Calculus: Analytic Geometry and Calculus, with Vectors

326 Functions, graphs, and numbers

We now prove Theorem 5.27 the first part of which says that if f is

continuous over an interval ao 5 x < bo and f' (x) > 0 when ao < x < bo,

then f is increasing over the interval ao < x < bo. Let ao < x1 < x2

bo. The mean-value theorem guarantees existence of a number x* such

that xi < x* < x2 and

f(x2) - f(xi) = f'(x)(x2- XI)-
But f'(x) > 0 and (X2 - x1) > 0, so f(x2) - f(xi) > 0. Thus f(x2) >
f (xi) and f is increasing. For the second part of Theorem 5.27, every-
thing is the same except that f' (x*) < 0 and f is decreasing.
The following theorem expresses the fact that if a function f is con-
tinuous over a closed interval a 5 x < b, then it is uniformly continuous
over the interval.
Theorem 5.58 If f is continuous over a closed interval a -<_ x < b, then
to each e > 0 there corresponds a S > 0 such that

I f (xi) - f(x2) I < e

whenever a<xi<b,a<X2<b,and 1x2-x1I <S.

While neater proofs of this theorem can be given in advanced calculus

after more mathematics has been digested, there is virtue in knowing that

it is possible to base a proof upon a straightforward application of the

Dedekind axiom 5.43. The bookkeeping by which we inch along toward

the answer is really very elementary, and students who have the patience

to see that this is so are very likely to become the leading scientists of the

future. Let e be a given positive number. Let a number x be placed in

the set 4 if x 5 a and also if a < x < b and there is a positive number S

such that If(X2) - f (xi) I < e whenever a <--_ xl S x, a < x2 < x, and

Ix2 - x1I 5 S. Let B contain each number x not placed in A. Let t be

the partition number. Clearly, a < < b. Since f is continuous at a,

we can choose a positive number Sl such that I f(x) - f(a)I < e/2 when-

ever a 5 x < a + S. Then

If(X2) - f(xi)I = ILf(xz) - f(a)] - [f(xi) - f(a)] I

< If(x2) - f(a)I + If(xi) -f(a)I < + 2 < e

whenever a < xi 5 a + 51, a < X2 <= Sl, and Ix2 - xiI<--_Sl. Therefore,

a + S1 belongs to I and it follows that t > a. Our next step is to prove

that Z = b by obtaining a contradiction from the assumption that

a < < b. Suppose, then, that a < < b. Since f is continuous at ,

we can choose a positive number 52 such that a < - S2 < + S2 < b

and If(x) - f(t) I < e/2 whenever Ix - fl < 62. Moreover, since - S2

must belong to -4, we can choose a positive number S3 such that as < S2

and If(X2) - f (xi) I < e/2 whenever a < XI G - 82j a < x2 <

r

S2,

and Ix2 - xiI < S3. Now suppose that a < xi < + S2, a < X2 < t