Vitali's Convergence Theorem

Theorem
Let $U$ be an open,  connected subset of $\C$.

Let $S \subseteq U$ contain a limit point $\sigma$.

Let $\sequence {f_n}_{n \mathop \in \N}$ be a normal family of holomorphic  mappings $f_n : U \to \C$.

Let $\sequence {f_n}_{n \mathop \in \N}$ converge to some holomorphic mapping $f : U \to \C$ at $\sigma$.

Then $f_n$ converges uniformly to $f$ on all compact subsets of $U$.

Proof
there exists some compact subset $K$ of $U$ such that $f_n$ does not converge uniformly to $f$ on $K$.

Consider $K^* := K \cup \set \sigma$.

From Subsets Inherit Uniform Convergence, $f_n$ does not converge uniformly to $f$ on $K^*$.

From Uniformly Convergent iff Difference Under Supremum Norm Vanishes, the above is equivalent to:
 * $\exists \epsilon > 0 : \forall N \in \N : \exists n \ge N : \norm {f_n - f}_{K^*} \ge \epsilon$

where $\norm {\cdot}_{K^*}$ denotes the supremum norm over $K^*$.

From Finite Union of Compact Sets is Compact, $K^{*}$ is compact.

Since $\sequence {f_n}$ is a normal family, there is some subsequence $\sequence {f_{n_r} }$ of $\sequence {f_n}$ and some mapping $g \in \map {\mathcal H} U$ such that:
 * $\sequence {f_{n_r} }$ converges uniformly to $g$ on $K^*$.

Further:

From the Identity Theorem, $f$ and $g$ agree on $U$.

From Uniformly Convergent iff Difference Under Supremum Norm Vanishes:
 * $\exists N \in \N: r \ge N \implies \norm {f_{n_r} - f}_{K^*} < \epsilon$

This contradicts the result that:
 * $\forall N \in \N: \exists n \ge N: \norm {f_n - f}_{K^*} \ge \epsilon$

Hence the result, by Proof by Contradiction.