Hypothesis (CH), these differences vanish and
one can give an explanation for the unusually strong analogies between the theory
of the countable dimensional subspaces of RN and the theory of the first category
subsets of RN: These two collections are dual in the sense of Erdo¨s and Sierpin´ski.
Chapter 19 of [17] gives a nice exposition of the following classical theorem:
1
2 LILJANABABINKOSTOVAANDMARIONSCHEEPERS
Theorem 1 (Erd¨os-Sierpin´ski Duality Theorem). Let X be a set of cardinality
ℵ . Let I and J be σ-complete ideals on X such that:
1
(1) X =A B where A and B a disjoint sets with A∈I and B ∈J.
S
(2) cof(I)=ℵ =cof(J).
1
(3) For each I ∈I there is an S ⊂X\I such that |S|=ℵ and S ∈I.
1
(4) For each J ∈J there is an S ⊂X \J such that |S|=ℵ and S ∈J.
1
Then there is a bijective function f :X −→ X which is its own inverse, such that
for each set E ⊆X we have
E ∈I ⇔f[E]∈J.
A classicaltheoremof Tumarkin[28] is animportanttoolinprovingthe duality
between countable dimensional sets and first category sets of RN under CH:
Theorem 2 (Tumarkin). In a separable metric space each n-dimensional set is
contained in an n-dimensional G -set.
δ
Corollary 3. Assume the Continuum Hypothesis. There is a bijection f :RN −→
RN such that f =f−1 and for each set E ⊂RN, E is countable dimensional if, and
only if, f[E] is first category.
Proof: InTheorem1takeI tobeCD,theidealofcountabledimensionalsubsets
of RN, and J to be M, the ideal of first category subsets of RN. By Hurewicz’s
theoremthatRN is notcountabledimensional,CD isaσ-complete idealonRN. By
the Baire category theorem M is a σ-complete ideal on RN.
SinceRNisseparable,letDbeacountabledenseset. ThenDiszero-dimensional.
By Theorem 2 there is a zero-dimensional G set A⊃D. But then B = RN\A is
δ
a first category set, and so (1) of Theorem 1 is satisfied.
Theorem2alsoimpliesthateachcountabledimensionalsubsetofRNiscontained
in a countable union of zero-dimensionalG sets, and so CD has a cofinalfamily of
δ
cardinality 2ℵ0. Since M has a cofinal family of cardinality 2ℵ0, C.H. implies that
(2) of Theorem 1 holds. Evidently (3) and (4) also hold. (cid:3)
3. Hurewicz sets.
Some strong analogies between the theory of the ideal of countable dimensional
subsets of RN and the theories of the ideal N of Lebesgue measure zero subsets
of the real line, and the ideal M of first category subsets of uncountable complete
separable metric spaces emerge when we consider the following features: Call a
subset of RN a Hurewicz set if it is uncountable, but its intersection with every
zero-dimensional subset of RN is countable. Hurewicz sets are first category: Take
a countable dense set D ⊂ RN. Since D is zerodimensional it is, by Theorem 2,
contained in a zerodimensional dense G set E. Since a Hurewicz set meets E in
δ
only countably many points the Hurewicz set is first category. We shall see below
that Hurewicz sets are first category in a strong sense.
The notion of a Hurewicz set is analogousto the notions of a Lusin set and of a
Sierpinski set: A subset of R is a Lusin set if it is uncountable but its intersection
with any first category subset of R is countable. A subset of the real line is a
Sierpinski set if it is uncountable but its intersection with each Lebesgue measure
zero set is countable.
Mahlo[15]andLusin[14]independentlyshowedthattheContinuumHypothesis
implies the existence of a Lusin set of cardinality 2ℵ0, and Sierpinski [26] showed
WEAKLY INFINITE DIMENSIONAL SUBSETS OF RN 3
that the Continuum Hypothesis implies the existence of a Sierpinski set of cardi-
nality 2ℵ0. It is known that the existence of a Lusin set of cardinality 2ℵ0 does
not imply the Continuum Hypothesis, and that the existence of a Sierpin´ski set of
cardinality 2ℵ0 does not imply the Continuum Hypothesis. But Rothberger [25]
provedthat the simultaneousexistence ofboth a Lusinsetof cardinality2ℵ0 anda
Sierpinskisetofcardinality2ℵ0 isequivalentto2ℵ0 =ℵ1. W.Hurewicz[10]showed
that the existence of a Hurewicz set is equivalent to the Continuum Hypothesis.
SomedeeperpropertiesoftheLusinsetsandthe Sierpin´skisetsaretiedupwith
the algebraic properties of the real line viewed as a topological group.
Theorem 4 (Galvin-Mycielski-Solovay). If L ⊂ R is a Lusin set, then for each
first category set M ⊂R, L+M 6=R.
Galvin,MycielskiandSolovay[6]provedasignificantlystrongertheorem: Strong
measure zero sets are characterized as sets whose translates by first category sets
donotcoverthe realline. Theorem4followsfromthis sinceSierpin´skiprovedthat
Lusin sets are strong measure zero. Pawlikowski [19] proved the counterpart for
Sierpin´ski sets:
Theorem 5 (Pawlikowski). If S ⊂ R is a Sierpin´ski set, then for each Lebesgue
measure zero set N ⊂R, S+N 6=R.
WeexpectthatasimilarstatementistrueaboutHurewiczsetsinthetopological
group (RN,+):
Conjecture 1. If H ⊂RN is a Hurewicz set, then for each countable dimensional
set C ⊂RN, H +C 6=RN.
We have a partial result on this conjecture: Call a subset of RN strongly count-
able dimensional if it is a union of countably many closed, finite dimensional sub-
sets. This notion was introduced by Nagata [16] and Smirnov [27]. Every strongly
countable dimensional set is countable dimensional, but not conversely. Thus, the
σ-ideal SCD generated by the strongly countable dimensional subsets of RN is a
proper subideal of CD. Indeed, every strongly countable dimensional set is of first
category in RN, and thus SCD is also a subideal of the σ-ideal of first category
subsets of RN. Call a subset of RN compactly countable dimensional if it is a union
of countably many compact finite dimensional sets. The σ-deal KCD generated by
the compact finite dimensional sets is a proper subideal of SCD. We shall show
Theorem 6. If H is a Hurewicz subset of RN, then for every compactly countable
dimensional set C ⊂RN, H +C is a first category subset of RN.
Sincetheunionofcountablymanyfirstcategorysetsisfirstcategory,Theorem6
followsdirectlyfromLemma8below. ThisLemmausesanotherimportantclassical
result:
Theorem 7 (Hurewicz, Tumarkin). In a separable metric space the union of
countably many closed, n-dimensional sets, is an n-dimensional set.
Lemma 8. If H ⊆RN is a Hurewicz set and C ⊂RN is a compact n-dimensional
set, then H +C is a first category subset of RN.
Proof: Let D ⊂ RN be a countable dense set. Then D − C is a union of
countably many closed n-dimensional sets, so by Theorem 7 it is n-dimensional.
4 LILJANABABINKOSTOVAANDMARIONSCHEEPERS
By Theorem 2 choose open sets U ⊃ U ⊃ ··· ⊃ U ⊃ ··· ⊃ D −C such that
1 2 k
∞ U is n-dimensional. Then put Y = {x ∈ RN : x−C ⊆ ∞ U }. Then Y
Tk=1 k Tk=1 k
is dense in RN. Note that Y is a G -set: Consider a k, and an x ∈ Y. Since C is
δ
compact there is an open neighborhood V (x) of x such that V (x)−C ⊂U . But
k k k
then V = ∪ V (x) is an open set containing Y, and V −C ⊂ U . It follows
k x∈Y k k k
that Y =∩ V .
k<∞ k
Next, put
∞
X =H ∩(\ Uk).
k=1
Since H is a Hurewicz set and ∞ U is n-dimensional, X is a countable set.
Tk=1 k
Define Z = Y \ (X + C). Since X + C is a union of countably many closed,
n-dimensional sets, it is an n-dimensional F -set and thus first-category in RN.
σ
Consequently Z is co-meager. But
Z ={x∈Y :(x−C)∩H =∅}⊆Y \(H +C).
Thus, H +C is meager. (cid:3)
One can show that if there are Hurewicz sets, then there are ones satisfying the
property in Conjecture 1. There are several well-studied classes of weakly infinite
dimensional subsets of RN which are not countable dimensional. In Example 1 we
give, under CH, an example of a set S in a very restrictive class of weakly infinite
dimensional spaces (but not countable dimensional), and a Hurewicz set H, such
that S+H =RN. The following is an unresolvedweaker instance of Conjecture 1:
Problem 1. Is it true that if C is a strongly countable dimensional set in RN and
H is a Hurewicz set, then H +C 6=RN?
4. New classes of weakly infinite dimensional spaces.
Manyexamplesininfinitarydimensiontheorybelongtoclassesofweaklyinfinite
dimensionalspacesintroducedin[2]. OurexamplesinconnectionwithConjecture1
are also in these classes. These classes are defined by applying selection principles
to specific types of open covers. We now introduce these covers and selection
principles.
Classes of open covers
For a given space X the symbol O denotes the collection of all open covers of
X. The following special types of open covers are relevant to our discussion:
• Ofd: This is the collection of all open covers U such that there is for each
finite dimensional F ⊂X a U ∈U with F ⊆U, but X is not a member of
U. (Thus, we are assuming X is not finite dimensional.)
• Ocfd: This is the collection of all open covers U such that there is for each
closed finite dimensional F ⊂ X a U ∈ U with F ⊆ U, but X is not a
member of U. (Thus, we are assuming X is not finite dimensional.)
• Okfd: This is the collection of all open covers U such that there is for
each compact finite dimensional F ⊂ X a U ∈ U with F ⊆ U, but X is
not a member of U. (Thus, we are assuming X is not compact and finite
dimensional.)
• Ok: This is the collection of all open covers U such that there is for each
compact F ⊂X aU ∈U withF ⊆U,but X is nota memberofU. (Thus,
we are assuming X is not compact.)
WEAKLY INFINITE DIMENSIONAL SUBSETS OF RN 5
• Ω: This is the collection of ω covers of X. An open cover U of X is an
ω-cover if X is not a member of U, but for each finite subset F of X there
is a U ∈U with F ⊆U.
• Γ: Thisisthecollectionofγ coversofX. AnopencoverU ofX isaγ-cover
if it is infinite and every infinite subset of U is a cover of X.
• Ogp: This is the collection of groupable open covers: An open cover U of
a space X is groupable if there is a partition U = F into finite sets
Sn∈N n
F that are disjoint from each other, such that for each x∈X there is an
n
N with x∈ F for all n≥N.
S n
We have the following inclusion relations among these classes of open covers:
Ofd ⊂Ocfd ⊂Okfd ⊂Ω⊂O; Ok ⊂Okfd; Γ⊂Ω.
It is also worth noting that if X is a separable metric space then each U ∈Okfd
has a countable subset V ∈Okfd. The same is true about Ok.
Three selection principles.
Let A and B be families of sets. The selection principle, S (A,B) states:
1
Foreachsequence(A :n∈N)ofelementsofAthereisasequence
n
(B : n ∈ N) such that for all n we have B ∈ A , and {B : n ∈
n n n n
N}∈B.
The selection principle S (A,B) states:
fin
Thereisforeachsequence(A :n∈N)ofmembersofAasequence
n
(B : n ∈ N) of finite sets such that for each n we have B ⊂ A
n n n
and B ∈B.
Sn∈N n
The selection principle S (A,B) states:
c
Foreachsequence(A :n∈N)ofelementsofA,thereisasequence
n
(B :n∈N) whereeachB is a refinementofthe collectionofsets
n n
A ,eachB isapairwisedisjointfamily,and B isamember
n n Sn∈N n
of B.
InourcontextthefamiliesAandB willbeclassesofopencoversofatopological
space. The selection principle S (A,B) has the following monotonicity properties:
1
If A⊂C and B ⊂D, then we have the following implications:
S (C,B)⇒S (A,B); S (A,B)⇒S (A,D).
1 1 1 1
Replacing S with S or S in these implications give corresponding facts for the
1 fin c
other two selection principles.
Specialinstancesofthesethreeselectionprinciplesappearinclassicalliterature.
For example: S (O,O) is known as Rothberger’s property and was introduced in
1
the1938paper[24]. S (O,O)isknownasMenger’spropertyandwasintroduced
fin
in the 1925paper [7]. S (O,O) is known as property C, and was introduced in the
c
1978paper[1]. S (O,O)isaselectiveversionofBing’spropertyofscreenability[3],
c
thus also known as selective screenability. Selective screenability is an important
propertyininfinitarydimensiontheory. LetO denotethecollectionofopencovers
2
U with |U| ≤ 2. Then S (O ,O) is equivalent to Alexandroff’s notion of weakly
c 2
infinite dimensional. A space whichis notweakly infinite dimensionalis saidto be
strongly infinite dimensional.
The class S1(Okfd,O)
6 LILJANABABINKOSTOVAANDMARIONSCHEEPERS
MetrizablespaceswithpropertyS1(Okfd,O)seemimportant. First: Theorem17
of [2] shows that for separable metric spaces S1(Ofd,O)⇒Sc(O,O). Monotonicity
propertiesofthe selectionprinciplesgive: S1(Okfd,O)⇒S1(Ocfd,O)⇒S1(Ofd,O).
Thus for metrizable spaces S1(Okfd,O) implies the important property Sc(O,O).
Second: The following lemma shows that these spaces have the classical Menger
property S (O,O):
fin
Lemma 9. S1(Okfd,O)⇒S1(Ok,O)⇒Sfin(O,O).
Third: Severalcentralexamplesfromthetheoryofinfinitedimensionalspacesare
inthis class. Forexample: R.Pol’sexample in[22]whichshowsthata weaklyinfi-
nitedimensionalspaceneednotbecountabledimensionalisintheclassS1(Okfd,O).
The preservation of infinitary dimension properties by products is not well un-
derstood yet. Whenever a new class of infinite dimensionality is defined it is of
interest to know how this class behaves under various mathematical constructions,
like products. We now make some remarks about products by spaces in the class
S1(Okfd,O), using recent examples constructed by E. Pol and R. Pol in [21]:
Proposition 10. Assume Martin’s Axiom.
(1) Finite powers of sets in S1(Okfd,O) need not be in Sc(O,O).
(2) The product of a space in S1(Okfd,O) with the space of irrational numbers
need not be weakly infinite dimensional.
To see this consider the following: In the proof of Proposition 4.1 of [21] E.
Pol and R. Pol construct, using Martin’s Axiom, spaces E and E which have
0 1
S1(Okfd,O). To see that these two spaces have this property we consider elements
oftheargumentgiveninpart(6)oftheproofof[21],Proposition4.1: Fori∈{0,1}
the spaces E are of the form E∪S where
i i
• S is the union of countably many compact, finite dimensional subspaces,
i
and
• For each open set W ⊃Si in Ei, |Ei\W|<2ℵ0.
Consider Ei. Let (Un : n < ∞) be a sequence in Okfd. Write Si as a union of
compactfinite dimensionalsubspacesSi ⊂Si ⊂···⊂Si ⊂···. Foreachn choose
1 2 n
a U ∈ U with Si ⊆ U . Then W = U is an open set
2·n−1 2·n−1 n 2·n−1 Sn<∞ 2·n−1
containing Si, and so Ei \W has cardinality less than 2ℵ0. But Martin’s Axiom
implies that a separable metric space of cardinality less than 2ℵ0 has Rothberger’s
property S (O,O) (Theorem 5 of [13]). Thus, choose for eachn a U ∈U such
1 2·n 2·n
that E \W ⊆ U .
i Sn<∞ 2·n
ItfollowsthatthespaceX constructedthereasthefreeunionofE andE still
0 1
has S1(Okfd,O). Recall that a metric space (X,d) has the Haver property if there
is for eachsequence (ǫ :n<∞) ofpositive realnumbers a sequence (U :n<∞)
n n
such that each V is a pairwise disjoint family of open sets, each of d-diameter less
n
than ǫ , and U is a cover of X. Metrizable spaces with property S (O,O)
n Sn<∞ n c
havetheHaverpropertyinallequivalentmetrics. Since[21]showsthatX2 hasthe
Menger property but under some metric X2 does not have the Haver property, it
follows that X2 does not have property S (O,O).
c
Toseetheseconditem: InTheorem6.1of[21]itispointedoutthatthe product
of the space E with the subspace of irrationalnumbers of the real line is not even
i
S (O ,O), that is, is strongly infinite dimensional.
c 2
The class S1(Okfd,Ω)
WEAKLY INFINITE DIMENSIONAL SUBSETS OF RN 7
S1(Ok,Ω) was considered in Section 2 of [18]. Theorem 2.1 of [18] implies:
Lemma 11 (Pansera, Pavlovic). A space has S1(Ok,Ω) if, and only if, it has
S1(Ok,O) in all finite powers.
Proposition 12. IfaspacehaspropertyS1(Okfd,Ω)thenithaspropertySfin(Ω,Ω).
Proof: Let X be a space with S1(Okfd,Ω). By monotonicity properties of S1(·,·)
X then has S1(Ok,Ω). Then by Lemma 11 Xn has S1(Ok,O) for all finite n. By
Corollary 9 Xn has S (O,O) for all n. By Theorem 3.9 of [11] this is equivalent
fin
to: X has S (Ω,Ω). (cid:3)
fin
Usingstandardtechniquesonecanprove: IfaspacehasthepropertyS1(Okfd,Ω),
then it has the property S1(Okfd,O) in all finite powers. We don’t know if the
converse is true:
Problem 2. For metrizable space X is it true that if each Xn has property
S1(Okfd,O) then X has S1(Okfd,Ω)?
It islikely thatthe spacesE0 andE1 of[21] mentionedaboveareinS1(Okfd,Ω),
but we have not examined this. The space X which is the free union of E and E
0 1
isanexampleofaspaceinS1(Okfd,O)butnotinS1(Okfd,Ω). Thus,underMartin’s
Axiom, S1(Okfd,Ω) is a proper subclass of S1(Okfd,O). Below we construct under
CH an example (Example 1) of a space which has the property S1(Okfd,Ω) and
which is not countable dimensional.
The class S1(Okfd,Γ)
S1(Ok,Γ)wasexplicitlyconsideredin[4],Section3,andin[5],Section3,wherethe
second implication of Corollary 13 is noted.
Corollary 13. S1(Okfd,Γ)⇒S1(Ok,Γ)⇒Sfin(Ω,Ogp).
Theorem 12 of [4] shows that each σ-compact space has the property S1(Ok,Γ).
ThisisnottrueforS1(Okfd,Γ): TheHilbertcubeisnotweaklyinfinitedimensional,
but is a σ-compact space. Combining [5] Theorem 10 and [5] Proposition 13 gives
the following result:
Lemma 14 (Di Maio, Kocinac, Meccarriello). If X has S1(Ok,Γ), then for all n
also Xn has this property.
The Hurewicz covering property is a strengthening of the Menger property
S (O,O)andisdefinedasfollows: TopologicalspaceX hastheHurewiczcovering
fin
property if: Foreachsequence(U :n<∞)ofopencoversofX thereisasequence
n
(V :n <∞) of finite sets such that for each n, V ⊆ U , and for each x∈ X, for
n n n
all but finitely many n, x∈ V . It was shown in [12] that the Hurewicz covering
S n
proprty is equivalent to the selection principle S (Ω,Ogp).
fin
Corollary 15. If a space X has property S1(Okfd,Γ) then it has Sfin(Ω,Ogp) in
all finite powers.
Using the technique in the proof of [2] Theorem 17 and using the techniques in
[2] Lemma 6 one obtains the following two results:
Theorem 16. S1(Ofd,Ogp)⇒Sc(O,Ogp).
Lemma 17. Ifaspace X has S1(Okfd,Γ), thenit has Sc(O,O) inall finitepowers.
8 LILJANABABINKOSTOVAANDMARIONSCHEEPERS
Proof: Let X be a space with property S1(Okfd,Γ). By Lemma 13 it has property
S1(Ok,Γ), and by Lemma 14 it has S1(Ok,Γ) in all finite powers, and thus the
Hurewiczpropertyinallfinitepowers. ThenbyCorollary13of[2],allfinitepowers
of X has the property S (O,O). (cid:3)
c
Let X be an infinite dimensional space which satisfies S1(Okfd,Γ). By [2] and
Corollary13,allfinitepowersofX havetheHaverproperty,andindeed,theproduct
of X with any space having the Haver property again has the Haver property. It
followsthat X×P has the Haverproperty. We don’t know if this product must be
selectively screenable:
Problem 3. If X has S1(Okfd,Γ) then does X×P have Sc(O,O)?
5. Examples.
We now describe two infinite dimensional examples that will be used to demon-
strate a number of facts in connection with the preceding sections. In both cases
we are interested in showing properties of Hurewicz sets, and thus we are confined
to assuming the Continuum Hypothesis.
A ZFC+CH example of S1(Okfd,Ω).
Write: R := {x ∈ RN : (∀j > n)(x(n) = 0)} and R := ∞ R . Standard
n ∞ Sn=1 n
arguments prove Lemma 18 and Corollary 19 below.
Lemma 18. Assume G⊃R is an open subset of RN and I , ··· , I are closed
∞ 1 n
intervalsofpositivelength. Foreachf ∈RN thereareanm>nandclosedintervals
I , ··· , I of positive length such that
n+1 m
I × ··· ×I × RN ⊆G−f.
1 m
Corollary 19. If (G :n<∞) is a sequenceof G subset of RN suchthat each G
n δ n
containsR ,andif(f :n<∞)issequenceofelementsofRN,then ∞ (G −f )
∞ n Tn=1 n n
contains a Hilbert cube.
Example 1 (CH). There are subsets X and Y of RN such that:
(1) X and Y satisfy S1(Okfd,Ω);
(2) X \R and Y \R are Hurewicz sets;
∞ ∞
(3) X and Y have S1(Okfd,Ω) and X ×Y not.
(4) X ∪Y is S1(Okfd,O), and not S1(Okfd,Ω).
Fix the following enumerations:
• (fα :α<2ℵ0), the list of all elements of RN
• (Fα :α<2ℵ0), the list of all finite dimensional Gδ subsets of RN
• (Lα :α<2ℵ0), the list of all Gδ subsets of RN containing R∞
• ((Unα :n<ω):α<2ℵ0) where for each α each element of R∞ is in Unα for
all but finitely many n, and each Uα is open in RN.
n
For each α<2ℵ0 the set Tα =Tn<∞(Sm≥nUmα) is a Gδ set containing R∞.
Since G = T ∩L is a dense G set which contains R . Thus G ∩(G −f )
0 0 0 δ ∞ 0 0 0
contains, by Corollay 19, a homeomorphic copy of the Hilbert cube. Choose
x ∈(G ∩(G −f ))\((R ∪F )∪((R ∪F )−f ))
0 0 0 0 ∞ 0 ∞ 0 0
WEAKLY INFINITE DIMENSIONAL SUBSETS OF RN 9
and fix y ∈ G with x = y − f . Then define S1(0) = {n : x ∈ U0} and
0 0 0 0 0 0 0 n
S2(0)={n:y ∈U0}.
0 0 n
Assume we have 0 < α < ω , and that we have selected for each γ < α an x
1 γ
and a y and defined sets S1(γ) and S2(γ), ν ≤γ such that:
γ ν ν
(1) δ <γ <α⇒S1(δ)=S2(δ)=ω;
γ γ
(2) γ ≤ν <δ <α⇒
(a) S1(δ)⊆∗ S1(δ); S2(δ)⊆∗ S2(δ)
γ ν γ ν
(b) S1(δ), S1(ν)⊆S1(0); S2(δ), S2(ν)⊆S2(0) are all infinite
γ γ γ γ γ γ
(3) γ ≤δ <α⇒{n:x ∈Uγ}⊇S1(δ); {n:y ∈Uγ}⊇S2(δ)
δ n γ δ n γ
(4) γ ≤ δ < α ⇒ {x , y } ⊆ G \(R (∪ F ) {x : ν < δ} {y : ν <
δ δ γ ∞S ν≤δ ν S ν S ν
δ});
(5) δ <α⇒x =y −f .
δ δ δ
Towards selecting x and y we consider two cases:
α α
Case 1: α=β+1
For δ <α define S1(δ)=ω =S2(δ). For γ ≤β we define
α α
∞
Tγ1 = \( [ Umγ),
n=1 m≥n,m∈S1(β)
γ
∞
Tγ2 = \( [ Umγ),
n=1 m≥n,m∈S2(β)
γ
and for α we define
∞
Tα1 =Tα2 = \( [ Umα).
n=1 m≥n
and
Gα = \(Lγ ∩Tγ1∩Tγ2).
γ≤α
Then G is a G set containing R and so by Corollary19 the set G ∩(G −f )
α δ ∞ α α α
contains a Hilbert cube. But the set
Bα =(R∞[(∪γ≤αFγ)[{xγ :γ <α}[{yγ :γ <α})−({fγ :γ ≤α}∪{0})
is countable dimensional and thus does not contain a Hilbert cube. Choose x ∈
α
(G (G −f ))\B and fix y ∈G with x =y −f .
αT α α α α α α α α
Then (4) holds at α. Since for γ < α we have x ∈ T1 and y ∈ T2 the sets
α γ α γ
S1(α) = {n ∈ S1(β) : x ∈ Uγ} and S2(α) = {n ∈ S2(β) : y ∈ Uγ} are infinite.
γ γ α n γ γ α n
Similarly the sets S1(α)={n:x ∈Uα} and S2(α)={n:y ∈Uα} are infinite.
α α n α α n
Case 2: α is a limit ordinal.
For γ < α choose infinite sets F1 and F2 so that for all β < α, F1 ⊆∗ S1(β) and
γ γ γ γ
F2 ⊆∗ S2(β).
γ γ
For γ ≤β we define
∞
Tγ1 = \( [ Umγ),
n=1 m≥n,m∈F1
γ
∞
Tγ2 = \( [ Umγ),
n=1 m≥n,m∈F2
γ
10 LILJANABABINKOSTOVAANDMARIONSCHEEPERS
and for α we define
∞
Tα1 =Tα2 = \( [ Umα).
n=1 m≥n
and
Gα = \(Lγ ∩Tγ1∩Tγ2).
γ≤α
SinceG isaG setcontainingR ,Corollary19impliesthatthesetG ∩(G −f )
α δ ∞ α α α
contains a Hilbert cube. The set
Bα =(R∞[(∪γ≤αFγ)[{xγ :γ <α}[{yγ :γ <α})−({fγ :γ ≤α}∪{0})
is countable dimensional and does not contain a Hilbert cube.
Choose x ∈(G (G −f ))\B and fix y ∈G with x =y −f .
α αT α α α α α α α α
Then (4) holds at α. Since for γ < α we have x ∈ T1 and y ∈ T2 the sets
α γ α γ
S1(α) = {n ∈ F1 : x ∈ Uγ} and S2(α) = {n ∈ F2 : y ∈ Uγ} are infinite.
γ γ α n γ γ α n
Similarly the sets S1(α)={n:x ∈Uα} and S2(α)={n:y ∈Uα} are infinite.
α α n α α n
Thiscompletesthe descriptionofthe processforchoosingx andy forα<ω .
α α 1
Finally define X =R {x :α<ω } and Y =R {y :α<ω }.
∞S α 1 ∞S α 1
Claim 1: If U is a collection of open subsets of RN which contains an infinite
subset V, each infinite subset of which covers R , then there are countable sets A
∞ X
and A such that V is an ω cover of X\A and of Y \A .
Y X Y
Proof of Claim 1: For let V be such a subfamily of U. We may assume that V
is countable. Thus, we may assume that for some α, fixed from now on, V is (Uα :
n
n<ω), asenumeratedbefore the constructionfX andY. PutA ={x :γ ≤α}
X γ
and A ={y :γ ≤α}.
Y γ
Consider any finite subset F ⊂ X \A . We may write F = {x ,··· ,x }
X γ1 γn
where we have α < γ < ··· < γ . From the construction of X and Y we have
1 n
S1(γ )⊆∗ ···⊆∗ S1(γ ). Choose n large enough that
α n α 1
S1(γ )\n⊆···⊆S1(γ ).
α m α 1
Then for k ∈S1(γ ) we have F ⊂Uα. A similar argument applies to Y \A .
α m m Y
Claim 2: X and Y both satisfy the selection principle S1(Okfd,Ω).
ProofofClaim2: WeproveitforX. TheproofforY issimilar. Let(U :n<∞)
n
be asequenceoffamiliesofopensetsofRN suchthateachisanelementofOkfd for
X. Write N = Z where each Z is infinite and Z ∩Z = ∅ whenever we
Sk<∞ k k k m
have k 6=m.
Starting with Z , choose for each n ∈ Z a U ∈ U such that every infinite
1 1 n n
subset of {U : n ∈ Z } covers R . Choose a countable set X ⊂ X such that
n 1 ∞ 1
{U :n∈Z } is an ω-cover of X \X .
n 1 1
Continuing with Z , choose for each n ∈ Z a U ∈ U such that every infinite
2 2 n n
subsetof{U :n∈Z }coversX ∪R . ChooseacountablesetX ⊂X suchthat
n 2 1 ∞ 2
{U :n∈Z } is an ω-cover of X \X . Note that we may assume X ∩X =∅.
n 2 2 2 1
With X , ··· , X selected countable subsets of X, and for each j ≤n, U ∈U
1 n k k
selectedfor k ∈Z so that (U :k∈Z ) is anω-coverof X\X , and every infinite
j k j j
subset of(U :k ∈Z ) coversX ∪···∪X ∪R . Choose U ∈U for k ∈Z
k j 1 j−1 ∞ k k n+1
so that each infinite subset of {U :k ∈Z } is a cover of R ∪X ∪···X and
k n+1 ∞ 1 n
byClaim1chooseacountableX ⊂RN suchthat{U :k ∈Z }is anω-cover
n+1 k n+1
of X \X .
n+1