Analysis with an introduction to proof 5th by steven lay ch03

Chapter 3
The Real Numbers

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Section 3.3, Slide 1


Section 3.3
The Completeness Axiom

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Section 3.3, Slide 2


There is one additional axiom that distinguishes
completeness axiom.

from

. It is called the

Before presenting this axiom, let’s look at why it’s needed.

When we graph the function f (x) = x2 – 2, it appears to cross the horizontal axis
at a point between 1 and 2.But

does it really?

How can we be sure that there is a “number” x on

f (x)

2
the axis such that x – 2 = 0?

2
f (x) = x – 2

3

It turns out that if the x-axis consists only of rational
2

numbers, then no such number exists.

1

–2

That is, there is no rational number whose square is 2.

1

2

x

–1

In fact, we can easily prove the more general result

p

that

is irrational (not rational) for any prime

number

p.

Recall that an integer

p > 1 is prime iff its only

divisors are 1 and p.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.3, Slide 3


Theorem 3.3.1
Let p be a prime number. Then

p

Proof: We suppose that

p

If

pis not a rational number.

is rational and obtain a contradiction.

p

is rational, then we can write

= m / n, where m and n are integers
Then

with no common factors other than 1.

p n2 = m2, so p must be a factor of m2.

2
Now the prime factored forms for m and m have exactly the same prime factors,
so p is a factor of m.
so that n

2
2
  = k   p.

That is, m = k

 p for some integer k.

But then p

2
2 2
 n = k   p  ,

2
Thus p is a factor of n , and as above we conclude that
p is also a factor of n.

Hence p is a factor of both m and n, contradicting the fact

that they have no common factors other than 1. ♦

In Section 2.4 we learned that there are uncountably many irrational numbers.
Thus, if we were to restrict our analysis to rational numbers, our “number line”
would have uncountably many “holes” in it.

It is these holes in the number line

that the completeness axiom fills.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.3, Slide 4


In order to state the completeness axiom for

, we need some preliminary definitions.

Upper Bounds and Suprema

Definition 3.3.2

Let S be a subset of

. If there exists a real number m such that m ≥ s for all s ∈ S,

then m is called an upper bound of S, and we say that S is bounded above.
If m ≤ s for all s ∈ S, then m is a lower bound of S and S is bounded below.
The set S is said to be bounded if it is bounded above and bounded below.
If an upper bound m of S is a member of S, then m is called the maximum
(or largest element) of S, and we write m = max S.
Similarly, if a lower bound of S is a member of S, then it is called the minimum
(or least element) of S, denoted by min S.

While a set may have many upper and lower bounds, if it has a maximum or a
minimum, then those values are unique (Exercise 6).

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Section 3.3, Slide 5


Example 3.3.3

(a) Let S = {2, 4, 6, 8}.

min

0

max

•

•

•

•

2

4

6

8

lower bounds

10

upper bounds

2
Then S is bounded above by 8, 9, 8.5, π , and any other real number greater than

or equal to 8.Since

8 ∈ S, we have max S = 8.

Similarly, S has many lower bounds, including 2, which is the largest of the
lower bounds and the minimum of S.
It is easy to see that any finite set is bounded and always has a maximum and
a minimum.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.3, Slide 6


Example 3.3.3
(b) The interval [0, ∞) is not bounded above.
min
no maximum

[
–2

0

2

4

6

8

no upper bounds
lower bounds

It is bounded below by any nonpositive number, and of these lower bounds,
0 is the largest.

Since

0 ∈ [0, ∞), 0 is the minimum of [0, ∞).

(c) The interval (0, 4] has a maximum of 4, and this is the smallest of the upper bounds.
max
no min

(
0

lower bounds

]
2

4

upper bounds

It is bounded below by any nonpositive number, and of these lower bounds, 0 is the largest.
Since 0 ∉ (0,1], the set has no minimum.

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Section 3.3, Slide 7


Example 3.3.3

(d) The empty set ∅ is bounded above by any m ∈

.

Note that the condition

m ≥ s for all s ∈ ∅ is equivalent to the implication “if s ∈ ∅, then m ≥ s.”
This implication is true since the antecedent is false.
Likewise, ∅ is bounded below by any real m.

Definition 3.3.5

Let S be a nonempty subset of

. If S is bounded above, then the least upper bound

of S is called its supremum and is denoted by sup S. Thus m = sup S iff
(a) m ≥ s, for all s ∈ S,
(b) if

That is, m is an upper bound of S.

m′ < m, then there exists s′ ∈ S such that s′ > m′.
Nothing smaller than m is an upper bound of S.

If S is bounded below, then the greatest lower bound of S is called its infimum
and is denoted by inf S.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.3, Slide 8


{

Let T = q ∈

}

:0 ≤ q ≤ 2 .

Does T have a supremum?

sup T =

If we think of T as a subset of the real numbers, then

2But

is not a rational number.

So T does not have a supremum in

When considering subsets of
least upper bound.

2.
.

, it has been true that each set bounded above has had a

This supremum may be a member of the set, as in the interval [0,1],

or it may be outside the set, as in the interval [0,1),
exists as a real number.

It depends on the context.

but in both cases the supremum

This fundamental difference between

and

is the basis

for our final axiom of the real numbers, the completeness axiom:

Every nonempty subset S of

that is bounded above has a least upper bound.

That is, sup S exists and is a real number.

It follows readily from this that every nonempty subset S of

that is bounded below

has a greatest lower bound. So, inf S exists and is a real number.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.3, Slide 9


Theorem 3.3.7
Given nonempty subsets A and B of

, let C denote the set C = {x + y : x ∈ A and y ∈ B}.

If A and B have suprema, then C has a supremum and sup C = sup A + sup B.
Proof: Let
Thus

If z ∈ C, then z = x + y for some x ∈ A and y ∈ B.

sup A = a and sup B = b.

z = x + y ≤ a + b, so a

+

By the completeness axiom,

b is an upper bound of C.

C has a least upper bound, say sup C = c.
least upper bound of C, we have
To see that a + b ≤ c, choose any ε

c≤

We must show that c Since
= a + cb.is the

a + b.

> 0.

Since a = sup A,

of A, and there must exist x ∈ A such that x > a – ε .

exists y ∈ B such that
But,

Similarly, since b = sup B, there
Combining these inequalities, we have

y > b – ε .

That is, a + b <

x + y ∈ C and c = sup C, so c > a + b – 2ε .

Thus by Theorem 3.2.8,

a – ε is not an upper bound

a + b ≤ c.

x + y > a + b – 2ε .

c + 2ε for every ε > 0.

Finally, since c ≤ a + b and c ≥ a + b, we conclude

that c  = a + b. ♦
B

A

y

x

•
a – ε

•

A+B

•

•

a

b – ε

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

•

•

•

b

a + b – 2ε

x+y

•

•

c

Section 3.3, Slide 10

Theorem 3.3.8
Suppose that D is a nonempty set and that

x, y ∈ D, f (x) ≤ g
Furthermore,

 ( y),

then

f : D → 

and g : D → 

f (D) is bounded above and g (D) is bounded below.

f

sup  (D)

≤ inf g

 (D).

Proof: Given any z ∈ D, we have f (x) ≤ g

Thus f (D) is bounded

 (z), for all x ∈ D.

It follows that the least upper bound of

above by g (z).

That is, sup

. If for every

f (D) is no larger than g

Since this last inequality holds for all z ∈ D, g

f (D) ≤ g (z).

bounded below by sup

Thus the greatest lower bound of g

f (D).

than sup

f (D). That

is, inf g

 (D) ≥ sup

•
g

sup f (D)

•
lower bound of g(D)

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

 (D) is no smaller

z

f

least upper bound of f (D)

 (D) is

f (D). ♦

D

f (D)

 (z).

inf g(D)

g(z)

•

•

g (D)

upper bound of f (D)
greatest lower bound of g(D)

Section 3.3, Slide 11


Theorem 3.3.9

The Archimedean Property of

The set

of natural numbers is unbounded above in

.

Proof:
If

were bounded above, then by the completeness axiom it would have a least upper

bound, say

sup

Thus there exists an n in

Since m is a least upper bound, m – 1 is not an upper bound of

= m.

such that n > m – 1.

this contradicts m being an upper bound of

But then n +and
1 > m, since n + 1 ∈

.

,

.♦

The Archimedean Property is widely used in analysis and there are several equivalent
forms (which are also sometimes referred to as the Archimedean Property).
We state them and establish their equivalence in the following theorem.

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Section 3.3, Slide 12

Theorem 3.3.10
Each of the following is equivalent to the Archimedean property.
(a) For each

z∈

, there exists an n ∈

such that n > z.

(b) For each x > 0 and for each y ∈
(c) For each x > 0, there exists an n ∈

, there exists an n ∈

such that nx > y.

such that 0 < 1/n < x.

Proof:
We shall prove that Theorem 3.3.9  (a)  (b)  (c)  Theorem 3.3.9, thereby
establishing their equivalence.
Theorem 3.3.9  (a)
If (a) were not true, then for some

z0 ∈

we would have n ≤ z0 for all n ∈

But then z0 would be an upper bound of

.

, contradicting Theorem 3.3.9.

Thus the Archimedean property implies (a).
(a)  (b)
Given

x > 0 and y ∈

, let z =

y /x. Then there exists n ∈

such that n > y/x,

so that n x > y.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 3.3, Slide 13


Theorem 3.3.10
Each of the following is equivalent to the Archimedean property.
(a) For each

, there exists an n ∈

z∈

such that n > z.

(b) For each x > 0 and for each y ∈
(c) For each x > 0, there exists an n ∈

, there exists an n ∈

such that nx > y.

such that 0 < 1/n < x.

Proof:
We shall prove that Theorem 3.3.9  (a)  (b)  (c)  Theorem 3.3.9, thereby
establishing their equivalence.
(b)  (c)
Given
Since n ∈

Then n

x > 0, take y = 1 in part (b).

,

 x > 1, so that 1/n < x.

n > 0 and also 1/n > 0.

(c)  Theorem 3.3.9
Suppose that
That is, n ≤ m for all n ∈

were bounded above by some real number, say m.

.

This contradicts (c) with

Let k = m + 1.
x = 1/k .

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Then

n ≤ k – 1 < k and 1/n > 1/k for all n.

Thus (c) implies the Archimedean property. ♦

Section 3.3, Slide 14


In Theorem 3.3.1 we showed that

pis not rational when p is prime.

We are now in a position to prove there is a positive real number whose square is p,

thus illustrating that we actually have filled in the “holes” in

.

Theorem 3.3.12
2
Let p be a prime number. Then there exists a positive real number x such that x = p.
Proof:
Let

S = {r ∈

2
: r > 0 and r < p}.

Furthermore, if r ∈ S, then

Since
2
r <

2
p < p , so r <

p.

p > 1, 1 ∈ S and S is nonempty.

Thus S is bounded above by p,
Let x = sup S.

and by the completeness axiom, sup S exists as a real number.

2
It is clear that x > 0, and we claim that x  = p.

To prove this, we shall show that

2
2
neither x < p nor x > p is consistent with our choice of x.

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Section 3.3, Slide 15


Theorem 3.3.12
2
Let p be a prime number. Then there exists a positive real number x such that x = p.
Proof:
Let

S = {r ∈

2
Suppose that x <

2
: r > 0 and r < p} and x = sup S.

•
x

•

( x+ )

2

1 2
n

p.

•
p

We want to find something positive, say 1/n, that we can add on to x and still
have its square be less than p. This will contradict x as an upper bound of S.

We have (

p − x2
1
<
.
2x + 1
n

n∈

such that

2
 p – x )/(2x + 1) > 0, so that Theorem 3.3.10(c) implies the existence of some

The text shows where this estimate comes from.

2

Then

1
2x 1
1
1

2
2
x
+
=
x
+
+
=
x
+
2

x
+

÷

÷
n
n n2
n
n

1
≤ x 2 + ( 2 x + 1) < x 2 + ( p − x 2 ) = p.
n
It follows that x + 1/n ∈ S, which contradicts our choice of x as an upper bound of S.

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Section 3.3, Slide 16


Theorem 3.3.12
2
Let p be a prime number. Then there exists a positive real number x such that x = p.
Proof:
Let

S = {r ∈

2

Now suppose that x >

2
: r > 0 and r < p} and x = sup S.

•

•

( x− )

p

1 2
m

p.

•
x

2

We want to find something positive, say 1/m, that we can subtract from x and still
have its square be greater than p. This will contradict x as the least upper bound of S.

2
We have (x –

 p)/(2x) > 0, so there exists an m ∈

such that

2

Then

x2 − p
1
<
.
2x
m

1
2x 1
2x

2
2
2
2
x
−
=
x
−
+
>
x

−
>
x
−
(
x
− p ) = p.

÷
2
m
m
m
m


This implies that x – 1/m > r, for all r ∈ S, so x – 1/m is an upper bound of S.
Since x – 1/m < x, this contradicts our choice of x as the least upper bound of S.
2
2
Finally, since neither x < p nor x > p is a possibility, we conclude by the

2

trichotomy law that in fact x = p. ♦

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Section 3.3, Slide 17

Theorem 3.3.13

The density of

in

If x and y are real numbers with x < y, then there exists a rational number r such
that

x < r < y.

Proof: We begin by supposing that x > 0.
there exists an n ∈
not difficult to show

Using the Archimedean property 3.3.10(a),

such that n > 1/(

That is, n

 y – x).

(Exercise 9) that there exists m ∈

such that m – 1 ≤

n

 x + 1 < n y.

Since n

 x > 0, it is

 x < m.

But then m ≤ n

 x + 1 < n y, so that n x < m < n y. It follows that the rational number

r = m/n satisfies

x < r < y.

Finally, if x ≤ 0, choose an integer k such that k > |

 x |.

Then apply the argument

above to the positive numbers x + k and y + k.
If q is a rational satisfying
x + k < q < y + k, then the rational r = q – k satisfies x <

r < y. ♦

It follows easily from this that the irrational numbers are also dense in the real numbers
(Theorem 3.3.15). That is, if x and y are real numbers with x < y, then there exists an

irrational number w such that

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x < w < y.

Section 3.3, Slide 18