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)