Strongly meager sets and subsets of the plane by

156 (1998)

Strongly meager sets and subsets of the plane

by

Janusz P a w l i k o w s k i (Wrocław)

Abstract. Let X ⊆ 2

. Consider the class of all Borel F ⊆ X × 2

with null vertical sections F

, x ∈ X. We show that if for all such F and all null Z ⊆ X, S

x∈Z

F

is null, then for all such F , S

x∈X

F

6= 2

. The theorem generalizes the fact that every Sierpi´ nski set is strongly meager and was announced in [P].

A Sierpiński set is an uncountable subset of 2

which meets every null (i.e., measure zero) set in a countable set. Such sets may not exist, but they do, e.g., under the Continuum Hypothesis.

A strongly meager set is a subset of 2

whose complex algebraic sum with null sets cannot give 2

.

Answering a question of Galvin I proved in [P] that every Sierpi´ nski set is strongly meager (see [M] and [P] for more about Galvin’s question).

Since Sierpiński sets may not exist, this result is somewhat defective. Here we prove its “absolute” version, which seems to be of independent inter- est. The paper is an elaboration of the Note given at the end of [P]. A different elaboration, using “small sets” of [B] and closely following my lecture at Cantor’s Set Theory meeting, Berlin 1993, is given in [BJ] in Section 5 (repeated in [BJ1]). There is, however, a major gap in the expo- sition in [BJ], namely, in the proof of Lemma 5.5, where one really needs a sort of Kunugui–Novikov theorem (see the proof of Lemma 4 below).

Also it seems reasonable to avoid “small sets” because they do not form an ideal.

Let X ⊆ 2

. Consider the class of all Borel F ⊆ X ×2

with null vertical sections F

, x ∈ X. If for all such F , S

x∈X

F

is null, resp. 6= 2

, we say that X ∈ Add, resp. X ∈ Cov. (See [PR] for an explanation of this notation.) Here Borel means relatively Borel. It is useful to remember that for a Borel F ⊆ X × 2

, the function x 7→ µ(F

) is Borel. In particular, any Borel

1991 Mathematics Subject Classification: 04A15, 03E15.

Partially supported by KBN grant 2 P03A 011 09.

[279]

subset of Y × 2

, Y ⊆ X, with all vertical sections null, extends to a Borel subset of X × 2

with all vertical sections null.

Theorem. Suppose that every null subset of X ⊆ 2

is in Add. Then X ∈ Cov.

The theorem implies that every Sierpi´ nski set is strongly meager as fol- lows. Suppose X ⊆ 2

is a Sierpi´ nski set. Let D ⊆ 2

be null. We want to see that D + X 6= 2

. Consider

F = [

x∈X

{x} × (D + x).

Clearly F is a Borel subset of X × 2

and all its vertical sections are null.

Also, every null subset of X, being countable, is in Add. So, by the theorem, S

x∈X

F

6= 2

. But D + X = S

x∈X

F

.

Notation. Given a set K and A ⊆ 2

, let µ(A) be the measure of A in the product measure arising from assigning to each point in {0, 1} weight 1/2. Note that if K is finite, then µ(A) = |A| · 2

.

For K ⊆ L ⊆ ω and A ⊆ 2

let [A] = {t ∈ 2

: t|L ∈ A} and let A|K = {t|K : t ∈ A}. Likewise, for A ⊆ (2

)

let [A] = {ht

, . . . , t

i ∈ (2

)

: ht

|L, . . . , t

|Li ∈ A}.

Clearly µ(A) = µ([A]). Note that any clopen subset of 2

can be written as [A] for some A ⊆ 2

. Also for A ⊆ 2

and m > n, [A] = [B], where B = {τ ∈ 2

: τ |n ∈ A}.

For σ ∈ ω

of length n + 1 , let σ

be σ|n.

We use the following abbreviations:

∃

— there exist infinitely many,

∀

— for all but finitely many, W

n

— T

m

S

n>m

, V

n

— S

m

T

n>m

.

For F ⊆ X × T and x ∈ X let F

= {t ∈ T : hx, ti ∈ F }. Likewise, for F ⊆ X × S × T , x ∈ X, s ∈ S, let F

= {t ∈ T : hx, s, ti ∈ F }, etc. In particular, if F ⊆ X × (2

)

, t

, . . . , t

∈ 2

, then

F

= {ht

, . . .i ∈ (2

)

: hx, t

, . . . , t

, t

, . . .i ∈ F }.

Let F [X] = S

x∈X

F

. If F has all sections F

, x ∈ X, null, we say that F is X-null.

The following simple lemma is crucial.

Lemma 1. Let every null subset of X be in Add. Suppose F ⊆ X × (2

)

is Borel and X-null. Then, given null Y ⊆ X, there exist t ∈ 2

and Borel

null Z ⊆ X \ Y such that F

, x 6∈ Z, are null.

P r o o f. Let

G = {hx, ti ∈ X × 2

: µ(F

) > 0}.

Then G is Borel and X-null. By Fubini’s theorem find t ∈ 2

\ G[Y ] such that

Z = {x : hx, ti ∈ G}

is null. (This is possible because G[Y ] is null.)

We shall need the following property B, which is a Borel version of prop- erty H of Hurewicz (for more see [FM], [PR]):

X ⊆ 2

has property B if, given for each n ∈ ω a Borel cover {U

}

of X, there exist k

’s such that X ⊆ V

n

S

k≤kn

U

.

It is not hard to see that we can use increasing covers in this definition and write X ⊆ V

n

U

. Also, easily, X has property B iff for any Borel function f : X → ω

, f [X] is dominated. (Y ⊆ ω

is dominated if there exist z ∈ ω

such that ∀

n y(n) < z(n), for all y ∈ Y .) Moreover, it is enough to consider only f for which all f (x) are increasing.

The following lemmas are well known.

Lemma 2. If all null subsets of X ⊆ 2

have property B, then X has property B.

P r o o f. Let {U

}

, n ∈ ω, be increasing Borel covers of X. Find k

’s with µ

(X \ U

) < 2

. Let Z = X \ V

n

U

. Then Z is null, so it has property B. Thus Z ⊆ V

n

U

for some l

’s. It follows that X ⊆ V

n

U

.

Lemma 3. (1) If A ⊆ 2

is null then for any sequence {ε

} of positive reals there exists an increasing sequence {a

} ∈ ω

together with sets B

⊆ 2

of measure ≤ ε

, n ∈ ω, such that A ⊆ W

n

[B

].

(2) If a

∈ ω and B

⊆ 2

, n ∈ ω, are such that P

n

µ(B

) < ∞, then A = W

n

[B

] is null. If moreover K ⊆ ω is such that P

n

µ(B

)·2

< ∞, then also A|(ω \ K) is null (in 2

).

P r o o f. We prove the first part, the second is straightforward. Given null A ⊆ 2

and ε > 0, we can cover A by an open set of measure < ε/2

, which next can be split into disjoint clopen sets. In this way we can find clopens C

, i ∈ ω, such that A ⊆ W

i

C

and P

i

µ(C

) < ε.

Suppose now that {ε

} is a sequence of positive reals. Use the above to find clopens C

, i ∈ ω, such that A ⊆ W

i

C

and P

i

µ(C

) < ε

. Next find an increasing sequence {i

} such that P

i≥i_n

µ(C

) < ε

. Finally, let A

= S

i<i0

C

and for n > 0 let A

= S

in−1≤i<in

C

. Then µ(A

) < ε

and A ⊆ W

n

A

. Each A

, being clopen, is of the form [B

] for some B

⊆ 2

.

We can easily arrange that a

> a

.

It follows from Lemma 3 that, given {b

} ∈ ω

and null A ⊆ 2

, there is an increasing {a

} ∈ ω

such that

∀

n |a

∩ K| ≤ b

⇒ A|(ω \ K) is null.

Lemma 4. Suppose that F ⊆ X × 2

is Borel X-null. Let {ε

} be a sequence of positive reals. Then there exist for each n a countable Borel partition U

of X together with integers a

and sets A

⊆ 2

of measure

≤ ε

, U ∈ U

, such that

F ⊆ _

n

[

U ∈Un

U × [A

].

If additionally X has property B, we can require that for some increasing {a

} ∈ ω

all a

, U ∈ U

, equal a

.

P r o o f. This is a parametrized version of the first part of Lemma 3. We indicate the main steps.

For any ε > 0 there are Borel sets W

and clopens C

, i ∈ ω, such that F ⊆ W

i

W

× C

and for all x ∈ X, P

i∈K(x)

µ(C

) < ε, where K(x) = {i : x ∈ W

}.

This follows from the following facts:

• any Borel subset of X × 2

with open vertical sections can be written as a union of countably many disjoint sets of the form W × C, W Borel, C clopen (a theorem of Kunugui and Novikov, see [K]);

• for any Borel B ⊆ X × 2

and ε > 0, there exists a Borel A ⊆ X × 2

, B ⊆ A, such that vertical sections of A are open and µ(A

\ B

) < ε (the sets with such a covering property form a monotone family that includes all finite unions of Borel rectangles).

Note that the function x 7→ P

i∈K(x)\j

µ(C

) is Borel. (Because the function µ(C

)1

that takes µ(C

) on W

and 0 outside is Borel and P

i∈K(x)\j

µ(C

) = P

i≥j

µ(C

)1

(x).) It follows that for any δ > 0 we can find a countable Borel partition U of X and numbers j

∈ ω, U ∈ U, such that on each U the mapping x 7→ K(x) ∩ j

is constant and P

i∈K(x)\jU

µ(C

) < δ.

Using this find Borel sets U

and integers j

, σ ∈ ω

, such that

• U

= X, j

= 0,

• U

is partitioned into U

’s, k ∈ ω,

• if |σ| > 0, then j

< j

and x 7→ K(x) ∩ [j

, j

) is constant on U

,

• if |σ| > 0, then on U

, X

i∈K(x)∩[j_σ∗,jσ)

µ(C

) < ε

.

If |σ| > 0, let

B

= [

i∈K(x)∩[j_σ∗,j_σ)

C

, x ∈ U

.

This is a clopen set of measure < ε

, so we can find a

∈ ω and A

⊆ 2

of measure < ε

such that B

= [A

].

Now just note that _

i

W

× C

⊆ _

n>0

[ {U

× [A

] : σ ∈ ω

}.

Up to some enumeration, we are done.

The following lemma is a version of Miller’s [M1] result that additivity of measure is below number b. (See also [PR].)

Lemma 5. Add ⊆ B.

P r o o f. Let Y ∈ Add. Let Y 3 y 7→ y ∈ ω

be Borel with all y’s increasing. Define F ⊆ Y × 2

by

t ∈ F

⇔ ∃

n ∀i < n t(y(n) + i) = 0.

Then F is Borel and Y -null, so A = F [Y ] is null. Use Lemma 3 to find an increasing sequence {a

} such that

∀n |K ∩ a

| ≤ n(n − 1)/2 ⇒ A|(ω \ K) is null.

We claim that {a

} dominates all y’s. Indeed, suppose that ∃

n y(n) ≥ a

. Consider

K = [

{[y(n), y(n) + n) : y(n) ≥ a

}.

Then ∀n |K ∩ a

| ≤ n(n − 1)/2 (we take to K below a

at most n − 1 intervals).

It follows that A|(ω \ K) is null. This is a contradiction because F

⊆ A and F

|(ω \K) is 2

. (Any element of 2

can be extended to an element of 2

which on infinitely many intervals [y(n), y(n)+n) is constantly zero.)

Proof of theorem. By Lemmas 2 and 5, X ∈ B. Let F ⊆ X × 2

be Borel X-null. We seek a point outside F [X]. Let Q = {t ∈ 2

: ∀

n t(n) = 0}.

Enlarging F if necessary we can assume that for all x, F

= F

+ Q. Use Lemma 4 to find an increasing {a

} ∈ ω

together with a sequence {U

} of countable Borel partitions of X such that for some A

⊆ 2

, U ∈ U

, of measure ≤ 2

,

F ⊆ _

n

[

U ∈Un

U × [A

].

Let B

be A

for the unique U ∈ U

that covers x.

Say that σ

, . . . , σ

∈ 2

have a diagonal in A ⊆ 2

if for some n

≤ . . . ≤ n

≤ n,

σ

|a

∪ σ

|[a

, a

) ∪ . . . ∪ σ

|[a

, a

) ∈ A.

Say that t

, . . . , t

∈ 2

have a diagonal in A ⊆ 2

if t

|a

, . . . , t

|a

do.

Define E ⊆ X × (2

)

by

ht

, t

, . . .i ∈ E

⇔ ∃k ∃

n t

, . . . , t

have a diagonal in B

. Claim 1. E is Borel and X-null.

P r o o f. Let

B

(k) = {hσ

, . . . , σ

i ∈ (2

)

: σ

, . . . , σ

have a diagonal in B

}.

Then |B

(k)| ≤ 2

2

· (1 + n)

. Indeed, there are ≤ (1 + n)

possi- ble sequences n

, . . . , n

, and for each sequence we have |B

| times 2

possible choices for hσ

, . . . , σ

i.

So

µ([B

(k)]) = |B

(k)|/2

≤ 2

(1 + n)

. It follows that

µ(E

) ≤ X

k

Y

m

X

n≥m

2

(1 + n)

= 0.

Claim 2. There exist {t

} ⊆ 2

and a Borel partition {X

} of X such that each

E

, x ∈ X

, k ≥ i, is null.

P r o o f. Apply Lemma 1 with Y = ∅ and F = E to find t

∈ 2

and Borel null X

⊆ X such that the following sets are null:

E

, x ∈ X

, E

, x 6∈ X

. Next apply Lemma 1 to Y = X

and

F = [

x∈X₁

{x} × E

∪ [

x6∈X1

{x} × E

to get t

∈ 2

and Borel null X

⊆ X \ X

such that the following sets are null:

E

, x ∈ X

,

E

, x ∈ X

,

E

, x 6∈ X

∪ X

.

Similarly find t

∈ 2

and Borel null X

⊆ X \ (X

∪ X

) such that the following sets are null:

E

, x ∈ X

, E

, x ∈ X

, E

, x ∈ X

,

E

, x 6∈ X

∪ X

∪ X

, etc. Finally, set X

= X \ S

i>0

X

.

It follows from Claim 2 that for x ∈ X

, ht

, t

, . . .i 6∈ E

.

Otherwise we would have E

= (2

)

for some k.

Thus for x ∈ X

and k ≥ i,

∀

n t

, . . . , t

have no diagonal in B

. For all k and n let

V

= [

i>k

X

∪ [

i≤k

{x ∈ X

: ∀m ≥ n t

, . . . , t

have no diagonal in B

}.

Then for all k, V

’s form an increasing Borel cover of X. By X ∈ B, there is an increasing sequence {n

} such that

X ⊆ ^

k

V

.

Let

t = t

|a

∪ t

|[a

, a

) ∪ t

|[a

, a

) ∪ . . . . Claim 3. t 6∈ F [X].

P r o o f. Fix x ∈ X

. Since ∀

k x ∈ V

, for all sufficiently large k ≥ i and n ≥ n

, t

, . . . , t

have no diagonal in B

. Hence,

t

|a

∪ t

|[a

, a

) ∪ . . . ∪ t

|[a

, a

) 6∈ B

, and thus

t

|a

∪ [

k>i

t

|[a

, a

) 6∈ _

n

[B

].

It follows that t 6∈ F

.

Note. We have really proved that if X ∈ B has all its null subsets in Cov,

then X ∈ Cov. The crucial Lemma 1 goes through because if Y ∈ Cov, then

for all Borel Y -null F ⊆ Y × 2

, µ

(2

\ F [Y ]) = 1. (Otherwise we could

find in F [Y ] a perfect set P of positive measure. Then D = F ∩ (Y × P )

would be a Borel Y -null subset of Y × P such that D[Y ] = P . This would

yield a similar subset of Y × 2

.)

Note also that if X ∈ B ∩ Cov, then 2

\ F [X] contains a perfect set for all X-null F ⊆ X × 2

. (It is enough to require in Lemma 4 that A

’s have measure ≤ 2

and consider B

= {σ ∈ 2

: ∃τ ∈ A

σ|K = τ |K}, where K is a fixed co-infinite subset of ω such that ∀n |a

\ K| ≤ n. Then B

is a subset of 2

of measure ≤ 2

. If X × {t} avoids W

n

S

U

U × [B

], then {s ∈ 2

: s|K = t|K} is a perfect set disjoint from F [X].)

We cannot drop B in the above remark. If we add ω

random reals to a model of CH then the ground model reals constitute a counterexample. (Use the fact that a random real does not add a perfect set of random reals.)

We cannot require in the theorem that X ∈ Add. It is enough to take for X a Sierpi´ nski set and for F the diagonal in X ×X. There is however no ZFC example for this. Indeed, suppose the Dual Borel Conjecture holds, i.e. all strongly meager sets, hence also all Cov sets, are countable. Suppose also that every uncountable set has an uncountable null subset. (Both assumptions are true when ω

Cohen reals are added to a model of CH, see [C].) If all null subsets of X are in Cov, then they are all countable by the first assumption.

So X has no uncountable null subsets, and thus X itself is countable by the second assumption. It follows that X ∈ Add.

Suppose that all null subsets of X are in Cov. Does it follow that X is strongly meager? We have the following partial result:

Proposition. Let X have property B. Let D = W

k

[B

], B

⊆ 2

, where L

⊆ ω, k ∈ ω, are pairwise disjoint. Suppose for every finite F ⊆ 2

, D + (X ∩ (D + F )) 6= 2

. Then D + X 6= 2

.

P r o o f. Choose t

∈ 2

and inductively t

∈ 2

so that t

6∈ D + (X ∩ (D + {t

, t

, . . . , t

})).

Then for all x ∈ X,

∀

n x 6∈ D + t

.

Indeed, if x ∈ D + t

and m > n, then t

6∈ D + x, so x 6∈ D + t

. Let

U

= {x : x 6∈ D + t

⇒ ∀m ≥ k x|L

6∈ B

+ t

|L

}.

Then for all n, U

’s form an increasing Borel cover of X. So X ⊆ V

n

U

for some increasing sequence {k

}. Then for all x,

∀

n ∀k ≥ k

x|L

6∈ B

+ t

|L

(remember that ∀

n x 6∈ D + t

).

So for all x,

∀

n ∀k ≥ k

t

|L

6∈ B

+ x|L

.

Let t ∈ 2

be such that for all n, [

k_n≤k<k_n+1

t

|L

⊆ t.

Then for all x,

∀

n ∀k ∈ [k

, k

) t|L

6∈ B

+ x|L

. It follows that ∀x ∈ X t 6∈ D + x.

Note. Let small mean small in the sense of Bartoszy´ nski [B]. Suppose that every small subset of X has property B. Suppose also that for every small Y ⊆ X and small D ⊆ 2

, Y + D 6= 2

. Then for every small D ⊆ 2

, X + D 6= 2

. (Every null set is a union of two small sets. So, if all small subsets of X have property B, then all null subsets of X have property B, thus X itself has property B. Also, a union of finitely many translates of a small set is small.)

References

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[BJ] T. B a r t o s z y ń s k i and H. J u d a h, Borel images of sets of reals, Real Anal. Ex- change 20 (1995), 1–23.

[BJ1] —, —, Set Theory: on the Structure of the Real Line, A. K. Peters, Wellesley, Mass., 1995.

[C] T. C a r l s o n, Strong measure zero and strongly meager sets, Proc. Amer. Math.

Soc. 118 (1993), 577–586.

[FM] D. H. F r e m l i n and A. M i l l e r, On some properties of Hurewicz , Menger and Rothberger, Fund. Math. 129 (1988), 17–33.

[K] A. K e c h r i s, Classical Descriptive Set Theory, Springer, 1995.

[M] A. W. M i l l e r, Special subsets of the real line, in: Handbook of Set-Theoretic Topology, K. Kunen and J. E. Vaughan (eds.), Elsevier, 1984, 203–233.

[M1] —, Additivity of measure implies dominating reals, Proc. Amer. Math. Soc. 91 (1984), 111–117.

[P] J. P a w l i k o w s k i, Every Sierpiński set is strongly meager, Arch. Math. Logic 35 (1996), 281–285.

[PR] J. P a w l i k o w s k i and I. R e c ł a w, Parametrized Cichoń’s diagram and small sets, Fund. Math. 147 (1995), 135–155.

Department of Mathematics University of Wrocław Pl. Grunwaldzki 2/4 50-384 Wrocław, Poland

E-mail: pawlikow@math.uni.wroc.pl

Received 10 June 1997