Difference between normal and compact families of holomorphic functions. I'm doing an exercise in complex analysis and I'm not sure if I'm confusing some ideas:
First I have to prove that the family of automorphisms of the unit disk is normal in $\mathcal{H}(D(0,1))$. So I used Montel's theorem and done.
Then I'm asked if it is a compact family. I think I can disprove it by finding a sequence of automorphisms with no convergent subsequence (converging to an automorphism). For example:
$ \phi_n=e^{i\theta}\frac{z-\alpha_n}{1-z\overline{\alpha_n}}$ with $\alpha_n=(1-\frac{1}{n})$
And finally I'm asked if $\mathcal{H}(D(0,1))$ is compact. What can I use in this case? Is there any boundedness property I should use?
If you could give me some information about the differences between normal families and compact ones it'd be great.
 A: Let $U \subset \mathbb C$ be an open set and $(X, d)$ be a complete metric space, $C(U, X)$ be the vector space of all continuous functions
from $U$ into $X$.
$F \subset C(U,X)$ is relatively compact if and only if any sequence of functions $f_n \in F$ has a subsequence which converges uniformly on compact subsets to an element of $C(U, X)$. If $F$ is relatively compact, then the closure of $F$ in $C(U, X)$ is compact.
$F \subset C(U,X)$ is called normal if any sequence of functions $f_n\in F$ has a subsequence which converges uniformly on compact subsets to an element of $C(U, X)$ or, possibly, to a constant map from $U$ into $\partial X$.
For automorphisms, since it cannot be the constant map case, it's compact, but for general analytic functions on $\mathbb D$, it's not hard to construct a family which is not relatively compact, for example, let $F = \{f_n(z) = z + n\; \text{for}\; n \in N \;\text{and} \;z \in \mathbb D\}$. Then since the orbit of $0$ is unbounded in $\mathbb C$, $F$ is not relatively compact in $C(\mathbb D, \mathbb C)$ (Arzela-Ascoli).
A: What is relevant here is the following:
Claim: Let $f_n$ be a convergent sequence of automorphisms of the disc, then its limit must be an automorphism of the disc.
Proof:
Injectivity of $f:=\lim f_n$ follows from Hurwitz's theorem.
Now let us consider $g_n:=f_n^{-1}$. Since they are still automorphisms of the unit disc, Montel's theorem implies that $g_n$ has a convergent subsequence $g_{n_k}\to g$. It is easy to see that $f\circ g=z$ and this implies that $f$ is surjective.
So the set of automorphisms is sequentially compact in $\mathcal{H}(D(0,1))$ equipped with the topology of the compact-open convergence. As this space is metrizable, the set of automorphisms is compact.
