Principles of Mathematics in Operations Research

160 11 Continuity

Proof. Let g = f'^1 : Y -> X. Show that V closed set C in X, g~l{C) is a

closed set in Y: 5 _1(C) = (/~^1 )~^1 (C) — /(C), since X is compact. Hence,

/(C) is closed, thus g~l{C) is closed. D

Remark 11.2.4 If compactness is relaxed, the theorem is not true. For ex-

ample, take X — [0,2n), with d\ metric. Y — {(x, y) £ R^2 : x^2 + y^2 = 1} with

d% metric.

f :X<->Y, f(t) = (cos t, sin t).

f is one-to-one, onto, continuous. However /_1 is not continuous at P =

(0,1) = /(0). If we let e = 7r, suppose there is a 8 > 0 9 V(x,y) £ Y wiift

d 2 {{x,y), (1,0)) < 8, then we have

\r^1 (x,y)-r^1 (l,0)\<e.

However, for {x,y) 9 ^ < f~^1 (x,y) < 2ir (S = V%), we have

\f-\x,v)-r

1

M\>\>*-

Thus, we do not have

\f~\x,y) - r\l,0)\ < e = 7T V(x,y) €Y B d 2 [(x,y), (1,0)] < J.

11.3 Uniform Continuity

Definition 11.3.1 Let (X, dx), (Y, dy) be two metric spaces, f : X M- Y.
We say f is uniformly continuous on X if

\/e > 0, 38 > 0 3 Vp, <? € X with dx(p, q) < 8, we have dY(f(p), f(q)) < e.

Remark 11.3.2 Uniform continuity is a property of a function on a set,
whereas continuity can be defined at a single point. If f is uniformly continuous
on X, it is possible for each e > 0 to find one number 5 > 0 which will do for
all points p of X. Clearly, every uniform continuous function is continuous.

Example 11.3.3

f{x,y) = 2x+\, E = {(x,y)eR^2 :l<y<2}.

Let us show that f is uniformly continuous on E. Let e > 0 be given. Suppose
we have found 8 > 0 whose value will be determined later. Let p = (x,y),q =
(u,v) G E be such that d 2 (p,q) < 8, Show \f(x,y) — f(u,v)\ < e: d 2 (p,q) <
8 =>\x-u\ <8 and \y - v\ < 8 =*> \f(x,y) - f(u,v)\ = \2x+-^ -2u- -^\ <

2\x-u\ + {^-^)<28 +

we have \f(x,y) - f(u,v)

{y-y)(v+y) = 28+ l«-»jl»+»l. Since I"-»JIP+"I < 48,

(vy)^2

< 65 = s. Hence, one can safely choose 8 = | > 0.