Why does mapping to unit disk and close to infinity imply non-normalcy of family of iterates?

Question

I am reading Devaney's "Introduction to Chaotic Dynamical Systems", the chapter on the exponential family Prop. 9.1 I don't really understand how he arrives at the conclusion that if for $x\in\mathbb R$ and $U$ any neighborhood of $x$ you can find points $z_1,z_2\in U$ such that there is an $n\in \mathbb N$ for which $|E^n(z_1)|<1$ and $|E^n(z_2)|$ is as large as you wish, then $\{E^n\mid n\in\mathbb N\}$ is not a normal family. If someone could tell me what property or equivalence of normal families he is using I would be grateful.

Answer 1

By definition of normality, every subsequence of $\{E^n \ | \ n \in \mathbb{N}\}$ must have a, possibly finer, subsequence that either converges uniformly on compact subsets or converges uniformly to $\infty$.

By his remarks, given $U$ a neighborhood we may construct a subsequence $\{E^{n_j} \ | \ j \in \mathbb{N}\}$ such that for each $j$ there exists $z_{1_j}, z_{2_j} \in U$ such that $|E^{n_j}(z_{1_j})| < 1$ and $|E^{n_j}(z_{2_j})| >j$. We can actually take $z_{2_j} = x \in \mathbb{R}$ for all $j \in \mathbb{N}$, since this point is in particular compact and converges to $\infty$, then the subsequence we constructed cannot have a subsequence that converges uniformly on compact sets. This means that the only chance of it being normal is that it has a subsequence that uniformly converges to $\infty$, but there's always a point such that its image is in the unit disk, so that can't happen either.