Combinatorics of open covers (IV): Subspaces of the Alexandroff
double of the unit interval
Marion Scheepers
1
Abstract
2
We show that:
1. Strong measure zero subsets of the unit interval can be characterized
by a Ramseyan partition relation – Theorem 8;
2. The cardinal number suggested by Theorem 3.8 of [11] lies between
cov(M) and unif(SM Z);
3. The answer to Question 4.25 of [11] is “No”.
Selection hypotheses occur commonly in mathematics, and usually have
games associated with them. Often, the associated game characterizes the selection hypothesis. When this happens the game is a powerful combinatorial
tool for analysing the combinatorial properties of the basic objects involved in
the selection hypothesis.
We illustrate this here for a special case of the following selection hypothesis:
Let X be an infinite set and let A and B be collections of subsets of X. Then
the symbol S1 (A, B) denotes the selection hypothesis that for every sequence
(An : n ∈ N) with terms from A there is a sequence (bn : n ∈ N) such that
for each n bn ∈ An , and {bn : n ∈ N} ∈ B. The symbol G1 (A, B) denotes the
game which is associated with this selection hypothesis. This game is played
as follows: Players ONE and TWO play an inning per positive integer. In the
n–th inning ONE chooses an element On of A, and TWO responds by choosing
an element Tn of On . A play O1 , T1 , O2 , T2 , . . . is won by TWO if {Tn : n ∈ N}
is a member of B; otherwise, ONE wins.
The example we use is motivated by concepts introduced by Tkachuk in his
paper [11]. Let (X, τ ) be a topological space which is at least T3 12 . Define the
following collections of families of open sets:
D is the family consisting of U ⊂ τ such that ∪U is a dense subset of X.
K is the set of those U ∈ D such that X = ∪{U : U ∈ U}.
O is the collection of all open covers of X.
In [11] the symbol Ω∗ denotes the game G1 (K, D) and the symbol Θ∗ denotes
G1 (O, D). In Theorem 3.8 of [11] Tkachuk proves, assuming the Continuum
Hypothesis, that ONE has a winning strategy in G1 (K, D) on any space of uncountable cellularity. Thus the Continuum Hypothesis implies that if TWO has
1 Research
supported by NSF grant DMS 95 - 05375
and phrases: Borel Conjecture, Rothberger property, strong measure zero set,
partition relation, ω–cover, infinite game, winning strategy.
2 Keywords
1
a winning strategy in G1 (K, D), then the space must have countable cellularity.
But in Theorem 2.18 he shows that the Continuum Hypothesis is not needed for
the result about TWO. Naturally, one now wonders if the Continuum Hypothesis (or any other additional axiom) is needed for Theorem 3.8. This is asked
in Question 4.25.
After establishing a connection with Borel’s strong measure zero sets (Theorem 2 and Theorem 8), we explain how these matters are related to Theorems
2.17 and 3.8 of [11], and give a strengthening of Theorem 3.8 (Theorem 9). Then
we establish a connection with a property introduced by Rothberger (Theorem
10). This information puts us in a position to answer Question 4.25 negatively.
Strong measure zero sets of real numbers.
Let I be the closed unit interval. A subset X of the real line R has strong
measure zero if there is for every sequence (n : n ∈ N) of positive real numbers
a sequence (Jn : n ∈ N) of nonempty open intervals such that each Jn has
length at most n , and X ⊆ ∪n∈N Jn . Since X ⊆ I has strong measure zero if,
and only if, (X + Q) ∩ I has strong measure zero, we may confine our attention
to dense subsets of I. This concept was introduced in [1] where Borel observed
that every countable set of real numbers has this property. He conjectured
Borel Conjecture: Each strong measure zero set is countable.
A few years later Sierpiński showed that the Continuum Hypothesis implies
that Borel’s Conjecture is false. This was shortly before Gödel proved the
consistency of the Continuum Hypothesis. By 1976 Laver proved in [8] the
consistency of the Borel Conjecture. Thus, the Borel Conjecture is not decidable
by classical mathematics. (All these consistency results presuppose that classical
mathematics is consistent, something we assume here without further ado.)
Now consider the following subspace of the Alexandroff double of I (see for
example Exercise 14B of [12]). Let X be a dense subset of I. Then T (X) is
I × {0} ∪ X × {1}. For A ⊆ I and for x ∈ I we write Ai for A × {i} and xi for
(x, i), i ∈ {0, 1}. The family
B := {U0 ∪ ((U ∩ X)1 \ {x1 }) : U open in I and x ∈ U ∩ X} ∪ {{x1 } : x ∈ X}
is a basis for a topology on T (X). In this topology T (X) is compact and T4 .
We use the following well–known theorem of Lebesgue (Theorem 22.5 in [12])
to analyse when the selection hypothesis S1 (O, D) is valid for T (X).
Theorem 1 (Lebesgue Covering Lemma) For each finite open cover of a
compact metric space there is a δ > 0 such that every set of diameter at most δ
is a subset of some member of the cover.
A δ associated like this with a cover is said to be a Lebesgue number for it.
2
Theorem 2 For X a dense subset of I, the following are equivalent:
1. X has strong measure zero.
2. T (X) satisfies selection hypothesis S1 (O, D).
Proof : 1 ⇒ 2: Let (Un : n ∈ N} be a sequence of open covers of T (X). After
making the necessary refinements, we may assume that for each n each element
of Un which has nonempty intersection with I0 is of the form U0 ∪ ((X ∩ U )1 \
F (U )), where U is a nonempty interval open in I, and F (U ) is a finite subset
of X1 . Since I is compact we find for each n a finite set F2n of open intervals
of I such that
a F2n covers I and
b For each U ∈ F2n , (U0 ∪ ((U ∩ X)1 \ F (U )) is in U2n .
For each n let δ2n be a Lebesgue number for F2n . Since X has strong measure
zero we find for each n an open interval J2n ⊆ I of length at most δ2n such that
X ⊆ ∪n∈N J2n . For each n choose U 2n ∈ F2n with J2n ⊆ U 2n , and let F 2n be the
corresponding finite subset of X1 such that V2n := U02n ∪ ((U 2n ∩ X)1 \ F 2n ) ∈
U2n . For each n choose V2n−1 ∈ U2n−1 such that the sequence (V2n−1 : n ∈ N)
2n
covers ∪∞
. Since X is dense in I, the set {Vn : n ∈ N} belongs to D for
n=1 F
T (X).
2 ⇒ 1: Let (n : n ∈ N) be a sequence of positive real numbers. For each n let In
be the set of all intervals open in I and of length at most n . For each n define
Un := {J0 ∪ (J ∩ X)1 : J ∈ In }. Each Un is an open cover of T (X). Applying
the hypothesis S1 (O, D) we find for each n a Vn ∈ Un such that {Vn : n ∈ N}
is in D. But each Vn is of the form J0n ∪ (J n ∩ X)1 , where J n is an interval
of length at most n . But then the sequence (J n : n ∈ N) is an appropriate
sequence of intervals covering X. 2
Next we show that for T (X) the game G1 (O, D) characterizes S1 (O, D).
Let X be a subset of I. In [4] the authors define the following game on X
– let SM Z(X) denote this game: Players ONE and TWO play an inning per
positive integer. In the n–th inning ONE chooses a positive real number n ,
then TWO chooses an open interval Jn ⊆ I of length at most n . TWO wins
a play 1 , J1 , 2 , J2 , . . . if X ⊆ ∪n∈N Jn ; otherwise, ONE wins. Theorem 3 of [4]
states:
Theorem 3 (Galvin, Mycielski, Solovay) Let X be a subset of I.
1. ONE has a winning strategy in SM Z(X) if, and only if, X is not of strong
measure zero.
2. TWO has a winning strategy in SM Z(X) if, and only if, X is countable.
3
Thus the Borel Conjecture is equivalent to the assertion that for each X ⊆ I,
SM Z(X) is determined.
Theorem 4 For X a dense subset of I the following are equivalent:
1. ONE has a winning strategy in SM Z(X).
2. ONE has a winning strategy in G1 (O, D) on T (X).
Proof : 1 ⇒ 2: Let F be ONE’s winning strategy in SM Z(X). Define a strategy
G for ONE in G1 (O, D) on T (X) as follows: With 1 = F (∅), put
G(∅) = {U0 ∪ (U ∩ X)1 : U an interval open in I, length(U ) ≤ 1 }.
If TWO chooses T1 ∈ G(∅), look at T1 = U01 ∪ (U 1 ∩ X)1 . Compute F (U 1 ) = 2 ,
say, and put
G(T1 ) = {U0 ∪ (U ∩ X)1 : U an interval open in I, length(U ) ≤ 2 }.
If TWO chooses T2 ∈ G(T1 ), look at T2 = U02 ∪(U 2 ∩X)1 , compute F (U 1 , U 2 ) =
3 , and set
G(T1 , T2 ) = {U0 ∪ (U ∩ X)1 : U an interval open in I, length(U ) ≤ 3 },
and so on.
To see that G is a winning strategy, look at a G–play
G(∅), T1 , G(T1 ), T2 , G(T1 , T2 ), T3 , . . . .
For each n let J n be the open interval of I for which Tn = J0n ∪ (J n ∩ X)1 . Then
the sequence
F (∅), J 1 , F (J 1 ), J 2 , F (J 1 , J 2 ), J 3 , . . .
is an F –play of SM Z(X), so won by ONE. This means that X is not covered
by the J n ’s, and so X1 is not covered by the Tn ’s. But then {Tn : n ∈ N} is not
in D.
2 ⇒ 1: Let F be ONE’s winning strategy in G1 (O, D) on T (X). Since refining
the sets ONE played to “smaller” ones has no advantage for TWO, we may
assume that in each inning F requires that ONE plays an open cover of T (X)
consisting of elements of the basis B; for that matter we may assume that
{{x1 } : x ∈ X} is always a subset of all moves by ONE.
We now define a strategy G for ONE of SM Z(X). Let ≺ be a well-order of
I. To define G(∅), ONE’s first move, first look at
F (∅) = {U0 ∪ (X ∩ U )1 \ {x1 } : x ∈ U ∩ X, U ∈ I 1 } ∪ {{x1 } : x ∈ X},
where I 1 is a cover of I by intervals open in I. Choose a finite subset F 1 of I 1
which covers I, and let δ1 be a Lebesgue - number for F 1 . Define G(∅) = δ1 .
4
If TWO of SM Z(X) now chooses an interval T1 of length at most δ1 , ONE
determines G(T1 ) as follows: First, choose a U 1 ∈ F 1 with T1 ⊆ U 1 . Let x1 be
the ≺–least element of U 1 ∩ X, and let S11 be the set U01 ∪ (U 1 ∩ X)1 \ {x11 },
a legitimate response by TWO in G1 (O, D) on T (X), and let S21 be {x11 }, a
legitimate response by TWO to F (S11 ). Then compute
F (S11 , S21 ) = {U0 ∪ (U ∩ X)1 \ {x1 } : x ∈ (U ∩ X), U ∈ I 2 } ∪ {{x1 } : x ∈ X},
where I 2 is a cover of I by intervals open in I. Choose a finite subset F 2 of
I 2 which covers I, and let δ2 be a Lebesgue number for this cover. Define
G(T1 ) := δ2 .
If TWO of SM Z(X) now chooses an interval T2 of length at most δ2 , ONE
determines G(T1 , T2 ) as follows: First, choose a U 2 ∈ F 2 with T2 ⊆ U 2 , and let
x2 be the ≺–first element of U 2 ∩ X. Let S12 be the set U02 ∪ (U 2 ∩ X)1 \ {x21 },
a legitimate response by TWO in G1 (O, D) on T (X), and let S22 be {x21 }, a
legitimate response by TWO to F (S11 , S21 , S12 ). Then compute
F (S11 , S21 , S12 , S22 ) = {U0 ∪(U ∩X)1 \{x1 } : x ∈ (U ∩X), U ∈ I 3 }∪{{x1 } : x ∈ X},
where I 3 is a cover of I by intervals open in I. Choose a finite subset F 3 of
I 3 which covers I, and let δ3 be a Lebesgue number for this cover. Define
G(T1 , T2 ) := δ3 , and so on.
To see that this G is a winning strategy for ONE in SM Z(X), consider
a G–play G(∅), T1 , G(T1 ), T2 , G(T1 , T2 ), T3 , . . .. A recursive computation based
on the definition for G from F shows that there are sequences U 1 , U 2 , U 3 , . . .
and x1 , x2 , . . . such that if for each n we set S1n = U0n ∪ (U n ∩ X)1 \ {xn1 } and
S2n = {xn1 } then:
1. U n is such that Tn ⊆ U n ;
2. xn is the ≺–least element of U n ∩ X;
3. F (∅), S11 , F (S11 ), S21 , F (S11 , S21 ), S12 , F (S11 , S21 , S12 ), S22 , F (S11 , S21 , S12 , S22 ), . . .
is a play of G1 (O, D) on T (X).
Since F is a winning strategy for ONE, the play in 3 is lost by TWO; this means
that ∪n∈N (S1n ∪ S2n ) is not dense in T (X). Since each of the points xn1 is in this
union, this means that the intervals U n left some other point of X uncovered.
But then 1 implies that TWO also lost the G–play of SM Z(X). 2
Corollary 5 For a dense subset X of I, the following are equivalent:
1. T (X) has property S1 (O, D).
2. ONE has no winning strategy in G1 (O, D).
5
Proof : Theorems 2, 3 and 4. 2
Theorem 6 For a dense subset X of I, the following are equivalent:
1. TWO has a winning strategy in SM Z(X).
2. TWO has a winning strategy in G1 (O, D) on T (X).
Proof : 1 ⇒ 2. If TWO has a winning strategy in SM Z(X) on X, then by
Theorem 3 X is a countable set. But if X is a countable dense subset of I then
TWO has an easy winning strategy in the game G1 (O, D) on T (X).
2 ⇒ 1. Let F be a winning strategy for TWO in G1 (O, D) on X. Define a
strategy G for TWO in SM Z(X) as follows:
When ONE makes first move 1 > 0, TWO translates this first as a move
O1 := {U0 ∪ (U ∩ X)1 : U an interval of length ≤ 1 open in I} for ONE of
G1 (O, D) on T (X), and computes F (O1 ), say it is U01 ∪ (U 1 ∩ X)1 . Then TWO
plays G(1 ) = U 1 .
When ONE next moves 2 > 0, TWO translates this as a move O2 :=
{U0 ∪ (U ∩ X)1 : U an interval of length ≤ 2 open in I} for ONE of G1 (O, D)
on T (X), and computes F (O1 , O2 ), say it is U02 ∪ (U 2 ∩ X)1 . Then TWO plays
G(1 , 2 ) = U 2 , and so on.
It is left to the reader to check that G is winning for TWO. 2
For collections A and B of subsets of an infinite set S and for positive integers
n and k, the symbol A → (B)nk denotes the statement:
For each A ∈ A and for each f : [A]n → {1, . . . , k} there is a B ∈ B
and an i ∈ {1, . . . , k} such that on [B]n f is constant and of value i.
B is said to be homogeneous of color i for f . Written in this notation Ramsey’s
famous theorem asserts that if A is the collection of infinite subsets of N, then
for all k and n in N, A → (A)nk holds.
An open cover U of a topological space is an ω-cover if X is not a member of
U, and for each finite subset F of X there is a U ∈ U with F ⊂ U . This concept
was introduced in [5] where they prove the important fact that every ω–cover
of a space has a countable subset which is an ω–cover if, and only if, every
finite power of the space has the Lindelöf property. They call spaces with the
property that each ω–cover has a countable subset which is an ω-cover –spaces.
Since T (X) is compact, it has the property that every ω–cover has a countable
subset which is still an ω–cover. We use this without further mention. We
shall also use the symbol Ω to denote the collection of all ω–covers of a space.
The collection of ω–covers of an infinite space always satisfies: for each k ∈ N,
Ω → (Ω)1k . This makes these sorts of covers convenient for Ramsey-theoretic
considerations. The property S1 (O, D) is equivalent to another which lends
itself better to Ramsey-theoretic arguments.
6
Theorem 7 A space has property S1 (O, D) if, and only if, it has property
S1 (Ω, D).
Proof : S1 (O, D) implies S1 (Ω, D) since Ω ⊆ O. Suppose now that the space
satisfies S1 (Ω, D), and let (Un : n ∈ N) be a sequence of open covers of the space.
Let (Yn : n ∈ N) be a partition of N into infinitely many infinite, pairwise disjoint
sets. For each n define
Vn = {Ui1 ∪ . . . ∪ Uik : i1 < . . . < ik are in Yn and for j ≤ k, Uij ∈ Uij }.
If some Vn has an element which is a dense subset of the space we are done.
Otherwise, each Vn is in Ω, and we can apply S1 (Ω, D) to find a selector for the
original sequence of Un ’s. 2
Theorem 8 For a dense subset X of I the following are equivalent:
1. X has strong measure zero.
2. The space T (X) satisfies: For each k, Ω → (D)2k .
Proof : 1 ⇒ 2: Let U be an ω–cover of T (X), let k be a positive integer,
and let f : [U]2 → {1, . . . , k} be given. We may assume that U is countable.
Let (Un : n ∈ N) enumerate it bijectively. Recursively construct a sequence
((Un , in ) : n ∈ N) such that
1. U1 = {Un : n > 1 and f ({U1 , Un }) = i1 } ∈ Ω;
2. For each n, Un+1 = {Um : m > n + 1 and f ({Un+1 , Um }) = in+1 } is in Ω.
. This is done by repeatedly using the partition relation Ω → (Ω)1k . Next, define
Cj := {Un : in = j}. Then for each n, Un = (Un ∩ C1 ) ∪ . . . ∪ (Un ∩ Ck ) partitions
the ω–cover Un into k classes. Applying Ω → (Ω)1k once more, we find for each
n a jn such that Un ∩ Cjn is an ω–cover of X. Since the Un ’s form a descending
sequence, we may assume that all jn are equal to a fixed j.
Let (Unm : m ∈ N) be an enumeration of Cj using the subscripts of the
original enumeration of U, and for each m put Vnm = Unm ∩ Cj . We are now
going to play the game G1 (O, D) to extract an appropriate homogeneous set for
f . Define a strategy F for ONE as follows: Let m1 be minimal with Unm1 ∈ Vn1 ,
and let ONE’s first move be F (∅) := Vnm1 . If TWO responds with T1 =
Unm2 ∈ F (∅), then we have nm2 > nm1 . ONE’s move now is F (T1 ) := Vnm2 . If
TWO responds with Unm3 ∈ F (T1 ), then nm2 < nm3 , and ONE’s move will be
F (T1 , T2 ) := Vnm3 , and so on.
Since X is of strong measure zero, Theorems 3 and 4 imply that F is not a
winning strategy for ONE. Consider an F –play F (∅), T1 , F (T1 ), T2 , F (T1 , T2 ), T3 , . . .
which is lost by ONE. The set of Tr ’s is in D. For each r we have Tr = Unmr ,
and Tr+1 ∈ Vnmr . This implies that the set of Tr ’s is homogeneous of color j
7
for f .
2 ⇒ 1: By Theorems 2 and 7 it suffices to show that 2 implies that T (X) has
property S1 (Ω, D). For this we use an idea from [7]. Let (Un : n ∈ N) be a
sequence of ω–covers of T (X). If one of these contains an open set which is
dense in T (X), we have nothing more to do. Thus we may assume that no
Un has an element dense in T (X) and is countable; enumerate it bijectively as
n
n
(Um
: m ∈ N). Define V to be {Un1 ∩ Um
: m, n ∈ N} \ {∅}. Then V is an ω–cover
of T (X) and has no dense subset of T (X) as member. For each element of V
n
choose once and for all a representation as Un1 ∩ Um
. Define f : [V]2 → {1, 2} by
1 if n1 = n2
n1
n2
f ({Un11 ∩ Um
, Un12 ∩ Um
}) =
1
2
2 otherwise
Apply 2 to find W ⊂ V which is in D and is homogeneous for f . Sets which are
homogeneous of color 1 for f are not in D since each of these is a refinement of
the same Un1 and no element of U1 is dense in T (X). Thus, W is homogeneous
of color 2 for f . But W refines the set of second terms of elements of W, and
these second terms can all be taken from distinct Un ’s. This provides us with a
selector in D for the original sequence of Un ’s. 2
[11], Theorems 2.18 and 3.8.
To see how our preceding results fit in with [11], we outline “Tkachuk’s algorithm” for associating subsets of I with a T3 12 space. For this outline refer to
the following diagram:
X
-
-
β(X)
E(β(X))
6
- T (S(A))
ZA
6
YA
8
6
- S(A)
In this diagram, X is a given T3 12 –space, β(X) is its Stone–Čech compactification, and E(β(X)) is the absolute (also known as projective cover) of X.
These three spaces have equal cellularity. E(β(X)) is (compact and) extremally
disconnected, X is dense in β(X), and there is an irreducible, continuous (and
thus closed) function from E(β(X)) onto β(X).
For A an arbitrary family of cardinality ≤ 2ℵ0 of nonempty pairwise disjoint
clopen subsets of E(β(X)), YA = ∪A and ZA = YA . Since ZA is compact and
extremally disconnected (properties inherited from E(β(X))), it is the Stone–
Čech compactification of YA . This explains A and the positions of YA and ZA
in the diagram.
With X and A as above, S(A) is any dense subset of I of same cardinality as
A: list it bijectively as S(A) = {xA : A ∈ A}. The function g : YA → T (S(A))
defined by g(z) = xA
1 whenever z ∈ A is continuous. Then g extends to a
continuous function h : ZA → T (S(A)). Since the range of h is dense and
compact, h is a continuous surjection.
To summarize how game-theoretic information propagates along this diagram, consider the following statements for ONE. Each implies the succeeding
one. We have seen that ONE.a and ONE.b are equivalent. Also ONE.d and
ONE.e are equivalent because E(β(X)) is extremally disconnected. ONE.b⇒ONE.c
follows from Lemma 2.6 (ix) of [11], ONE.e⇒ONE.f follows from Theorem 2.17
(ii) of [11], and ONE.f⇒ ONE.g follows from Lemma 2.6(iv) of [11]. The remaining implication is easy to prove.
ONE.a ONE has a winning strategy in SM Z(S(A).
ONE.b ONE has a winning strategy in G1 (O, D) on T (S(A)).
ONE.c ONE has a winning strategy in G1 (O, D) on ZA .
ONE.d ONE has a winning strategy in G1 (O, D) on E(β(X)).
ONE.e ONE has a winning strategy in G1 (K, D) on E(β(X)).
ONE.f ONE has a winning strategy in G1 (K, D) on β(X).
ONE.g ONE has a winning strategy in G1 (K, D) on X.
Now let unif(SM Z) be the minimal cardinality for a set of real numbers which
does not have strong measure zero. The Borel Conjecture as well as the Continuum Hypothesis implies that unif(SM Z) = ℵ1 , but it is also consistent that
this is larger than ℵ1 . We return to this point below. We now find the following
sharpening of Theorem 3.8 of [11]:
Theorem 9 On any T3 12 –space with cellularity at least unif(SM Z) ONE has a
winning strategy in the game G1 (K, D).
9
Proof : Let A in the above “algorithm” be a family of unif(SM Z) pairwise
disjoint nonempty clopen subsets of E(β(X)), and let S(A) be any dense subset
of I which is not of strong measure zero. By Theorem 3 ONE has a winning
strategy in the game SM Z(S(A)), and so by the preceding remarks ONE has
a winning strategy in G1 (K, D) on X. 2
In the following list, each statement is implied by the one succeeding it. We
have seen that TWO.a is equivalent to TWO.b; also TWO.d and TWO.e are
equivalent because E(β(X)) is extremally disconnected. TWO.c⇒TWO.b follows from [11], Lemma 2.6(vii); TWO.d⇒TWO.c follows from Lemma 2.6(v) of
[11]; TWO.f⇒TWO.e follows from Theorem 2.17(i) of [11], and TWO.g⇒TWO.f
follows from Lemma 2.6(iii) of [11].
TWO.a TWO has a winning strategy in SM Z(S(A)).
TWO.b TWO has a winning strategy in G1 (O, D) on T (S(A)).
TWO.c TWO has a winning strategy in G1 (O, D) on ZA .
TWO.d TWO has a winning strategy in G1 (O, D) on E(β(X)).
TWO.e TWO has a winning strategy in G1 (K, D) on E(β(X)).
TWO.f TWO has a winning strategy in G1 (K, D) on β(X).
TWO.g TWO has a winning strategy in G1 (K, D) on X.
Using this, one now argues as follows to prove Theorem 2.18 of [11]. Let X be a
T3 12 –space such that TWO has a winning strategy in G1 (K, D) on X. Then for
A any infinite family of pairwise disjoint nonempty clopen subsets of E(β(X)),
and for S(A) any dense subset of I having the same cardinality as A, TWO has
a winning strategy in the game SM Z(S(A)). Theorem 3 implies that S(A),
and hence A, is countable.
Rothberger’s property and [11], Question 4.25.
Theorem 9 suggests the following cardinal number, denoted j:
j is the least κ such that on any T3 12 –space with cellularity at least
κ, ONE has a winning strategy in the game G1 (K, D).
We have proved that j ≤ unif(SM Z). Question 4.25 of [11] now becomes the
question:
Is j = ℵ1 ?
10
In this section we shall give a lower bound for j, which shows that the answer
to Question 4.25 is No.
Our lower bound is related to the property S1 (O, O) for sets of real numbers.
Rothberger introduced this property in [9]. Rothberger showed that if a set of
real numbers has property S1 (O, O), then it has strong measure zero. Rothberger showed in [10] that the Continuum Hypothesis implies the existence of
a strong measure zero set which does not have property S1 (O, O). Since every
countable set has property S1 (O, O), the Borel Conjecture implies that strong
measure zero sets have property S1 (O, O).
Let cov(M) denote the least κ such that the real line is the union of κ
first category sets. It is implicit in Rothberger’s work that cov(M) is the least
cardinality of a set of real numbers not having property S1 (O, O) – see for
example Theorem 5 of [3] on the matter. Though the property that every finite
power of a set X have property S1 (O, O) is stronger than that X has property
S1 (O, O), it is true that the minimal cardinality of a set X of reals, not all of
whose finite powers has property S1 (O, O), is cov(M) – [7], Theorem 4.8.
Recall that a space is weakly Lindelöf if each open cover contains a countable
subset which is in D. Let us say that a space is weakly K–Lindelöf if every
element of K contains a countable subset which is in D.
Theorem 10 Let κ be an infinite cardinal number. Then the following are
equivalent:
1. κ < cov(M).
2. For every T3 –space which is weakly K–Lindelöf and has π–weight κ, ONE
has no winning strategy in the game G1 (K, D).
3. For every T3 –space which is weakly Lindelöf and has π–weight κ, ONE
has no winning strategy in the game G1 (O, D).
Proof : We must show that 1 ⇒ 2 and 2 ⇒ 1. These implications for 3 are
proved in a similar but slightly easier way.
1 ⇒ 2: Assume 1. Let a weakly K–Lindelöf T1 –space X with π(X) = κ be
given. Consider a strategy F for ONE in the game G1 (K, D). Since X is weakly
K–Lindelöf, each move by ONE contains a countable subset which is in D.
Construct a family (Uν : ν ∈ <ω N) of open subsets of X as follows: (Un : n ∈
N) is an enumeration of a countable subset in D, contained in ONE’s first move,
F (∅). If TWO chose Un1 in the first inning, then (Un1 ,n : n ∈ N) enumerates a
countable element of D contained in ONE’s move F (Un1 ). If TWO now chooses
Un1 ,n2 , then (Un1 ,n2 ,n : n ∈ N) enumerates a countable element of D contained
in ONE’s move F (Un1 , Un1 ,n2 ), and so on.
Next, let B be a π-basis of cardinality κ for X and use the family of Uν ’s
just constructed to assign to each B in B a closed nowhere dense subset NB of
N
N as follows:
NB = {f ∈ N N : (∀n)(B ∩ Uf dn+1 = ∅)}.
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Since κ is less than cov(M) the union of the NB ’s does not cover N N (since the
latter is homeomorphic to the set of irrational numbers); let f be an element
not in any of the NB ’s, and write nk for f (k). Then the play
F (∅), Un0 , F (Un0 ), Un0 ,n1 , F (Un0 , Un0 ,n1 ), . . .
is won by TWO, showing that F is not a winning strategy for ONE.
3 ⇒ 1: Let X be a set of real numbers of cardinality κ, and let P R(X) be the
Pixley-Roy space over X. Then P R(X) is a T1 –space with countable cellularity,
and thus weakly K–Lindelöf, and π–weight κ. By 2 ONE has no winning strategy
in G1 (K, D) on P R(X). This implies that P R(X) has property S1 (K, D), and
thus property S1 (O, D). Theorem 5 A of [2] implies that every finite power of
X has property S1 (O, O). We have shown that 2 implies that whenever a set of
real numbers has cardinality κ, then all its finite powers have property S1 (O, O).
By Theorem 4.8 of [7] κ is less than cov(M). 2
The π–weight of a space is at least as big as the cellularity of the space.
The important role played by π–weight in the preceding theorem raises the
question whether cellularity is really the cardinal function, say φ, giving rise
to the phenomenon that whenever X is a space with φ(X) larger than a fixed
κ, then ONE has a winning strategy in G1 (K, D). Maybe the only reason why
cellularity has this property is because it raises π–weight, and π–weight is really
the function responsible for the phenomenon? This is not the case. Let c
denote 2ℵ0 . Then the power 2c of the two-point discrete space is separable, thus
of countable cellularity, but has π–weight c. Since the space is separable, TWO
has a winning strategy in G1 (K, D). Thus, large π–weight does not explain the
phenomenon. On the other hand countable cellularity per se is of no benefit to
TWO. Let X be a set of real numbers of cardinality cov(M), which does not
have property S1 (O, O). Then P R(X) has countable cellularity and π–weight
cov(M), and Theorem 5 A of [2] implies that ONE has a winning strategy in
the game G1 (O, D), and thus in G1 (K, D).
Corollary 11 cov(M) ≤ j ≤ unif(SM Z).
Proof : The cellularity of a space is no larger than its π–weight. Thus, Theorem 10 implies that for some spaces with cellularity less than cov(M) (namely
those whose π–weight is also below cov(M)), ONE has no winning strategy in
G1 (K, D). This by itself at first glance does not give the required lower bound; we
must verify that there are actually for all regular infinite κ < cov(M) spaces of
cellularity κ where ONE does not have a winning strategy in the game G1 (K, D).
Moreover, only uncountable κ’s require discussion. Probably the easiest examples are as follows: If κ < cov(M) is a regular uncountable cardinal number, let
D(κ) be the discrete space of cardinality κ. Then β(D(κ)), the Stone–Čech compactification of D(κ), has π–weight and cellularity equal to κ, and is extremally
disconnected, thus weakly K–Lindelöf. 3 According to 1 ⇒ 2 of Theorem 10
3I
am grateful to Professor Jack Porter who brought these facts to my attention.
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ONE has no winning strategy in the game G1 (K, D). 2
Corollary 12 The answer to Question 4.25 of [11] is “No”.
Proof : Martin’s Axiom and Corollary 11 imply that j = 2ℵ0 . But Martin’s
Axiom plus the negation of Continuum Hypothesis is consistent. 2
This leaves us now with the question whether the inequalities in Corollary
11 are sharp. The mere fact that not all strong measure zero sets have property
S1 (O, O) does not rule out that cov(M) = unif(SM Z) – Rothberger’s example
was obtained for ℵ1 = 2ℵ0 . Should it be true that cov(M) = unif(SM Z), then
it would be true that j = unif(SM Z). But in [6] the authors show that it is
consistent that ℵ1 = cov(M) < unif(SM Z) = ℵ2 = 2ℵ0 .
Problem 1 Could j be different from both unif(SM Z) and cov(M)?
I suspect that the answer is “yes”; it would be more interesting to determine
if j is one of the well-studied cardinal numbers related to structures of cardinality
≤ 2ℵ0 .
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