Julia set ($J(f)$)

For a rational map $f$ of degree of a rational map of the Riemann sphere at least two, the Julia set isthe points where the iterations of a map fail to form a normal family on every neighbourhood. It is nonempty, closed, a completely invariant set of a rational map, and perfect. The repelling periodic points are dense in the Julia set.

For $f(z)=z^2$, the Julia set is the unit circle. Every neighbourhood of a point on that circle meets points whose iterations of a map tend to zero and points whose iterations of a map tend to infinity. These two incompatible local limits prevent normality there. The circle separates the two Fatou components.

A Julia set need not enclose a nonempty interior region: the Chebyshev polynomial Julia set is an interval, and other rational maps have a Julia set equal to the whole Riemann sphere. For a polynomial function, distinguish the Julia set from the filled Julia set, which includes every nonescaping point and has the Julia set as its boundary.^[ref]

Let $R,S$ commute. On an open set where $\{R^n\}$ is equicontinuous, so is $\{S\circ R^n\}$, by the rational map is Lipschitz in the chordal metric bound. Since this equals $\{R^n\circ S\}$, equicontinuity descends through the open map $S$: near $S(x)$ every nearby value has a preimage near $x$, and the corresponding estimates hold for all $n$. Thus $S(F(R))\subseteq F(R)$. A degree-at-least-two Julia set contains at least three points, all omitted by the iterates of $S$ on $F(R)$. The three-value Montel theorem gives $F(R)\subseteq F(S)$. Exchanging the maps proves equality of their Fatou and Julia sets. The degree hypothesis excludes Möbius examples with only one Julia point.

For a polynomial $P$ of degree at least two, the filled Julia set $K(P)$ consists of the points with bounded forward orbit. Choose an escape radius $B$ so that $|P(z)|>2|z|$ for $|z|>B$. Then $K(P)={\bigcap }_{n\ge 0}(P^n{)}^{-1}(\{|z|\le B\})$ is compact, and its complement is the basin of infinity. On the interior all iterates are bounded by $B$, so they form a normal family; on the complement they converge locally uniformly to infinity. At a boundary point, normality would give a subsequential limit equal to infinity on an open escaping part and hence identically infinity, contradicting the uniform bound at a nearby point of $K(P)$. Therefore $\partial K(P)=J(P)$.

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 $|E|\ge 3$.

Let $f$ be a rational map of degree of a rational map of the Riemann sphere at least two. Its nonempty Julia set $J$ is infinite: a finite completely invariant closed set of a rational map lies in the Fatou set. Since $J$ is compact, its set $J^{\prime }$ of accumulation points is nonempty. It is closed, contained in $J$, and completely invariant; local openness of a nonconstant holomorphic map gives the inverse-image direction.

Apply the completely invariant closed set of a rational map classification to $J^{\prime }$. Its alternative with at most two points would put the nonempty set $J^{\prime }$ in the Fatou set, contradicting $J^{\prime }\subseteq J$. Hence $J\subseteq J^{\prime }$, so $J=J^{\prime }$ and the Julia set has no isolated points. This supplies the perfect property without assuming it in advance.^[ref]

For every $d\ge 2$, choose a $d$th root of unity $r\ne 1$ with $d|1-r|>1$, put $a=1-r$, and setIts only critical points are $0$ and infinity, and both land on the repelling fixed point $r$. The postcritically finite map has no possible Fatou component, so its Julia set is the whole sphere.

The degree-$d$ Chebyshev polynomial has Julia set the interval $[-1,1]$ after the standard normalization. Its Fatou set is the connected complement of that interval in the Riemann sphere, so it has exactly one Fatou component.