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

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

Chapter 3 The Real Numbers

Section 3.3 The Completeness Axiom

There is one additional axiom that distinguishes from It is called the

completeness axiom. 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

2

1 3

– 1

But does it really?

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

the axis such that x2 – 2 = 0?

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

numbers, then no such number exists

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

In fact, we can easily prove the more general result

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.

p

If is rational, then we can write = m / n, where m and n are integers

with no common factors other than 1.

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

that they have no common factors other than 1 ♦

Thus p is a factor of n2, and as above we conclude that

p is also a factor of n.

Theorem 3.3.1

Let p be a prime number Then is not a rational number.

Proof: We suppose that is rational and obtain a contradiction.

Now the prime factored forms for m and m2 have exactly the same prime factors,

so p is a factor of m That is, m = k  p for some integer k. But then p  n2 = k 2 p 2,

so that n  2 = k 2 p.

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

p p

p

p

If m ≤ s for all s ∈ S, then m is a lower bound of S and S is bounded below.

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.

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).

0 2 4 6 8 10

Example 3.3.3

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

upper bounds

max

lower bounds

min

Then S is bounded above by 8, 9, 8.5, π2, 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

– 2 0 2 4 6 8

Example 3.3.3

(b) The interval [0, ∞) is not bounded above

no upper bounds

no maximum

lower bounds

min

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

upper bounds

max

lower bounds

no min

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

Example 3.3.3

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

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

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, That is, m is an upper bound of S.

(b) if 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

but in both cases the supremum

exists as a real number.

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

But is not a rational number So T does not have a supremum in

When considering subsets of , it has been true that each set bounded above has had a

least upper bound 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),

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.

2

By the completeness axiom,

C has a least upper bound, say sup C = c.

•

b

•

b – ε

•

a

•

a – ε

Since a = sup A, a –  ε is not an upper bound

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

Since c is the

least upper bound of C, we have c ≤ a + b

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 sup A = a and sup B = b If z ∈ C, then z = x + y for some x ∈ A and y ∈ B

Thus z = x + y ≤ a + b, so a + b is an upper bound of C.

We must show that c = a + b.

To see that a + b≤ c, choose any ε > 0.

Similarly, since b = sup B, there

exists y ∈ B such that y > b – ε Combining these inequalities, we have x + y > a + b – 2ε

But, x + y ∈ C and c = sup C, so c > a + b – 2ε That is, a + b < c + 2 ε for every ε > 0.

Thus by Theorem 3.2.8, Finally, since a+ b ≤ c c ≤ a + b and c ≥ a + b, we conclude

that c   = a + b ♦

•

c

•

x

•

y

•

x + y

•

a + b – 2ε

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

bounded below by sup f (D).

Thus f  (D) is bounded

above by g  (z).

Theorem 3.3.8

Suppose that D is a nonempty set and that f  : D →  and g : D →   If for every

x, y ∈ D, f (x) ≤ g  (  y), then f (D) is bounded above and g  (D) is bounded below

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

It follows that the least upper bound of f  (D) is no larger than g  (z).

That is, sup f (D) ≤ g  (z).

Thus the greatest lower bound of g  (D) is no smaller

than sup f (D). That is, inf g  (D) ≥ sup f (D) ♦

D

•

z

sup f (D)

•

inf g(D)

g(z)

upper bound of f (D)

least upper bound of f (D)

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

and since n + 1 ∈ ,

this contradicts m being an upper bound 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 = m. Since m is a least upper bound, m – 1 is not an upper bound of .

Thus there exists an n in such that n > m – 1. But then n + 1 > m,

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

Theorem 3.3.10

(a) For each z ∈ , there exists an n ∈ such that n > z.

Each of the following is equivalent to the Archimedean property

(b) For each x> 0 and for each y ∈ , there exists an n ∈ such that nx  >  y.

(c) For each x> 0, there exists an n ∈ 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.

Theorem 3.3.10

(a) For each z ∈ , there exists an n ∈ such that n > z.

Each of the following is equivalent to the Archimedean property

(b) For each x> 0 and for each y ∈ , there exists an n ∈ such that nx  >  y.

(c) For each x> 0, there exists an n ∈ 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 x > 0, take y = 1 in part (b).

(c)  Theorem 3.3.9

Suppose that were bounded above by some real number, say m.

Then n  x > 1, so that 1/n < x.

Since n ∈ , n > 0 and also 1/n > 0.

That is, n ≤ m for all n ∈ Let k = m + 1. Then n ≤ k – 1 < k and 1/n > 1/k for all n.

This contradicts (c) with x = 1/k Thus (c) implies the Archimedean property ♦

Thus S is bounded above by p,

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

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

Let p be a prime number Then there exists a positive real number x such that x2 = p.

Proof:

Let S = {r ∈ : r > 0 and r2 < p}. Since p> 1, 1 ∈ S and S is nonempty.

Furthermore, if r ∈ S, then r2 < p < p2, so r < p.

Let x = sup S.

It is clear that x To prove this, we shall show that > 0, and we claim that x2 =  p.

neither x2 < p nor x2 > p is consistent with our choice of x.

p

Theorem 3.3.12

Let p be a prime number Then there exists a positive real number x such that x2 = p.

Proof:

Let S = {r ∈ : r > 0 and r2 < p} and x = sup S. Suppose that x2 < 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)/(2x + 1) > 0, so that Theorem 3.3.10(c) implies the existence of some

n ∈ such that The text shows where this estimate comes from.

Then

n

x +

2

1

p x x n

− +

<

2

2

2 1

x

n

Theorem 3.3.12

Let p be a prime number Then there exists a positive real number x such that x2 = p.

Proof:

Let S = {r ∈ : r > 0 and r2 < p} and x = sup S. Now suppose that x2 > p.

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

•

x2 p

We have (x2 –  p)/(2x) > 0, so there exists an m ∈ such that

Then

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.

Finally, since neither x2 < p nor x2 > p is a possibility, we conclude by the

trichotomy law that in fact x2 = p ♦

m

x −

2

1

x x m

p

−

<

2

2

Since n x > 0, it is not difficult to show (Exercise 9) that there exists m ∈ such that m – 1 ≤ n x < m.

Then apply the argument

above to the positive numbers x + k and y + k.

Using the Archimedean property 3.3.10(a),

there exists an n ∈ such that n > 1/( y – x).

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

That is, n  x + 1 < n y.

But then m ≤ n  x + 1 < n y, so that n x < m < n y It follows that the rational number

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

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 x < w < y.