Calculus and Elementary Analysis

1.2 Supremum and Infimum

Integers, Rational Numbers and Real NumbersCompleteness of Real NumbersSupremum and Infimum Characterizations of Sup and of Inf

23/4/18 Calculus and Elementary Analysis

1.2.1 Integers, Rational and Real Numbers

0, 1, 2, 3, Integers

Notations 0,1,2,3, Natural Numbers

, , 0, gcd , 1p

p q q p qq

Rational Numbers

This condition means that p and q have no common factors.

Theorem There are no rational numbers r such that r2=2.

Real Numbers

Real numbers consist of rational and of irrational numbers.

Notation

23/4/18 Calculus and Elementary Analysis

Integers, Rational and Real Numbers

Assume the contrary.

This means that p2=2q2.

Proposition There are not rational numbers r such that r2=2.

Proof

Then there are positive integers p and q such that p2/q2=2 and p and q do not have common factors.

Hence the area of a square with side length p is twice the area of the square with side length q. Furthermore p and q are smallest such numbers since they do have common factors.

Graphically:Now p and q are integers so that B=p2=2q2=2G. I.e. the area B of the large brown square is twice the area G of the green square.

p

q

GB

23/4/18 Calculus and Elementary Analysis

Integers, Rational and Real NumbersTheorem There are not rational numbers r such that r2=2.

Proof (cont’d)

p

q

p-q

A

A

I

Furthermore, p and q are smallest integers such that the area B of the large brown square is twice the area G of the smaller green square, B=2G.

p

qNow move a copy of the green square to the upper right hand corner of the larger square.

The two squares marked by A in the picture have the same area A. Furthermore, by the assumptions, I=2A. This is impossible, since B and G were the smallest squares with integer side lengths such that B=2G.

The intersection I of the two copies of the green square is the square I with the area I.

23/4/18 Calculus and Elementary Analysis

1.2.2 Upper and Lower Bounds

We say that a number is an for the set if

upper bo

un

:

d

.

M A

a A a M

Let A be a non-empty set of real numbers. Definition

A number is an for the set if

lower bound

: .

m A

a A a m

A set A need not have neither upper nor lower bounds.

The set A is bounded from above if A has a finite upper bound.

The set A is bounded from below if A has a finite lower bound.

The set A is bounded if it has finite upper and lower bounds.

23/4/18 Calculus and Elementary Analysis

1.2.3 Bounds of Integers, Rationals and RealsObservations

Any non-empty bounded set of integers has the largest (and the smallest) element which is also an integer.

1

2 Consider the set of rational numbers r such that r2 ≤ 2.

This set is clearly bounded both from the above and from below. Observe that the set of rational upper bounds for the elements of this set does not have a smallest element. This follows from the previous considerations showing that is not rational.2

This means that the set of rational numbers is not complete in the sense that the set of rational upper bounds of a bounded set does not necessarily have the smallest rational upper bound.

23/4/18 Calculus and Elementary Analysis

1.2.4 Supremum Let , , be bounded from above.A A

Definition

The set A has finite upper bounds. An important completeness property of the set of real numbers is that the set A has a unique smallest upper bound.

The smallest upper bound of the set A is called the supremum of the set A.

Notation sup(A) = the supremum of the set A.

Example 1 3 7Let , , , 1 2 , 0 . Then sup 1.

2 4 8nA n n A

Completeness of Real Numbers

23/4/18 Calculus and Elementary Analysis

Infimum

Let , , be bounded from below.A A

Definition

The set A has finite lower bounds. As in the case of upper bounds, the set of real numbers is complete in the sense that the set A has a unique largest lower bound.

The largest lower bound of the set A is called the infimum of the set A.

Notation inf(A) = the infimum of the set A.

Example 3 5 9Let , , , 1 2 , 0 . Then inf 1.

2 4 8nA n n A

23/4/18 Calculus and Elementary Analysis

1.2.5 Characterization of the Supremum (1)

Let , , be bounded from above. This is equavalent to

saying that sup .

A A

A

Assume that sup . Then since is an upper bound of ,

the condition 1) is trivially satisfied.

s A s AProof

To prove the condition 2) assume the contrary, i.e.

assume that there is an 0 such that there are no elements

such that .

a A

s a

If this is the case, then also is an upper bound for the set .2

But this is impossible, since is the smallest upper bound for .

s A

s A

1) : and sup

2) 0 : such that

a A s as A

a A s aTheorem

23/4/18 Calculus and Elementary Analysis

Characterization of the Supremum (2)

1) : , and sup

2) 0 : such that s- .

a A s as A

a A a s

To prove the converse, assume that the number

satisfies the conditions 1) and 2). We have to show

that sup .

s

s A

By the condition 1), is an upper bound for the set .s A

Theorem

Proof

Cont’d

If sup , then sup , and sup 0.

By the condition 2), such that s- .

s A s A s A

a A a s

By the definition of the number this means that sup .

This is impossible, hence sup .

a A

s A

For a non-empty set A:

23/4/18 Calculus and Elementary Analysis

1.2.6 Characterization of the Infimum

1) : , and inf

2) 0 : such that m<a<m+ .

a A a mm A

a A

Theorem

The proof of this result is a repetition of the argument the previous proof for the supremum.

For a non-empty set A:

23/4/18 Calculus and Elementary Analysis

1.2.7 Usage of the Characterizations Assume that sup , and let 2 2 | .A A a a A

sup 2 2sup .A A

Let sup . Then : . Hence

: 2 2 sup 2 2 2sup .

m A a A m a

a A m a A m A

Example

Let 0. Apply the above characterization for sup replacing

by to conclude: such that a- . Then m- 2 . 2 2

implying sup 2 2 2sup .

A

a A a m a m

A m A

Claim

Proof of the Claim

1

2

1 2and

sup 2 2sup .A A