Proof.
Since
S
is bounded below, by the Completeness Axiom there exists
w
= inf
S
.
If
w
2
S
then we are done. So assume
w
62
S
, in order to derive a contradiction. We apply
the two properties of the inf: first,
w
n
for all
n
2
S
. Since
w
62
S
, in fact we have the
slightly stronger
w < n
for all
n
2
S
,
and
w
+ 1 is not a lower bound for
S
, so there exists
k
2
S
⇢
Z
for which
k < w
+ 1
.
Since
S
has no minimum element,
k
is not the smallest element of
S
, so there must be
j
2
S
with
j < k
. Putting these all together,
w < j < k < w
+ 1
.
In particular, 0
< k

j <
(
w
+ 1)

j <
(
w
+ 1)

w
= 1, but
k

j
2
Z
and the distance
between any two integers is at least one! So this is impossible, and we conclude that
S
has
a minimum element.
Applied to subsets of the natural numbers this property is called the “wellordering
principle”:
Corollary 1.13
(WellOrdering Principle)
.
Any nonempty subset of
N
has a minimum ele
ment.
The Corollary follows from the previous Theorem, since
N
is bounded below (by 1), and
so any nonempty subset of
N
is bounded below.
We’re now ready to prove the Density Theorem.
10
Proof of the Density Theorem.
First we choose the denominator of
r
=
m/n
. By the Archimedean
Property, there exists
n
2
N
with
n >
1
(
b

a
)
, that is
1
n
<
(
b

a
)
.
(1.2)
To find the numberator, consider the set
S
=
{
k
2
Z
:
k > na
}
. Since
S
⇢
Z
and
S
is
bounded below by
na
, by Theorem 1.12
S
contains a minimal element
m
, and so
m
2
S
and
m

1
62
S
. This implies:
m > na
and
m

1
na.
(1.3)
Putting (1.2) and (1.3) together we get:
a <
m
n
=
m

1
n
+
1
n
a
+
1
n
< a
+ (
b

a
) =
b.
Then the conclusion holds with
r
=
m/n
.
11
2
Sequences
A
sequence
of real numbers is a function
f
:
N
!
R
; for each counting number
n
2
N
we associate to it a real number
f
(
n
) =
x
n
. There are many di
↵
erent ways of denoting the
sequence. Here are a few which you may see:
(
x
1
, x
2
, x
3
, . . .
) = (
x
n
) = (
x
n
)
n
2
N
= (
x
n
:
n
2
N
)
.
The text also writes
X
= (
x
n
), using a single capital letter for the sequence as a whole.
Example 2.1.
(a) The function
f
can be explicitly given. For instance,
x
n
=
2
n
2
+1
n
2
,
✓
2
n
2
+ 1
n
2
◆
n
2
N
=
✓
3
,
9
4
,
19
9
,
33
16
, . . . ,
◆
.
(b)
(
x
n
) = ( (

1)
n
)
n
2
N
= (

1
,
1
,

1
,
1
,

1
, . . .
)
.
(c) We may define sequences by iteration. Let
g
:
R
!
R
be a given realvalued function.
Choose an initial value
x
1
2
R
and then define the sequence iteratively,
x
n
+1
=
g
(
x
n
)
,
n
= 1
,
2
,
3
, . . .
This is a natural way to define sequences, for example to approximate solutions to equa
tions. For a more specific example, take
g
(
x
) =
1
2
(
x
+
2
x
)
, and generate the sequence
(
x
n
)
by iteration:
x
1
= 2
and
x
n
+1
=
1
2
✓
x
n
+
2
x
n
◆
,
n
= 1
,
2
,
3
,
. . . .
The first few values are:
(
x
n
) = (2
,
1
.
5
,
1
.
41
¯
6
,
1
.
414215686
. . . ,
1
.
414213562
. . . , . . .
)
Later, we will prove that this sequence converges to
p
2
.
Note that a sequence is not the same thing as a set. A sequence is an infinite ordered
list of numbers. In a sequence the same number may appear several times, and changing
the order of the elements of a sequence creates an entirely di
↵
erent sequence. A set has no
order, and there is no point in repeating the same value several times. Taking Example (b)
above,
{
x
n
:
n
2
N
}
=
{

1
,
+1
}
is a set with two elements. The sequence (
x
n
) is not the
same thing. If we let
{
y
n
}
n
2
N
= (

1
,

1
,
1
,
1
,

1
,

1
,
1
,
1
, . . .
), this is a
di
↵
erent sequence
than
{
x
n
}
n
2
N
, yet it takes values in the same set
{
y
n
:
n