# Is the set of non-escaping points in a Julia set always totally disconnected?

I am looking for examples of transcendental entire functions $$f:\mathbb C\to \mathbb C$$ such that the set of non-escaping points in the Julia set of $$f$$ is not totally disconnected. I denote this set $$J_r(f)=\{z\in J(f):f^n(z)\not\to\infty\}$$ because in some interesting cases it is the same as the "radial Julia set'' of $$f$$.

For two types functions, including the exponential family of $$\exp(z)-2$$, I have recently shown that $$J_r(f)$$ is not only totally disconnected, but is zero-dimensional in the topological sense: my paper. The Julia sets of these functions have a relatively simple "Cantor bouquet'' structure which was used in the proofs.

Question. Can $$J_r(f)$$ contain non-degenerate or unbounded connected sets?

If $$f$$ has order $$<1/2$$ then there is a sequence $$r_k\to\infty$$ with the property that $$\min_{|z|=r_k}|f(z)|>r_k.$$ Restricting $$f$$ on $$\{ z:|z| we obtain a polynomial-like map in the sense of Douady and Hubbard. If $$J_k$$ is the Julia set of this map, then evidently $$J_k\subset J(f)$$, and the points of $$J_k$$ are not escaping. Now, if $$f$$ has an attracting cycle, then for large $$k$$, $$J_k$$ contains the boundary of the attraction domain of this cycle, which is a continuum. Thus $$J(f)$$ contains a continuum consisting of non-escaping points.
• Examples of functions of order less than $1/2$? Plenty. Take an appropriate power series, for example. To arrange an attracting cycle, multply on $z^2$. – Alexandre Eremenko Feb 10 at 1:47