1550251515-Classical_Complex_Analysis__Gonzalez_

Functions. Limits and Continuity. Arcs and Curves 147

Proof Let A be any open set in W. We have

By the continuity of g, g-^1 (A) is open in E (note that it could be the

empty set), and by the continuity off, the set f-^1 (g-^1 (A)) is open in D.

By Theorem 3.7, go f is continuous in D.

Theorem 3.9 Under a continuous function the image of any .connected
set is connected.

Proof Let f: E -t f(E) be continuous and suppose that E is connected.

Assume that f(E) =AU B, where A and B are open [relatively to f(E)]

and disjoint, i.e., An B = 0. Then we have

E = f-^1 (A) U f-^1 (B)

where f-^1 (A) and f-^1 (B) are open (by Theorem 3.7). Furthermore,
f-^1 (A) n f-^1 (B) = 0. Since E is connected, either f-^1 (A) = 0 or
f-^1 (B) = 0, which implies that either A = 0 or B = 0. Hence f(E)
is connected.

Remark If f is continuous and f(E) = G is connected, it may happen

that f-^1 (G) is not connected. For example, if E = E 1 U E 2 , where E 1

and E 2 are disconnected, and f(E 1 ) = f(E2) = G, then f(E) = G and

f-^1 (G) = E 1 U E 2. This is illustrated in Fig. 3.11 with G = {x: x =
Re z, z E E 1 or z E E 2 }.

Theorem 3.10
set is compact.

Under a continuous function the image of any compact

/ /

Proof Let f: E -t f(E), f continuous, and suppose that E is compact.
Let Q be a· covering of f(E). by open sets A._ Then the sets f-^1 (A)

y

E1

~ E = E 1 U E 2

I E 2 I

~

I I

: I

I I

0 G = f(E) x

Fig. 3.11