= Solution
Let $J'$ 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 that
$$
f^{-1}(J')=J'.
$$
Thus $J'$ is completely invariant. If $J'$ were a proper subset of $J(f)$, part b would imply $|J'|\leq2$ and $J'\subseteq F(f)$. Since $J'\subseteq J(f)$, this forces $J'=\varnothing$.
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 $J'\ne\varnothing$. This contradiction proves that $J'=J(f)$ and hence the <Julia set has no isolated points>.
Back to article page
