Entire function whose escaping set has only bounded path-components

You need to know: Set ${\mathbb C}$ of complex numbers, absolute value $|z|$ of $z\in{\mathbb C}$ , limits, function $f(z)$ in complex variable z, its derivative $f'(z)$ , polynomials, notation $f^n(z)$ for $f(f(...f(z)...))$ where f is applied n times, bounded subsets of ${\mathbb C}$ , path (or curve) connecting points $x \in {\mathbb C}$ and $y \in {\mathbb C}$ .

Background: A function $f:{\mathbb C}\to {\mathbb C}$ is called entire if $f'(z)$ exists for all $z\in {\mathbb C}$ . An entire function f is called transcendental if it is not a polynomial. An escaping set of a transcendental entire function f is the set $I(f)=\{z\in {\mathbb C} : \lim\limits_{n\to\infty}|f^n(z)|=\infty\}$ . A path-component of $I(f)$ is the set of all $y\in I(f)$ which can be connected by a path $P \subset I(f)$ to any fixed $x\in I(f)$ .

The Theorem: On 24th April 2007, Günter Rottenfusser, Johannes Rückert, Lasse Rempe, and Dierk Schleicher submitted to arxiv a paper in which they proved the existence of a transcendental entire function f whose escaping set $I(f)$ has only bounded path-components.

Short context: The study of limiting behaviour of sequence $f^n(z)$ for transcendental entire function f goes back to 1926 paper of Fatou. This is an example of dynamical system. In 1989, Eremenko, formalising a question of Fatou, asked whether the escaping set $I(f)$ of any transcendental entire function f has the property that every point $z\in I(f)$ can be joined with $\infty$ by a curve in $I(f)$ . The Theorem answers this question in negative.