Real and Complex Analysis (Rudin)

272 REAL AND COMPLEX ANALYSIS
m = max {I z - IX I: Z E K}.
I p(z) - f(z) I < 11m for all z E K. Then
Suppose P is a polynomial, such that
I (z - IX)P(Z) - 11 < 1 (z E K). (1)
In particular, (1) holds if z is in the boundary of V; since the closure of V is
compact, the maximum modulus theorem shows that (1) holds for every
z E V; taking z = IX, we obtain 1 < 1. Hence the uniform approximation is
not possible.
The same argument shows that none of the IXi can be dispensed with in
Theorem 13.6.
We now apply the preceding approximation theorems to approximation
in open sets. Let us emphasize that K was not assumed to be connected in
Theorems 13.6 and 13.7 and that n will not be assumed to be connected
in the theorem which follows.
13.9 Theorem Let n be an open set in the plane, let A be a set which has one
point in each component of S2 - n, and assume f E H(n). Then there is a
sequence {Rn} of rational functions, with poles only in A, such that Rn - f uniformly
on compact subsets ofn.
In the special case in which S2 - n is connected, we may take A = {oo}
and thus obtain polynomials P n such that P n - f uniformly on compact subsets
ofn.
Observe that S2 - n may have uncountably many components; for instance,
we may have S2 - n = {oo} u C, where C is a Cantor set.
PROOF Choose a sequence of compact sets Kn in n, with the properties specified
in Theorem 13.3. Fix n, for the moment. Since each component of
S2 - Kn contains a component of S2 - n, each component of S2 - Kn contains
a point of A, so Theorem 13.6 gives us a rational function Rn with poles
in A such that
1
I Rn(z) - f(z) I < - n
(z E KJ. (1)
If now K is any compact set in n, there exists an N such that K c Kn for
all n ~ N. It follows from (1) that
which completes the proof.
1
I Rn(z) - f(z) I < - n
(z E K, n ~ N), (2)
IIII