AN ABSTRACT OF THE THESIS OF
Arthur Eugene Olson, Jr.
for the
(Name)
Date thesis is presented
M.A. in
(Degree)
Mathematics
(Major)
June 23, 1966
Title NEW THEOREMS AND EXAMPLES FOR FREEDMAN'S
DENSITY SPACES
Abstract approved
Redacted for Privacy
(Major profesor)
Allen Freedman defined a density space to be the ordered pair
(S, °S
and
'
)
where
S
is a certain kind of seinigroup called an s -set
is a particular type of family of finite subsets of
a fundamental family on
S
called
S.
New theorems for s -sets are presented, with emphasis on the
generation of s -sets from subsets of the positive real numbers.
Several theorems are proved concerning density inequalities for a
special class of fundamental families called discrete fundamental
families.
A
list
of density spaces and the density
satisfy is included.
properties they
NEW THEOREMS AND EXAMPLES
FOR FREEDMAN'S DENSITY SPACES
by
ARTHUR EUGENE OLSON, JR.
A THESIS
submitted to
OREGON STATE UNIVERSITY
in partial fulfillment of
the requirements for the
degree of
MASTER OF ARTS
June 1967
APPROVED:
Redacted for Privacy
Associate Professor of Mathematics
In Charge of Major
Redacted for Privacy
Chairman of Department of Mathematics
Redacted for Privacy
Dean of Graduate School
Date thesis is presented
Typed by Muriel Davis
June 23, 1966
ACKNOWLEDGMENTS
The author wishes to express his gratitude to
Professor Robert Stalley for his help in the writing
of
this thesis.
TABLE OF CONTENTS
Chapter
I
II
Page
INTRODUCTION
BACKGROUND
2. 1
2. 2
s -sets
Fundamental Families
2.3 Density
III
DENSITY SPACES: THEOREMS AND EXAMPLES
Definitions and Preliminary Theorems
for Some s -sets of Positive Real
Numbers
3. 2 Some s -sets Generated by Strictly Increasing Sequences of Positive Real
Numbers
3. 3 Some s -sets Generated by Strictly Decreasing Sequences of Positive Real
Numbers
3. 4 Some s -sets Generated by Other Subsets
of the Positive Real Number
3. 5 Other Examples of s -sets
3. 6 Construction of Fundamental Families
3.
IV
DISCRETE FUNDAMENTAL FAMILIES
1
5. 2
5. 3
VI
Discrete Fundamental Families in General
Purely Discrete Fundamental Families
Singularly Discrete Fundamental Families
Nested Singularly Discrete Fundamental
Families
THE TRANSFORMATION PROPERTIES
5.
The Landau -Schnirelmann Inequality
_
and T -1
The Schur Inequality and T -2
The Correspondence Between T -1 and T -2
DENSITY SPACES AND PROPERTIES
Examples and Properties
6.2 Comments
6.
4
6
9
12
1
4. 1
4. 2
4. 3
4. 4
V
1
1
BIBLIOGRAPHY
12
16
21
29
32
33
38
38
40
41
44
60
60
62
63
65
65
67
70
NEW THEOREMS AND EXAMPLES
FOR FREEDMAN'S DENSITY SPACES
CHAPTER I
INTRODUCTION
In 1930, L. Schnirelmann [ 9, 10] introduced the following
density for a subset
of the positive integers:
A
number of positive integers in the set A
Then the Schnirelmann density of A
a
(1)
Definition
integers. The sum of
I
n
Let A
1. 1.
and
B
AvBv{a+bl
Now
A, B, and
let
a,
ß,
and y
C = A + B
be two subsets of the positive
A + B
aeA, beB}
is the set
,
.
denote the Schnirelmann densities of
respectively. Some of the results which
then y
(2)
If
(3)
y > a+ ß - aß (E.
(4)
If a+ ß< 1, then y >
(5)
y >
min
{
1, a
=
1
(Schnirelmann [ 10] ).
Landau [ 7] and Schnirelmann [ 10] ).
+ í3}
n.
n>1}.
-
have been obtained are:
a+ ß > 1,
be the
which do not exceed
B, denoted by
and
A
A(n)
is given by
{A(n)
=
Let
ß/(1-
a) (I. Schur
(H. Mann
[
[
11] ).
8] and F. Dyson
[
2]
.
2
In 1965, A. Freedman [ 3, p.
1
generalized the notion of
]
Schnirelmann density to arbitrary sets as follows:
Definition
X
of
1. 2.
S, and a
Let
S
finite subset
number of elements in the set
empty finite subsets of
with respect to ,
,
S.
be an
D
arbitrary set. For a subset
of
S,
let
X(D)
Lets be
Xr-D.
denote the
any family on non -
Then the density of a subset
of
A
S,
is
a
=
gQb
A( G)
S( G)
1%
I
Ge ,6C }
Then Freedman [ 3, p. 8-15] developed a general theory for
density by introducing two sets of axioms. The first set of axioms
requires
to be a certain type of abelian semi -group which is
S
called an s -set. The second set of axioms gives structure to the
family ,6C
pair
(S, °
family on
which is then called a fundamental family on
,
)
S
where
S
is an s -set and
T
The
S.
is a fundamental
is called a density space. Freedman [ 3, p.
51
-103]
was able to extend many of the results which have been obtained
for positive integers, including (2), (3), (4), and (5),
In this thesis we present new theorems useful in the construc-
tion of s -sets and study a special class of fundamental families
called discrete fundamental families. In Chapter II we state those
definitions and results of Freedman which are utilized in our work.
3
Freedman
s -sets
[
3, p. 28 -40]
also provided a list of examples of
and fundamental families. In Chapter III we make a detailed
study of methods for constructing s -sets, particularly those which
are subsets of the positive real numbers.
We
also study methods
for constructing fundamental families.
In Chapter IV we study a special class of fundamental families
which Freedman [ 3, p. 101-103] calls discrete fundamental families.
In Chapter
ties
(3) and (4)
V
we investigate the relationships between inequali-
for density spaces and two transformation properties
defined by Freedman [ 3, p. 43 -44] ..
In Chapter VI we
list ten examples
of density spaces and
summarize the results which have been obtained for each of them.
Throughout this thesis various unsolved problems are stated.
Such problems are clearly marked.
4
CHAPTER II
BACKGROUND
In this chapter we state those definitions and theorems from
Freedman'
of the
s
thesis
3, p. 8 -78] which
[
are used in the remainder
thesis. Proofs of these theorems are found in Freedman'
s
thesis.
s-sets
2.1.
Throughout this section, unless otherwise indicated,
non -empty subset of an abelian group
denoted by
+
and the identity element by
Definition
y> x)
(or
E
of
S
x
For
2. 1.
whenever
Definition
y
For
2. 2.
with respect to
S
and
x
y - x
for which y-<x
The set
The operation in
G.
x
E
E
or y
y
in
is a
S
G
O.
G,
we write
x.< y
S.
let
S,
=
x.
L(x)
We
call
denote the set of all
L(x)
the lower set
S.
is called an s -set whenever the following three
axioms are satisfied.
Axiom s -1.
S
Axiom s -2.
0
is closed under
¿
S .
is
+.
5
Axiom s -3.
L(x)
is finite for each
It is easy to see the set
Now we look at some
I of
x
S.
e
positive integers is an s -set.
theorems which enable us to construct
new s -sets from given ones.
Theorem
T
is an
s
If
2. 1.
is a closed subset of an s -set
T
then
S,
-set.
For example, the set of even positive integers is an s -set.
Definition
denote by
of
the set
X°
contained in a group G,
X
XU {0}
where
we
is the identity element
0
G.
Definition
for each
X
For a set
2. 3.
Let
2. 4.
a set
S e A,
be a non -empty index set.
A
X6
Consider,
contained in an abelian group
be the set of all functions
defined on
f
A
G.
Let
which satisfy the
following two properties:
(i) f(ó)
e
for each
X°
(ii) the set of
We
call
X
{Só
I
S e
2. 2.
A}
,
for which
ó
the product of the
Theorem
S = II
ó e A
If
Só
f( 5) # 0
X5
is non -empty and finite.
and write
is an s -set for each
X =TI {X5
S e
is an s -set. Addition is defined by
(f1 + f2) (ó)
=
f1(ó)
+
f2(6).
I
óe
t }.
o, then
6
Theorem
Let
2. 3.
be an s -set and
T
abelian group. Then the set
S =
an s -set where addition on
(x,y)
=
{
(x, y)
X
I
G, y
is
T}
is defined by
S
(x'
+
GX T
be a finite
G
,y')
(x+ x', y+ y').
_
Fundamental Families
2. 2
Before listing the fundamental family axioms we must define
some terminology used in the axioms.
Definition
by
the set of all
U(x)
Note that
Definition
ment
x
Let
2. 5.
X
E
x
be an
S
y
S
E
s -set
such that
is never a member of
2. 6.
Let
X
and let
U(x).
is called a maximal element of
Definition
2. 7.
X
s -set
write
F\A
for the set of all elements in
=
4
Max(X).
let ,P
S,
denote the family of all non -empty finite subsets of
We
X ìU(x)
if
X
An ele-
S.
is denoted by
For an arbitrary set
Denote
S.
E
x-<y.
be a subset of an
The set of all maximal elements of
x
a(S)
e
=
S.
F
and not in
A.
Definition
and let
F
2. 8.
Let
be a set in
1
1
.
be an
arbitrary subfamily
An element
x
E
F
of
is called a
1)
.
corner element of F
if either
F
set of all corner elements of F
Let
=
{x}
-.
\
or
F
is denoted by
called a fundamental family on
S
{x}
e
°
.
The
F *.
be an s -set. A non -empty family
S
7
IC V (S)
is
if the following four axioms are
satisfied:
Axiom f -l. For each
Axiom f -2.
1
is a set in
1
S
there is an F
-p
E
J
with
x
E
F.
The union of any non -empty finite subfamily of
1
The intersection of any non -empty subfamily of
1
,
provided the intersection is non - empty.
Axiom f -4. If F
Definition
space whenever
on
E
.
Axiom f -3.
is a set in
x
then Max (F)
°
E
The ordered pair
2. 9.
is an s -set and
S
C
F *.
(S,1) is called
Y is
a density
a fundamental family
S.
The following theorem is useful in the actual construction of
fundamental families for
Theorem 2.4.
to each
x
e
S,
s -sets.
Let
be an
S
let B(x)
arbitrary s -set. Corresponding
be a subset of
S
satisfying the follow-
ing three conditions:
(i) x
E
B(x),
(ii) B(x)C L(x),
(iii) if
y
E
B(x),
then B(y) C B(x).
8
Let
1B
{FIFE
=
a(S),
x
is a fundamental family on
J' on
family
there exists a function B(x)
S,
Let
Definition 2. 10.
J
set of
such that
F
determined by
Theorem
2, 5.
Let
1B
E
F
1.
E
Then
s -set
be an
S
(ii), and (iii) of Theorem
,
Cheo set of
°B determined by
[
[
the intersec-
x]
x] is called the Cheo
and let
Then
2. 4.
is equal to
x,
satisfy
B(x)
[
x]
,
the
B(x).
define a special fundamental family as follows:
Definition
the fundamental
Theorem
we have
satisfying condi-
x.
conditions (i)
We
JB
J.
=
Denote by
xE S.
x
Then
.
Conversely, given any fundamental
S.
tions (i), (ii), and (iii) such that
tion of all
B(x) C F}
implies
F
E
family
2. 6.
be an
s -set.
1B with
B(x)
Let
2. 11.
S
We denote by /C=
=
for each x
L(x)
For any fundamental family
°
(S)
E
on an s -set
S.
S
`xC 5C D
The following theorem enables us to construct new fundamental
families from given ones,
We
will consider this problem in more
detail in the next chapter.
Theorem
2. 7,
Let
{
16
fundamental families on an s -set
fundamental family on
S.
I
S
E
S.
t} be a non -empty class of
Then
n
{
eg6I
5
E
A}
is a
9
2.3 Density
(S,5)
In the remainder of this chapter let
arbitrary
be an
density space.
Definition 2.12. Let
(possibly empty) subset
of
elements in the set
n D.
Definition
A
2. 13.
Let
Definition
C- density of A
with respect to
2. 14.
D
For any finite
be the number
is non -empty, let
5)
is the Cheo set of
J
o
arbitrary subset
q(X, D)
be an
=
I
of
S.
of
S.
The
is
I
with respect to
c
x]
If
X(D)
gib { q(A, F) F
=
Let A
d (A,
[
we let
be an
A
d(A,1)
where
S
S.
X(D) /S(D).
be the quotient
K- density of
of
D
X
be a subset of
X
E}.
arbitrary subset
The
is
[x])IxES},
gib
{q(A,
J
determined by
x.
K- density generalizes the density defined by B. Kvarda [ 6]
and C- density generalizes the density used by L. Cheo
Kasch [ 5]
.
s -set
1
]
and F.
Both of these densities reduce to Schnirelmann density
when the density space is
the
[
(I,
of positive integers.
X), where
we
recall that
I
denotes
It is immediate from the definitions
10
that
0 <
d(A,°)
<
and
1
Definition 2.15.
sum A
+ B
of
and
A
We
also write
yc
=
C = A + B,
a
c
bi a
+
< 1.
and
B
be subsets of
a=
e
are arbitrary subsets of
d(B,°),
ß=
p
c
dc (B,
=
c
Theorem 2.8.
y >
Theorem
2, 9.
If a+ ß> 1, then
Theorem
2. 10,
y
max
{
whenever
[
4),
y
x]
,
x
and
then
[
Theorem
d(C,5).
and
c-> max {ac,c
If
1=
y
1.
=
ßc}
c
.
26S) and
x]
y
are elements of
n [y ]
2. 12.
If
S
a
c
+ ß
c
'1
-
> 1,
ACS.
then y c
=
such that
xd [y] and
is empty.
J
1.
is separated if,
is a separated fundamental family,
then K- density and C- density are identical. That is,
each
=
a,131.
Definition 2.16. A fundamental family
I
and
S.
dc(C,J')
Theorem 2.11.
y
The
S.
B}.
E
B
d(A,°),
(A,5),
c
A, b
and
A
dc
=
5)
is the set
B
In all that follows
write
dc (A,
Let A
AL)Bk,_){a
We
0 <
a= a
for
11
Freedman
[
ties, which we call
Definition
D =
if
F r\U(x)
T1[ D]
and
For each F
T1[ D]
7 v {4
- x
{y
=
=
if
[
x]\ F
T2[ D]
and
T2[ D]
is in
°
Theorem
and
2. 13.
y
I
x
E
- yI y
{
=
S
E
E
and
F
S
x
e
Fo,
let
°
is
Then
D }.
and
x
F
and
E
D\{x} }.
for every x
{4 }
If
1
E
for every
}
Definition 2. 18. For each
D
respectively.
T -2
and
T -1
2. 17.
is in
defines two transformation proper-
3, p. 43 -44]
E
°
T-1
Fo.
v {4
let
} ,
J
Then
is T-2
F.
and
7(S) is both
is an s -set, then
T -1
T -2.
Freedman
[
3, p. 65 -78]
uses properties
and
T -1
T -2
to obtain several results including the following:
Theorem 2.14. If
Theorem
y > r3/(1 -
°
is
T -1,
2. 15.
If
T is
T -2
2. 16.
If
is
T-1,
then
and
y > a+ ß -
a+ p<
1
,
aß.
then
a).
Theorem
then
y
c-> c
+ ß-
c
ß'
12
CHAPTER III
THEOREMS AND EXAMPLES
DENSITY SPACES:
In Section 2.
1
several ways to obtain new s -sets from
we saw
given ones. In this chapter we develop theorems which are useful
in constructing s -sets, particularly s -sets which are subsets of the
set of positive real numbers. We also study the problem of con-
structing fundamental families.
3. 1.
Definitions and Preliminary Theorems
for Some s -sets of Positive Real Numbers
Let
R+
denote the set of all positive real numbers.
Definition 3.
1.
Let
A
be any subset of
R +.
the set of all finite linear combinations of elements of
tive integral coefficients.
Then we say
Let
A
S
be
with posi-
is the set generated by
S
A.
CR+ and
x
set generated by A. Each way in which
x
Definition
3. 2.
Let
A
is called a representation of
represented in
A
x
in
A.
e
We
if it can be generated by
S,
where
S
is generated by
say
A
x
is the
A
is finitely
in only a finite
number of ways. The sum of the coefficients of a representation is
called the length of the representation.
13
Theorem
Let
3, 1.
A
the set generated by A.
A
C R+
Then
y-<x
is
iff a representation of
y
x, y
can be extended to a representation of
finite, non -zero number of elements of
Proof: By Definition
x - y
have
iff there is a representation of
S
E
2. 1, we
S,
e
in
x
of elements of
Theorem
in A
in
by adding a
A
to it.
A
x
iff
x - y
x - y
in
A, which is
y
equivalent to saying that a representation of y
tended to a representation of x
where
S
and
in
E
But
S.
can be ex-
A
by adding a finite number
to it.
A
3, 2,
Let
set generated by A.
A
Then
C R+
L(x)
and
x
E
S,
where
is the
S
is finitely repre-
is finite iff x
sented in A.
Proof: Assume
is finitely represented in
x
the number of distinct representations of
the length of the longest one.
x
in
Let
A.
and let
A
Then there can be at most
n
be
be
h
dis-
nh
tinct elements of A occurring in representations of x in A.
be the set of elements of
Let A '
.
of
x
course
in
A.
Let
A
occurring in representations
be the number of elements in
m
A'
.
Of
m< nh. Let
m
m
b,a.
B = {b b =
i=
1
,
a.
E
A'
,
integers,
b.non-negative
i
, b. - h}
<
i=
1
.
14
Then
y
E
is a finite set. Now let
B
and so by Theorem 3.
S,
h -1 or
of length
y E B.
1,
y
E
L(x). If y i x, then y -<x and
any representation of
less and consists entirely
of
in
y
is
A
elements of A'. Thus
we also have y e B. Hence, L(x) CB and L(x) is finite.
If y = x,
Assume x is not finitely represented in A. We show that
the set
A'
of
elements occurring in representations of x in
is finite and obtaining a contradic-
is infinite by supposing that A'
tion. If A'
is finite there is a least element a o A'. Let h
E
x
ao
be the greatest integer part of
A
A
Each representation of x in
or less. Therefore, we can construct only
must be of length h
a finite number of different
representations
tradicts our assumption that
x
x
of
in
This con-
A.
is not finitely represented in A.
x
Therefore, A' is infinite. If
some representation of
.
a e A' and a
in
By adding the
A.
occurs in
x, then a
rest
that repre-
of
sentation to the element a, we can extend a to a representation of
x
in
A
by adding only a finite number of elements of
Hence, by Theorem 3.
we have
L(x)
1,
a <x.
we have
infinite. Hence,
L(x)
Therefore,
A
a
finite implies
E
x
to
L(x)
a.
and
is
finitely represented in A and the proof is complete.
Definition 3.3. Let
{an}
of distinct elements of
AC R+.
A
If
there exists a sequence
such that
is called an accumulation point of A.
n
-
lim a n
oo
=
a,
then a
15
Theorem
by
If
A.
3. 3.
Let
A
C R +.
Let
S
be the set generated
has no accumulation points, then
A
Proof: Axioms s -1 and s -2 are immediate.
we must show that
s -3
x
E
Since
S.
A
{a
=
I
aEA, a-<x or a
=
is finite. Let
x}
Consider any
ao be the
0
smallest
Let h be the greatest integer part
.
Then each representation of
.
a
S.
e
verify Axiom
in the interval (0, x]. Therefore, the set
A
element (real number) in A'
of
x
To
has no accumultion points, there are only a finite
number of elements of
A'
is finite for all
L(x)
is an s -set,
S
x
in
A
has length
h
or
0
less. However, we can construct only a finite number of different
linear combinations of elements of A'
exceeding
h,
because A'
with coefficient sums not
is a finite set and
h
is finite.
There-
is finitely represented in A, and by Theorem 3. 2, we
fore,
x
have
L(x)
finite. Therefore,
S
is an s -set and the proof is
complete.
If A
Theorem
is finite, it has no accumulation point, and so by
3. 3, we have
S
is an s -set.
Hence, we have proved the
following corollary.
Corollary
by
A.
If
A
3. 4.
Let
A
is finite, then
+. Let
C R +,
S
S
is an s-set.
be the set generated
16
Generated by Strictly Increasing
Positive Real Numbers
of
Sequences
s -sets
Some
3. 2.
Lemma 3.
Let
5.
fan}
be a
n
sequence of positive real numbers.
ments in
and let
{an}
n
strictly increasing convergent
Let
denote the set of ele-
A
be the set generated by
S
Then
A.
there are no strictly decreasing convergent sequences in
Proofs Since
A
al,
has a least element
the maximum length of the representations of
x
E
which is positive,
in A, for any
is finite. Hence, the proof is by induction over the maximum
S,
of the sum of the coefficients of
A.
x
S.
Let
linear combinations of elements of
denote the set of elements generated from A
S
P
linear combinations with coefficient sum not exceeding
When
quence in
p
=
have
S1
=
A.
P.
strictly decreasing se-
Any
must begin with an element in
S1
However,
1, we
by
{an}
,
say
a..
J
{an}
is strictly increasing and hence exactly
j -1
A
are less than a.. Therefore, there exist
no
elements of
J
strictly decreasing infinite sequences in
S1
=
A.
For our induction step, we assume that there are no strictly
decreasing convergent sequences in Sk,
Let
x
E
R +.
We
creasing sequences in
lim an
n--
=
a.
k(k
for some
> 1).
must show that there are no strictly deSk
+
which converge to
x.
Let
1
From our induction hypothesis, we know that there
17
are no strictly decreasing sequences in
none converging to
Ex,
E
a
and
,
and none converging to
x - a,
each positive integer
x
+
E
), (x - a, x - a
x
for each positive integer
1
such that the open
i
such that
a - ai
(1)
<
min
=
E
{
),
E
and
a
Ex,
,
E
i > J.
By
in
(
E2, .
. . ,
EJ
,
1
+ E.)
i
1
J
> 0
2 Ea
} .
Sk
in the open
+1
+ E).
1), we have
(x, x
a.
-
Let
proceed to show that there are no elements of
interval (x, x
x
a.,
i
(x
a, we can find an integer
for all
Ea
2
+
contain no elements of Sk.
i
converges to
Ian}
n
Since
Sk
for
x - a.
i
intervals (x,
We
x,
Therefore there exist positive numbers
i.
for each positive integer
E.
which converge to
Sk
-<
E
x
Therefore, there are
.
no
since there are no elements of Sk
E),
+
E
Hence, in order for an element of
Sk
to be in
+1
elements of
in
(x, x
+
(x, x+
Ex).
E),
there must be an element of Sk in at least one of the intervals
(x - a., x - ai
+
because all elements in
E),
3.
are of the form y
+
where
ai
y
E
There are no elements of Sk
i(1< i
i > J,
and so
have
<
because by (1),
J),
then
a - a.
x - ai
x - a.
i
+
+
E
E
<
<
2 Eaa
<
i
x -
x -
E
.
a+2Ea+
a+
1
ca.
a
Ei
Sk
in
and
a.I.
I.
for all
Since
<i
<
J).
a-
aii
<
x+
If
2
Ea
a
,
1
2 Ea by (1), we
Hence (x - a., x - a.+
i
for
+ E)
3.
i(1
E<
Sk
A.
E
(x - a., x - a.
Therefore, x+
E.
not also in
Sk+
+1
i
E)
C (x - a, x - a+
E
),
a
18
which contains no elements of
ments of
(x - ai, x - ai
in
Sk
We have shown
(x, x
Sk+1
+
Sk.
Therefore, there are no elefor
+ E)
i >
J.
that there are no elements of
in
Sk
+l
Hence, there are no strictly decreasing sequences in
c).
which converge to
is arbitrary, so the proof
But x
x.
is complete.
Theorem
Let
3. 6.
fad
n
be a strictly increasing convergent
sequence of positive real numbers.
ments in
fan}
n
and let
S
denote the set of ele-
Let A
be the set generated by
Then
A.
S
is an s -set.
Proof: Axioms s -1 and s -2 are immediate.
s
-3, we must show that
that there is an xo
3. 2, we know
that
over, since
fad
is
n
al
<
xo
a1
ai for all
i(i
is finite for each
such that
S
e
x
L(x)
L(xó
To
x
e
a
> 1).
S.
Assume
is not finite. By Theorem
is not finitely represented in
0
verify Axiom
A.
More -
strictly increasing sequence, we have
Let
be the greatest integer part of
h
Then each representation of xo
in
A
has length
h
or
less.
Let
{
an4}
n
be the set of all elements of
A,
listed in strictly
increasing order, which occur in at least one representation of
Then
{a
n
contains an infinite number of elements. If not,
would be finitely represented in
A,
x
x
0
o
since each representation has
.
19
or less and there would be only finitely many elements
length
h
of
available.
A
Consider the sequence
because each
sequence
{x
n
*
{an}
{an
}
o
}
Each
.
Therefore,
Axiom s -3 holds.
Theorem 3.
Theorem
Hence,
6
o
S
The
.
n
}
is
3. 7.
L(x)
But now we have a
S,
which contradicts
is finite for each
x
E
S
and
is an s -set and the proof is complete.
S
can be generalized to the following theorems.
Given a finite number of strictly increasing
A
be the set
elements which occur in at least one of the sequences. Let
be the set generated by
s
n
-a)
convergent sequences of positive real numbers, let
of
*
an E
{
h--00
strictly decreasing convergent sequence in
3. 5.
o
-
lim a n = a. Then lim (x o
= x -a,
n
o
09
is an infinite subsequence of the convergent sequence
Let
and hence converges to the same limit.
Lemma
x
is strictly decreasing, because
- an }
n
strictly increasing.
since
á
- an
is in at least one representation of x
an
o
{x
A.
Then if
S
S
is non -empty, it is an
-set.
Theorem
3. 8.
Given a countably infinite collection of strictly
increasing convergent sequences of positive real numbers, let
A
be the set of elements which occur in at least one of the sequences.
Let
S
be the set generated by
A.
Then
S
is an s -set if the
20
following condition holds:
(
for each x
*)
E
R+,
the interval
contains elements
(0,x]
from only a finite number of the sequences.
The proof of Theorems 3.
We
3. 6.
7
and 3.
8
parallel that of Theorem
omit the details of these proofs but indicate how they may
In each case we
be constructed from the proof of Theorem 3. 6.
first show that there are
in
no
strictly decreasing convergent sequences
which uses a proof paralleling that of Lemma 3.
S,
show that
S
is an s -set.
The basic difference is that at each step
we must work with a finite number of sequences instead of
We only
Then we
5.
just one.
need to work with a finite number of sequences, even in the
proof of Theorem
3. 8,
because given any
x
E
all but a finite
R +,
number of the sequences occur completely to the right of x
on
the real line. In Theorem 3. 8, this follows immediately from con-
dition ( *). In fact, if we do not have ( *), the set
an
s
is not always
S
-set. For example, if the first elements of the sequences form
the strictly decreasing sequence
{
1n }
,
an s -set is not generated,
as we shall see in Theorem 3.15.
We
can generalize Theorem 3.
Theorem
3. 9.
7
to the following theorem.
Given a finite number of strictly increasing
convergent sequences of positive real numbers, let
of elements which occur in at
A
be the set
least one of the sequences. Let
a
21
be the maximum of the set of limits of the given sequences.
B
be a finite or countably infinite set of positive
Let
real numbers
having no accumulation point and subject to the condition that
b > a
for each
Then
S
is an
b
Let
B.
e
be the set generated by
S
A
v B.
s -set.
By properly placing more elements in the set
Proof:
B,
we
can create a finite or countably infinite number of strictly increasing
convergent sequences satisfying condition
such a set of additional elements
erated by
AL)BuB'
know
is an s -set.
s
S'
-set, because
3. 3.
S
B'
.
(
of Theorem 3. 8.
*)
Let
Call
be the set gen-
S'
By Theorem 3. 8, or Theorem 3. 7, we
.
Therefore, by Theorem
is a closed subset of S'
2. 1,
S
is an
.
Some s -sets Generated by Strictly Decreasing
Sequences of Positive Real Numbers
Now we
are ready to examine strictly decreasing convergent
sequences of positive real numbers. The word convergent can be
omitted, because all strictly decreasing sequences of positive real
numbers converge. The next lemma is similar to Lemma
Lemma 3. 10.
Let
gent) sequence of positive
of elements in
{an}
n
be a strictly decreasing (conver-
{an}
real numbers. Let
and let
3. 5.
S
A
denote the set
be the set generated by A.
Let
22
lim a n = a. If a > 0, then there are no strictly increasing con vergent sequences in S.
n-.
Proof: Since
tions of
in
x
the maximum length of the representa-
a > 0,
for any x
A,
e
is finite. Hence, the proof is
S,
by induction over the maximum of the sum of the coefficients of
linear combinations of elements
elements generated from
sum not exceeding
When
p
=
Let
A.
S
S1
However,
fan}
p.
we have
1,
Any
A.
S1 =
strictly increasing se{an}
,
say
a..
is strictly decreasing and hence exactly j -
n
A
denote the set of
P
by linear combinations with coefficient
must begin with an element in
quence in
elements of
A
of
are greater than a..
1
Therefore, there exist
no
J
strictly increasing convergent sequences in
S1 =
A.
For our induction step, we assume that there are no strictly
increasing convergent sequences in Sk,
Let x
E
R+
.
We
creasing sequences in
for some
k(k >
1
).
must show that there are no strictly in-
Sk
+1
which converge to
x.
From an
induction hypothesis, we know that there are no strictly increasing
sequences in Sk which converge to
and none converging to
x - a.
for each positive integer
fore, there exist positive numbers
positive integer
i
none converging to
,x,
E
x
,
e
a
,
and
such that the open intervals
i.
x -
There-
ci for each
(x -
Ex ,
x
x),
a,
23
(x - a - Ea, x - a),
integer
Since
E.
n
a. - a
{Ex, El
.
,
,
EJ
2
Ea }
.
in
(x -
E ,
in the open
+l
x).
E,
By (2), we have
Sk
0
1
,
proceed to show that there are no elements of Sk
interval (x -
J>
Let
i > J.
E2
,
we can find an integer
a,
for all
Ea
2
min
=
E
<
for each positive
x - a.)
,
Sk.
converges to
an
(2)
We
(x - a. -
contain no elements of
i
such that
and
E
- E.
x
Therefore, there are
<
x), since there are
Sk
elements of
elements of Sk in
no
Hence, in order for an element of
no
to be in
+1
(x - Ex, x).
E,x),
(x -
there
must be an element of Sk in at least one of the intervals
because all elements in
(x - a. - E, x - a.)
are of the form
y + ai
where y
There are no elements of
i(1
< i <
J),
in
Sk
because by (2),
<
E
and
Sk
e
a.
E
Sk
A.
for
(x - ai - E, x - a.)
i(1<
for all
E
not also in
Sk+
i <
J).
If
i-
then
i > J,
and so
ai - a
x - a -
Ea <
a
<
2
x - ai 1
x - a - Ea <x -ai -E.
a
i
have
Therefore,
Ea .
1
2
E
in
(x - a. -
We have shown
(x -
E,
x).
.
Hence, (x
which contains no elements of
ments of Sk
a
E,
Sk.
X
- a.)
Since
x
E
+
-<
ai - a
<
x
+
2
Ea
,
22 Eaa by (2), we
-a -Ea,
-a.i -E, x- a.)(x
i
a
x -a)
Therefore, there are no elefor
i >
J.
that there are no elements of
Sk
+1
in
Hence, there are no strictly increasing sequences in
24
which converge to
Sk+1
But
x.
is arbitrary, so the proof
x
is complete.
This proof is valid only when
zero; that is, when
clear, but suppose
n
The reason may not be immediately
a > 0.
lim a
n -P-00 n
is bounded away from
fan}
=
0.
Then for any positive integer
there are no strictly increasing sequences in Sk
converging to x.
However, there may be strictly increasing sequences in
verging to
all
n( n
x.
obtain such a case by taking
We
and letting
> 1),.
S
A
bounded away from zero, because then given any
and
Sk
such that for all
K > 0
identical in the interval (0, x]
Theorem
Let
3. 11.
{an}
{
x
A
If
A.
a > 0,
we must show that
that there is an xo
3. 2, we know
that
E
x
S
0
L(x)
{an}
then
n
R+
there
S
L(xo)
lim a
n--c,0 n
and let
S
S
=
a.
be the
is an s -set.
is finite for each
such that
is
.
Proof: Axioms s -1 and s -2 are immediate.
s -3,
an}
be a strictly decreasing (con -
denote the set of elements in
set generated by
The
A.
we have
k > K,
vergent) sequence of positive real numbers, where
Let
E
con-
S
1n } for
= {
be the set generated by
induction proof remains valid, however, as long as
exists an integer
k,
To
x
E
verify Axiom
S.
Assume
is not finite. By Theorem
is not finitely represented in A.
25
Moreover, since
we have
length
n
}
is strictly decreasing and lim a
n-. oo n
for all
a < a,
1
x
of
a
0 <
integer part
{a
ó
i(i
Let
> 1).
h
x
a> 0,
greatest
be the
Then each representation of
=
has
in A
o
or less.
h
Let
{an
n
be the set of all elements of A, listed in strictly
}
decreasing order, which occur in at least one representation of x 0 .
{a}
n
Then
contains an infinite number of elements. If not, x o
would be finitely represented in
representation has
or less and there would be only finitely many elements
length
h
of
available.
A
A, since each
*
*
Consider the sequence
*
{x
*
o
- an
n
o
}
.
Each x o - an
is in at least one representation of
because each ann
sequence
{x
- an
is strictly increasing, because
}
strictly decreasing.
x
o
E
.
{an}
S
The
is
-
lim (x o - a n ) = x o - a, since
o0
is an infinite subsequence of the convergent fan} and hence
n
{an}
n
We have
n
converges to the same limit. But now we have a strictly increasing
convergent sequence in
fore,
L(x)
Hence,
S
S,
which contradicts Lemma 3. 10.
is finite for each
x
e
S
and Axiom
s -3
There-
holds.
is an s -set and the proof is complete.
Theorem 3.
11
Theorem 3.
12.
can be generalized to the following theorems.
Given a finite number of strictly decreasing
(convergent) sequences of positive real numbers, let
A
be the
26
set of elements which occur in at least one of the sequences. Let
be the set generated by A.
S
has a positive limit point and
If each of the given sequences
S
is non -empty, then
S
is an
s -set.
Theorem
Given a countably infinite collection of strictly
3. 13.
decreasing (convergent) sequences of positive real numbers, let
A
be the set of elements which occur in at least one of the sequences.
Let
be the set generated by
S
A.
Then
is an s -set if the
S
following conditions hold:
(i)
for each x
E
the interval (0,x]
R+,
contains elements
from only a finite number of the sequences,
(ii) each of the given sequences has a positive limit point.
The proofs of Theorems 3. 12 and 3. 13 parallel that of
Theorem
3, 11.
how they may be
We
omit the details of these proofs but indicate
constructed from the proof of Theorem
3. 11.
In
each case we first show that there is no strictly increasing convergent sequence in
3.10.
S,
which uses a proof paralleling that of Lemma
Then we show that
S
is an s -set.
The basic difference is
that at each step we must work with a finite number of sequences
instead of just one.
We only
need to work with a finite number of
sequences, even in the proof of Theorem
x
E
R +,
3. 13,
because given any
all but a finite number of the sequences occur completely
27
to the right of
x
on the
real line.
In Theorem 3.13
this follows
immediately from condition (i). In fact, if we do not have condition
(i), the set
is not always an s -set.
S
For example, if the first
elements of some of the sequences form the strictly increasing sequence
is
{ 1
+
1
1n
-
{ 1
}
and if one of the strictly decreasing sequences
an s -set is not generated, as we shall see in Theorem
} ,
3. 17.
We
can generalize Theorems 3.12 and 3.13 to the following
theorem.
Theorem
3. 14.
Given the conditions of either Theorem 3.12
or Theorem 3. 13, let
the given sequences.
a
Let
be the minimum of the limit points of
B
be a finite set of positive real num-
bers in the open interval (0,a).
AFB,
Then
S
Let
be the set generated by
is an s -set.
Order the elements in
Proof:
S
B
in decreasing order.
the last element (the smallest) in this finite sequence
b.
We
Call
can
extend this strictly decreasing finite sequence of positive reals to a
strictly decreasing infinite sequence
of positive
real numbers which
converges to 2 > 0 by inserting appropriate elements from the
interval (2 , b) into the sequence. Call these newly inserted
elements
B'
.
Let
S'
be the set generated by
A
L.)BB'
The conditions of Theorem 3. 12 or Theorem 3. 13 still hold.
.
28
Therefore,
is an s -set. Hence, by Theorem
S'
is an s -set, because
is a closed subset of S'
S
Suppose we have a set
lim a
n
co
n
{an}
Then sometimes
= O.
the set
S
.
generated by the elements of a
S
strictly decreasing sequence
2, 1,
S
of
real numbers and that
is an
s -set
and sometimes it
is not as we see from the following two theorems.
Theorem
{ñ}
and
If
3. 15.
is the set generated by
S
First, note that
Proof:
write
-1 qE
r
=
P
{
S
where p
r
and
=
p
q
e
r,
r.
s -set.
S.
Hence,
because
L(r)
S
=
0.
We can
are positive integers.
q
positive rational numbers. The value of L(r)
any positive rational
is not an
is strictly decreasing and lim -1
n-.con
}
n
then
A,
Consider any positive rational number
Therefore,
A.
is the set of elements in the sequence
A
Now
is the set of all
is not finite for
is the set of all positive
rationals less than or equal r. Therefore, Axiom s -3 fails and
S
is not an s -set.
{
tn}
Theorem 3.
16.
where
t
0 <
<
set generated by A,
First, note that
and that
limoo to
n__.
=
0.
If A
1
is the set of elements in the sequence
and
then
{tn}
t
S
is transcendental, and
S
is the
is an s -set.
is strictly decreasing since
0 < t < 1,
29
Proof: Axioms s -1 and s -2 are immediate. To show that
Axiom s -3 holds, consider any
x
and assume we have two
S
E
distinct representations of x in A.
Thus,
m
tk
X =
k=
and
,
1
m
x
ckt k
=
,
kLLL=1
k=
where the bk
for some k
( 1
and
<
ck
1
are non -negative integers,
k < m) ,
and
m
=
max
{
k bk
I
+
bk
ck
> 0
ck
} .
Then
m
x - x
0 =
=
bktk-- m
L cktk
k=1
and so
k=
1
m
ck)
=
k=
1
satisfies a polynomial equation in one unknown and with
t
integral coefficients not all zero. Hence,
tradiction. Therefore,
x
Theorem
L(x)
3. 2, we have
element of
3. 4.
tk,
S.
t
is algebraic, a con-
has a unique representation in
is finite.
But
x
Therefore Axiom s -3 is valid and
A.
is an arbitrary
S
is an s -set.
Some s -sets Generated by Other Subsets
of the Positive Real Number
Suppose we have a set
S
By
generated by the elements of a
30
strictly increasing and a strictly decreasing sequence of positive
real numbers. Then sometimes
is an s -set and sometimes it
S
is not as we see from the following two theorems.
Theorem
If
3. 17.
of the two sequences
and
1
{1 +
n= 1, 2, ...
I
is the set generated by
S
Note that
while
{ 1 +
1
{
1
-
1n
n(n >2),
{1
--
We
can write
by Theorem 3. 2,
(1
2 =
1
+
-n
+
e
1n )
n
A
and
for each integer
1n )
-
+ (1
n
1
-
is infinite, Axiom
s
-3
1
1.
1n
E
for each
A
is not finitely represented in A.
2
L(2)
n= 2, 3, ... },
is not an s -set.
S
is strictly decreasing and converges to
the element
(
is strictly increasing and converges to
}
Therefore, since
-
> 2).
then
A,
and
}
}
n
Proof:
n(n
is the set of elements in at least one
A
fails, and
S
Hence,
is not
an s -set.
Theorem 3.18. If
the two sequences
dental, and
S
{ 1
1n
+
A
{1
- tn}
} ,
and
is the set of elements in at least one of
where
S
0< t
<
1
and
t
is the set generated by
is transcenA,
then
is an s -set.
Note that
while
{
1
+
1
n
}
{
1
- tn}
is strictly increasing and converges to
is strictly decreasing and converges to
Proof: Axioms s -1 and s -2 are immediate.
Axiom
s -3
holds, consider any
x
E
S
1.
To show
that
and assume we have two
1
31
distinct representations of x in A. Thus,
m
x
bk
=
k=
(
1
- tk)
gk(1+),
+
k=1
1
and
X
m
)
=
k=
where the
m
or
=
gk
max {k
k=
and
hk
#
,
1
are non -negative integers,
hk
for some
ck+ gk+ hk>
I
hk(1+k)
+
1
bk, ck, gk,
ck
bk
ck(1-tk)
k (1 < k < m),
0
and
} :
Then
m
0= x
-x=
(bk-ck)(1-t
k=
m
k
(gk -
) +
k=
1
hk)(1+k)
1
m
-bk)tk+ ) [(gk-hk)(1+k)+(bk-ck)
k=
and so
t
k=
1
=
bk
for all
k(1
<
k < m).
the elements from the sequence
sentation of x
Let
,
1
satisfies a polynomial equation in one unknown and with
rational coefficients. Since t
ck
]
in
A.
is transcendental, we must have
Therefore,
{
1
- tn}
x
uniquely determines
which occur in any repre-
32
x'
bk(1
_
k=
Since
x
-tk).
1
uniquely determines
x'
we know that
,
and
x
both have the same number of representations in A.
by Theorem 3. 2, we have
is finite iff
L(x)
However, any representation of
from the sequence
a positive limit.
sequence
{ 1
+
{1 +
Now the
Theorems 3.
3
set
S'
through 3.
S
Thus,
L(x - x'
is an s -set.
provide us with several sufficient
18
conditions for s -sets of positive real numbers.
of
is finite.
generated by the elements of the
is finite and
L(x)
)
which is strictly decreasing and has
},
n
Therefore,
contains only elements
is an s -set by Theorem 3. 11.
1n }
Therefore,
is finite.
1
x - x'
L(x - x'
x - x'
A
corresponding set
theorems can be listed for negative real numbers.
Unsolved Problem. Give necessary and sufficient conditions
for a subset of the positive (negative) real numbers to generate an
s -set.
3. 5.
Other Examples of s -sets
The most useful s -set in our study of density spaces is the set
I of
positive integers with the usual addition. If, in Theorem
we set
Ss
=
I
for each
b
E
A
,
then
2. 2,
)
33
S
which Freedman
[
=
II
{I
I
SE A},
3, p. 28] denotes by
is an s -set. We will
In,
use this example in Chapter VI.
We
can construct an unlimited number of different s -sets using
Theorems
2.
1
through
2. 3 and 3. 3
through 3. 14. All of Freedman'
examples [ 3, p. 28 -30] follow from Theorems
lary
2. 2, 3. 3,
and Corol-
3. 4.
Unsolved Problem.
Construct an s -set which is essentially
different from those we can construct with present theorems.
Construction of Fundamental Families
3. 6.
Theorem
2. 4
is very useful in the construction of fundamental
families for s -sets. For example, if we are given the s -set I,
we can
construct a fundamental family on
B(x)
{1, 2,
=
...
,
x}
for all
x-e I.
I
by letting
Here we obtain the density
(I,7() which is the one Schnirelmann [ 9, 10) worked with.
space
Furthermore, if we let
density space
Section
(I,z).
B(x) _ {x}
Here
2
for all
and
rj
x
E
I
we obtain the
are defined as in
2. 2.
From this point on, whenever we define B(x)
in order to
obtain a particular density space, it is to be understood that condi-
tions
(i), (ii), and (iii)
of Theorem 2. 4 are satisfied.
s
34
Theorem
2. 7
states that the intersection of any non -empty
class of fundamental families on an s -set
is itself a fundamental
S
family. If we replace intersection by union, the theorem fails as we
see in the following example on the
Let
1B
4B 2
1,3}
if
{
x
otherwise.
e
3,
}
be defined by
if x
=
3,
_
{
{2, 3}
x=
{
=
B2(x)
J=
I.
1
{2,3}
Let
-set
be defined by
B1(x)
Let
s
otherwise.
x}
1 B1 v J B2
-
5B2 C 5, but
Then
{1,3}
3}
11,3}
n {2, 3}
e
1BI C 5 and
_ {3}
¿
5.
Therefore,
5
does not satisfy Axiom f -3 and is not a fundamental family on I.
Definition 3.
4 [
4, p. 189]
.
A
lattice is a partially ordered
set in which any two elements have a least upper bound and a great-
est lower bound.
A
lattice is complete if any subset has a least up-
per bound and a greatest lower bound.
Freedman
[
3, p. 24]
proved the following theorem.
Theorem 3. 19. The class of all fundamental families on an
s
-set
S
forms a complete lattice with respect to the partial
35
ordering by set inclusion.
Given an s -set
°B
and two fundamental families
S
and
1
(5B2
on
defined by
S
have seen that
v1B2
B1
However, Theorem 3.
family
19
5
mental families
Theorem
J B1 v
Let
families, on an s -set
5,,,
=
is not always a fundamental family.
C
B2
of a fundamental
-5
5,, C
with the property that
3. 20.
tively and let
respectively, we
B2(x)
assures the existence
such that
J*
and
B1(x)
4B1
B1
JB2
and
,2µb
{
,
}.
°
v " C ?.
B2-
be two fundamental
and
B2(x)
respec-
Then
°,,,
is defined
defined by B1(x)
S,
for all funda-
B1
by
where
B(x)
B(x)
=
for all x
B1(x) ,-B2(x)
Proof: First, we show that B(x)
(ii), and (iii) of Theorem 2. 4.
for all
x
E
S
we have
,
x
E
Since
B(x)
y
E
B(x)
B1(y)
B(y)
x
=
E
=
B1(x)
B1(y)
is satisfied.
B(x)
=
B1(x)
=
and
B1(x)
,m
x
E
B2(x)
Therefore,
B2(x).
B1(x)nB2(x)CB1(x)CL(x)
Hence, condition (ii) is satisfied. Furthermore, if
S.
B1(x)
E
B1(x)
condition (i) is satisfied. Now B(x)
for all
satisfies conditions (i),
x
=
S.
E
,
B2(x),
and
we have y
B2(y)
n B2(y) C B1(x)
B2(x).
,
B2(x)
Therefore, the family
B2(x),
E
and
B1(x)
y
E
B2(x).
Thus,
Therefore,
=
B(x)
°IB,
and condition (iii)
defined by
is a fundamental family on
S.
36
B(x)
Now
and
1B
l-C JB,
fined, and so
and
C B1(x)
B2(x)
E
and
JB
1
v cpJB
C_
5B.
fined. Therefore,
,,
B(x)
by the way
,
B1(x)
=
1B
n B2(x)
/
SB1
and
,
S
are de-
°B
B1(x)
Arp
Ç5,
E
1
/B
and
are de2
1
for each x
J,;,
E
`TB
2
However,
1B C /-e . Therefore,
cp
1B2 C °5B C ` * and so
L)
=
Jk
vB
E
In C
1
also have
We
by Axiom f -3 for fundamental families.
all x
Thus,
2
.
B1
B
,.p
by the way
B11
C B2(x).
B(x)
B(x)
E
S,
e?,,, for
implies
- -
and the proof is
complete.
We conclude the
chapter by showing that if B1(x)
determine fundamental families, then B1(x)
v
always define a fundamental family.
s -set
B1(x)
=
Let the
12,4} if x
=
4,
=
3,
{
1, 3}
if x
{
x}
otherwise.
2(x)
B2(x)
=
4,
=
{
x
otherwise.
}
Then
B1(x)
L.)
B2 (x)
=
if x
{
2, 3, 4}
{
1, 3}
if
{
x}
otherwise.
x
=
=
does not
be I and let
and
{3,4} if x
and
4,
3,
B2(x)
37
However,
I
B1(x)
1/4..)B2(x)
because condition (iii)
that
3
E
B1(4)
does not define a fundamental family on
of Theorem 2. 4 fails.
To see this note
uB2(4), but B1(3) cB2(3) ,$ B1(4) uB2(4).
38
CHAPTER IV
DISCRETE FUNDAMENTAL FAMILIES
In this chapter we study discrete fundamental families, and
find restrictions needed for various density properties to hold.
We
also state some problems which have come up in this study and remain unanswered.
4. 1.
Discrete Fundamental Families in General
Freedman
of order
n
[
3, p. 1011
as follows:
Definition 4.
discrete of order
(i)
°S.
defines a discrete fundamental family
A
1.
if
n
J
fundamental family
T
for some
Recall that
satisfies -the following two conditions:
x
E
S, we have
S([ x]
)
< n
with equality holding
x.
S(
[
x]) is the number of elements in
which is the same as the number of elements in
is separated, we know by Theorem
C- density are identical.
2. 12
S
,
,
knowing that a
S
B(x).
ç[ x]
,
Since
that K- density and
Therefore, whenever we are working with
discrete fundamental families we can say that the set
a
is
S
is separated,
(ii) for each
J
on an s -set
A
has density
is both the K- density and the C- density of A.
39
Freedman
[
Theorem 4.
discrete of order
of
Y
S
proves the following theorem:
3, p. 102 -103]
or
1
(5,1
Let
1.
2.
with densities
Let A, B,
a, ß
and
,
1
be a density space where
)
_
and
C
=
be subsets
A + B
respectively.
y
is
Then
>min{1, a+ß}.
The following example shows that the preceding theorem fails
for discrete fundamental families of order
for each
n
n(n
> 3).
In fact, Theorem 2. 8, which is true for all density spaces, gives
Let the s -set be
the strongest result.
B(x)
and let
I
{1,3}
ifx=
3,
{1,3,4}
ifx=
4,
{2,5}
ifx=
5,
{2,5,6}
ifx=
6
{11,12,...,10+n}
ifx
10+ n (n >3),
=
{
=
otherwise.
x}
It can be verified in a straightforward manner that
B(x)
conditions (i), (ii), and (iii) of Theorem 2.4 and hence
a fundamental family on I.
separated. Also
I
(
[
let
A
Since
=
B =
{
)
1,
q(B, [ x] ),
n
=
J"
1,2,7,8,9,... }.
q(A, [ x] ),
1B
is
Likewise it can be verified that
10 + n]
Therefore, by Definition 4.
satisfies
and
[
(
x]
) <
is discrete of order
Then
and
I
vB is
n for all x e I.
C =
q (C, [ x]
{
)
n.
Now
1,2,3,4,7,8,9,... }.
are all minimized
40
at
x
=
6,
we have
a=
ß= y =
3
.
Hence, Theorem 2,
8
gives the strongest valid result.
In the
rest
restrictions on dis-
of this chapter we place various
crete fundamental families so that we can prove results which are
stronger than that given by Theorem
For some of these re-
2. 8.
sults it may be possible, of course, to weaken our additional
restrictions somewhat without changing the conclusions.
A
con-
jecture of such a result is included later in the form of an unsolved
problem,
4. 2.
Purely Discrete Fundamental Families
Definition 4,
2,
A
1
fundamental family
is purely discrete of order
if
n
irS
on an s -set
S
satisfies the following two
conditions:
(i)
1
(ii) if
is discrete of order
y
[
E
Definition 4.
and
x]
2
y
,
x,
n,
then
[
y] _
{
y }.
restricts c5 severely. However,
we prove
the following theorem which is useful later.
Theorem 4.
Let
2,
purely discrete of order
Proof: Let
density of A.
A
(S,
n.
1)
Then
be a density space where
y >
be any subset of
min
{
S,
and let
1, a+ p
From condition (ii) of Definition 4.
J
is
} .
a be the
2, we
conclude
41
that for any x
a)
[
(b)
[
(
or
x]
E
_
x]
_
either
S,
{
x}
{
,
...
x1,
xi
,
x
where
}
x.]
[
_
{x.} for
J
j(1
<j <i
- 1).
a=
0
if x ¿ A.
j.
Thus, if
In case (a), we have
for some
x. ¿ A
if
all
J
In case (b), we have a=
a
0
we must have
0,
3
q (A,
x]
[
sities
i (2
a, ß
i
<
) =
q (A,
and
,
y >
max { a ,
[
x]
i -1
) =
or i -1
for all
i
then by Theorem 2. 8,
a
ß
} =
or
is
ß
min
0, 1,
0,
1, a+ ß
{
then by Theorem 2. 9, we have
is
1,
If
a= aal
b(2
<b <n),
for some
a(2
<
If either
}.
a+
min {
bbl for
ß=
some
y
=
and
a < n)
or
a
=
1
1
,
1, a + p
}
ß
ß
}.
then
a+
ß=
a-1
b-1
a
b
Thus, by Theorem 2. 9, we have
4. 3.
Therefore, the only den-
.
can take on are
y
If either
< n).
we have
or
1
y
=
>
1
1=
1
-
2
2
=
min
{
1.
.
Singularly Discrete Fundamental Families
°
Definition 4. 3. A fundamental family
singularly discrete of order
n
T satisfies
if
on an
s -set
is
S
the following two
conditions:
(i)
°
(ii)
S
is discrete of order
(
[
x])
=
i
n.
for at most one x
e
S
where
i
=
2, 3,
...
,
n
42
Theorem 4.
Let
3.
I)
(S,
singularly discrete of order
be a density space where J is
y > min
Then
3.
1, a+ (3).
{
Proof: Consider the following cases:
Case I:
Let
JB be defined by
{xi,x2,x3} if x
B(x)
_
{x
where
x2-<x3.
Case II: Let
otherwise,
}
Then
is purely discrete of order
vB
by Theorem 4. 2, we have
y > min
{
x1'.x2 <x3 and
order
3
Let
B (x)
x1x2-<x3.
where
0, 3
,
2
,
Theorem
3 ,
and
2. 8, we
x3,
=
x5,
x4, x5}
if
{
x}
otherwise,
x4 <x5.
x
Then
is purely discrete
B
Y
>
min
x
=
x3,
if x
=
x2,
{
{
xi, x2, x3
{
xi, x2
{
x}
if
}
}
have
3).
otherwise.
The only densities possible for
1.
1,
be defined by
1B
=
=
{
and, by Theorem 4. 2, we have
Case III:
and,
be defined by
1B
B(x) _
of
3
1, a+ ß }.
{x1,x2,x3} if x
where
x3,
=
If either
y >
max
a=
{
or
0
a, P}
=
ß = 0,
min
{
a or
ß
then by
1, a +ß }.
If
are
43
then by Theorem
a+ P> 1,
2. 9, we
2
1
(or a=
If
a=
are in both
xl -< x2 <x3,
x2 - (x2 - x1)
#
x2 - x1
Thus,
imply
S
a=
E
x2 - x1
x2- x1
xi
E
B,
#
x2
so
elements of
,
x1
+
ß=
and
x2
=
2
(x2 - x1)
a= 3
,
+
and so
are the only two eleE
y >
2
E
min
=
A; x3
B;
E
B, hence in
and
x2 - x1
xi
and
A
a
B
{1, a+ß }.
and
C.
and again
S,
1
#
Therefore,
x3.
=
x2
E
x2 - x1
Since
C.
and so
C
y
case where
and
because
is transitive.
A, B; x3
we have
x1
=
.< 11
x3
A
Since
xl. x2 <x3,
Therefore,
C.
and all other
A, B
x2 - x1 ..<x2
x2 - x1
are in both
S
are in
x1ux2<x3
1, a+ (3).
{
hence in C.
>>
and
x2
'
x3
x2 -
since
then x2
,
We do not have the
where
E
x2 - x1
B(x)
min
=
1
and
Also
S.
E
However,
xl< x2 -<x3,
S
B,
x3
because
A
all other elements of
Again since
and
A
Therefore,
S.
E
not in A.
3
and
x1
(x2 - x1)
If
A, B
x2 - x1
and
x2 - x1
q'
we have
=
x2
ments of
then x2
,
S
=
).
1
ß=
elements of
1
ß=
,
y
a= ß= 3
The only cases left to consider are
ß=
have
=
1
°
B
>
x3
6
=
E
A.
C C,
min
{
Now
all
1, a+ ß }.
is defined by
{
x1, x2, x3}
if
x
=
x3,
{
xi , x4
if
x
=
x4,
{
x}
}
B
E
otherwise,
xi -<x4 because here
*
is not
separated and hence not discrete. Therefore the proof is complete.
44
Indications are that we can prove similar theorems for singu4
larly discrete fundamental families of
5, 6, and so on. How-
ever, each successive proof seems to require a greater number of
cases and we have found no generalization of the proof for n
-
Unsolved Problem. Prove (or disprove) that if
a density space where
then y
min
>
{
1,
a+
=
is
(S, °4.)
7 is singularly discrete of order
3,
n,
ß }.
Before going on, we should show that
y > min
the strongest valid result for a density space
is singularly discrete of order
n.
(S,
5
{
1, a+ ß
)
is
}
where DT
Consider the following example
where the s -set is I:
Let
vB
be defined by
{iJ
B(x)
{
where
n > 2.
Then
2
Y =
n
C
=
{
y
=
if
1
-<x -<n,
otherwise,
(I, 5B)
1,2,n+1,n+2,...
and so
-<i<x}
-
x}
singularly discrete of order
Then
1
_
is a density space where
n.
Let
},
Therefore
A
=
B
=
{
a=ß
1, n+1, n+2,
=
is
`7B
...
} .
1 and
a+ P.
4.4. Nested Singularly Discrete Fundamental Families
Definition 4.4. A fundamental family
is nested singularly discrete of order
n
if
B
JB
on an s -set
S
is defined by
45
{x.
B(x)
i
4
=
{
and
x --<x
1
x
. .
2
n
x
1
I
- -m }
< i <
if x
=
x
(1<m<
- - n)
m
.
,
otherwise,
}
,
Case III of Theorem 4.
which is the most difficult case to
3,
prove, concerns nested singularly discrete fundamental families of
order
3.
Theorem 4.4. Let
(S,
is nested singularly discrete of order 4.
The inequality
y >
min
5
be a density space where
)
{ 1,
a
+ ß}
result according to the example at the end
Then
min { 1, a
y >
+
P}
is the strongest possible
of Section 4, 3.
Before
proving Theorem 4. 4, we prove two lemmas.
Lemma A.
Let
(S,
eS)
be a density space where
vB
is the nested singularly discrete fundamental family of order 4 defined by
x2, x3, x4}
{
B(x)
a > 0
and
x1
a
E
and
B,
Proof: Since
=
x4,
x1, x2, x3}
if x
=
x3,
{
xl,
if x
=
x2,
{
x}
x2}
otherwise,
where x1 <x2- .x3-<x4. Let
with densities
if x
{
=
ß
A
and
B
be subsets of
respectively and let
then x2
E
C
=
A + B.
S
If
C.
x1 -<x2-.x3-<x4,
.
we have
x2 - x1
E
S.
46
< x2
x2 - x1
Now
x2. ßx3 -<x4
x2 - x1
x2 - x1 # x2,
Now
x2 - x1
x1
so
B
e
#
- (x2 - x1)
x3,
and x2 - x1
(x2 - x1)
Lemma B.
Let
x1
+
=
°B)
(S,
x2
x1
=
".<"
and since
can be missing from
or x4
x3,
x2,
because x2
E
Hence,
S.
is transitive we have
Since
x4.
#
a> 0, only
Therefore,
A.
X2
- X1
E
A.
C.
e
be a density space where
(4B
is
the nested singularly discrete fundamental family of order 4 defined
by
B(x)
if x
=
x4,
{
x1, x2, x3}
if x
=
x3,
{
x1, x2
if x
=
x2,
{
x}
_
with densities
a and
a > 0, x1
and x2
E
B,
Proof: Since
-
x1, x2, x3, x4}
x11 x2 .<x3 <x4.
where
X3
{
X1 e
x3 - (x3 - x2)
Likewise,
x3 -X1
x2
=
x3 -
ßx4.
E
Let A
(x3 - x1) - (x3 - x2)
Therefore, either
x3 -
B
be subsets of
S
C = A + B.
If
and so
x3 - x2
#
e
x2
we have
or
and
x3 - x1
E
S,
#
x3 - x2
x3
x4.
and
because
and so
S,
x3 - x2
x2-<x3-<x4 because
x3 - x2 # x3
x2<x3 -xl,
x2 - x1
C.
x3 -
Now
S.
e
and so
S,
=
e
<x2 <x3 -<x4,
x1.x3.1x4,
But
and
then x3
B,
x2 - xi
and
S,
x1
otherwise,
respectively and let
ß
e
}
x3 - x2 # x3 - x1.
x3 - x1 # x2.
Since
a> 0,
47
x3
can be missing from
x2, x3, or x4
only
-x2EA or
either case,
x3 - x1
But x1
A.
E
and
B
E
1
x3
Therefore, either
A.
x2
so in
B,
E
C.
E
Proof of Theorem 4.4: By an appropriate selection of nota-
°
tion,
is defined by
xi , x2, x3, x4}
{
{x1,x2,x3}
B(x)
We
,
(i) a=
ß=
et
The only densities possible for
1.
If either a=
0
y >
max
min
{
f3
ß=
1
a= ß= 4
hence in
C.
we have
x2
In
e
and so
Case (ii); If
all other elements of
S
3
,
2
E
=
} =
y
1
=
1
1, a+ ß
If
} .
min {1, a+
=
1
1
ß=
,
/ A, B;
ß-
,
ß }.
1
,
3
2
x3
are in both
S
then
P= 0,
{
(iii) a=
,
3
.
I
A, B;
A
and
B,
Therefore, by Lemma A,
B.
min
{
then x2
are in both
or
a and
Then the cases which
(vi) a=
,
x1
y >
a= p=
ß
then x2
,
particular,
C,
,
a< P.
4, (ii) a= ß=
,
a
have
2. 9, we
and all other elements of
A, B;
x2,
otherwise,
(v) a=
,
Case (i): If
x4
=
x}
may select our notation so that
(iv) a=
x3,
{
then by Theorem
remain are
=
if x
0,
a+ ß> 1,
if x
x1, x2}
x1.<x2ux3.<x4.
are
x4,
{
4' 3' 2' 3' 4' and
by Theorem 2. 8, we have
P
=
=
L
where
_
if x
A
1, a+ ß }.
e'
A, B;
and
B,
x3 ¿A, B;
hence in
and
C.
48
Again since
y
x1
>= min
2
{
a+
1,
we have
B,
E
x4
hence in
C.
Since
E
x4
Since
B
C
B
hence in
C.
are in
x3
A
y
x2
E
B
and
x4
E
ACC,
=
1
>
6
=
min
may not be in
=
and
B,
Lemma A.
>
1
x3
E
=
E
B
;
x4
5
E
A
E
C C,
min
=
1,
{
¿A,B;
and
x3
we have
B,
E
x3
A
by Lemma A.
C
min
1, a+ p }.
{
x4
A
E
x2
E
may or
x4
;
are in
S
and
A
by Lemma
C
then all elements of
x4
EA;
S
¿B; and all
x4
Since
C.
by Lemma B.
S
The other possi-
a+ (3).
hence in
B,
C
B,
and
A
Since
x2
are in
C
x1
E
B
B
E
CC
and so
1, a+ 131.
{
a-
;
p =
,
4
x2 ¿ A, B
B
12
>
then all all elements of
Case (v): If
one of which is
x1
and
we have
and
x2
1 A, B;
then there are two possibilities,
,
x3 WA;
x2EB
;
2
13=
3
=
are in both
S
and all other elements of
other elements are in
y
1
y
,
3
CC
B
E
x2
a=
Since
and so
C
bility is
;
Hence,
then x2 IA, B; x3
3'
we have
B,
x2 ¿ A, B;
may not be in
Since
E
we have
C,
one of which is
A.
ß=
,
x1
Case (iv): If
B,
1
a=
all other elements of
B;
E
by Lemma A.
C
E
p }.
Case (iii): If
x4 ¿A;
x2
;
1
then there are two possibilities,
2'
x3EB;
x3 ¿A
x4 ¿ A
and all other elements of
hence in
C.
Since
x3
The other possibility is
Since
E
B
xi
C. C,
x2 d A
;
we have
we have
y >
x2
E
B;
x3
are in both
S
B,
E
of
4
may or
x4
;
x2
=
A, B
E
min
;
by
C
{
A
1,
a+ p I.
x4 ¿A, B
;
49
and all other elements of
Since
C.
Since
x2
x1
E
and
B
BCC,
E
both
;
x4 é A
A and
we have
x3
C
1
a=
x4
;
4
B
E
32
p=
,
;
min
E
by Lemma B.
C
1, a+ p }.
{
then x2
,
Since
C.
x1
Since
are in
S
=
4
x3
hence in
B,
A
(I
x2
;
E
and all other elements of
by Lemma B.
then all elements of
3
y >
and
A
we have
B,
E
hence in
B,
E
x2
we have
Case (vi): If
x3 dA, B
are in both
S
x2
y
=
1
;
are in
S
x2
BCC
E
and so
C
and
B
E
B
E
B,
and x4
>
122
BCC,
E
=min
{1, a +ß }.
The proof of Theorem 4. 4 is complete.
Now we examine the density space
nested singularly discrete of order
unable to prove that
y >
min
{
(I, °
)
Y>
1, a+ ß }.
(1
[
7]
and Schur
is
However, we prove
y > a+ p - a p
when a+ p<
ßà)
7
proofs depend on the nested property of
proofs given by Landau
°41
Even here, we have been
n.
below both the Landau -Schnirelmann inequality,
and the Schur inequality,
where
[
11 ]
1.
,
The
and are based on
for the density space
(I,¡/-).
Definition 4.
x
E
S.
Then x
5.
Let (S,
1B)
be a density space and let
is called a singleton if
The following definition for
X(x)
B(x)
_
2. 4.
.
is used in the remainder
of this chapter and is not to be confused with the
Theorem
{x}
B(x)
defined in
50
Definition 4.
Let
6.
be a subset of
X
denotes the number of integers in
Then
I.
X(x)
which do not exceed
X
The following theorem is useful in proving Theorem 4.
x.
6
and
4. 7.
Theorem 4.
Let
5.
(I,
1)
be a density space where
nested singularly discrete of order
and
I
in I
be the density of A.
a
which are greater than
Let
If
1.
- T(x)
a= gib {A(x)
x - T(x)
J
Proof: Since
a
)
=
1
a
Since
7
I
.
for all
x
gib {q(A,
=
be any subset of
be the set of singletons
{1}
XE I
A
CA,
then
} .
is discrete, we have
Now the set of all singletons,
so q(A, [ x]
T
T
gib {q(A, [x] )I xE
=
Let
n.
E
T
v{
1
I
} .
is contained in
},
T{i}.
x
I
x
E
I , xEl
Tu{ 1}}= {xiI
then using
n,
<i <n}
Thus,
(2)
and
[x])IxeI, xi Tu {1 } }.
is nested singularly discrete of order
{
A,
Therefore,
the notation of Definition 4. 4, we have
(1)
5 is
a= min {q(A,[xi] )I
2 <
i < n )
.
.
51
Hence
for each
,
we have
i ( 2 < i < n),
xi]
[
_ {x1,
x2,
... xi}
.
.
Since
is the number of singletons from
T(xi)
2
through xi, then
3.
T(xi)
(3)
=
(xi -
=
x. - i
=
x.
i
-I([x.])
i
whether
x1
singletons in
=
or
1
i
Let
x1 > 1.
singletons are in A,
be the number of non-
R
which do not exceed
A
- (i - 1)
1)
x. (2 < i < n).
Since all
we have
A(xi) - (T(xi)
(4)
R.
+ 1) =
However, since the non - singletons in A which do not exceed
form the set
{
x. 12
J
then
-< j < i },
A([ xi]
)
(5)
v{x2,
=
A ({x1}
=
A ({x1})+ A
=
1+R
. . .
,
xi})
(x2,...
,
xi)
Combining (4) and (5), we obtain
A([ x.]
(6)
Equations
(3) and (6)
)
=
A(xi) - T(x.).
yield
A([x, ]
q(A,
[
x. ]
)
=
1([ xl]
A(xi) - T(xi)
)
1
)
x. - T( xi)
x.
i
52
where
2 <
and so by (2), we have
i < n,
A(x.) - T(x.)
a
=
min
i
{
i
- i -< n
2 <
xi - T(x.)
T(xi)
and by (1),
(
a= gib
7)
If
and
x
E
Tv{1},
u
E
x-
and there exists a
whenever
T L,{1}
If no such
y
such that
y
=
x,
A(x) - T(x)
x - T(x)
}.
y < x, y d T
v
T(x)
=
A(y) - T(y)
y - T(y)
x - 1,
and so
1.
=
Therefore, by (7), (8), and (9), we have
a
=
- T(x)
gib { A(x)
x - T(x)
x
e
I}
and the proof is complete.
Theorem 4.
6.
Let
(I,
)
be a density space where
is nested singularly discrete of order
{1 } ,
then
y < u < x,
exists then A(x)
(9)
xiTv{1}
A(x-1) - T(x-1)
x-1 - T(x-1)
A(x) - T(x)
x - T(x)
(8)
I,
x
I
T(x)
n.
Then
ci5e
y > a +ß a ß
53
Proof: Recall that
X
If
be the densities of
a,ß,y
A
=
then
I,
a= y
construct integers
0<-
<b
a
1
Let
is in
b1 +
<
<b 2
a2
=
A
+
be subsets of
B
respectively.
al
+
1
...
a
<
k-1
< b
k-1
a
<
k
Therefore,
A.
We
<
m
be the least positive integer missing from
be the least integer greater than
1
I,
where
b.
<
In general, let
A.
A, B, C
B
and the theorem follows.
1
and
a.
1
Let
as follows.
=
and
A
least one positive integer m is not in
we assume that at
A.
Let
which do not exceed x.
I
and
denotes the number of integers in
X(x)
a.
+
al
+
which
1
be the least integer greater
1
than b. -1 + 1 which is not in A and let b. + 1 be the least
integer greater than ai + 1 which is in A. This process ter -
minates when we reach ak
exist or
We
hence in
b
<
and find that either
m
bk
does not
k-> m.
can assume that all singletons are in both
Otherwise,
C.
immediate. Let
greater than
1,
T
a=
or
0
f3
=
0,
and
B,
and the theorem is
be the set of all singletons in
as in Theorem 4.
A
I
which are
5.
We have
C(m) > A(m)
and so
+
B
(bi - al)
+ . . . +
B(bk-1
-
ak_i) + B(m - ak)
,
54
(10)
C(m) -T(m) >A(m)-T(m)
-
m-T(m)
m-T(m)
B(b -a )+... +B(b
1
k-1
1
-a k-1 )+B(m-ak )
m - T(m)
+
By Theorem 4. 5, we have
q,
-
A(m) - T(m)
m -T(m)
Hence, there exists a non -negative constant
a+ a
m
Also by Theorem 4.
ß<
such that
have
5, we
-
(12)
m
A(m) - T(m)
m - T(m)
=
B( m
a
- ak) - T( m - ak)
B(m
<
-
m - ak - T(m - ak)
- ak)
m - ak
and
(13)
ß<
-
B(b. - a.) - T(b -a.)
i
i
i 1
b. - a. - T(b. - a.)
i
Combining (10) ,
i
i
(1 1)
,
(12)
i
,
<
B(b. - a.)
-
Cm)T(m,) >a+án+ ß
m-T(m)
m-T(m)
But
A(m)
ak - (b1
- al)
i
b. - a.
i
i
'
i
=
1, 2
k-1.
and (13) , we obtain
(14)
=
i
...
{(b
-a 1 )+...(bk-l-ak-1)+(m-ak)}
- (bk- - ak -1)'
so (14) becomes
.
55
C(m) -T(m)
m-T(m)
>
a+a
=
a+am
=
- T(m)
a}am+ ß-ß{ A(m)
m - T(m)
>
-
a+a +ß -ß { a+ a
m
m
=
a+ß
m+
+
R
m-T(m) {m-A(m)
}
m-T(m) m- T( m)+ T( m) -A(m)}
}
}
-aß+am 1-ß}
{
>a+ß - aß
because
a
m
is non -negative and
C(m) - T(m)
m - T(m)
(1 5)
for all
If
m
- 1.
-
me A
- aß
ß
and all integers less than
C(m) - T(m)
m - T(m)
=
1>a+
-
Otherwise, we find the largest integer
(17)
+
WA.
(16)
in A.
Therefore,
ß<
m
ß
m'
are in
A, we have
- aß
less than m
Then
C(m) - T(m)
m - T(m)
>
-
0(m') - T(m'
m' - T(m' )
)
-> a+ ß
- a
ß
.
and
56
From (15), (16), and
(17) we can conclude that
C(x) - T(x)
x - T(x)
for all
x
E
I
>
a
-
+
- aß
ß
Therefore,
.
gib
{
C(x) - T(x)
x - T(x)
and so by Theorem 4. 5, we have
(
X
a+ß -a ß,
I} >
E
y > a +ß
-a ß and the proof is
complete.
Theorem 4.
Let
7.
(I,
4')
nested singularly discrete of order
y>
ß
/(1
X,I
sets of I,
X(x)
denotes the number of integers in
which do not exceed
and
greater than
1,
are in both
A
T
B,
5.
We
hence in
C.
and so
Hence, we can assume that
is missing from
C.
A, B, C
Let
Y(1
Otherwise,
-a)>ß
=
A
I
+
B
which are
y
=
1,
a=
0
or
we have
and the theorem follows.
and so, at least one integer
y < 1,
x1, x2,
be sub-
B
can assume all singletons
and the theorem is immediate. If
y>a+ß= ay+ß,
and
be the set of all singletons in
as in Theorem 4.
and
Let A
x.
be the densities of
a ,ß ,y
respectively. Let
P= 0,
then
a +ß< 1,
If
n.
is
- a).
Proof: Recall that
the set
J
be a density space where
...
be the integers missing from
57
C.
Let
x
x. - x.
(18)
for any
B
i (i > 1).
x. - b. áA
1
j
> B (x.) - B(x. )
i
i-1
+
A(x. - x.
i
i-1 -1),
Now
b2, ... b
b
xi -b.< xi -x.
xi-xi-1
t
j(1 < j < p).
-
J-
1
1
=
For any h
1
1
j<P)
Therefore,
Hence,
- p.
p+xi-xi-1
i
i-1
we know that
also know that
We
for each j(1<
xi - xi-1 -
x. ¿C,
Since
.
P
- -
for each
A(xi - xi-1-1)
h,
1
we show that
B(x.) - B(x ) is the number of integers
i
i -1
which lie in the interval (xi xi] . Assume there are P
such integers
0 <
-
i-1
1
in
First,
= O.
o
-p
- B(xi)i -B(xi-1
>
such that xh ¿C,
h
xh
-h>
B(xh)+
A(x.-x.
i
-1).
i-1
i-1
However,
C(xh)
=
xh - h
a<
-
and by Theorem 4. 5, we have
A(m) - T(m)
m -T(m)
for any positive integer m.
<
-
-xi-1-1).
we can sum (18) from
obtaining
(19)
i
A(m)
m
Therefore, (19) becomes
1
to
58
h
C (xh)
>
B(xh)
Lj a(xi
+
i=
=
B ( xh) + a
=
B(xh)
+
- xi-1 -
1)
1
(
xh - h)
a C (xh)
Hence,
(20)
- a) C(xh) >B(xh)
(1
Therefore, by (20) and Theorem 4.
(1- a) C(xh) - T(xh)
(21)
0 <
1
- a<
1
so
(1
have
B(xh) - T(xh)
-
xh - T( xh)
But
5, we
.
xh - T( xh)
- a)T(xh) < T(xh).
ß
Hence, (21) becomes
C(xh) - T(xh)
(1 - a
>
xh - T(xh)
ß,
and so
C ( xh) - T(xh)
(22)
If
(23)
>
m
E
A
ß
-
xh - T ( xh)
(1
- a)
and all integers less than
C(m) - T(m)
m - T(m)
_
=
1
>
are in A, we have
m
ß
(1
- a)
'
59
Otherwise, we find the largest integer
A.
less than m and in
m'
Then
C(m) - T(m)
m - T(m)
(24)
>
-
C(m' ) - T(m' )
m' - T(m' )
>
-
ß
(1 -
a)
From (22), (23), and (24) we can conclude that
C(x) - T(x)
x - T(x)
for all
x
E
I.
Therefore,
gib
{
C(x) - T(x)
x - T(x)
ß
>
-
(1
- a)
XEI}? (1ßa)
,
I
and so by Theorem 4. 5, we have
y > ß /(1 - a)
and the proof
is complete.
Unsolved Problem. Study discrete fundamental families of
infinite order. Such a family is defined by replacing condition (ii)
of Definition 4.
1
by " the set
The density space
(I,
family of infinite order.
)
{
S
(
[
x])
(
x
E
S
}
is unbounded. "
is an example of a discrete fundamental
60
CHAPTER
V
THE TRANSFORMATION PROPERTIES
In this chapter we study some relations between the two
formation properties
T -
and
1
trans-
and the Landau -
T - 2,
Schnirelmann and Schur inequalities.
The Landau -Schnirelmann Inequality and T -1
5. 1.
Let
(S, °
be any density space.
)
Theorem
that the Landau -Schnirelmann inequality,
S
is
5
imply that
y > a +ß - a
In this section we show that
T -1.
is
2. 14
ß,
tells us
holds if
-a ß does not
y > a +ß
In order to show this we prove the fol-
T -1.
lowing theorem which is of independent interest.
Theorem
Let
5. 1.
is discrete of order
fined by
B(x)
Proof:
Then
Let
JB =i),
subsets of
I
Now let
by
B(x)
_ {x}
Then
n.
{x}
=
(I, 1B)
1
and so
is
B
I;
T -1
for all x
T -1
iff
that is, iff n
T -1
B(x)
_ {x}
"B
=
B
is de-
1.
for all x
e
I.
because all non -empty
=D
"B
be
E
be defined by
`SB
vB
is
`B
for all x
are in
be a density space where
E
I.
and assume that
c4)B
is not defined
Then there exists an integer
61
k(k
and an integer
> 2)
x(1 < x < k -
for all
Suppose
tion
if
2. 17,
and
T1 [D]
and
1)
B(k+1)
=
x
and
=
1
{k} ¿
=
i(1 << i < k -
a4B is not
is not a singleton.
Thus,
Then, using the notation of Defini-
} .
1
F
=
°B v {(i)}
{
1,k
+
1
}
by the way
,
As our induction step we assume that
ton for an arbitrary integer
Suppose
F
=
T1
[D]
{ 1
,
k
=
B(k
+
j +
{k
+
1
+ j +
}
j(j
{k
) =
we have
,
j } I °gB
is not
Thus,
1
{(I)
T -1,
is not a singleton for any
we have
,
a contradiction.
T -1,
D=
{
k
+
1
}
is defined.
°4B
Therefore,
B(k
B(k
+ 1)
is not a single-
+ j)
> 1).
+ j +
1
Then, if
{k + j
D =
},
}.
by the way
a contradiction.
j(j
B(x) _ {x}
i, k} C B(k).
{
{k+
such that
1)
+
1
}
x
=
and
1
and
cjB
is defined.
Therefore,
B(k+ j)
Hence, only the first
> 1).
k -
1
positive integers are singletons.
Now each
must contain at least one singleton, because
B(x)
the smallest integer in each
or more of the first
of the
B(x).
k -1
is a singleton.
B(x)
Therefore, one
integers must occur in infinitely many
Let y be any such integer.
Then y
co
E
n
i=
for some strictly increasing sequence
{xi }.
Since
B(x.)
i
1
B
is a
discrete fundamental family it must be separated. Hence, since
y
E
B(xi)
n B(xi +1)
Therefore,
and
xi
<
xi +1
,
we must have
B(xi)
CB(x.
i
.
62
B(xl) C B(x2)
which contradicts the fact that
Therefore,
. . .
1B
is defined by
IB
C
C B(xi) C
. . .
is discrete of order
for all
B(x) _ {x}
x
e
n.
I
and
the proof is complete.
Let
of
order
2.
(I,B)
be any density space where *TB
By Theorem 4.
1
,
we have
By Theorem 5. 1, we know that
y > a+ ß - a ß
y >
min
{ 1
,
is discrete
a+13}> a +ß - a p..
vB is not T -1. Therefore,
does not assure us that ,1B is T-1.
Unsolved Problem.
pothesis of Theorem
2. 14
p
Replace the
property in the hy-
T -1
with a weaker condition which still im-
plies the Landau -Schnirelmann inequality.
5. 2.
The Schur Inequality and
Let
(S,
that if El is
)
T -2
be any density space.
T -2,
then y
> (3/(1
Theorem
- a)
2. 15
whenever
tells us
a+13< 1.
In this section we show that the Schur inequality does not imply
Consider the following example on the s -set
Let
IB
be defined by
B(x)
=
{
2, 4 }
if x
=
4,
{
1, 5}
if x
=
5,
{x}
otherwise.
I.
T -2.
63
If, using the notation of Definition 2. 18, we let
empty, then
B
that if
To see this we note
y > min
{
T2[D]
and
5
F
be
a+ ß<
a+ ß}
1,
=
1
_ {4} ¿
then by Theorem 4.
,
a+ ß> ay
1, we
have
ß,
+
is discrete of order 2. Thus,
vsB
=
{0. Hence,
JB v
T -2. However, the Schur inequality holds in (I,
and
D = {1, 5}
is not
since
x
y
-ay>ß:,
and so
ßs(1 - a).
y >
Unsolved Problem. Replace the
pothesis of Theorem
2. 15
T -2
property in the hy-
with a weaker condition which still implies
the Schur inequality.
5. 3.
The Correspondence Between
Property
T -2
does not imply property
the following example on the
Let
alB
T -1 and T -2
as we see by
T -1
s -set I.
be defined by
{1,2,} if x=
B(x)
2,
=
otherwise.
{x}
Using the notation of Definition 2. 18, the only choices allowed for
D
are
for
Thus,
{0,
T2[ D]
°B
{1,
are
is
and hence is not
2
},
and
{4}
and
{x }.
{
1),
T -2,
However,
T -1 by
Theorem
Hence, the only possibilities
which are both in
1B v {0.
JB is discrete
order
5. 1.
of
2
64
It seems unlikely that T -1 could imply
by Theorem 2.15,
ever,
T -1
T -1
T -2
because then,
would imply the Schur inequality.
How-
incorporates the characteristics of density spaces for
which the Landau- Schnirelmann inequality hold, and the Landau -
Schnirelmann and Schur inequalities are dissimilar.
Unsolved Problem.
a density space which is
Prove that
T -1
but not
T -1
T -2,
implies
T -2
or find
65
CHAPTER VI
DENSITY SPACES AND PROPERTIES
list several density spaces and summarize
In this chapter we
properties which they satisfy.
We
are particularly interested in
the properties Freedman [ 3, p. 43 -46] has defined and the in-
equalities made famous in the study of Schnirelmann density on the
set of positive integers.
Examples and Properties
6. 1.
The properties we study are
T -1, T -2,
separation, the Schur
inequality, the Landau -Schnirelmann inequality, and y > min {1, a +ß}.
We
consider both K- density and C- density.
We
examine the follow-
ing density spaces which have been of particular interest in our work.
These are only representatives of a much longer list which could be
developed.
.
where
order
Example
5
l':
(S,
ß).
This is the density space
is discrete of order
Example
2:
(S, °
Example
3:
(S,
(S,
1),
1.
)
where
°'S is discrete of order 2.
5)
where
°
Example 4: (S, 5')
where
is purely discrete of order n.
Example
where °
is singularly discrete of
n.
5:
(S,
°,
)
is discrete of order
n(n >2).
66
Example
of
(I,
Example
7:
(S,
X)
Example
8:
(I,
)
order
cj
where
6:
is nested singularly discrete
n.
.
_
This is the density space for which
Schnirelmann density was first studied.
Example
3.5 and n>
9:
(In,
/)
where
is defined as in Section
1.
Example
(I,
10:
{
B(x)
B)
where
2, 4,
...
,
x}
B
is defined by
if x is even,
_
{x}
A
In
if x is odd.
clear way to summarize the results for these ten examples
is with a chart like the one we have included at the end of this
chapter. Each box in the chart represents one of the properties and
one density example. A
YES
in the box means that the property
under consideration is always true for that example, a NO means
it is not always true, a
UKN
means that to our knowledge the
truth of the result is unknown, and a CNJ means that the truth of
the result is unknown but we conjecture a
YES.
Many boxes also
contain letters which identify the person or persons who first verified
the results. The results in boxes containing no letters are easily
verified or are not attributed to any individual. The letter code for
the chart is as follows:
67
S =
Schnirelmann [ 9, 10]
L
=
Landau
H
=
Schur
[ 11 ]
M
=
Mann
[ 8]
D
=
Dyson[2]
C
=
Cheo
[
1
[ 7]
]
K =
Kvarda [ 6]
F
=
Freedman
A
=
The author
[
3]
If a number occurs in a box, then a comment concerning the
sults in that box is found following that number in Section
6. 2,
re-
6. 2.
Comments
The number preceding each comment in this section identifies
it with the box or boxes in the chart which contain that number.
1.
These results follow from Theorem 4.
2.
The example following Theorem 4.
1
1.
shows that all of the
listed density inequalities fail to hold in general for Example
We
3.
These results follow from Theorem 4,
4.
These results hold for
n
conjecture that they hold for all
<
3,
2.
3, by Theorems 4.
n.
5.
These results follow from Theorem 4,
6.
These results follow from Theorem 4. 7.
6.
1
and 4. 3,
68
7.
and 4. 4,
These results hold for
We
n < 4, by
Theorems 4,
conjecture that they hold for all
1, 4, 3,
n.
8.
These results follow from Theorem
2, 13.
9.
This result follows from Theorems
2. 13
and 2. 14,
10.
These results follow from Theorems 2.13 and 2.15.
11.
Example
9
is a particular case of Example
result is known to fail.
12.
These results are easily verified.
7
for which this
69
Example
Property
1
2
YES
NO
4
3
6
5
8
7
9
10
NO NO NO NO YES YES YES NO
F -8
T -1
T
YES NO
NO NO NO NO YES YES YES YES
F -8
T -2
Separation
YES YES YES YES YES YES NO YES NO
a+ ß-
YES YES NO YES n
F -1 F -1 A -2 A -3
>
Y
a
ß
ES
YES
YES YES YES YES YES
CNJA
-5 F -9 L,S
K
12
A -4
a
>
c_>
Yc-
c
- a
+ßc
cßc
YES
YES YES NO YESh <3 YES UKN YES UKN YES
F -1 F -1 A -2 A -3
L, S
12
CNJA -5
A -4
ß/ (1 -a)
when a} ß<
Y
YES YES NO YES n3S YES YES YES YES YES
F -1 F -1 A -2 A 3
6 F -10
12
H F -10
>
1
CNJ
A -4
Y
> ß
ß
c-
when a
y
-
/ (1 - a
c
c
+13
c
<
)
c
1
>min {lea +ß}
{
1
ac+
YE
-
CNJ
A -4
YES UKN YES UKN YES
A -6
12
H
'
i
YES YES
YES YES NO YES n< 3 <4 UKN YES UKN YES
n
F -1 F -1 A -2
- --
- - Yc > min
YES YES NO YES
F -1 F -1 A -2 A -3
-
-
A
-3
-
- -
CNJ CNJ
A -4 A -7
M,D
YES YES
YES YES NO YES n<
NO YES NO
3
ßc} F -1 F -1 A -2 A -3 - n< 4 11 M,D C
,;,NJ CNJ
A -4 A -7
-
12
YES
12
70
BIBLIOGRAPHY
1.
Cheo, Luther.
2,
Dyson, F.
3.
Freedman, Allen R. A general theory of density in additive
number theory. Ph. D. thesis. Corvallis, Oregon State
University, 1965. 125 numb. leaves.
4.
Jacobson, Nathan. Lectures in abstract algebra. Princeton,
Van Nostrand, 1951. 217 p.
5.
Kasch, Friedrich. Wesentliche Komponenten bei Gitter punktmengen. Journal fur die reine und angewandte
Mathematik 197:208 -215. 1957.
6.
Kvarda, Betty. On densities of sets of lattice points.
Journal of Mathematics 13:611-615. 1963.
7.
Landau, Edmund. Die Goldbachsche Vermutung und der
Schnirelmannsche Satz. Nachrichten von der Gesellschaft
der Wissenschaften zu Göttingen, Mathematisch- physikalische Klasse, Fachgruppe II, 1930, p. 255 -276.
8.
Mann, Henry B. A proof of the fundamental theorem on the
density of sums of sets of positive integers. Annals of
Mathematics, Ser. 2, 43:523 -527. 1942.
9.
Schnirelmann, L. Über additive Eigenschaften von Zahlen.
Annales d'Institut polytechnique, Novocerkask 14: 3 -28. 1930.
On the density of sets of Gaussian integers.
American Mathematical Monthly 58:618 -620. 1951.
theorem on the densities of sets of integers.
Journal of the London Mathematical Society 20:8 -14. 1945.
A
Pacific
10.
Über additive Eigenschaften von Zahlen.
Mathematische Annalen 107 :649 -690. 1933.
11.
Schur, I. Über den Begriff den Dichte in der additiven
Zahlentheorie. Sitzungsberichte der Preussischen Akademie
der Wissenschaften, Physikalisch- mathematische Klasse,
1936, p. 269 -297.
.