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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.