Proof, [Lemma III.1.10, Con78]. First suppose that $\infty \not\in P$. Let $A \subset \complex_{\infty} \setminus K$ be the collection of elements such that the lemma holds^[1].

Let $a_{0} \in \complex \setminus K$, then there exists $V \in \cn_{\complex}(a)$ compact such that $V \cap K = \emptyset$. In which case, the function

is continuous, and hence uniformly continuous by Proposition 6.4.5. Thus $1/(z - a) \to 1/(z - a_{0})$ uniformly on compact sets as $a \to a_{0}$, and if $a_{0} \in \ol{A}$, then $a_{0} \in A$ as well. Therefore $A$ is a closed subset of $\complex_{\infty} \setminus K$.

Now, let $a_{0} \in A \cap \complex$, then by Proposition 8.2.5, $r = d(a_{0}, K) > 0$. Let $a \in B_{\complex}(a_{0}, r)$, then for each $z \in K$,

Since $\sup_{z \in K}|a - a_{0}|/|z - a_{0}| < 1$,

where the convergence is uniform on $K$. Therefore $B_{\complex}(a_{0}, r) \subset A$ as well.

Finally, for each $a \in \complex \setminus K$ with $|a| > \sup_{z \in K}|z|$,

where the convergence is uniform on $K$, so there exists a neighbourhood of $\infty$ that is contained in $A$.

Since $P \cup \bracs{\infty}\subset A$, and $A$ is an open and closed subset of $C_{\infty} \setminus K$, $A$ is a union of connected components of $\complex_{\infty} \setminus K$ by Lemma 5.12.5. Given that $P$ intersects every connected component of $\complex_{\infty} \setminus K$, the lemma holds for all $a \in \complex \setminus K$.$\square$