Julia set

Remarks

Example: for $f(z) = z^2$, $J(f)$ is the unit circle. On $\abs{z} < 1$ the iterates $z^{2^n}$ tend to $0$ locally uniformly, and on $\abs{z} > 1$ they tend to $\infty$, so both regions lie in $F(f)$. Every neighbourhood of a point with $\abs{z} = 1$ contains points whose orbits tend to $0$ and points whose orbits tend to $\infty$. So a limit of a subsequence of iterates would be discontinuous there, and the point is in $J(f)$. Non-example: $z = 1/2$ is not in $J(z^2)$, because the disc $\abs{z} < 3/4$ lies in the Fatou set. For degree at least $2$, $J(f)$ is nonempty, compact and perfect, it satisfies $f^{-1}(J(f)) = J(f)$, and it is the closure of the set of repelling periodic points. The same definition, with $\C$ in place of $\hat{\C}$, is used for transcendental entire functions.