30 1. The Real Numbers

(c) Since a ≤ sup A and b ≤ sup B for all a ∈ A, b ∈ B, we have

a + b ≤ sup A + sup B for all a ∈ A, b ∈ B.

It follows that

sup(A + B) ≤ sup A + sup B.

Let x be any number less than sup A+sup B. We claim that there are elements

a ∈ A and b ∈ B such that

(1.5.3) x a + b.

Once proved, this will imply that no number less than sup A + sup B is an upper

bound for A + B. Thus, proving this claim will establish that sup(A + B) =

sup A + sup B.

There are two cases to consider: sup B finite and sup B = ∞. If sup B is

finite, then x − sup B sup A, and Theorem 1.5.4 implies there is an a ∈ A with

x − sup B a. Then x − a sup B. Applying Theorem 1.5.4 again, we conclude

there is a b ∈ B with x − a b. This implies (1.5.3) and proves our claim in the

case where sup B is finite.

Now suppose sup B = ∞. Let a be any element of A. Then x−a sup B = ∞

and so, as above, we conclude from Theorem 1.5.4 that there is a b ∈ B satisfying

x−a b. This implies (1.5.3), which establishes our claim in this case and completes

the proof.

Sup and Inf for Functions. If f is a real-valued function defined on some set

X and if A is a subset of X, then

f(A) = {f(x) : x ∈ A}

is a set of real numbers, and so we can take its sup and inf.

Definition 1.5.8. If f : X → R is a function and A ⊂ X, then we set

sup

A

f = sup{f(x) : x ∈ A} and inf

A

f = inf{f(x) : x ∈ A}.

Thus, supA f is the supremum of the set of values that f assumes on A and

infA f is the infimum of this set. They themselves may or may not be values that

f assumes on A. If supA f is a value that f assumes on A, then it is called the

maximum of f on A. Similarly, if infA f is a value assumed by f somewhere on A,

then it is called the minimum of f on A.

Example 1.5.9. Find supI f and infI f if

(a) f(x) = sin x and I = [−π/2,π/2);

(b) f(x) = 1/x and I = (0, ∞).

Solution: (a) The function sin x takes on all values in the interval [−1,1) on

I but does not take on the value 1. Thus, infI f = −1 and supI f = 1. In this case,

infI f is a value assumed by f on I, but supI f is not.

(b) The function 1/x takes on all values in the open interval (0, ∞). Thus,

infI f = 0 and supi f = ∞ in this case. Neither one of these extended real numbers

is a value taken on by f on I.