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**](https://www.researchgate.net/publication/338980999_The_topological_dimension_of_radial_Julia_sets). 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?



Is the set of non-escaping points in a Julia set always totally disconnected?
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 $\{ z:|z|<r\_k\}$ 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.


