ProofWiki:Sandbox

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

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

Let $\mathcal H \left({U}\right)$ be the space of mappings holomorphic on $U$.

Let $\left\langle{ f_n }\right\rangle_{n \mathop \in \N}$ be a normal family of  mappings contained in $\mathcal H \left({U}\right)$.

Suppose that $f_n \left({ \sigma }\right) \to f \left({ \sigma }\right)$.

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

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

Consider $K^{*} := K \cup \left\{ { \sigma } \right\}$.

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:
 * $\exists \epsilon > 0 : \forall N \in \N : \exists n \geq N : \left\Vert{ f_n - f }\right\Vert_{K^{*}} \geq \epsilon$

where $\left\Vert{ \cdot }\right\Vert_{K^{*}}$ denotes the supremum norm over $K^{*}$.

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

Since $\left\langle{ f_n }\right\rangle$ is a normal family, there is some  subsequence $\left\langle{ f_{n_r} }\right\rangle$ of $\left\langle{ f_n }\right\rangle$ and some mapping $g \in \mathcal H \left({ U }\right)$ such that:
 * $\left\langle{ f_{n_r} }\right\rangle$ 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:
 * $\exists N \in \N : r \geq N \implies \left\Vert{ f_{n_r} - f }\right\Vert_{K^{*}} < \epsilon$

This contradicts the result that:
 * $\forall N \in \N : \exists n \geq N : \left\Vert{ f_n - f }\right\Vert_{K^{*}} \geq \epsilon$

Hence the result, by Proof by Contradiction.