Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 157 / 1

A family $\mathcal{F}$ of holomorphic or meromorphic functions on a domain $U$ is a normal family if every sequence in $\mathcal{F}$ has a subsequence converging locally uniformly in the spherical metric to a meromorphic function or to infinity. Montel theorem states that a family of meromorphic functions omitting three fixed points of the Riemann sphere is normal; for plane-valued holomorphic functions, two omitted values suffice.

Suppose first that $|E|\ge 3$. Complete invariance implies that every iterate maps $\hat{\mathbb{C}}\setminus E$ into itself. Hence the family $\{f^n\}$ omits the same three points of $E$ on this open set. By Montel theorem,so

If $E$ consists of one or two points, complete invariance makes $f$ permute those points and makes every preimage of them remain in $E$. Some iterate fixes each point and is totally ramified there. In a local coordinate it therefore has the form $w\mapsto a{w}^m+O(w^{m+1})$ with $m\ge 2$, so the point is superattracting for that iterate and belongs to the Fatou set. This proves the completely invariant closed set of a rational map dichotomy.

Let $J^{\prime }$ be the set of accumulation points of $J(f)$. It is closed. Since a rational map is open and has finite local degree, images and preimages of convergent sequences of distinct Julia points show thatThus $J^{\prime }$ is completely invariant. If $J^{\prime }$ were a proper subset of $J(f)$, part b would imply $|J^{\prime }|\le 2$ and $J^{\prime }\subseteq F(f)$. Since $J^{\prime }\subseteq J(f)$, this forces $J^{\prime }=\varnothing$.

On the other hand, $J(f)$ is infinite: if it had at most two points, applying part b to the completely invariant set $J(f)$ would put it inside the Fatou set. Every infinite compact subset of the sphere has an accumulation point, so $J^{\prime }\ne \varnothing$. This contradiction proves that $J^{\prime }=J(f)$ and hence the Julia set has no isolated points.