East-West J. of Mathematics: Vol. 1, No 2 (1999) pp. 171-178
SOLUTIONS OF TWO PROBLEMS
CONCERNING ADDITIVE PROPERTIES OF
STRONG MESURE ZERO SETS
Andrzej Nowik
Inst. of Mathematics, University of Gdańsk, Wita Stwosza 57, 80 – 952 Gdańsk, Poland
e-mail: [email protected]
Abstract
CH implies that there exists perfectly meager and strong measure
zero set X such that X + X is not totally imperfect.
For every meager set X player TWO has a winning strategy in the
game G(X, SM Z).
Most notation used in this paper will follow [1]. The σ - ideal of meager sets
in X will be written as M(X).
A set X⊆2ω is perfectly meager (X ∈ AF C) if P ∩ X ∈ M(P ) for every
perfect set P ⊆2ω . A set X⊆2ω is strong measure zero iff for every F ∈ M,
X + F = {x + f : x ∈ X, f ∈ F } = 2ω . A set X⊆2ω is strong first category iff
for every F ∈ N , X + F = {x + f : x ∈ X, f ∈ F } = 2ω .
Let SM Z denote the ideal of strong measure zero sets and SF C the collection of strong first category sets. For an ideal F on a set X we define
add(F ) = min{|C| : C⊆F ∧ ∪C ∈ F }.
For X, Y ⊆ω define X⊆∗ Y iff Y \ X is finite. A family {Xα : α < θ}⊆[ω]ω
is called a tower if Xα ⊆∗ Xβ for α ≥ β. We define cardinal number t as the
size of the smallest maximal tower in [ω]ω .
Theorem 1 Assume CH. Let A ∈ [ω]ω be a set with |ω \ A| = ω. Then
there exists a perfectly meager and strong measure zero set X⊆2ω such that
2A ⊆X + X, where 2A = {x ∈ 2ω : x|(ω \ A) ≡ 0}.
Partially supported by the KBN grant 2 P03A 047 09
Key words and phrases: Strong measure zero set; Perfectly meager set; Additive games
AMS Mathematics Subject Classification: 03E20, 28E15, 90D44
171
172
Solutions of two problems ...
Proof. Let (An )n<ω be a partition of ω into infinite disjoint sets. Assume also
that A0 = A. Consider the following perfect set defined in the article [4] (see
[4], proof of the Theorem 1):
D((An )n<ω ) = {< x, y >∈ (2ω )2 : ∀i ∈ ω,
x|Ai ≡ 1 ∨ y|Ai ≡ 1 }
This set has the following property:
Lemma 1 Suppose M ∈ M(D((An )n<ω ) and let z ∈ 2A0 be an element such
that z(a0 ) = z(a1 ) = 1, where a0 and a1 is the first and the second element of
A0 , respectively. Then there exists < x, y >, < x , y >∈ D((An )n<ω ) \ M such
that (x + x )|A0 = z and x|(ω \ A0 ) = x |(ω \ A0 ).
Proof of the Lemma 1. We adopt the following notation:
B<x ,y > = {< x, y >∈ 2ω × 2ω : x ⊆x ∧ y ⊆y},
where x ∈ 2<ω , y ∈ 2<ω , and dom(x ) = dom(y ). Let M ⊆ n<ω Fn , Fn
are closed, nowhere dense sets such that Fn ⊆Fn+1 . We recursively define two
sequences:
<< xn , yn >: n < ω >
<< xn , yn >: n < ω >
of elements of 2<ω such that
1.
2.
dom(xn ) = dom(xn ) = dom(yn )
∀n ∈ ω,
xn ⊆xn+1 ; xn ⊆xn+1 ; yn ⊆yn+1 ;
∀n ∈ ω,
3.
∀n < ω, ∀i ∈ dom(xn )
∀n < ω,
4.
5.
∀n < ω,
i ∈ A0
i ∈ A0
⇒ xn (i) + xn (i) = z(i)
⇒ xn (i) = xn (i)
yn |A0 ≡ 1
B<xn ,yn > ∩ D = ∅ ∧ B<xn ,yn > ∩ D = ∅
6.
∀n < ω,
7.
∀0 < n < ω,
n odd ⇒ B<xn ,yn > ∩ Fn = ∅
n even ⇒ B<xn ,yn > ∩ Fn = ∅
A. Nowik
173
Construction of the sequences xn , yn , xn :
Let dom(x0 ) = dom(x0 ) = dom(y0 ) = [0, a1]. Define:
0 if i = a0
x0 (i) =
for i ∈ [0, a1]
1 if i = a0
x0 (i)
=
0
1
if i = a1
if i =
a1
for i ∈ [0, a1]
y0 (i) = 1 for i ∈ [0, a1].
Let n be odd.
We extend the sequence < xn−1 , yn−1 > to sequence < xn , yn > such that
the following condition will hold: B<xn ,yn > ∩ D((An )n<ω ) = ∅ and B<xn ,yn > ∩
Fn = ∅. Define:
if i ∈ A0
xn (i)
for i ∈ dom(xn )
xn (i) =
xn (i) + z(i) if i ∈ A0
We apply now the condition to the sequence xn−1 , xn−1 and we obtain
that xn is an extension of the sequence xn−1 . It is not difficult to see that the
other conditions from our construction remain true.
Let n > 0 be even.
We extend the sequence < xn−1 , yn−1 > to < xn , yn > such that the
following condition will hold: B<xn ,yn > ∩D((An )n<ω ) = ∅ and B<xn ,yn > ∩Fn =
∅. Define
if i ∈ A0
xn (i)
xn (i) =
for i ∈ dom(xn )
xn (i) + z(i) if i ∈ A0
Applying the condition to the sequences xn−1 , xn−1 we conclude that xn
forms an extension of xn−1 . It is easy to verify the other conditions of this
construction.
At the end we put
x=
xn
n<ω
y=
yn
n<ω
x =
n<ω
xn
We conclude from the condition that (x + x )|A0 = z and x|(ω \ A0 ) = x |(ω \
A0 ). This completes the proof of the Lemma 1.
174
Solutions of two problems ...
Lemma 2 Assume that CH holds. Let A ∈ [ω]ω be a set such that |ω \ A| = ω.
Then there exists X1 ⊆2ω perfectly meager and strong measure zero such that
{z ∈ 2ω : z|(ω \ A) ≡ 0 ∧ z(a0 ) = z(a1 ) = 1}⊆X1 + X1 ,
where a0 and a1 denotes the first and the second element of A, respectively.
Proof of Lemma 2. Suppose that (Mα )α<ω1 is an enumeration of all meager
sets in the relative topology of D((An )n<ω ). Suppose also that (zα )α<ω1 is an
enumeration of all elements of the set {z ∈ 2ω : z|(ω \ A) ≡ 0 ∨ z(a0 ) = z(a1 ) =
1}.
We will construct using the Lemma 1 sequences << xα , yα >; < xα ; yα >:
α < ω1 > by the following description: Having constructed << xα , yα >; <
xα ; yα >: α < θ > for θ < ω1 we choose
< xθ , yθ >, < xθ , yθ >∈ D((An )n<ω )\
Mα ∪{< xα , yα >; < xα , yα >: α < θ}
α<θ
such that xθ + xθ = zθ .
df
It is obvious that the set L = {< xα , yα >, < xα , yα >: α < ω1 } is indeed a
Luzin set in the relative topology of D((An )n<ω ). Therefore a projection π1 [L]
of the set L on the first coordinate is a strong measure zero set. We conclude
from the Corollary 1 in the article [4] that this projection is perfectly meager
set.
It is evident that π1 [L] + π1 [L] contains the set (zα )α<ω1 . We define
X1 = π1 [L]. It is easy to see that X1 has the desired properties. This finishes
the proof of the Lemma 2.
From the Lemma 2 we easy obtain the conclusion of the Theorem 1.
To deduce Theorem 1 from the Lemma 2, define
df
X = X1 ∪ (X1 + e0 ) ∪ (X1 + e1 ) ∪ (X1 + e0,1 )
where
e0 (i) =
e1 (i) =
e0,1 (i) =
0
1
0
1
if i = a0
if i = a0
0
1
if i = a1
if i = a1
if i ∈ {a0 , a1 }
if i ∈ {a0 , a1 }
It follows easily that 2A ⊆X + X.
The Theorem 1 is no longer true if we deal with towers. In fact, we have
the following easy theorem:
A. Nowik
175
Theorem 2 Suppose that X = (xα )α<t and Y = (yβ )β<t are towers in [ω]ω .
Then the algebraic sum X + Y is a perfectly meager set.
Proof. Let P ⊆2ω be a perfect set. Let (pn )n<ω be a dense set in P such that
∀n < ω, pn ∈ P ∩ [ω]ω .
Lemma 3 Suppose a ∈ [ω]ω . Then there is θ < t such that a ∈ [xθ ∪ yθ ]∗ ,
where [A]∗ = {B ∈ [ω]ω : B⊆∗ A}.
Proof of Lemma 3. Choose θ1 < t such that a ⊆∗ xθ1 . Thus |a \ xθ1 | = ω.
Choose θ > θ1 , θ < t such that (a \ xθ1 ) ⊆∗ yθ . Therefore
|a \ (xθ1 ∪ yθ )| = ω
and then
|a \ (xθ ∪ yθ )| = ω.
Lemma 4 Suppose θ < t and θ < α, β < t. Then xα + yβ ∈ [xθ ∪ yθ ]∗ ∪ [ω]<ω .
Proof of Lemma 4. We have: xα + yβ ⊆xα ∪ yβ ⊆∗ xθ ∪ yθ , This completes
the proof of the Lemma 4.
Following the Lemma 3 take θ < t such that
∀n < ω, pn ∈ [xθ ∪ yθ ]∗.
Thus P ∩ [xθ ∪ yθ ]∗ ∈ M(P ), therefore from the Lemma 4 we obtain:
P ∩ (xα )α>θ + (yβ )β>θ ∈ M(P )
Then:
X + Y ⊆(xα )α>θ + (yβ )β>θ ∪ X + (yβ )β≤θ ∪ Y + (xα )α≤θ .
It is easy to conclude that the last two sets from this equation are perfectly
meager (because add(AF C) ≥ add (M) ≥ t, but towers of hight t are perfectly
meager from the result in [3]). This ends the proof of the Theorem 2.
In [5] and [6] M. Scheepers introduced the following additive game:
Definition 1 Given a set X⊆2ω and a family X, Player ONE chooses a set
X0 ∈ X, Player TWO chooses a x0 ∈ 2ω , Player ONE chooses a set X1 ∈ X,
Player TWO chooses a x1 ∈ 2ω , and so forth; Player TWO wins the play if
and only if
1. The sequence (xn )n<ω is convergent, say to x∗ .
2. ( n<ω Xn ) ∩ (X + x∗ ) = ∅.
176
Solutions of two problems ...
Denote this game by G(X, X).
In [6] the author showed that if the family of sets X and set X have the
following property:
∀Y ∈ X, X + Y ∈ s0
then the player TWO has a winning strategy.
From the fact (see [2]) that for every strong first category set X⊆2ω and
for every strong measure zero set Y ⊆2ω , X + Y ∈ s0 we obtain:
Corollary 1 For every strong measure zero set X⊆2ω the player TWO has a
winning strategy in the game G(X, SF C)
It is easy to obtain the following ”dual” version of this theorem: Suppose
X⊆2ω is a strong first category set, then in the game G(X, SM Z) player TWO
has also a winning strategy. We will show something more.
Theorem 3 For every meager set X⊆2ω player TWO has a winning strategy
in the game G(X, SM Z).
Notice that this theorem solves Problem 2.1 from [5]. Notice also that the
”dual” question about null and strong first category sets (see Problem 2.2 from
[5]) remains open.
Problem 1 Is it consistent with ZFC that SF C is σ - ideal and there exists a
measure null set N ⊆2ω such that Player TWO has not a winning strategy in
the game G(N, SF C)?
Lemma 5 Consider the following auxiliary additive game
G (X, X): Given a set X⊆2ω and a family X, Player ONE chooses a set
X0 ∈ X, Player TWO chooses a t0 ∈ ∪s∈[ω]<ω 2s , Player ONE chooses a set
X1 ∈ X, Player TWO chooses a t1 ∈ ∪s∈[ω]<ω 2s such that dom(t0 )∩dom(t1 ) =
∅, and so forth:
• ONE
X0
X1
X2
• TWO
t0
t1
t2
Player TWO wins the play if and only if
Xn = ∅
∀t ∈ 2ω , (∀i, t|dom(ti ) = ti ) ⇒ (t + X) ∩
n<ω
If Player TWO has a winning strategy in the game G (X, X), then Player TWO
has a winning strategy in the game G(X, X).
Take X ∈ M. From the characterization of meager sets we may assume
without loss of generality that X = {x ∈ 2ω : ∀n < ∞, x|In = z|In }, where
z ∈ 2ω and {In }n<ω are pairwise disjoint, finite intervals in ω.
Fix any bijection φ : ω × ω → ω. Consider k–th move of Player ONE:
A. Nowik
• ONE
• TWO
177
C0
C1
t0
...
t1
Ck
. . . tk−1
where dom(ti ) ∩ dom(tj ) = ∅ for i = j and ti : dom(ti ) → 2, dom(ti )⊆ω. Since
(k)
Ck ∈ SM Z, for each l ∈ ω there exists a sequence sl ∈ 2<ω such that
(k)
|sl | > max Iφ(k,l)
(1)
and
Ck ⊆
L<ω l>L
(k)
[sl ].
We define a function
tk :
(2)
Iφ(k,l) → 2.
l<ω
by the following condition:
(3)
(k)
tk |Iφ(k,l) = sl |Iφ(k,l) + z|Iφ(k,l)
for each l < ω.
It is not hard to see that we have described a winning strategy for Player
TWO in this game. Indeed, take t ∈ 2ω such that
∀i,
(4)
t|dom(ti ) = ti .
Fix n < ω and x ∈ Ck where k < ω.
There exists l < ω such that
(k)
x ∈ [sl ] ∧ max Iφ(k,l) > n.
(5)
Thus:
(x + t)|Iφ(k,l)
= x|Iφ(k,l) + t|Iφ(k,l)
(k)
= sl |Iφ(k,l) + t|Iφ(k,l)
(k)
= sl |Iφ(k,l) + tk |Iφ(k,l)
= z|Iφ(k,l) .
from (1)(5)
from (4)(2)
from (3)
Therefore ∃r < ∞, (x + t)|Ir = z|Ir then t + x ∈ X. From this we
conclude that Player TWO has a winning strategy in the game G (X, SM Z).
From the Lemma 5 we deduce that the Player TWO has a winning strategy in
the game G(X, SM Z). This finishes the proof.
178
Solutions of two problems ...
References
[1] T.Bartoszyński H.Judah, ” Set Theory. On the Structure of the Real
Line”, A.K. Peters, Wellesley, Massachusets, 1995.
[2] A. Nowik, M. Scheepers, T. Weiss. The algebraic sum of sets of real
numbers with strong measure zero sets, Journal of Symbolic Logic 63(1)
(1998), 301-324.
[3] S. Plewik Towers are universally measure zero and always of first category, Proceedings of the AMS 119 (3) (1993), 865-868.
[4] I. Reclaw, ”Products of perfectly meagre sets”’ Proceedings of the AMS,
112 (4) (1991), 1029-1031.
[5] M. Scheepers, ”The Boise problem book”,
http://www.unipissing.ca/topology/.
[6] M. Scheepers, Additive properties of sets of real numbers and an infinite
game, Quaestiones Mathematicae 16 (1993), 177-191.