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 $|f'(z)| > 1$ )

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.