1.5. Sup and Inf 29

Thus, 2/3 is an upper bound for A. If x 2/3, then = 2/3 − x is positive, and

the Archimedean property implies we can choose n large enough that

1

2n + 1

1

n

.

Then

x

2

3

−

1

2n + 1

=

4n − 1

6n + 3

for such an n, which means that x is not an upper bound for A.

We conclude that 2/3 is the least upper bound for A – that is, sup A = 2/3.

By the previous theorem, the set of all upper bounds for A is the interval [2/3, ∞).

Example 1.5.6. If A =

n2

n + 1

: n ∈ N , find sup A and the set of all upper

bounds for A.

Solution: Long division yields

n2

n + 1

= n − 1 +

1

n + 1

≥ n − 1.

Then the Archimedean property implies that there are no upper bounds for A,

since, for every x ∈ R, there is an n ∈ N for which n − 1 is larger than x. Thus, the

set of upper bounds for A is the empty set and sup A = ∞.

Properties of Sup and Inf. The next theorem uses the following notation con-

cerning subsets A and B of R:

−A = {−a : a ∈ A};

A + B = {a + b : a ∈ A, b ∈ B};

A − B = {a − b : a ∈ A, b ∈ B}.

Theorem 1.5.7. Let A and B be non-empty subsets of R. Then

(a) inf A ≤ sup A;

(b) sup(−A) = − inf A and inf(−A) = − sup A;

(c) sup(A + B) = sup A + sup B and inf(A + B) = inf A + inf B;

(d) sup(A − B) = sup A − inf B;

(e) if A ⊂ B, then sup A ≤ sup B and inf B ≤ inf A.

Proof. We will prove (a), (b), and (c) and leave (d) and (e) to the exercises.

(a) If A is non-empty, then there is an element a ∈ A. Since inf A is a lower

bound and sup A an upper bound for A, we have inf A ≤ a ≤ sup A.

(b) A number x is a lower bound for the set A (x ≤ a for all a ∈ A) if and

only if −x is an upper bound for the set −A (−a ≤ −x for all a ∈ A). Thus, if

L is the set of all lower bounds for A, then −L is the set of all upper bounds for

−A. Furthermore, the largest member of L and the smallest member of −L are

negatives of each other. That is, − inf A = sup(−A). This is the first equality in

(b). If we apply this result with −A replacing A, we have − inf(−A) = sup A. If

we multiply this by −1, we get the second equality in (b).