When reconciling two different definitions of "normal family of analytic functions", Ahlfors gives the lemma:
If a sequence of meromorphic functions on a region $\Omega$ converges in the sense of spherical distance, uniformly on every compact set, then the limit function is meromorphic or identically to $\infty$.
I am attempting to fill in gaps in my understanding of his proof.
In summary, his argument as I understand it is:
1) The limit function $f$ is continuous in the spherical metric from uniform convergence.
2) If $f(z) \ne \infty$, then by by Weierstrass's theorem $f$ is analytic near $z$.
3) If $f(z) = \infty$, then by by Weierstrass's theorem $1/f$ is analytic near $z$, and hence $f$ is meromorphic.
It's the bold part of step 3) that I'm wrestling with. Should it instead read ...and hence $f$ is meromorphic or identically equal to $\infty$?
If this is the case, then does one then show that not identically infinite implies every singularity is an isolated pole, by using connectedness argument such as:
Define $A$ to be the set of points in the region around which there is a punctured disk on which $f$ is finite.
Define $B$ to be the set of points contained in a disk on which $f = \infty$.
Use continuity and Uniqueness/Identity to show $A$ and $B$ both open, disjoint, and cover the region, etc...
Thank you.
Suppose the set $\{f\ne \infty\}$ is nonempty. This set is evidently open. Hence, $E:=\{f = \infty\}$ is closed (relative to $\Omega$). On the other hand, every point of $E$ is either an isolated point (if $1/f$ turns out to be nonconstant in a neighborhood) or an interior point (if $1/f\equiv 0$ in a neighborhood).
Suppose $E$ has an interior point, say $a$. Let $b\in \Omega\setminus E$. Connect $a$ to $b $ by a polygonal curve $\gamma$ lying in $\Omega$. We can perturb each link in $\gamma$ so it does not meet the aforementioned set of isolated points. But now we have a contradiction: $\gamma$ goes from the interior of $E$ to its exterior without hitting the boundary.
Conclusion: $E$ has empty interior; thus it consists of isolated points only.