1549901369-Elements_of_Real_Analysis__Denlinger_

262 Chapter 5 • Continuous Functions

Since A is closed, Theorem 3.2.19 says that L E A. Thus, f is continuous

at L. Hence, by the sequential criterion for continuity,

Thus, lim [f(xnk) - f(Ynk)] = lim f(xnk) - lim f(Ynk) = 0. However,
k->oo k->oo k->oo
'<In EN, lf(xn) - f(Yn)I ~ c. Contradiction.

Therefore, f is uniformly continuous on A. •

*SEQUENTIAL CRITERIA FOR UNIFORM CONTINUITY

*Theorem 5.4.8 If f is uniformly continuous on A then, for all Cauchy se-

quences {xn} in A, {f(xn)} is a Cauchy sequence.

Proof. Suppose f is uniformly continuous on A and {xn} is a Cauchy
sequence in A. Let c > 0. Since f is uniformly continuous on A , :J 8 > 0 3
Vx,y EA,

Ix -yl < 8 '* lf(x) - f(y)I < c. (9)

Since { xn} is a Cauchy sequence, :J n 0 E A 3

m, n >no =* lxm -Xnl < 8

'* lf(xm) - f(xn)I < c by (9).

Therefore, {f (xn)} is a Cauchy sequence. •

The following example is an appli cation of this theorem.

Example 5.4.9 The function f(x) =sin(~) is not uniformly continuous on

(0, 1).

2

Proof. '<In E N, let Xn = -. Then {xn} is a Cauchy sequence in (0, 1)

nn

since it converges. But {f (xn)} is the sequence

1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, .. ·,

which does not converge, so {f (xn)} is not a Cauchy sequence. Therefore, by

Theorem 5.4.8, f is not uniformly continuous on (0, 1). D

*Theorem 5.4.10 If A is a bounded set, then a function f : A --+ JR. is uni-

formly continuous on A <:::?for all Cauchy sequences {xn} in A, {f(xn)} is a

Cauchy sequence.

*An asterisk with a theorem, proof, or other item in this chapter inidcates that the item is

optional and can be omitted, especially in a one-semester course.