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# Julia set

Let $U$ be an open subset of the complex plane and let $f\colon U\to U$
be analytic. Denote the $n$-th iterate of $f$ by $f^{n}$, i.e. $f^{1}=f$
and $f^{{n+1}}=f\circ f^{n}$. Then the *Julia set* of $f$ is the
subset $J$ of $U$ characterized by the following property: if $z\in J$
then the restriction of $\{f^{n}\mid n\in\mathbb{N}\}$ to any neighborhood
of $z$ is not a normal family.

It can also be shown that the Julia set of $f$ is the closure of the set of repelling periodic points of $f$. (Repelling periodic point means that, for some $n$, we have $f^{n}(z)=z$ and $|f^{{\prime}}(z)|>1$.)

A simple example is afforded by the map $f(z)=z^{2}$; in this case, the Julia set is the unit circle. In general, however, things are much more complicated and the Julia set is a fractal.

From the definition, it follows that the Julia set is closed under $f$ and its inverse — $f(J)=J$ and $f^{{-1}}(J)=J$. Topologically, Julia sets are perfect and have empty interior.

## Mathematics Subject Classification

28A80*no label found*

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## Comments

## filled julia set

it would probably make sense to define the filled Julia set in this article. Also a long time ago I started raising an interesting question on Julia set defined on different fields. There are situations where ”Julia sets” defined on fields that are still characteristic 0 but become empty. Here is an interesting discussion about this: http://mathoverflow.net/questions/81054/julia-sets-using-other-fields