148 Chapter^3
are open, and
Q* = {f-^1 (A): A E Q}
is a covering of E. Since E is compact, there is a finite number of sets in
Q* covering E. Suppose that
E c f-^1 (A1) U f-^1 (A2) U · · · U f-^1 (An)
Then it follows that
f(E) c Ai U A2 U · · · U An
so f ( E) is compact.
Corollary 3.2 Let f: E --+ JR, f continuous, E compact. Then f(z),
z EE, is bounded and attains its maximum and minimum on E.
Definition 3.12 Let (X, d) and (Y, d') be two metric spaces. A function
f: X --+ Y is said to be uniformly continuous on X iff for every e > 0 there
is a 6 > 0 (depending only one) such that d'(f(x 1 ),f(x 2 )) < e for all pairs
of points x 1 , x 2 E X satisfying d( xi, x 2 ) < 6.
In particular, the definition applies to the complex plane (C, d) or to
any subset thereof (with the induced metric).
Theorem 3.11 A continuous function on a compact set E C X is
uniformly continuous.
Proof Since f is continuous on E, given e > 0 for each x E E there is a
6x > 0 such that if x' E N5.,(x) n E, then d'(f(x), f(x')) <^1 / 2 e. Consider
the open covering of E by the neighborhoods
{ N(l/2)8., ( X)} xEESince Eis compact, there is a finite subcovering, i.e., there are points x 1 ,
x2, ... , Xn E E such that
nEC LJ N(1/2)8.,, (xi)
i=lLet 6 = min(1/ 2 6x.,i = l, .. .,n), and let x,x' EE with d(x,x') < 6. For
some i in {1,2,. .. ,n} we havex E N(l/2)8.,, (Xi)Since d(x, x') < 6 :S %6x; it follows thatd(x', x;) :S d(x', x) + d(x, Xi)<^1 / 2 6x; +^1 / 2 6x; = 6x;