THE ALGEBRA OF SEMIGROUPS OF SETS
MATS AIGNER, VITALIJ A. CHATYRKO and VENUSTE NYAGAHAKWA
Abstract
We study the algebra of semigroups of sets (i.e. families of sets closed under finite unions) and its applications. For eachn > 1 we produce two finite nested families of pairwise different semigroups of sets consisting of subsets ofRnwithout the Baire property.
1. Introduction
An interesting extension of the familyMof all meager subsets of the real line R, as well as the familyO of all open subsets ofR, in the familyP(R)of all subsets ofRis the family Bp of all sets possessing the Baire property. The property is a classical notion which is related to the thesis of R. Baire. Recall thatB∈Bpif there are anO∈Oand anM ∈Msuch thatB=OM.
It is well known that the familyBp is aσ-algebra of sets invariant under homeomorphisms of the real lineR, and the complementBpC=P(R)\Bpof BpinP(R)is not empty (for example, each Vitali setSofR([7]) is an element ofBpC). Moreover, there are elements ofBpCwith a natural algebraic structure (see [4] for subgroups of the additive groupR, which are elements ofBpC).
In [2] Chatyrko and Nyagahakwa looked for subfamilies of the familyBpC which have some algebraic structures. They proved that the familyV1of all finite unions of Vitali sets ofRand its extensionV2which elements are all sets of the typeAB, whereA ∈ V1andB ∈ M, are semigroups of sets (i.e.
families of sets closed under finite unions) invariant under translations of the real lineR and consisting of zero-dimensional subsets ofBpC. Furthermore, Chatyrko and Nyagahakwa extended the result to the Euclidean spacesRn, wherenis any positive integer.
In this paper we pay attention to the algebra of semigroups of sets. We look at the behavior of semigroups of sets under several operations. Then we suggest some applications. First, we show that the results from [2] can be obtained by the use of the theory. Moreover, we can suggest many different semigroups of sets inBpC. After that for eachn >1 we produce two finite nested families of
Received 30 December 2012, in final form 20 September 2013.
pairwise different semigroups of sets consisting of subsets ofRn without the Baire property.
2. Auxiliary notions
Recall that a non-empty setS is calleda semigroupif there is an operation α:S×S →S such thatα(α(s1, s2), s3)=α(s1, α(s2, s3)). The semigroup S is calledabelianifα(s1, s2)=α(s2, s1).
LetXbe a set andP(X)be the family of all subsets ofX. In the paper we will be interested in subsetsS ofP(X)such that for eachA, B∈S we have A∪B ∈ S. It is evident that such a family of sets is an abelian semigroup with respect to the operation of union of sets (in brief,a semigroup of sets).
LetA ⊂ P(X). PutSA = {∪i≤nAi : Ai ∈ A, n∈ N}. Note thatSA is a semigroup of sets. Recall that a setI ⊂ P(X)is called an ideal of sets if I is a semigroup of sets and if A ∈ I and B ⊂ A then B ∈ I. Put IA = {B ∈P(X): there isA∈SA such thatB ⊂ A}. Note thatIA is an ideal of sets.
Forx ∈Rdenote byTx the translation ofRbyx, i.e.Tx(y) = y+xfor eachy∈R. IfAis a subset ofRandx ∈R, we denoteTx(A)byAx.
The equivalence relationE onRis defined as follows. For x, y ∈ R, let xEy iffx −y ∈ Q, where Q is the set of rational numbers. Let us denote its equivalence classes byEα,α ∈ I. It is evident that|I| = c(continuum), and for eachα ∈ I and eachx ∈ Eα,Eα = Qx. Let us also note that every equivalence classEα is dense in R. Recall ([7]) thata Vitali set of Ris any subsetSofRsuch that|S∩Eα| =1 for eachα∈I, and each Vitali set neither possess the Baire property inRnor it is measurable in the sense of Lebesgue.
For other notions and notations we refer to [3] and [6].
3. Semigroups of sets and ideals of sets
LetA,B ⊂ P(X). Put A ∪B = {A∪B : A ∈ A, B ∈ B},A B = {AB:A∈A, B ∈B}andA∗B = {(A\B1)∪B2:A∈A;B1, B2∈B}. However,A∩Bdenotes the intersection ofA,B, i.e. the family of common elements ofA,B.
It is evident thatA∪B =B∪AandAB =BA. SinceA∪B = (A\B)∪B =(B\A)∪A, we haveA∪B ⊂A∗BandA∪B ⊂B∗A. Moreover, ifA,B are both semigroups of sets or both ideals of sets then the familyA ∪Bis of the same type.
On the other hand as we will see in the following examples in general for given semigroups of setsA,B the familiesAB,A ∗B,B∗A do not need to be semigroups of sets and none of the statementsAB ⊆A∪B, AB ⊇A∪B,AB ⊆A∗B,AB ⊇ A∗B,A∗B ⊆B∗A
needs to hold. Moreover, one of the familiesA∗B,B∗Acan be a semigroup of sets while the other is not.
Example3.1. Let|X| ≥2 andAbe a non-empty proper subset ofX. Put B = X\A,A = {A, X}andB = {B, X}. Note thatA = SA,B = SB
and the familiesA ∪B = {X},A B = {∅, A, B, X},A ∗B = {B, X}, B ∗A = {A, X}are semigroups of sets. Moreover, none of the following inclusionsA B ⊆ A ∪B, A B ⊆ A ∗B, A ∗B ⊆ B ∗A and B∗A ⊆A∗B holds.
Example3.2. LetX= {1,2,3,4},A1= {1,3},A2= {2,4},B1= {1,2}, B2= {3,4},C= {1,4},D= {2,3},A = {∅, A1, A2}andB = {∅, B1, B2}. Note thatSA = {∅, A1, A2, X}andSB = {∅, B1, B2, X}. Moreover, we have SA ∪SB = {∅, A1, A2, B1, B2,{1}−,{2}−,{3}−,{4}−, X}(hereY−denotes the complement of a setY in the setX),SASB = {∅, A1, A2, B1, B2, C, D, X}andSA∗SB = SB ∗SA =P(X)\ {C, D}. It is easy to see that the inclusionsSA ∗SB ⊆ SA SB andSA ∪SB ⊆ SA SB do not hold.
We note also that none of the familiesSASB,SA∗SBandSB∗SA is a semigroup of sets. In fact,A1, D∈SASB butA1∪D=4−∈/SASB, and{1},{4} ∈SA∗SB but{1} ∪ {4} =C /∈SA∗SB.
Example 3.3. Let X = {1,2,3,4,5,6,7,8,9}, A1 = {1,2,4,5,7,8}, A2 = {2,3,5,6,8,9}, B1 = {1,2,3,4,5,6}, B2 = {4,5,6,7,8,9}, A = {A1, A2},B = {∅, B1, B2}. Note thatSA = {A1, A2, X}andSB = {∅, B1, B2, X}. First we will show that the familySA∗SBis not a semigroup of sets. It is enough to prove that the setC=((A1\B1)∪∅)∪((A2\B2)∪∅) /∈SA∗SB. Note thatC=(A1\B1)∪(A2\B2)= {2,3,7,8}. Assume thatC∈SA∗SB. ThusC=(S1\S2)∪S3for someS1∈SA andS2, S3∈SB. Since|C| =4, we haveS3= ∅. LetS1=A1. Then|S1\S2|is either 2 (ifS2isB1orB2), 0 (if S2=X) or 6 (ifS2= ∅). We have a contradiction. IfS1=A2, we also have a contradiction by a similar argument as above. Assume now thatS1=X. Then
|S1\S2|is either 3 (ifS2isB1orB2), 0 (ifS2=X) or 9 (ifS2= ∅). We have again a contradiction that proves the statement.
Further note thatSB∗SA = {A1, A2,{1}−,{3}−,{7}−,{9}−, X} =SA∪ SB. Hence, the familySB∗SA is a semigroup of sets.
Proposition3.4.LetS be a semigroup of sets andI be an ideal of sets.
Then the familyS ∗I is a semigroup of sets.
Proof. In fact, letSi ∈ S andIi, Ii ∈ I, i = 1,2. Proceed as follows:
U =((S1\I1)∪I1)∪((S2\I2)∪I2)=(S1\I1)∪(S2\I2)∪(I1∪I2). Put I2=I1∪I2and continue:U =((S1∩I1−)∪(S2∩I2−))−−∪I2=((S1∩I1−)−∩ (S2∩I2−)−)−∪I2=((S−1 ∪I1)∩(S2−∪I2))−∪I2=((S1−∩S2−)∪(S1−∩I2)∪
(S2−∩I1)∪(I1∩I2))−∪I2. PutI1=(S1−∩I2)∪(S2−∩I1)∪(I1∩I2)and note thatU =((S1−∩S2−)−∩I1−)∪I2=((S1∪S2)∩I1−)∪I2=((S1∪S2)\I1)∪I2. It is easy to see thatS1∪S2∈S andI1, I2∈I. Hence,U ∈S ∗I.
Let(X, τ)be a topological space andM(X,τ)be a family of meager subsets of(X, τ). It is easy to see that the familyτ is a semigroup of sets andM(X,τ)
is an ideal of sets (in fact, σ-ideal of sets). The family B(X,τ) of sets with the Baire property is defined as the familyτ M(X,τ). It is well known that τ M(X,τ) = τ ∗ M(X,τ). In fact, this equality is a particular case of the following general statement.
Proposition3.5.LetS be a semigroup of sets andI be an ideal of sets.
Then
(a) S ∗I =S I ⊃S ∪I =I ∗S ⊃S; (b) (S ∗I)∗I =S ∗I,I ∗(I ∗S)=I ∗S.
Proof. (a) Note that for any setS ∈ S and for any setI ∈ I we have SI = (S\I)∪(I \S) ∈ S ∗I, S∪I = S(I \S) ∈ S I, S∪I = (I \S)∪S ∈ I ∗S andS = S∪ ∅ ∈ S ∪I. Thus,S ∗I ⊃ S I ⊃ S ∪I ⊃ S andI ∗S ⊃ S ∪I. Observe also that for any sets S1, S2∈S and any setsI1, I2∈I we have(S1\I1)∪I2=S1I ∈SI, whereI =((I1∩S1)\I2)∪(I2\S1)and(I1\S1)∪S2 ∈S ∪I. Thereby, S ∗I ⊂S I andI ∗S ⊂S ∪I.
(b) LetS∈SandI1, I2, I3, I4∈I. Observe that(((S\I1)∪I2)\I3)∪I4= (S\(I1∪I3))∪((I2\I3)∪I4)∈S ∗I. Hence,(S∗I)∗I ⊂S ∗I. The opposite inclusion is evident.
LetI1, I2, I3∈I andS1, S2, S3, S4∈S. Note that(I1\((I2\S1)∪S2))∪ ((I3\S3)∪S4)=((I1\((I2\S1)∪S2))∪(I3\S3))∪S4=I∪S4∈I ∗S, whereI =(I1\((I2\S1)∪S2))∪(I3\S3). Hence,I ∗(I ∗S)⊂I ∗S. The opposite inclusion is evident.
Corollary3.6.LetS be a semigroup of sets andI be an ideal of sets.
Then
(a) the familiesS I,I ∗S are semigroups of sets;
(b) (I ∗S)∗I =I ∗(S ∗I)=S ∗I. Proof. We will show only (b). Note that
(1) S ∗I =(S ∗I)∗I ⊃(I ∗S)∗I ⊃S ∗I; (2) S ∗I =(S ∗I)∗I ⊃I ∗(S ∗I)⊃S ∗I.
The following statement is evident.
Corollary3.7.LetI1,I2be ideals of sets. Then the familyI1∗I2is an ideal of sets. Moreover,I1∗I2=I2∗I1=I1I2=I1∪I2.
Example3.8. LetX= {1,2},A=X, B= {1}, C = {2},A = {A},B = {B}. Note thatSA = {A},IB = {∅, B},SA∗IB = {A, C}andIB ∗SA = {A}. Thus, in general, none of the following statements is valid:S∗I =I∗S, S ∗I ⊃I, the familyS ∗I is an ideal of sets orI ∗S is an ideal of sets, even ifS is a semigroup of sets andI is an ideal of sets.
The next statement is useful in the search of pairs of semigroups without common elements.
Proposition3.9 (See [2, Proposition 3.1]). LetI be an ideal of sets and A,B ⊂P(X)such that
(a) A ∩I = ∅;
(b) for each elementU ∈SA and each non-empty elementB∈B there is an elementA∈Asuch thatA⊂B\U.
Then
(1) for each element I ∈ I, each elementU ∈ SA and each non-empty elementB∈Bwe have(U ∪I)−∩B= ∅;
(2) for each elementsI1, I2∈I, each elementU ∈SAand each non-empty elementB∈Bwe have(U ∪I1)−∩(B\I2)= ∅;
(3) for each elements I1, I2, I3, I4 ∈ I, each elementU ∈ SA and each elementV ∈SBwe have(U\I1)∪I2=(V \I3)∪I4. i.e.(SA∗I)∩ (SB∗I)= ∅.
Proof. Our proof is very close to the proof of [2, Proposition 3.1].
(1) Assume thatU ∪I ⊃ Bfor some non-empty elementB ∈B. By (b) there isA∈ A such thatA⊂ B\U. Note thatA⊂ (U ∪I)\U ⊂ I. But this contradicts (a).
(2) Assume thatU∪I1⊃(B\I2)for some non-empty elementB ∈Band some elementI2∈I. Note thatU∪(I1∪I2)=(U∪I1)∪I2⊃(B\I2)∪I2⊃ B. But this contradicts (1).
(3) Assume that(U\I1)∪I2=(V \I3)∪I4for some elementsU ∈SA, V ∈SBandI3, I4∈I. IfV = ∅, then(U\I1)∪I2=I4and soU ⊂I1∪I4. But this contradicts (a). HenceV = ∅. Note that there is a non-empty element B ∈ B such thatB ⊂ V. Further observe that U ∪I2 ⊃ (U \I1)∪I2 = (V \I3)∪I4⊃B\I3. But this contradicts (2).
Example3.10 ([2]).
(a) The familyV of all Vitali sets ofRasA, the familyO of all open sets of Ras B and the family M of all meager sets of Ras I satisfy the
conditions of Proposition 3.3. Note thatSV =V1,SO =O,V1∗M = V2andO∗M =Bp (the notations are from the Introduction). Hence, V2∩Bp = ∅.
(b) Consider the Euclidean spaceRn for some n > 1. A Vitali set ofRn is any set S = n
j=1S(j), where S(j) is a Vitali set of R for each j = 1, . . . , n. The family Vn of all Vitali sets ofRn asA, the family Onof all open sets ofRnasBand the familyMnof all meager sets of RnasI satisfy the conditions of Proposition 3.3. LetV1nbe the family of all finite unions of Vitali sets ofRn,V2n = V1n∗MnandBpn be the family of all sets ofRnwith the Baire property. Note thatSVn = V1n, SOn =On,Bpn=On∗MnandV2n∩Bpn = ∅.
There is even a generalization of the result for the productsRn×RmS, whereRS is the Sorgenfrey line (see [3] for the definition).
4. Applications
In [2, Theorem 3.2] one can find the following statements about the families Vn,V1n,V2n, wheren≥1.
(i) Vn ⊂V1n⊂V2n⊂(Bpn)C.
(ii) For eachU ∈V1n, dimU =0, and for eachW ∈V2n, dimW ≤n−1.
(iii) The familiesVn,V1n,V2nare invariant under translations ofRn. (iv) The familiesV1n,V2nare semigroups of sets.
4.1. Two nested families of semigroups of sets
It follows easily from Corollary 3.1 and Proposition 3.2 that the familyMn∗V1n
is another semigroup of sets invariant under translations ofRnsuch thatV1n ⊂ Mn∗V1n ⊂ V2n. The following statement extends the variety of semigroups of sets without the Baire property based on the familyV1n.
Theorem 4.1. Let n > 1. Then there are two finite families {Ln,k}n−k=01, {Rn,k}n−k=10of pairwise distinct semigroups of sets invariant under translations of the Euclidean spaceRnsuch that
(a) for each0≤k≤n−2we haveLn,k ⊂Ln,k+1andRn,k ⊂Rn,k+1, (b) for eachL∈Ln,kandR ∈Rn,kwe havedimL≤kanddimR≤kand
there areL0 ∈Ln,kandR0 ∈ Rn,k such thatdimL0 = dimR0 = k, where0≤k≤n−1,
(c) for each 0 ≤ k ≤ n−1 we haveLn,k ⊂ Rn,k but Rn,k−1 does not containLn,k,
(d) Rn,n−1⊂V2nbutRn,n−1=V2n,Ln,n−1⊂Mn∗V1nbutLn,n−1=Mn∗ V1nandMn∗V1ndoes not containRn,0,V1n⊂Ln,0butV1n=Ln,0.
Proof. For each 0≤k < nlet us consider the familyFk of all closedk- dimensional subsets ofRn. Note that every familyFkis a semigroup of sets, and the inclusionIFk ⊂IFk+1holds for each 0≤k ≤n−2. Since every element ofIFn−1 is nowhere dense in the Euclidean spaceRn we haveIFn−1 ⊂ Mn. For each 0≤k < nputRn,k =V1n∗IFkandLn,k=IFk∗V1n. The point (a) is evident. It follows from Proposition 3.1 and Corollary 3.1 that the families Rn,k,Ln,k are semigroups of sets for each 0 ≤ k < n. It is also clear that the familiesRn,k,Ln,kconsist of sets which are invariant under translations ofRn and which have dimension dim ≤ k. Since for each Vitali setS ofRn the unionS∪Ik = (Ik\S)∪S = (S\Ik)∪Ik, whereIk is any subset of Rnhomeomorphic to thek-dimensional cube [0,1]k, belongs to both families Ln,k,Rn,kand dim(S∪Ik)=k, we have (b). Note that Proposition 3.2 implies the inclusion of (c), and (b) implies thatLn,k−1=Ln,k,Rn,k−1=Rn,kand that the familyRn,k−1cannot contain the familyLn,k. On the other hand for each Vitali setSofRnthe differenceS\ {p}, wherep ∈S, cannot belong to the familyMn∗V1nbut it belongs to the family Rn,0. Hence,Ln,k = Rn,l for each 0≤k, l < n−1. Note thatRn,n−1⊂V2n,Ln,n−1⊂Mn∗V1nand V1n⊂Ln,0. In order to finish the proof of (d) let us recall (see [2, Lemma 3.4]) that for each element U ∈ V1n there are elements V1, . . . , Vn ∈ V1 such thatU ⊂ n
i=1Vi. This easily implies that no element ofV1n can contain a countable subset of Rn consisting of points with rational coordinates. Thus the setCn∪S = (Cn \S)∪S ∈ Ln,0, whereCis the standard Cantor set of [0,1] andS is any Vitali set ofRn, is not an element ofV1n, and the set Qn∪S=(Qn\S)∪S∈Mn∗V1n, whereQis the set of all rational numbers ofRandSis any Vitali set ofRn, is no element ofRn,n−1. This completes the proof of (d).
4.2. Supersemigroups based on the Vitali sets
LetQbe a countable dense subgroup of the additive group of the real numbers.
One can consider the Vitali construction ([7]) with the groupQinstead of the groupQof rational numbers (cf. [4]). The analogue of a Vitali set with respect to the groupQwe will calla VitaliQ-selector of R. One can introduce in the same way as above a VitaliQ-selector ofRn, n≥1 and the corresponding families Vn(Q),V1n(Q),Mn∗V1n(Q), V2n(Q),Ln,k(Q),Rn,k(Q), where 0≤k < n. Note that similar statements as in part 4.1 are valid for the families.
LetF be the family of all countable dense subgroups of the additive group of the real numbers.
Set Vsup = {V : V ∈ V1(Q), Q ∈ F}, V1sup = SVsup and V2sup = V1sup∗M.
It is easy to see that
(i) for eachQ∈F we haveV21(Q)⊂V2sup.
(ii) V1sup,V2supare semigroups of sets invariant under translations ofR. (One can even show that for each Q ∈ F we have V1(Q) ⊆ Vsup but V1(Q)=Vsup, resp.V11(Q)⊆V1supbutV11(Q)=V1sup. We do not know if V21(Q)=V2supfor eachQ∈F.)
We will call the familyV1supthe supersemigroup of sets based on the Vitali sets.
Lemma4.2.For any setU ∈V1supand any non-empty open setOofRthere is a setV ∈Vsupsuch thatV ⊂O\U.
Proof. LetU = ∪ni=1Vi, whereVi ∈ V1(Qi)andQi ∈F. Note that the statement is valid whenQ1 = · · · =Qn(see [2, Lemma 3.1]). Now we will consider the general case. PutQ=n
i=1Qi =n
i=1qi :qi ∈Qi
and note thatQ∈F.
Claim4.3.For eachx∈Rwe have|Qx∩(O\U)| ≥1.
(In fact,|Qx∩(O\U)| = ℵ0.)
Proof. Forn=1 the statement evidently holds ([2, Lemma 3.1]).
Letn≥2. LetOi, i≤n, be non-empty open sets ofRsuch thatx+O1+
· · · +On = {x+x1+ · · · +xn:xi ∈Oi, i ≤n} ⊂O. For eachi ≤nchoose n+1 different pointsqi(j), j ≤n+1, ofOi ∩Qi.
Let nowQi = {qij :j ≥1},i=1≤n, andqij =qi(j), i ≤n;j ≤n+1.
Observe that for eachi ≤nand eachj1, . . . ,ji, . . . , jn(the notationameans thatais not there) the set{x+q1j1+ · · ·qik+ · · · +qnjn : k ≥1}consists of countably many different points (a coset ofQi) and only one of them belongs toVi.
Consider now an n-dimensional digital box B = {(j1, . . . , jn) : ji ≤ n+1, i ≤n}. Note that|B| =(n+1)n and call the elements ofBby cells.
Put in each cell(j1, . . . , jn)ofBthe sumx+q1j1+ · · · +qnjn.
Fixi ≤nand observe that each intervalI (j1, . . . ,ji, . . . , jn)= {(j1, . . . , k, . . . , jn) : k ≤ n+1}of cells contains at most one element ofVi. So the whole boxBcontains at most(n+1)n−1elements ofVi. Summarizing we have at mostn(n+1)n−1elements ofUin the boxB. Since(n+1)n> n(n+1)n−1 forn ≥2, there are pointspinB which are not elements ofU. But suchp must be elements of the setQx∩Oby our choice. The claim is proved.
Let us finish the proof of the lemma. For each equivalence classQxchoose a point from the setQx∩(O\U). The set of such points is a VitaliQ-selector V ofRsuch thatV ⊂O\U.
Theorem4.4.
(a) V2sup⊂BpC.
(b) for eachA∈V2supwe havedimA=0.
(c) for eachQ∈F we haveV21(Q)⊂V2sup.
(d) V2supis a semigroup of sets invariant under translations ofR.
Proof. (a) and (b) follow Lemma 4.1 and Proposition 3.3. (c) and (d) were observed in (i) and (ii) of this section.
Remark4.5.
(a) Considering different ideals of sets in the real lineR(the ideal of finite sets, the ideal of countable sets, the ideal of closed discrete sets, the ideal of nowhere dense sets, etc) we can produce many different semigroups of sets inBpCby the use of the operation∗and the semigroupsV11(Q), Q∈ F, andV1sup.
(b) Let us note that one can define supersemigroups of sets based on the Vitali sets inRn, n≥2, by a similar argument as above.
4.3. A nonmeasurable case
In [5] Kharazishvili proved that each elementU of the familyV1is nonmeas- urable in the Lebesgue sense. LetN be the family of all measurable sets in the Lebesgue sense on the real lineRandN0⊂N be the family of all sets of the Lebesgue measure zero. Recall that the familyN0is an ideal of sets (in fact, a σ-ideal). It follows from Propositions 3.1 and 3.2 that the familiesV1,N0∗V1
andV1∗N0are three different semigroups of sets invariant under translations ofRandV1 ⊂ N0∗V1 ⊂ V1∗N0. We have the following generalization of Kharazishvili’s result.
Proposition4.6.Each element of the familyV1∗N0is nonmeasurable in the Lebesgue sense.
Proof. In fact, letA∈V1∗N0and assume thatA∈N. By Proposition 3.2 there are anU ∈ V1and anN ∈N0such thatA=U N. It is known that ifA1, A2are sets such thatA1 ∈ N and the setA1A2is of the Lebesgue measure zero then the set A2 must belong to the family N (see [1]). But AU =(UN)U =N, henceU ∈N. This is a contradiction with [5].
SoA /∈N.
Question4.7. Is each elementUof the familyV1supnonmeasurable in the Lebesgue sense?
Acknowledgements. The authors would like to thank the referee for his (her) valuable comments.
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7. Vitali, G.,Sul problema della mesura dei gruppi di punti di una retta, Bologna 1905.
DEPARTMENT OF MATHEMATICS LINKOPING UNIVERSITY 581 83 LINKOPING SWEDEN
E-mail:mats.aigner@liu.se vitalij.tjatyrko@liu.se
DEPARTMENT OF MATHEMATICS NATIONAL UNIVERSITY OF RWANDA BUTARE
RWANDA
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