28 1. The Real Numbers

Solution: Clearly, inf A = −1 and sup A = 1. These are finite, sup A belongs

to A, but inf A does not.

Also, inf B = −∞ and sup B = 5. In this case, the inf is not finite. The sup is

finite but does not belong to B.

Since

n2

n + 1

≥

n

2

, the set C is unbounded, and so sup C = ∞. Also, we have

n + 1 ≤

n2

+

n2

=

2n2,

and so

1

2

≤

n2

n + 1

for all n ∈ N. Thus, 1/2 is a lower bound for C. It is the greatest lower bound,

since it actually belongs to C, due to the fact that

n2

n + 1

=

1

2

when n = 1. Thus,

inf C = 1/2.

Certainly 0 is a lower bound for the set D. It follows from the Archimedean

property (see Example 1.4.9) that there is no x ∈ F with x 0 which is a lower

bound for this set, and so 0 is the greatest lower bound. Thus, inf D = 0. Clearly,

sup D = 1.

If A is a set of numbers and sup A actually belongs to A, then it is called the

maximum of A and is denoted max A. Similarly, if inf A belongs to A, then it is

called the minimum of A and is denoted min A.

The following theorem is really just a restatement of the definition of sup, but

it may give some helpful insight. It says that sup A is the dividing point between

the numbers which are upper bounds for A (if there are any) and the numbers

which are not upper bounds for A. A similar theorem holds for inf. Its formulation

and proof are left to the exercises.

Theorem 1.5.4. Let A be a non-empty subset of R and let x be an element of R.

Then

(a) sup A ≤ x if and only if a ≤ x for every a ∈ A;

(b) x sup A if and only if x a for some a ∈ A.

Proof. (a) By definition a ≤ x for every a ∈ A if and only if x is an upper bound

for A. If x is an upper bound for A, then A is bounded above. This implies its sup

is its least upper bound, which is necessarily less than or equal to x.

Conversely, if sup A ≤ x, then sup A is finite and is the least upper bound for

A. Since sup A ≤ x, x is also an upper bound for A. Thus, sup A ≤ x if and only

if a ≤ x for every a ∈ A.

(b) If x sup A, then x is not an upper bound for A, which means that x a

for some a ∈ A. Conversely, if x a for some a ∈ A, then x sup A, since

a ≤ sup A. Thus, x sup A if and only if x a for some a ∈ A.

Example 1.5.5. If A =

4n − 1

6n + 3

: n ∈ N , find the set of all upper bounds for A.

Solution: Long division yields

4n − 1

6n + 3

=

2

3

−

1

2n + 1

≤

2

3

.