Show a normal family $\{f_n\} $ converges uniformly on compacts Let $\{f_n\}$ be a sequence of holomorphic functions on a domain $\Omega\subset \mathbb C$ which is bounded uniformly on compact subsets of $\Omega$. Let $\{z_k\}$ be a sequence of distinct points in $\Omega$.
with lim$_{k\to \infty} z_k = z_0 \in \Omega$. Assume that lim$_{n\to\infty} f_n(z_k)$ exists, for all $k$. Prove that
$\{f_n\}$ converges uniformly on compact subsets of  $\Omega$.
So far I can show that for disk $D$ centered at $z_k$, $\{f_n\}$ converges uniformly on $D$ if  $\{f_n^{(i)}(z_k)\}$ converges for all $i$. But I don't know how this will help in proving uniform convergence on arbitrary compacts. Any help will be appreciated.
 A: It suffices to show that there exists a holomorphic $f$ with the property that every subsequence of $\{f_n\}$ has a further subsequence which converges to $f$: if this were to hold but $f_n\not\to f$, then by Montel’s theorem we can pass to a subsequence which converges to something that is not $f$, however this is impossible since passing to yet another subsequence shows that this “not $f$” is indeed $f$.
Now by Montel’s theorem yet again, every subsequence of $\{f_n\}$ has a further subsequence which converges to something; but by our condition on $\{z_k\}$, we know that all these convergent subsequences must agree: this is a consequence of the identity principle (the set of points where they agree has an accumulation point).
A: If $(f_n)_n$ converges uniformly on each closed disk $D\subset \Omega$ then there exists a function $f$ on $\Omega$ such that $(f_n)_n$ converges uniformly to $f$ on any compact subset of $\Omega.$ 
Proof: (1). For each $p\in\Omega$ there is  a closed disk $D$ with $p\in D \subset \Omega,$ and $(f_n)_n$ converges uniformly on $D,$ so $(f_n)_n$ converges point-wise on $\Omega$ to a function $f.$ 
(2). Let $E$ be a non-empty compact subset of $\Omega.$ For each $p\in E$ let $D_p$ be a closed disk of positive radius, such that $p\in \text {int}(D_p)\subset D_p\subset \Omega.$
Now $C=\{\text {int}(D_p):p\in E\}$ is an open cover of $E,$ and $E$ is compact. So  let $F$ be a finite (non-empty) subset of  $E$ such that $E\subset  \cup_{p\in F}\text {int}(D_p).$ A fortiori, we have $E\subset \cup_{p\in F}D_p.$
For $p\in F$  let $\|f-f_n\|_p=\sup \{|f(x)-f_n(x)|:x\in D_p\}.$ Then $$\lim_{n\to \infty}\|f-f_n\|_p=0$$  which is to say that $(f_n)_n$ converges uniformly to $f$ on $D_p.$ 
Since $F$ is finite we have  $$\sup \{|f(x)-f_n(x)|:x\in E\}\leq \sup \{|f(x)-f_n(x)|: x\in \cup_{p\in F}D_p=$$ $$=\max_{p\in F}(\sup \{|f(x)-f_n(x)|:x\in D_p\})\leq    $$   $$\leq \sum_{p\in F}\sup \{|f(x)-f_n(x)|: x\in D_p\}=$$ $$=\sum_{p\in F}\|f-f_n\|_p.$$
Now $F$ is fixed and finite. And each of the terms, in the last summation above,  goes  to $0$ as $n\to \infty.$ QED. 
