146 Chapter 3Corollary 3.1 Any polynomial function P(z) = a 0 zn + a1zn-l +···+an
is a continuous function in (<C,d). A rational function R(z) = P(z)/Q(z)
is also continuous in (C, d), except at the points where Q(z) vanishes.
Proof The constant function h(z) = c, as well as the identity function
I(z) = z, are obviously continuous in C. It follows from Theorem 3.6 thatl(z) = z · z = z^2 , l(z) = z^2 · z = z^3 , and in general, l(z) = zn (n a positive
integer) are continuous in C. Then a repeated application of Theorem 3.6
yields the continuity of P(z), and from here that of P(z)/Q(z), except at
points z 0 where Q(z 0 ) = 0. However, if a z 0 is also a root of P(z) = 0
(with the same multiplicity), that z 0 is a removable discontinuity, as we
have seen in the case of the example (z^2 - 4)/(z - 2). A more detailed
discussion of the rational function will be given in Section 5.16.
The following rewording of the definition of continuous function at a
point is useful in generalizing the concept to topological spaces and in
proving Theorem 3. 7.
Definition 3.11 The function l: D ___.. C is said to be continuous at
a point a E D if for each neighborhood N 2 (J(a)) the set l-^1 (N 2 (J(a)))
contains a neighborhood N 1 (a).
Theorem 3. 7 Let l: D ___.. <C. Then l is continuous on D iff either:
- Given any open set A C C, the set l-^1 (A) C D is open, or
- Given any closed set B C C, the set l-^1 (B) C D is closed.
Proof (1) Assume that l-^1 (A) is open whenever A is open. Let z 0 be any
point of D and A an open neighborhood of l(z 0 ). Then l-^1 (A) is open and
zo E l-^1 (A), so that l-^1 (A) is a neighborhood of z 0. By Definition 3.11
it follows that l is continuous at z 0 • Next suppose that l is continuous on
D, and let A be open in <C. If z E l-^1 (A), A is a neighborhood of l(z), so
l-^1 (A) must contain a neighborhood of z. This means that l-^1 (A) is open.
(2) For the second part it suffices to note that for any set B C C we have
l-^1 (B') = [l-^1 (B)]'
So if l-^1 (B) is closed whenever B is closed, then [f-^1 (B)]' is open, and
so is l-^1 (B') where B' is open. This holds for any open set A, since we
can choose B = A', or B' = A. Therefore, by the first part of the proof
we conclude that l is continuous. Conversely, if l is continuous, and B
closed, B' is open, so l-^1 (B') is open. This implies that [f-^1 (B)]' is open,
or l-^1 (B) is closed.
Theorem 3.8 Let l: D ..Ebe a continuous function in D, and g: E ..
W be continuous in E. Then the composite function go l: D ___.. W is
continuous in D.