AN ABSTRACT OF THE THESIS OF Ronald Ralph Morgali for the M. S. (Name) Mathematics (Major) July 27, 1965 Date thesis is presented_ Title in (Degree) GAUSSIAN INTEGER ANALOGUES OF TWO THEOREMS OF MANN FOR POSITIVE INTEGERS Abstract approved (Major professor) Let A sum of the sets For n > 0 and B A and be two sets of non negative integers. B is the set C = A+B = {a+b|aeA, beB}. we denote the number of positive integers in the set which do not exceed n by C(n). is defined in a similar fashion. The notation A(n) n>k A(n) any positive integer missing from the set a , for the case in which for the case in which OeA, B(n) A is n is where H. B. Mann obtained a lower bound for and O^B, C, OeA and C(n), A. -where in terms of and OeB, n, and also leB. In this paper we prove Gaussian integer analogues of these two theorems of Mann. C n+1 is the smallest positive integer missing from the set B(n) and The Erdos density of the set gib k The GAUSSIAN INTEGER ANALOGUES OF TWO THEOREMS OF MANN FOR POSITIVE INTEGERS by RONALD RALPH MORGALI A THESIS presented to OREGON STATE UNIVERSITY in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE June 1966 APPROVED: 7f, Associate Professor of Mathematics In Charge of Major Charirman of Department of Mathematics Dean of Graduate School Date thesis is presented Typed by Carol Baker July 27, 1965 ACKNOWLEDGMENTS The author wishes to express gratitude to Dr. Robert D. Stalley for his patience and many helpful suggestions. TABLE OF CONTENTS Page CHAPTER I INTRODUCTION 1 DEFINITIONS AND PRELIMINARY RESULTS 5 CHAPTER II AN ANALOGUE OF MANN'S THEOREM CHAPTER III Q^ 12 KVARDA'S ANALOGUE OF MANN'S THEOREM <g 48 BIBLIOGRAPHY 57 APPENDIX 5 8 GAUSSIAN INTEGER ANALOGUES OF TWO THEOREMS OF MANN FOR POSITIVE INTEGERS INTRODUCTION Let A(n) A be a set of non-negative integers. denote the number of positive integers in than or equal to a n. The Erdos density [2] of For n > 0, A that are less A is denoted by let and is defined by IK A<n) where k is the smallest positive integer missing from there is no such k then a The sum of two sets is denoted by C = A+B A. If is undefined. A and B of non-negative integers and is defined by C = A+B = {a+b|aeA, beB) . Let B(n) and C(n) be defined in the same way as is defined. H. B. Mann proved the following two theorems. THEOREM a . If OeA, 0</B and 1 e B, then C(n) > an + B(n), for any positive integer n missing from the set C. A(n) THEOREM iO . If OeA, OeB, then C(n) > a (n+1) + B(n), for any positive integer n missing from the set In the case of Theorem 68 , C. Mann stated a weaker inequality. However, his proof actually establishes the inequality as given. For other proofs of these two theorems see [3] , [5] , [7] . A Gaussian integer is an ordered pair x and x . (x , x ) of integers The main purpose of this thesis is to prove Gaussian integer analogues of Theorems (X. and (3 Theorems 1 and 2 respectively. B. L. Kvarda [3] , which are called Theorem 2 was first proved by in her Ph.D. thesis. Then Kvarda generalized Theorem 2 to lattice points in Euclidean n-space [4] using a proof which is essentially the same as her earlier proof but which contains several improvements, and makes greater use of the language of Algebra. In presenting Theorem 2 in this paper we have specialized Kvarda's n-dimensional argument to Gaussian integers by taking n = 2, and at the same time have used the language of Algebra even more extensively. Our proof of Theorem 2 is quite different from Kvarda's original proof [3] . We prove four fundamental lemmas, which we call Lemmas A, B, C and D. Lemmas A, B and C are used in the proof of Theorem 1, and Lemmas rem 2. Lemma A B, C and D are used in the proof of Theo is new. In Chapter I we introduce definitions and establish results which are useful in later developments. Besides the definitions of "fundamental set" and "density" , this chapter contains the defini tions of two important order relations. In Chapter II we give the statement and proof of Theorem 1. Then we show that Theorem Q. 1. is a direct consequence of Theorem Attempts to produce a non-trivial example to show that equality can hold in the conclusion of Theorem 1 were not successful. In Chapter III we give the statement and proof of Theorem 2. Then we show that Theorem 2. © is a direct consequence of Theorem Finally, we give an example to show that the conclusion of Theo rem 2 is quite strong in the sense that equality can hold in non-trivial cases. Theorem (X may be obtained from Theorem GH by making a simple transformation on the set Actually Lim shows how Theorem $ rem (JL B, quite easily (see Lim[5]). may be obtained from Theo , but it is clear that the same method can be used to obtain Theorem (J from Theorem @ . By analogy it seems natural that Theorem 1 could be obtained from Theorem 2 by use of the same sort of transformation on the set B. Such is not the case. culty is encountered because the transformed set may not be Diffi contained in the universal set. Known proofs of Theorem 0 give proofs of Theorem Q . may be modified slightly to Again by analogy it seems natural that Theorem 1 could be proved by modifying slightly a proof of Theorem 2. Again such is not the case. In our proof of Theorem 2 we must replace Lemma D by Lemma A which in turn requires a new proof. CHAPTER I DEFINITIONS AND PRELIMINARY RESULTS The definitions and lemmas introduced in this chapter are used throughout the paper. The universal set Q is the set of all lattice points in the first quadrant of the Euclidean plane. called Gaussian integers. The elements of Q are More precisely, Definition 1.1. Q = {x | x = (x , x )| x , x > 0; x , x The notation Q ACQ means that the set A integral} . is a subset of in the weaker sense. The sum of two sets is defined in the usual way. Definition 1. 2. If A, BCQ, then C = A+B = {a + b | a e A, b e B} . The difference of two sets is given by Definition 1. 3. Note that belong to the set A-B={x|xeA, -A xc| B}. Also, is that subset of Q - A = Q - A. whose elements do not A. The following notation will be used for set elements. barred symbol, e. g. x, A will denote an arbitrary element of a subset of Q. A starred element, e. g. fixed element of some subset written in the form a+bi, Q. b*, will denote a special A Gaussian integer is sometimes where a and b are real integers. The following definition will impose a partial order relation on the elements of Q. Definition 1. 4. Suppose Then if and only if a ^ b if and only if a d! b LEMMA 1. -< a, beQ. and Let a <b a = (a , a ) and and a <b . b = (b ,b ). Also a -^ b~ a ^ b. The set Q is partially ordered with respect to " . PROOF. It can be verified directly from Definition 1. 4 that, 1) a -< a for every 2) a -< b and aeQ. Hence the relation is reflexive. b -< a implies a = b~ b -< c implies a -< c. Hence the rela tion is anti-symmetric. 3) a -< b and Hence the rela tion is transitive. The partial order relation of Definition 1. 4 is used to define a fundamental set. Definition 1.5. A finite, non-empty set R C Q is a fundamental set if and only if xeQ such that Let O^R and whenever reR then xeR for all 0 -^ x -^ T. *% be the class of all fundamental sets. LEMMA 2. PROOF. The class Let R .e J is closed under finite uni o n s . ''+! for i = 1, 2, • • • , n. Form the set n R :- \_j R.. Clearly this union is a finite, non-empty subset of Q. i=l Also, OrfR reR for some positive integer If since ' xeQ and O^R 0 -^ x X i r for each then k i. Now let r~eR. Then less than or equal to xeR . Hence xeR. n. There- AC fore Re ji and the proof is complete. Consider any finite, non-empty set M £ Q. Then by Lemma 1 the partial order relation of Definition 1. 4 holds in The set M M. contains elements that are maximal or minimal with respect to this partial order relation. The next definition gives the characteristic property of a maximal element. The definition for a minimal element is similar. Definition 1. 6. and x -{ y If by R . An element imply R e j1 xe M is maximal if and only if ye M x = y . then the set of maximal points of R is denoted The following definition gives a special kind of fundamental set. Definition 1.7. R If ~L Re J and the cardinality of # R is one then is called a Cheo set [l] . LEMMA 3. PROOF. A finite union of Cheo sets is a fundamental set. Since each Cheo set is a fundamental set the result follows immediately from Lemma 2. LEMMA 4. PROOF. If Let x, yeQ and x= (x , x ) x -< y and then y-xeQ-{0}. y = (y-.y.,). Since x.yeQ we have that all of these components are integral. Then x -«^ y implies x < y holding in at least one case. and Hence, these components are also integral. Hence x < y with strict inequality y - x = (y\ ~xi > y?~x?)> Moreover, an<^ (y -x ) + ,(y -x ) > 0. y - xeQ - {0} LEMMA 5. Let Re^1- If x,yeR and x-< y then y-xeR. PROOF. Since we have RC Q. y-xeQ-{0}by Lemma 4. Observe that y-x satisfies the inequality, 0-^y-x ^ y. Then y-xeR by Definition 1.5. The set Definition 1.8. y ~ (y, >y->)- Q can be ordered in another way. Suppose Then x <y x, yeQ. Let x = (x , x ) if and only if either x < y and or both x -- y and and x < y hold. Also, x < y if and only if x< y x ^ y. The ordering given by the relation just defined is usually called lexicographic ordering. Definition 1. 8 imposes a linear order relation on the set Q since in addition to satisfying Conditions 1, 2, and 3 of Lemma 1, it is also true that every pair of elements of Q are comparable. LEMMA 6. PROOF. If a -< b , Let a= (a , a ) 1 implies a < b and least one position. a <b a If a < b < b and the lemma follows. Now let then P, TCQ a < b~. and b= (b ,b ). £> 1 Then a-< b £- with strict inequality holding in at the lemma follows. If a = b then In either case the lemma is established. and suppose T is finite. Then we make the following definition. Definition 1. 9- Let P(T) denote the cardinality of the set P r~\ T. The statements of Theorem 1 in Chapter II and Theorem 2 in Chapter III involve the density of a set ACQ- The definition given here was first presented by Kvarda [3] as a natural extension of Erdos density defined in the Introduction. 10 Definition 1.10. of the set A If taken over all gib Re LEMMA 7. (3 is a proper subset of then the density J1 A(R) Q(R)+1 such that Suppose A(R) < Q(R). A, BCQ are the Kvarda densities of we have Q is a •- and A and A and A C B. B Then if a respectively, a < (3 . PROOF. Let (2= {R|Re^, @ - {R|Re ^i , B(R) < Q(R)}. A(R) < Q(R)} and Given ACB we show that (B C GL • Consider Re ^ . Then we have B(R) < Q(R). But ACB implies lows that Re^. - A(R) < B(R). Hence Thus (BC(2!K A(R) <r A(R) < Q(R). It fol Now> IK B(R) a This completes the proof. The final lemma presented in this chapter is LEMMA 8. and xe B then PROOF. Suppose A, B C Q and C = A+B. If y«/C y ~ x^A. Assume y - xe A. Then by the definition of the 11 set C (Definition 1.2) we have that But this contradicts y - x/A. WC (y - x) +xeC. Thus, which is part of the hypothesis. yeC. Hence 12 CHAPTER II AN ANALOGUE OF MANN'S THEOREM d Three of the four fundamental lemmas of this thesis are intro duced and proved in this chapter. These three lemmas are then used to prove Theorem 1. For easy reference Theorem 1 is stated here. However the proof of Theorem 1 must be deferred until after the three fundamental Lemmas A, B and C have been established. THEOREM1. l,ieB. Let R Let A, B C Q, OeA, be any fundamental set such that assume that for each b -^ "g . with beB r\ R there exists O^B and Q(R-C) > 1 geR-C and where Then C(R) > cQ(R) + B(R), where a is the Kvarda density of the set A. We now proceed to the development of the first fundamental lemma. Let A Definition 2. 1. and A set B be subsets of R is of Type I Q if with C = A+B. 13 1) Re "J , 2) Q(R-C) > 1, 3) for any beB ^R and any geR-C it follows that h-< g. In fact, whenever a subset of assume that If R "R" we then B ^R ^ $ and a unique element is determined in the following definition. Definition 2.2. Let B(R)> 1 that is made up of all and consider the subset of (b , b ) Ld 1 Ct 1 B r\ R such that b +b - max {x +x | (x , x ) e B r^ R} • L is denoted by is a fundamental set. B(R) > 1, b*e (B r^R) Q £ Then define b* to be the largest of these elements in the lexicographic ordering. If Q(R-C) > 1, then a unique element g*eR-C is determined in the following definition. Definition 2.3. of R-C Suppose that Q(R-C) > 1 that is made up of all (g , g ) g +g_ = max {y +y_ | (y , y_) eR-C} . 1 Ci 1 Cj 1 and consider the subset such that Define g* to be the largest Cj of these elements in the lexicographic ordering. Now we use the elements new sets. b* and g* to construct two 14 Definition 2.4. Let R be a Type I set with B(R) > 1. be the uniquely determined element of Definition 2. 3. Let g* We define G = {g-t-xjxeB r^R] . Definition 2. 5. Let R be a Type I set with B(R) > 1. be the uniquely determined element of Definition 2. 2. Let b* We define H = {y-b*|ycR-C} . The next lemma states that the two sets precisely one element in common, namely, LEMMA 9. Let G and H Let R G and have g*-b*. be a set of Type I with be the sets of Definitions H B(R) > 1. 2. 4 and 2. 5. Then GnH= {g*-b*} . PROOF. that We see that ue (G r> H) and prove that u - (1) where xeB ^R y ~ (Y-,'Yy)' 1 £> an<^ g*-b*e(G r>H). and g*-x - y eR -C. b* = (b ,b ). 1 Cm u =•" g*-b*. Hence we assume We have y - b*, Let g* ~ (g , g7), x = (x , x_), Then equation (1) becomes (g}, g2) - (x,x2) = (ylfy2) - (b ,b2). If we transpose, add, and equate components, then this equation becomes 15 •l+bl yi+Xl' ' + b '2 2 Yo +x. and (2) Addition of the equation in (2) yields g1+g2 +b:+b2 - Yl+y2 +Xl+x2 Since g* and sum from and x R-C +x i b* = b Cj are Gaussian integers with maximal component and B ^\ R +b . 1 Hence, respectively, since 5* then and b* y,+y9 = g,+g? 1 C* Y C* are the lexico Cm! graphically largest Gaussian integers with maximal component sum in R-C and B nR respectively, then (yry2) < (grg2) . (3) (x ,x2) < (b ,b2) From the inequalities (3) we have y < g and by the first of equations (2) we have y"i+xi - gi+xi - gi+bi = yi+xr Hence (4) y +x = g,+x, = g +b yY = g1 and so and Xj - b1 x .5 b and so 16 Now by inequalities (3) we have y _< g and x <b , and so by the second equation of (2) we have y2+X2 ^g2+X2 ^ g2+b2 = Y2+X2Hence y +x ' ; + b , +x '2 2 and so yz= g2 and x„, - b^ (5) Finally y~ g* by (4) and (5), and so, u~ g*-b* by (1). This com pletes the proof of Lemma 9LEMMA 10. G If R is a set of Type I with B(R) > 1, and is given as in Definition 2. 4, then GC(R-A). PROOF. Q(R-C)> 1. xeB r\R- Since R is of Type I it follows that Hence g* exists and g*eR-C. By Definition 2. 1, Part 3, we have g*-xeR follows from Lemma 5. g*-x^A by Lemma 8. LEMMA 11. Re J If g*-xeG then x -^ g*. Hence Also, g*^C and xe B imply Therefore GC(R-A). K R is of Type I with B(R)>1, and given as in Definition 2. 5, then hC(R-a). PROOF. b*eB r^R- where Since B(R) > 1 we have that Now the elements of yeR-C. and H b* exists and are of the form By Definition 2. 1, Part 3, we have y-b*, b*-<^y. H is 17 Hence y-b*eR follows from Lemma 5. imply y-b*<^A by Lemma 8. Also Therefore y^C and b*eB HC(R-A) . From the two preceding lemmas and set theoretic considera tions it follows that (G wH) C(R~A). Hence the following lemma has been established. LEMMA 12. G and H If R is of Type I with B(R) > 1, and the sets are given as in Definitions 2. 4 and 2. 5, then Q(R-A) > Q(G) + Q(H) - Q(G ^ H) . The next lemma is the first of the four fundamental lemmas. LEMMAA. If R Let A,BCQ is any set of Type I, with OeA, O^B and l,ieB. then C(R) > cQ(R) + B(R). PROOF. First we show that the Kvarda density of Definition 1. 10 is defined when the hypotheses of Lemma A are satisfied. If we review Definition 1. 10, we see that it is sufficient to prove A(R) < Q(R). that Since Q(R-C) > 1. x 7= 1, i because R is of Type I, we have from Definition 2. 1 Hence there exists xeR l,ieB Since and OeA. 1 -( x or i -<( x and possibly both. Suppose ilar if such that xeR x^C. Now we have either 1 -< x. (The proof is sim l-f\X and i-<x). Then x-1 e R by Lemma 5. But x- 1 ^A 18 since leB and x^C. Hence A(R)<Q(R). Now we return to the proof of Lemma A. Consider any fundamental set R. and non-zero number by Definition 1 . 5. Then Q(R) is a finite The elements of R are linearly ordered with respect to the lexicographic order relation of Definition 1. 8. Hence a unique element of R is determined in the following definition. Definition 2. 6. Let r* be the largest element of R with respect to the lexicographic ordering. Then the proof of the lemma proceeds by examining the cases r*eA and r*^A separately. The case r*eA is established with the help of a counting process developed by Kvarda (4). The case r*^A r*. requires use of the properties of the unique element Case 1. the sets G (r*eA), and H ner in which the sets Q(G) = B(R) 9, and Since exist. G and R is of Type I and (Definitions 2.4, H Q(H) = Q(R-C). 2.5). B(R) > 1, From the man are constructed it follows that Hence by Lemma 12 and Lemma we have Q(R-A) > B(R) + Q(R-C)-1. We add Q(R) Q(R-A) to each member and transpose the terms to obtain Q(R-C) and 19 Q(R)-Q(R-C) Since > Q(R)-Q(R-A)-1 + B(R). Q(R) - Q(R-C) = C(R), and Q(R) - Q(R-A) = A(R), we have by substitution, C(R) > A(R) - 1 + B(R). Before continuing we define a new fundamental set. Definition 2.7. R* = R - {r*} . By Definition 1.5, we have R*e \J . Also since r*eA, we have A(R) - 1 = A(R*) Q(R) - 1 = Q(R*). and First we see that C(R) > A(R*) + B(R). Next, we have established that that A(R*)<Q(R*). A(R) < Q(R), Thus, C(R) > a [Q(R*) +1] + B(R), and so, so it follows 20 C(R) Lemma A > cQ(R) + B(R). is established for Case 1. Case 2. (r*^A). We first prove that Definition 2. 1 we have that Q(R-C) > 1. xeR-C, and x^ B since because l,ieB. if i-^x). OeA. Suppose that Now x-leR Q(R-A)>2. Hence there exists We have that 1-< x. From 1-<. x or iAx (A similar argument holds by Lemma 5, and x-le/A since x^C and leB. Hence x-leR-A. Now by assumption r*eR-A. Since r* is by definition the largest element of R in the lexicographic ordering, and x-l<x where xeR, we have x-1 ^ r*. Therefore Q(R-A)> 2. Before continuing we introduce two new sets. Definition 2.7. From A = A w{r*}, Q(R-A) > 2 and CQ = AQ + B. it follows that A(R) + 1 <Q(R), and hence. AQ(R) < Q(R). Thus, Kvarda's density for A is defined. 21 Now we show C ^ R = C ^R, and hence also It follows immediately from the definitions of C and C (R) = C(R). C that (C r^R)C (CQ nR). Recall that r* is the largest Gaussian integer in to the lexicographic ordering. b + r*^R. ments of Thus, for every However, the only elements of C are of the form b + r*. C R "BeB with respect we have that that are not ele Hence, (CQ r^R)C (C ^R). Therefore we have that (CQ r^R) = (C ^R). Since sets A Q(R-C) > 1 B and By observing that C we also have Q(R-C)>1. Thus, the also satisfy the hypotheses of Lemma A. r*e A we can apply the first case of the proof of Lemma A to obtain, CQ(R) > aQQ(R) + B(R), where a is the Kvarda density of the set is replaced by C(R), we have A . When C (R) 22 C(R) > cQQ(R) + B(R). 'inally, AC A implies that a <a by Lemma 7. There fore, C(R) > oQ(R) + B(R). This completes the proof of Lemma A. An alternate proof, for the second case of Lemma A is given in the Appendix. Next we give some new definitions that are needed in order to state the second fundamental lemma. Let S be a non-empty set such that S= R'-R", where R' , R"e 31 • It follows that S is necessarily finite and O^S. note by "5., jel = {l, 2, ' ' ' ,u}, De the minimal points of S with respect to the partial order relation given in Definition 1. 4. Since S is a non-empty, finite set, at least one such minimal point can be found. set S The minimal points of S into subsets, not necessarily disjoint, as follows, Definition 2. 8. S.= {s|seS, 6. -< s}, Since, given any seS that are then used to decompose the 6. -4 s, J for jel. we can find at least one it follows that, _ S = w. TS. J el j 5.eS such 23 The minimal points of S are now used to construct new sets. Definition 2. 9. S! = {s-6.|se6., s -^ 6.} for jel. Also, we define S' = w. TS! . jel J Note that for each jel we have S!e J* . Hence S'eJ by Lemma 2. The second fundamental lemma is now introduced. LEMMA B. Let the non-empty set ence of two fundamental sets. tion 2. 9. Let the set S be given as a differ S' be given as in Defini Then Q(S') + 1 PROOF. < Q(S) . We define the set mapping \ on S., of Defini tion 2. 9, by the equation, SA1 = {s- (6.lf0)|s€S.}f jel, where set (1) S.. 6. = ( 6 , 6 ) . J J1 j2 Note that X 1 is a one-one mapping on the Then we also have by definition, S\_ = w. TSA. 1 J €I J 1 24 Note that S X is not a mapping on the set may have two distinct images in translate each set S. S since an element of S\ . The effect of X is to parallel to the x-axis until the left border of J S. coincides with the y-axis. J Now for each non-negative integer L , y we construct the "ray" defined by y L = {(x,y)| y fixed, (x, y) eQ} . y Let n = max {y I L r\S ^ $}, then we define y N={k|o<k<n, Note that the index set is true for the set N k integral }. is a non-empty finite set since the same S. From equation (1) we obtain, for any fixed Q(Ly ^S^) =Q(Ly ^(w^jSAj) )• Since intersection distributes over union we have Q(Ly ^S^) =Q( wJ€l(Ly nSAj) ). (2) Let jel be defined by Q(L ^S.\,)= max Q(L Y h' jel nSA.) Y J* yeN, 25 Also let k and k be defined by L ^S.\ = {(x,y)|(x,y)eQ, 0<x<k} v 1 1 and L ~S. \ ={(x,y)|(x,y)eQ, Ofxfkj} Jl Then k < k and it follows that for each (L But i Jl nS.\.)C(L y j i y r,S. \ Jj ! jel, ). is one of the values that the index j assumes, so that equation (2) becomes Q(L (3) Y ^S\_) = Q(L l Y r^S. \ ). Jx 1 Next, since the v-coordinate of an element in S. is not J changed by the mapping L y ^S.\. j i X. , = we have (L y nS.)\ j i . Thus equation (3) becomes (4) Q(L ^SXj) = Q((L r^S. )\j) Since of S- X. we have is a translation on S. andhence on any subset 26 Q(L (5) However, y nsSX) 1 = Q(L ^S. ) . y Jl (L ^S. ) £ (w.^ T(L ^S.)), y h~ ~ JeI y hence J Q(L r^S\.) < Q(w. T(Ly^S.) ). y i ~~ J€ f J Since intersection distributes over union we have Q(L r^SX ) < Q(L ^(w. TS.) ). y After replacing w. 6 TS. jel j Q(L (6) l - Y y with S J*i J we obtain r^SX) < Q(L r^S). l ~ Y From equation (6) it follows that n n Q(L (7) y ^SXJ < > \ Q(L ~ u y=0 y=0 y ^S) Since (L ^SXJ r\ (L pi q r>SXJ = $ i unless p = q and (L r^S) r\ (L p the equation (7) becomes r-,S) = $ q unless p = q, 27 <8) y e in y ^ S) >' QK.eN(Lv ^SX1» ^ Q( -yeN(Ly From the definitions of L and N, equation (8) becomes y Q(S\ ) < Q(S). (9) Next we define the transformation X_ 2 on the set S.X J 1 by the equation, (S.X )X = {s - 6. (i)|seS X }, J where 1 ^ 6. = (6 , 6 „) J jl J2 tion of the set (S.X ) J2 and jel . X . equations (1) through (9) for i Note that to the x-axis. similar to the translation (9') for J X_ 2 is a transforma- The translation X is very The equations that correspond to X have been named (1') through X2. From equation (1) we obtain <SV^<-jefVX2 From set theory we recall that the image of a union under a trans formation equals the union of the images. (!') Hence we have (SXJX7 = w. T(S.X )X 12 jel j 1 Now for each non-negative integer x 2 we construct the 28 !ray" L , defined by x L = {(x, y) Ix fixed, (x,y)eQ} x Let m = max {x IL 1 pS\, ^ $}, then we define x 1 M = {k | 0 < k < m, k integral }. Again the index set is a non-empty finite set. From equation (1' ) we obtain, for any fixed Q( Lx xe M, ^(S\.)^J = Q(Lx r^(w.j e T(S.X )X ) ). 12 lj!2 Since intersection distributes over union we have (2') Q(L ^ (SXJXJ = Q( w. T(L ^(S X )X ) ) X Let Also, let j k i. c jeix j l c be defined by Q(L ^(S. XJXJ = maxQ(L and k J2 1 2 j6l • X ^(S.XJX J 1 2). be defined by L ^(S.XJX = {(x,y)| (x,y)eQ, 0 < y < k} x j 1 2 - - and L ^(S. XJX = {(x,y)| (x,y)eQ, 0<y<k} x i I 2 — — 1 29 Then k <k and it follows that for each (L Since j eI r> (S.XJX )C (L x J 1 2 jel, MS. X )X ). x j 1 ^ we have, from equation (2') Q(L (3 x ^(SX)XJ-Q(L 12 x 0(S. X )X ) j 1 ^ Next, since the x-coordinate of an element in changed by the mapping X , J 1 ) is not we have L M(S.XJXJ= x (S.X Jl2 (L x 0S.\ )\ jl2 . Thus equation (3' ) becomes (4') Since S.X, J 1 Q(L X 2 X MSX )X ) = Q( (L ^S. X )X ). i L is a translation on XJ_1£ S.Xn and hence on any subset of J 1 we have (5') However, Q(L x MSXJXJ= Q(L 12 x r>S. X ). j (L ^S. XJC(w. T(L ^S.X )), X !„ 1 J €I X j 1 Q( L x 1 hence rMSXJXJ < Q(w.jelT(L x ^S.X )). 1 2 j 1 30 Since intersection distributes over union we have Q(L MSXJXJ < Q(L M ^. TS.X ) ). x After replacing (6') 12— w. T(S.XJ i e I J*1 Q(L x x with Jl 1 SX, 1 ^(SXJX_)< Q(L 1 2 jeljl — x we obtain ^SXJ 1 From equation (6') it follows that m m (7-) Y Q(Lx^(SX1X2))< Y Q(Lx^(SX1)). x:= x=0 0 Since (Lp r\ (SX )X ) n(L 1 L ^ (SX )X ) = $ unless p= q Q 1 L and (L p r\ SX ) r\ (L 1 r\ SX,) = <£> unless p q 1 equation (7') becomes (8') Q( w ,.(L ^(SXJXJ)< Q(w xeM x From the definitions of 12— L and M, x (9') Q((SX1)X2)< Q(SX1) . A.(L r-SX)). xeM x 1 equation (81) becomes 31 From inequalities (9) and (9') we obtain Q((SX1)X2) < Q(S). (10) From the definitions of X , X 1 and Definition 2. 9 we have L (SX^ = S' w {0} Hence Q((SX1)X2) = Q(S') + 1 (ID From the inequality (10) and equation (11) we obtain the final result. Q(S') + 1 < The proof of Lemma B Q(S) . is complete. The third fundamental lemma requires the definition of a special type of set. Definition 2. 10. A set S is of Type II if ; (1) There exists R' , R" , e 'J such that S = R' - R" , (2) B(S) > 1 (3) For any and Q(S-C) > 1, beB^S and any geS-C we have b -< g 32 LEMMA C. Let A, BCQ and C = A+B. If S is any set of Type II then, C(S) PROOF. Definition 2. 11. of B r^\ S > cQ(S) + B(S). First we need four definitions. Let S be a set of Type II and consider the subset that is made up of all (b , b ) b +b = max {x +xj (x , x ) e B r^ S} . such that Then define b* to be the largest of these elements in the lexicographic ordering. Definition 2. 12. of S-C Let S be a set of Type II and consider the subset that is made up of all (g ,g ) such that 1 g +g = max{y +y | (y , y )eS-C} . L Define g* to be the largest of these elements in the lexicographic ordering. Definition 2. 13. Let S be a set of Type II. Let uniquely determined elements of Definition 2. 12. g* be the We define G = {g*-x| x«B nS} . Definition 2. 14. Let S be a set of Type II. uniquely determined element of Definition 2. 11. Let b* be the We define H = {y-b*|yeS-C} . The proof of Lemma 9 is used to establish that 33 GnH= {g* - b*}. Before continuing with the proof of Lemma C we need two more lemmas. LEMMA 13. Let S be a Type II set, and let the set be the set as given in Definition 2. 9. If the set G S ' is given as in Definition 2. 13, then G C (S' -A) . PROOF. Since S is a Type II set the element By construction all of the elements of where that xe B r\S. G First we show that g* are of the form g*-xeS' exists. g*-x, and then we show g*-xfA. Since xeB r^S we have xeS. for some jel. Hence, J for some i we have 6. ^ J will consider the case x. First suppose that 6.-<(x. - We J 6. = x separately. J From Definition 2. 10, Part 3, we obtain 6. -^ g* since this order relation is transitive. follows that g*eS. by Definition 2. 8. Thus x-^g*. Since J by Definition 2. 9. Now J J since S'CS1. J Thus x - 6. -^ g*-6. J (g*-6.) - (x-5.)eS! 2. 9. J J since J If x= 6. .J g*-xeS' . g*eS x - 6.eS! J g*-6.eS! Hence, it and J implies J S'.ej>. Finally g*-xeS! implies g*-xeS' J <J thenweget g^xeS! directly from Definition J 34 Next we show that g*-xfA. Since g*eS-C we have in particular that g*/C. Sincewehave xe B, it follows that g*-xe/A by Lemma 8. Therefore g*-xe (S' -A) and the proof of Lemma 13 is com plete. LEMMA 14. Let S be a Type II set and let the set given as in Definition 2. 9- If the set H S' be is given as in Definition 2. 14, then HC(S' -A) . PROOF. Since uniquely determined. of the form y-b*, is a Type II set the element b* is By construction all of the elements of where and then we show that Since S yeS-C. First we show that H y-b*eS' y - b* /A. b*eB nS there exists a 6. such that 0-< 6.^b* J for some jel. are Assume that 6.^<b*. J We will consider the case J 6 = b* separately. j F y If 5.-<b* 3 then b*-6.eS! J From Definition 2. 10, Part 3, we have that and y-b*eS'.. S'e^. Thus j U we get Now b*-6.-<y-6. y-b*eS'. y-b*eS! so and hence by Definition 2. 9. J b*-^y. Hence (y-6 .)-(b*-6.) e S! y-b*eS'. J directly from Definition 2. 9. yeS. since If b* = 6. then J Therefore y-b*eS' Next we show that y-b*^A. Since yeS-C we have in 35 particular that y/C. Since we have b*eB it follows that y-b*</A by Lemma 8. Therefore y-b*e (S' -A) and the proof of Lemma 14 is com plete. Now we complete the proof of Lemma C. From Lemmas 13 and 14 we obtain (G WH) C(S'-A) . By set theory it follows that Q(S' -A) > Q(G) + Q(H) - Q(G ^ H). The construction of the sets identities: Q(G) = B(S), G Q(H)=Q(S-C), and and H gives the following Q(G ^ H) = 1. By substitution we have Q(S' -A) > B(S) + Q(S-C) -1 . Adding and Q(S) Q(S' -A) to each member and transposing the terms Q(S-C) we obtain Q(S)-Q(S-C) >Q(S)-Q(S' -A) -1 + B(S). Adding Q(S' )-Q(S' ) to the right member we obtain Q(S) - Q(S-C) >Q(S')-Q(S'-A) + Q(S) -Q(S')-1 + B(S). 36 Using the equalities, Q(S) - Q(S-C) = C(S) and Q(S')~Q(S'-A)= A(S'), we have after substitution, C(S) > A(S' ) + Q(S) - Q(S' ) - 1 + B(S) This last inequality can also be written C(S) >AqJs!^S1I )+1) +Q(S) -Q(S')-1 +B(S) Since S' e 0> and since a A(S' ) < Q(S' ) < A(S') Q(S')+1 by Lemma 13, we have ' Hence, C(S) > a (Q(S' )+ 1) + Q(S) - Q(S' )-l + B(S) . By Lemma B, we have Since by definition Q(S') + 1< Q(S), 0 < a < 1, and so 0 < Q(S)-Q(S')-1. we have C(S) > c[Q(S')+l] + cz[Q(S)-Q(S')-l] + B(S) . Finally, we obtain 37 C(S) > nQ(S) + B(S) . The proof of Lemma C is complete. The following theorem is an extension of Theorem (ji, to Gaussian integers. THEOREM 1. Let R Let A,BCQ, with be any fundamental set such that that for each beB r\R there exists OeA, O^B Q(R-C) > 1 geR-C where and l,ieB. and assume b -^ "g. Then C(R) > aQ(R) + B(R). PROOF. The argument is carried through by using complete induction on the number of elements in Let Q(R-C) = 1. Then R R-C. is a Type I fundamental set, which may be verified by checking Definition 2. 1. According to Lemma A the theorem holds for any set of Type I. Now assume that the theorem holds for any fundamental set R' that satisfies the hypotheses of the theorem and is such that Q(R' -C) < k, Q(R-C) = Let R-C. where k is some fixed integer and k > 2. Let k. %,,~g7, - ' ' >"g, denote the k Gaussian integers in Then we use these elements to construct k new sets. 38 Definition 2. 15. T.= {xIxeQ, 0-<x -< g.} for l Note that — T. i= 1,2, "•.k. i is a Cheo set for each i = 1, 2, • • • , k. Also, i note that since If l,ieB beT. we have for each B(R) > 1. i = 1,2, •••,k, and all beB^R then i R is of Type I (Definition 2. 1). Then Lemma A applies and the theorem is proved. Hence we may assume that our notation has been selected so that B(R-T ) > 0 Before continuing we need another definition. Definition 2. 16. Let B(R - (T J be the maximum index 1 w T £ J follows from the assumption that J < k-1, suppose J = k- B(R - w i=l w hich implies that there is a i = 1, 2, • • • , k. such that w • • • wT.) ) > 0. Next we prove the inequality, show j 1 < J < k-1. B(R-T ) > 0. The first part Hence 1 < J. Then T.) > 0, l beBnR and "E/ T. for each However, by the hypothesis of the theorem there To 39 exists a geR-C such that least one value of i, namely, the Gaussian integer that J b-<g. This means that that Cheo set geR-C. T. beT. for at determined by This contradiction gives the result <k-l. J LEMMA 15. If beB and beR- w T., i=l then b~e T x . J+1 J+ 1 PROOF. Suppose b</T . Then beR- w T. and so, i=l J+1 B(R - w T ) > 0, i=l which contradicts the choice of J as the maxi- i mum value of j for which J B(R - w T ) > 0. i=l One more lemma is needed before we continue with the proof of Theorem 1. LEMMA 16. Suppose determines the Cheo set g T is that element of . R-C that Then J+1 gJ+l* wi=l Ti PROOF. 1 < n < J, we have follows that then Suppose otherwise. gx,, °J+ 1 beT — n Now if by Lemma 15. J T 1 n such that By the definition of Cheo set it ^J+l£ Tn -^ g • Then for some beB J and Thus, since b~eR- w. ii =l T i is the Cheo J+1 40 set determined by ~gT , n transitive. Since g-,, . —( g , Therefore be w. J X i J T.. n b" -< g T . . . J t we have 1 The partial order is b -< "g . — Then n beT . toJ + 1 — , T . However, this contradicts the fact that i= 1 beR - u we have n i Thus, the proof of Lemma 16 is complete. i- 1 J Now we proceed with the proof of Theorem 1. Let ^q= w ]' j=l Since each set T. is a Cheo set it follows from Lemma 3 that J Consider the difference of the two fundamental sets W . If R - W R and is a Type II set (then Lemma C applies and we proceed to the final steps of the proof). Suppose R"wn is not a set of TyPe II- Then examination of Definition 2. 10 shows that only Part 3 fails to hold. 1) R-W is the difference of two fundamental sets. 2) There is a by the definition of R-W 0 beB(R-W ) J. Hence, since there is a B(R-W ) > 1. contains the Gaussian integer gT , . J+ 1 beB^T Also, the set by Lemma 16. Hence, Q((R-WQ)-C) > 1. Thus if R-W is not a set of Type II, then Part 3 of Defini tion 2. 10 fails to hold. "g". e[(R-W )-C] Hence, there exist such that 1l b~eR-Wn U and b -|< g. . (i.e. be/T. ) 11 Since beR-W implies 1\ beT by Lemma 15, we have U T 1 41 k "^ § ^y Definition 2. 15. J+1 _ From the definition of j such that i g. g.eW^ j 0 implies that i > J. Also, ^- J+1. Hence if integer we have that for every 1 < j < J. Thus "g. eR-W implies that W. 0 Hence R-Wn i b ^ gT+i > J+ 1 . is not a set of Type II, to construct the Cheo set T. . we use the Gaussian Then we let Wi^ wo^TV By Lemma 3 we have that If R-W W is a fundamental set. is a set of Type II then Lemma C applies and we proceed to the final steps of the proof. If R-W is not of Type II then examination of Definition 2. 10 reveals that while Parts land 2 of the definition are satisfied, Part 3 fails to hold. Thus we can find elements be R-W and ~g. e[(R-W )-C] X2 such that b -^ g. . As before beT by Lemma 15. By Definition 2. 15 we J+ 1 have b -^ g T + 1' Now i ^ J+1. g. eR-W X2 Hence implies i > J. Also, b -f«( g. implies "2 i > J+1 . We continue the construction by using the element construct the Cheo set T. . Lemma 3 we have that W£ X2 Then we let 'g. W._ ~ W, kj T. . 2 is a fundamental set. l X2 to By 42 Again we form the difference of two fundamental sets. R-W P If is not of Type II then the construction process may be re- eated. Eventually a set W must be constructed such that m R-W is a set of Type II. This follows since the indices i. m are J all distinct and are taken from a finite index set, namely, {J+2, J+3, • • • , k }. In the event that this index set is exhausted then we are left with S = R - [W ^ ( w T.)] It follows from the preceding remarks that S is a set of Type II. Hence we may assume that we have arrived at a set such that R-W W is of Type II. m Now W and for each e Tt m by Lemma 3. u y beBnW Also, we have there exists a geW m b-< "g. Thus W N ° -C Q(W m -C) > 1 - such that m is a fundamental set that satisfies the hypotheses m of the theorem. Finally 3 Q(W -C) < k m since fore by the induction hypothesis we have (1) C(W ) > cQ(W rn — ) •+ B(W m ). m By Lemma C we also have (2) C(R-W )>cQ(R-W m — ) + B(R-W m ). m gTil / W J+ 1 ' m . There- 43 The addition of (1) and (2) gives C(R) > gQ(R) + B(R). This completes the proof of Theorem 1. Now we establish Theorem $ as a corollary of Theorem 1. First we prove Lemma 17. LEMMA 17. Suppose ACQ- Assume that there is at least one Gaussian integer of the form (a, 0) , where integer, missing from the set and of a a > 1. Let the set A a is a positive A. Let (a ,aJeA if both (a^aJeQ A = {a J(a and a 0)eA}. If a is the Kvarda density is the Erdos density of the set A then -a. 1 PROOF. Let $ denote the class of all fundamental sets R which contain no imaginary Gaussian integers. First we prove ger missing from Then R e@ a A, and > a. If k Then n > k, let R ={(x, 0) | (x, 0)e Q, 1< x < n} . and A(R ) < Q(R ). So, A(R ) ^ 1T, A(R) A(R) > gib ^rrr, > R^ gib ^77^ n+1 Q(V+1 -R^Q(R)+1 n A(R)<Q(R) Hence • is the smallest positive inte n A(n) ViC J A(R)<Q(R) 44 gib n > Now we prove k a A(n) > a n+1 <a . Let R denote a fundamental set for which A(R3)<Q(R3). Let R ={(x, 0) | (x, 0) eR3} and R R = R -R L, J 3 "2' If R I R0, is empty, then R 1 and so " hypothesis. is non-empty and L A(R,) = A(R ) <Q(R ) = Q(R ), " --"-'2' Hence R ~v"3' -~»"3' is not empty. contrary to Now it suffices to show that A(RX) - Q(R1)+1 A(R3) Q(R3)+1 (1) ~v"2' > for then a = gib r4 A(R) Q(R)+1 not empty. (2) R 2 gib A(R) Q(R)+1 gib A(n) n+1 n>k A(R)<Q(R) A(R)<Q(R) If > is empty then (1) is trivial, so we assume We have, A(R3) = A(R^ + A(R2), Q(R3) - Q(R}) + Q(R2> and (3) Therefore, since A(R2) = Q(R2) A(R ) < Q(R_) we have R is ^ 45 A(R2) < Q(R1) Then it follows from (3) that A(R1)Q(R2) If we add A(R )Q(R ) < A(R2)Q(R1). to each member, we have A(R1)[Q(R1) + Q(R2)] < [A(RX) +A(R2)] QfRj). Hence, by (2) we have A(R1)Q(R3) < A(R3)Q(R1). Since A(R ) < A(R ), we have by addition that A(Ri)[Q(R3)+l] <A(R3)[Q(R1) +1] , from which (1) follows. This completes the proof of Lemma 17. Now we establish Theorem ^ 1. as a consequence of Theorem Assume the hypotheses of Theorem (2. , are sets of non-negative integers, n >1 and n^C. OeA, namely that O^B, leB, and B C = A+B, Then let A = {(a,0)|aeA} w {(x^xjIfx^xJeQ, x2>l}, B1 = {(b,0)|beB} w {(0,1)} , A 46 and C = A + B . The hypotheses for Theorem 1 are satisfied for the fundamental set R = {(x, 0)|(x, 0)eQ, 1 <x < n}, n — — and so, we have C.(R ) > cQ(R ) + B(R ) , In— where a n n is the Kvarda density of the set A . By Lemma 17 this inequality becomes C(n) > en + B(n) , which is the conclusion of Theorem & . Note that by Lemma 17 equality holds in Theorem 1 for any example constructed as follows. Let A and B denote sets of non-negative integers satis fying the hypotheses of Theorem Q . Let A = {(a,0)|aeA} u {(x^xj) (x^xJeQ, x2>l}, B1 = {(b, 0)| beB} w {(0,1)} , and C = A + B Let R = {(x, 0)|(x, 0)eQ, 1 <x < n}, 47 where n is a positive integer missing from the set C. Then the hypotheses for Theorem 1 are satisfied and equality will hold in the conclusion whenever the sets A and B are chosen so that equal ity holds in Theorem (L . See [5] for such examples. Theorem 1 is an extension of Theorem (X. to Gaussian inte gers. We note that the hypotheses of Theorem 1 "for each there exists geR-C where b" -( g " , beB ^R may be replaced by the stronger hypotheses "each maximal element of R is in R-C" . The analogy between this weaker theorem and Theorem CJL is stronger than the analogy between Theorem 1 and Theorem Qi . 48 CHAPTER III KVARDA'S ANALOGUE OF MANN'S THEOREM ® The last of four fundamental lemmas of this thesis is intro duced and proved. This lemma and fundamental Lemmas B and C of the previous chapter are then used to prove Theorem 2. This Theorem 2 is an extension of Theorem 00 THEOREM 2. Let R Let A, BCQ with 0eA,0eB, and C= A+B. be any fundamental set such that that for each . beB^R there exists Q(R-C) > 1 geR-C and assume such that b -^ g. Then C(R) > c[Q(R)+l] + B(R). PROOF. Definition 2. 1. From Chapter II we recall the following definition. A set R is Type I if (1) Re^ (2) (3) Q(R-C) > 1 For any beB r\R and any "geR-C it follows that b ~i I • Sets of Type I are used in the following fundamental lemma. 49 LEMMA D. If R Let A, B C Q with OeA, OeB, and C = A+B. is any set of Type I then C(R) > c[Q(R)+ 1] + B(R). Before continuing with the proof of Theorem 2 we prove Lem ma D. The proof proceeds with the examination of two cases . Case 1. Suppose C(R) = A(R) ). B(R) = 0. Then C(R)> A(R), (actually Hence, C{R)-Q(R)l\ (Q(R)+1) +B(R)" Since c —< ^TT, , we have, C(R) > a(Q(R)+l) +B(R). This L2(R)+1 completes the proof for Case 1. Case 2. b* and g* Suppose B(R) > 1. We recall the definitions of from Chapter II. Definition 2. 2. Let B(R) > 1 that is made up of all (b , b ) and consider the subset of such that b +b = max{x +x | (x , xJeB PR) . 1 L 1 L 1 B r\ R Then define b* to be the L largest of these elements in the lexicographic ordering. Definition 2. 3. Let Q(R-C) > 1 that is made up of all L 1 L R-C (g , g ) such that g +g = max{y +y | (y ,y ) eR-C}. 1 and consider the subset of 1 L Define g* to be the largest of 50 these elements in the lexicographic ordering. Again we construct the sets G = {g*-x xeB r~\ R} , and H = {y-b*| yeR-C}, if Definitions 2. 4 and 2. 5. As before we have (GwH)C(R-A) Now R-A neither that G g*fA 0^BnR. and Q(G ^ H) = 1. contains another element, nor in H. because Also, Thus OeB. g*fH namely g*eR-C. Since Hence because g*eR-A. y-b*-^y g* +g* = maxfy1+y2|(y1,y2)6R-C}, where g? g*, g*/C Next, g*^G since (gf.g|)- Again we have from set theory Q(R-A) > Q(G wH) + 1 which, by set theory, becomes Q(R-A) > Q(G) + Q(H) - Q(G ^ H) + 1. Q(R-A) > Q(G) + Q(H). it follows and g*|(GuH). By Lemma 9 we have that is in Hence 51 Since as before Q(G) = B(R) and Q(H) = Q(R-C), then Q(R-A) > B(R) + Q(R-C). Adding and Q(R) to both members and transposing the terms Q(R-C) Q(R-A) we obtain Q(R) - Q(R-C) > Q(R) - Q(R-A) + B(R). Using the identities, Q(R) - Q(R-C) = C(R) Q(R) - Q(R-A) = A(R), and we obtain C(R) > A(R) + B(R), or ClR)>Ag+1) +B(R)Since a<-^—^ , then C(R) > a(Q(R) + 1) + B(R). The proof of Lemma D is complete. Before returning to the proof of Theorem 2 we review the 52 definition of a Type II set and also Lemma C. Both were introduced in Chapter II. Definition 2. 10. A set S is of Type II if; (1) There exists R' , R" e ^ (2) B(S) > 1, and Q(S-C) (3) For any beB ^S LEMMA C. If S such that S = R' - R" , > 1, and any "geS-C we have b "{ \ is any set of Type II then, C(S) > oQ(S) + B(S) . We now return to the proof of Theorem 2. fundamental set such that beB^R there exists a Q(R-C) > 1 Let R be any and assume that for every "geR-C such that b -<( "g. Then we proceed by using an induction argument on the number of elements in the set R-C. Let 2. 1). Q(R-C) = 1. Then R is of Type I (check Definition According to Lemma D the theorem holds for any set of Type I. Assume the theorem holds for any fundamental set satisfies the hypotheses of the theorem and is such that where k B(R) = 0 is some fixed integer and then R k > 2. Let R' that Q(R'-C) < k, Q(R-C) = k. If is of Type I (check Definition 2.1, noting that Part 3 holds vacuously). Thus, the theorem follows by Lemma D. 53 Assume B(R)>1. Gaussian integers in R-C. Let g , g , * ' ' , g, denote the k Then as before we construct the Cheo sets of Definition 2. 15. T. = {x|xeQ, 0-<x-<;g.} for If beT. for each i = 1, 2, • • • , k and each i= l,2,---,k. beB ^ R then R i is of Type I (Definition 2. 1). In this case the theorem is proved by Lemma D. Hence we may assume also that our notation has been selected so that B(R-T ) > 0. Now that this assumption has been made we can refer back to the proof of Theorem 1. We repeat the part of the argument which runs from page 38 line 7 to page 42 line 17. there exists a set W such that R-W W is a set of Type II and m m the set Hence satisfies the induction hypothesis, m Thus, we have (1) C(W )>c(Q(W m — by the induction hypothesis. (2) C(R-W result, — ), m Also, we have )>cQ(R-W m by Lemma C. )+l) + B(W m ) + B(R-W m ) m The addition of inequalities (1) and (2) gives the final 54 C(R) > c(Q(R) + l) + B(R) . This completes the proof of Theorem 2. Now we show how Theorem (£) follows from Theorem 2. Assume the hypotheses of Theorem @ , namely that are sets of non-negative integers, n^C. A and OeA, OeB, C = A +B, n> 1, B and Then let B1 = {(b,0)|beB} , A = {(a,0)|aeA} w {(x1,x2| (Xl,x2)eQ, x£ > 1} , and C = A + B . The hypotheses of Theorem 2 are satisfied for the fundamental set R -'- {(x,0)| (x,0)eQ, 1 <x <n} Hence C.(R ) > g(Q(R )-fl) + B.(R ) , In— where a n In is the Kvarda desnity of the set inequality becomes C(n) > Ql(n) + B(n) , which is the conclusion of Theorem (8 A . By Lemma 17 this 55 Next an example is given to show that equality can hold in the conclusion of Theorem 2. Let A= {0, l,i, 1+i} w{(x,y)!(x,y)eQ, x > 5 or y > 2} , B = A, and C = A+B = {0, 1, 2, i, 1+i, 2+i} ^ A The hypotheses of Theorem 2 hold for the fundamental set R = {x | xeQ, 0 -< x ^< (3, 1)} . The sets A,B,C and R are displayed in the figure below. By computation C(R) - 5, a = 1/4, Q(R) - 7, B(R) = 3. Hence C(R) = 5 = -(7 + 1) + 3 - A ;tnnt ♦—•»—— A - c(Q(R)+l) + B(R). tnut, « • • t f O # • #__> 9 *-> —-#—©—> B C,R Note that by Lemma 17 equality holds in the exclusion of Theorem 2 for any example constructed as follows. 56 Let A and B denote sets of non-negative integers satis fying the hypotheses of Theorem (8 • Let A = {(a,0)|(a,0)eA} w {(xl!x2)|(x1,x2)eQ, *2>l}. B1 = {(b,0)| beB} , and Ci=Ai+Bi Let R= {(x, 0) | (x, 0)eQ, 1 < x < n}, integer missing from the set C. where n is a positive Then the hypotheses of Theorem 2 are satisfied and equality will hold in the conclusion of Theorem 2 whenever the sets A Theorem ® . See [5] and B are chosen so that equality holds in for such examples. Theorem 2 is an extension of Theorem $ gers. to Gaussian inte We note that the hypothesis of Theorem 2 "for each there is a geR-C where b -^ g" T5eB r\R may be replaced by the stronger hypothesis "each maximal element of R is in R-C. " The analogy between this -weaker theorem and Theorem (2 stronger than the analogy between Theorem 2 and Theorem $ is 57 BIBLIOGRAPHY 1. Cheo, Luther P. On the density of sets of Gaussian integers. American Mathematical Monthly 58:618-620. 2. Erdo's, Paul. sequences. 3. 1951. On the asymptotic density of the sum of two Annals of Mathematics 43:65-68. 1942. Kvarda, Betty Lou. Inequalities for the number of integers in a sum of sets of Gaussian integers. Ph.D. thesis. Corvallis, Oregon State University, 1962. 59 numb, leaves. 4. Kvarda, Betty Lou. An inequality for the number of elements in a sum of two sets of lattice points. Pacific Journal of Mathematics. (In press) 5. Lim, Yeam Seng. An inequality for the number of integers in the sum of two sets of integers. Master's thesis. Corvallis, Oregon State University, 1962. 6. Mann, Henry B. On the number of integers in the sum of two sets of positive integers. Pacific Journal of Mathematics 1:249-253. 7. 44 numb, leaves. Scherk, Peter. 1951. An inequality for sets of integers. Journal of Mathematics 5:585-587. 1955. Pacific APPENDIX 58 APPENDIX The algebraic properties of the two order relations defined in the first chapter can be used to give an alternate proof of the sec ond case in Lemma A. LEMMA A. If R is any set of Let A, BCQ, with OeA, O^B and l.ieB. Type I then, C(R) > cQ(R) + B(R) , where c is the Kvarda density of PROOF. Recall that the lexicographic ordering. Case 1. (r*eA). Case 2. (r*/A). r* A. is the largest element of in The two cases considered are, The proof proceeds as before. In this case r*eR-A. 11 we have, (G wH)C (R-A) . We have by assumption {r*)C (R-A) Next we establish that R By Lemmas 10 and 59 {r*} ^ (G wH) = *. Suppose In fact r*eG. r* = g* - x Lemma 6. Then implies that Suppose r* < y Hence, for some r*-^ g*. Thus But this contradicts the definition of graphically largest element of Since r* f*g* - x r*eH. Then B(R) > 1 we have that by Lemma 6. r*^H. R. Therefore r* < g* r* by as the lexico r*^G. r* = y - b* for some b*^ 0, xeB r^R- yeR-C. hence, r*-^ y . Thus Again this contradicts the definition of r*. Then by set theory we have Q(R-A) > Q(G wH) + Q({r*}). Thus, Q(R-A) > Q(G) + Q(H) - Q(G ^ H) + Q({r*}). By Lemma 9 we have Q(R-A) >Q(G) + Q(H). From the construction of the sets Q(G) = B(R) and G Q(H) = Q(R-C). and H it follows that Hence by substitution, we obtain Q(R-A) > B(R) + Q(R-C) . The addition of Q(R) to each member and the transposition of the terms Q(R-C) gives, Q(R-A) and Q(R) - Q(R-C) > Q(R) - Q(R-A) + B(R) 60 which, as before, reduces to, C(R) > A(R) + B(R) Since the Kvarda density is defined when the hypotheses of Lemma A are satisfied we have C(R) > aQ(R) + B(R)