Some properties of decompositions of a commutative group

ANNALES SOCIETATIS MATHEMATICAE POLONAE Series I: COM MENT ATIONES MATHEMATICAE XXVIII (1989) ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO

Serin I: PRACE MATEMATYCZNE XXVIII (1989)

Z.

(Katowice)

Some properties of decompositions of a commutative group

Abstract. Considering decompositions of a commutative group, we study some relations between the sets: A + A , А л - А ' , А 'Л - А , A — A , A — A ' and A — A , where A ' denotes the complement of A . Some examples which illustrate our results are also presented.

In this paper, G denotes a commutative group. For arbitrary subsets A and В of G we define the sum set and the difference set, respectively, in the following manner:

A + B = \a + b; a

A, b e B ), A — B — \a — b; a

A, beB ).

It is well known ([1], [2]) that if G is the additive group of all real numbers R then there exists a Lebesgue measurable and residual in R set A of the measure equal to zero such that A — A' contains no open interval.

1. The sum sets. We start with a simple lemma.

Le m m a.

Let G be a group divisible by two. For any subset A of G we have (A + A )u (A ' + A') = G.

P ro o f. Taking an хфА + А, we note that %

A'. Hence

A' + A' and the proof is completed.

R em ark 1. The assumption of the divisibility by two is essential in the above lemma.

To see that we take the additive group Z of all integers and the set A to be the set of all even numbers.

Th e o r e m 1.

Let G be a commutative group divisible by two. For arbitrary subset A of G we have A' + A — G or A + A = G or A' + A' = G.

P ro o f. Assume that A' + A ^ G . Then for every хфА' + А we get

x — A' a A' and x — A a A. Hence x — A' = A' and x — A = A, because G is a

commutative group and (x — A') u (x — A) = G. Putting A x := A — \ x and

Bx A '—\x , we note that A x = —A x, B x = —Bx and Bx = A\. We shall

show that A x+ A t = G or B x+ B x = G. Suppose that B x+ B x Ф G. For any

уфВх + B x we have y — B x a A x. It follows from symmetry of the sets A x and

250 Z. K o m in ek

B x with respect to zero that — y + Bx а А х and у + В х czAx. Therefore, В х+ В х cz А х+ А х. On account of Lemma we get A x + A x = G, which implies that A + A = G.

It is well known that there exists a subset A of the reals R such that int(>l + ^) = 'mi{A' + A') — 0 (here and in the sequel, int В, В czR, denotes the interior of B) as well as there exists a subset A of R with the property int (A + A) = int (A' + A) = 0 . We shall show the -existence of such a set with some special property.

R em ark 2. There exists a saturated non-measurable in the Lebesgue sense subset A of R (i.e., the inner Lebesgue measure of the sets A and A' are both equal to zero) and of the first Baire category such that mi{A + A)

= int (A' + A) = 0 .

P ro o f. Let Я be a saturated non-measurable in the Lebesgue sense basis of the first category of the space R over the field of all rationals Q (such a basis does exist; [3], Corollary 3, p. 261). Put

A: = (Q + H ) u ( Q - H ) .

Since Q — A c z A , thus (A' + A)rsQ = 0 which implies in t(A' + A) = 0 . Of course, int (A + A) = 0 , too. Hence and by a Steinhaus’ theorem [ 6 ], the inner Lebesgue measure of A is equal to zero. On the other hand, A' cr H' and therefore the inner Lebesgue measure of A' is equal to zero as well.

A topological analogue of saturated non-measurability of the set A reads as follows:

(*) if В is a set with the Baire property and В c A or В c A', then В is of the first category.

Similarly to Remark 2, we have

R em ark 3. There exists a set A a R which has property (*) and int {A + A) = \r)i(A' + A) = 0 .

P ro o f. We can repeat the proof of Remark 2 with H as a basis of the space R over Q having property (*) (see [3], Corollary 3 p. 261) and use a theorem of Piccard [5] instead of theorem of Steinhaus.

Th e o r e m 2.

Let A be an arbitrary subset of a commutative and divisible by two group G. Then A' + A = G if and only if for every y e G we have y - A Ф A - y .

P ro o f. Assume that A' + A = G and take a yeG . Then (2y — A ) n A '

Ф 0 and therefore у — А ф А — у. On the other hand, the condition

A' + A Ф G implies (see the proof of Theorem 1) the existence of an x e G

such that A —\ x = ? x — A which ends the proof.

Properties o f decompositions o f a commutative group 2 5 1

2. The difference sets. First we note that one always has А — А'

G, because 0 фА — А'. Miller in [4] showed, among others (using the method of transhnite induction), that there exists a subset A of R such that the conditions \ni(A —A) = 'mt(A'— A') = Ф hold. Now, we give also some example of such set A and we shall show that then one always has m i{A -A ') = 0 .

By E (C), C c / J , we denote the space spanned by C over Q, i.e., E(C):= {x = £ r aca; raeQ, caeC, the sum being finitej>.

Let Я be a basis of the space R over Q such that 1

H. Putting A: = E(H \\1\) + [Q

[2n,2n+l)],

• neZ

we observe that

A —A = £ (H \! l} ) + [<2\(2Z + l)],

where Z is the set of all integers. We shall show that (A — A) n (Е (Н \Щ ) + + 1) = 0. Suppose not. Then there exists u, veA such that u — v eE (H \\l]) + + \_Q\(2Z+Vf\ and и — v

E (H\ J1}) +1, which is impossible. So, int(A — A)

= 0 , because E(H\\1)) is a dense set in R. It follows from the fact that A' = A + 1 that int(A '-A ') = in t(A -A ') = 0 .

Theorem 3.

Let A be an arbitrary subset of a commutative group G with the properties A — A + G and А' — А'

G. Then A — A = A' — A' and A' = A + x iff хфА — А. If, moreover, G is a topological group then, in particular, the condition mt(A — A) = int( A '- A ') = 0 implies in t(A -A ') = 0 .

Proof. Assume that гфА — А and мгфА' — А'. Then we have z + A a A' and w + A ' ^ A . Hence A — A = A ' — A'. The condition x e A — A implies {x + A) n А

and therefore А'

A + x. If хфА — А then x + A a A' <= A — x.

By symmetry of A — A = A' — A' with respect to zero we have —хфА — А and, similarly as above, —x + A a A' c i + x. So, A' = A + x. This completes the proof.

3. Mixed case. It follows from Theorem 1 that if A + А

G and A' + A'

G then A + A' = G. A similar result holds true if А — А

G and А' —А'

G, Namely, we have the theorem.

Theorem 4.

Let G be an arbitrary commutative group divisible by two and A a G. I f A — А

G and A' — А' Ф G, then A + A — A' + A' = A' + A = G.

P roof. By Theorem

A - A = A ' - A ' and A' = A + x for each хф A —A. Moreover, it is easily seen that A = A' + x for each хфА — А. Hence A' + A' = A + A, because A + 2x = A' + x = A. On account of Lemma, A + A

= A' + A' = G. Now, A + A ' = A + A + x — G, too.

From the Theorem 4 we get immediately the following

— Commentationes Math. 28.2

252 Z. K o m in ek

Co r o l l a r y 1.

I f A + A

G or А' + А'

G or A'

А

G, then A

A

= G or A ' - A ' = G.

The next theorem describes some relations between the sets A — A and A' + A'.

Th e o r e m 5.

Let G be an arbitrary commutative group divisible by two and A c G. If А — А Ф G then A' + A' = G.

Proof. Similarly as in the proof of Theorem 3, we infer that A — A c A' — A'. Now, we assume that A' + A'

G. Then there exists a y e G such that

y

— A '

A which means that A'

A. Therefore, A'

A' с A — А

G.

By Theorem 4, A' + A' = G, a contradiction.

The following corollary is a direct consequence of Theorem 5.

Co r o l l a r y

2. Let V be a linear vector space. For every subset A of V either A or A' contains a basis of the space V.

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