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If a closed set $E$ satisfies $f^{-1}(E)=E$ for a rational map of degree at least two, then either $E$ has at most two points and lies in the Fatou set, or $J(f)\subseteq E$. The complement omits every point of $E$, so Montel theorem proves normality when $|E|\ge 3$.

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.