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

Serin I: PRACE MATEMATYCZNE XXVIII (1989)

### Z.

**K**

**o m in e k**

### (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 *

*g*

* A, b e B ), * *A — B — \a — b; a *

*g*

* 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.**

**L****e 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 %*

*x g*

*A'. Hence *

*x g*

*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.

**T****h 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.

**0**

**0 .**

### 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.*

**0**

**0 .**

**0 ,**

*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.

**T****h 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 *

**Ф 0**

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

**Ф**

*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' *

*Ф*

*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 *

*e*

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

*n*(J

*[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*

*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 .*

**T****heorem**** 3. **

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

*Ф*

*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 *

*Ф*

*0*

*and therefore А' *

*Ф*

*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 *

*Ф*

*G and * *A' + A' *

^{Ф }*G then A + A' = G. A similar result holds true if А — А *

^{Ф }*G and * *А' —А' *

*Ф*

*G, Namely, we have the theorem.*

**T****heorem**** 4. **

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

*Ф*

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

### P roof. By Theorem

**3,**

*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**

**C****o r o l l a r y**** 1. **

*I f A + A *

*Ф*

*G or А' + А' *

*Ф*

*G or A' *

*+*

*А *

*Ф*

*G, then A *

*—*

*A *

*= G or A ' - A ' = G.*

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

**T****h 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' *

*Ф*

*G. Then there exists a y e G such that *

*y *

*— A '*

*^*

*A which means that A' *

*a y —*

*A. Therefore, A' *

*—*

*A' с A — А *

*Ф*

*G. *

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

### The following corollary is a direct consequence of Theorem 5.

**C****o 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.*

**References**

**[1] ** **F. B a g e m ih l, ***S o m e s e ts o f s u m s a n d d iffe r e n c e s ,*** Michigan Math. J. 4 (1957), 289-290.**

**[2] ** **Z. K o m in e k , ***M e a s u r e , c a te g o r y , a n d th e s u m o f s e ts ,*** Amer. Math. Monthly 90 (October **
**(1983), 561-562.**

**[3] ** **M. K u c z m a , ***A n I n t r o d u c t i o n to th e T h e o r y o f F u n c tio n a l E q u a tio n s a n d I n e q u a litie s , *

**P.W.N. Uniwersytet Slqski, Warszawa-Krakow-Katowice 1985.**

**[4] ** **H. I. M ille r , ***S o m e d e c o m p o s itio n th r o r e m s f o r th e r e a l lin e,*** Radovi Matematicki 1 (1985), **
**31-37.**

**[5] ** **S. P ic c a r d , ***S u r les e n s e m b le s p a r fa its ,*** Mem. Univ. Neuchâtel 16 (1942).**

**[6] ** **H. S t e in h a u s , ***S u r les d is ta n c e s d e s p o in ts d e s e n s e m b le s d e m e s u r e p o s itiv e ,*** Fund. Math. 1 **
**(1920), 99-104.**