1774 CHAPTER 56. APPROXIMATION BY RATIONAL FUNCTIONS
Consider the rational function, Rk (z) ≡ Akwk−z where wk ∈ Vj, one of the components of
C\K, the given point of Vj being b j. By Lemma 56.1.8, there exists a function, Qk whichis either a rational function having its only pole at b j or a polynomial, depending on whetherVj is bounded such that
||Rk−Qk||K,∞ <ε
2M.
Letting Q(z)≡ ∑Mk=1 Qk (z) ,
||R−Q||K,∞ <ε
2.
It follows|| f −Q||K,∞ ≤ || f −R||K,∞ + ||R−Q||K,∞ < ε.
In the case of only one component of C\K, this component is the unbounded componentand so you can take Q to be a polynomial. This proves the theorem.
The next version of Runge’s theorem concerns the case where the given points are con-tained in Ĉ \Ω for Ω an open set rather than a compact set. Note that here there couldbe uncountably many components of Ĉ \Ω because the components are no longer opensets. An easy example of this phenomenon in one dimension is where Ω = [0,1]\P for Pthe Cantor set. Then you can show that R \Ω has uncountably many components. Nev-ertheless, Runge’s theorem will follow from Theorem 56.1.9 with the aid of the followinginteresting lemma.
Lemma 56.1.10 Let Ω be an open set in C. Then there exists a sequence of compact sets,{Kn} such that
Ω = ∪∞k=1Kn, · · · ,Kn ⊆ intKn+1 · · · , (56.1.10)
and for any K ⊆Ω,
K ⊆ Kn, (56.1.11)
for all n sufficiently large, and every component of Ĉ\Kn contains a component of Ĉ\Ω.
Proof: Let
Vn ≡ {z : |z|> n}∪⋃z/∈Ω
B(
z,1n
).
Thus {z : |z|> n} contains the point, ∞. Now let
Kn ≡ Ĉ\Vn = C\Vn ⊆Ω.
You should verify that 56.1.10 and 56.1.11 hold. It remains to show that every componentof Ĉ\Kn contains a component of Ĉ\Ω. Let D be a component of Ĉ\Kn ≡Vn.
If ∞ /∈ D, then D contains no point of {z : |z|> n} because this set is connected and Dis a component. (If it did contain a point of this set, it would have to contain the wholeset.) Therefore, D ⊆
⋃z/∈Ω
B(z, 1
n
)and so D contains some point of B
(z, 1
n
)for some z /∈
Ω. Therefore, since this ball is connected, it follows D must contain the whole ball andconsequently D contains some point of ΩC. (The point z at the center of the ball will do.)