1The real number system
IA Analysis I
1 The real number system
We are all familiar with real numbers — they are “decimals” consisting of
infinitely many digits. When we really want to study real numbers, while this
definition is technically valid, it is not a convenient definition to work with.
Instead, we take an axiomatic approach. We define the real numbers to be a set
that satisfies certain properties, and, if we really want to, show that the decimals
satisfy these properties. In particular, we define real numbers to be an ordered
field with the least upper bound property.
We’ll now define what it means to be an “ordered field with the least upper
bound property”.
Definition (Field). A field is a set
F
with two binary operations + and
×
that
satisfies all the familiar properties satisfied by addition and multiplication in
Q
,
namely
(i)
(
∀a, b, c
)
a
+ (
b
+
c
) = (
a
+
b
) +
c
and
a ×
(
b ×c
) = (
a ×b
)
×c
(associativity)
(ii) (∀a, b) a + b = b + a and a × b = b ×a (commutativity)
(iii) ∃0, 1 such that (∀a)a + 0 = a and a ×1 = a (identity)
(iv) ∀a ∃
(
−a
) such that
a
+(
−a
) = 0. If
a 6
= 0, then
∃a
−1
such that
a×a
−1
= 1.
(inverses)
(v) (∀a, b, c) a × (b + c) = (a ×b) + (a × c) (distributivity)
Example. Q, R, C, integers mod p, {a + b
√
2 : a, b ∈ Z} are all fields.
Definition (Totally ordered set). An (totally) ordered set is a set
X
with a
relation < that satisfies
(i) If x, y, z ∈ X, x < y and y < z, then x < z (transitivity)
(ii) If x, y ∈ X, exactly one of x < y, x = y, y < x holds (trichotomy)
We call it a totally ordered set, as opposed to simply an ordered set, when
we want to emphasize the order is total, i.e. every pair of elements are related to
each other, as opposed to a partial order, where “exactly one” in trichotomy is
replaced with “at most one”.
Now to define an ordered field, we don’t simply ask for a field with an order.
The order has to interact nicely with the field operations.
Definition (Ordered field). An ordered field is a field
F
with a relation
<
that
makes F into an ordered set such that
(i) if x, y, z ∈ F and x < y, then x + z < y + z
(ii) if x, y, z ∈ F, x < y and z > 0, then xz < yz
Lemma. Let F be an ordered field and x ∈ F. Then x
2
≥ 0.
Proof.
By trichotomy, either
x <
0,
x
= 0 or
x >
0. If
x
= 0, then
x
2
= 0. So
x
2
≥
0. If
x >
0, then
x
2
>
0
× x
= 0. If
x <
0, then
x − x <
0
− x
. So 0
< −x
.
But then x
2
= (−x)
2
> 0.
Definition (Least upper bound). Let
X
be an ordered set and let
A ⊆ X
. An
upper bound for
A
is an element
x ∈ X
such that (
∀a ∈ A
)
a ≤ x
. If
A
has an
upper bound, then we say that A is bounded above.
An upper bound
x
for
A
is a least upper bound or supremum if nothing
smaller that x is an upper bound. That is, we need
(i) (∀a ∈ A) a ≤ x
(ii) (∀y < x)(∃a ∈ A) a > y
We usually write
sup A
for the supremum of
A
when it exists. If
sup A ∈ A
,
then we call it max A, the maximum of A.
Example. Let
X
=
Q
. Then the supremum of (0
,
1) is 1. The set
{x
:
x
2
<
2
}
is bounded above by 2, but has no supremum (even though
√
2
seems like a
supremum, we are in Q and
√
2 is non-existent!).
max
[0
,
1] = 1 but (0
,
1) has no maximum because the supremum is not in
(0, 1).
We can think of the supremum as a point we can get arbitrarily close to in
the set but cannot pass through.
Definition (Least upper bound property). An ordered set
X
has the least upper
bound property if every non-empty subset of
X
that is bounded above has a
supremum.
Obvious modifications give rise to definitions of lower bound, greatest lower
bound (or infimum) etc. It is simple to check that an ordered field with the least
upper bound property has the greatest lower bound property.
Definition (Real numbers). The real numbers is an ordered field with the least
upper bound property.
Of course, it is very important to show that such a thing exists, or else we
will be studying nothing. It is also nice to show that such a field is unique (up
to isomorphism). However, we will not prove these in the course.
In a field, we can define the “natural numbers” to be 2 = 1 + 1, 3 = 1 + 2
etc. Then an important property of the real numbers is
Lemma (Archimedean property v1)). Let
F
be an ordered field with the least
upper bound property. Then the set {1, 2, 3, ···} is not bounded above.
Proof.
If it is bounded above, then it has a supremum
x
. But then
x −
1 is not
an upper bound. So we can find
n ∈ {
1
,
2
,
3
, ···}
such that
n > x −
1. But then
n + 1 > x, but x is supposed to be an upper bound.
Is the least upper bound property required to prove the Archimedean prop-
erty? It seems like any ordered field should satisfy this even if they do not have
the least upper bound property. However, it turns out there are ordered fields
in which the integers are bounded above.
Consider the field of rational functions, i.e. functions in the form
P (x)
Q(x)
with
P
(
x
)
, Q
(
x
) being polynomials, under the usual addition and multiplication. We
order two functions
P (x)
Q(x)
,
R(x)
S(x)
as follows: these two functions intersect only
finitely many times because
P
(
x
)
S
(
x
) =
R
(
x
)
Q
(
x
) has only finitely many roots.
After the last intersection, the function whose value is greater counts as the
greater function. It can be checked that these form an ordered field.
In this field, the integers are the constant functions 1
,
2
,
3
, ···
, but it is
bounded above since the function x is greater than all of them.