Let $f:\mathbb C \to \mathbb C$ be the complex exponential $$f(z)=e^z-2.$$ It is known that $J(f)$, the Julia set of $f$, is a uncountable collection of disjoint rays (one-to-one continuous images of $[0,\infty)$). It looks something like the figure below. The number $2$ is rather arbitrary; the dynamics of $f$ are practically identical if we replace $2$ with any number greater than $1$.

Let $E(f)$ be the set of all $0$-endpoints of these rays, i.e. the "endpoints" of $J(f)$.

Let $I(f)=\{z\in \mathbb C:f^n(z)\to\infty\}$. Here $f^2$ is the composition $f\circ f$, etc.

Let $\tilde E(f)=I(f)\cap E(f)$ be the set of *escaping endpoints* of $J(f)$.

It is known that $E(f)$ is completely metrizable.

**Question.** Is $\tilde E(f)$ completely metrizable?

We may independently consider the set of escaping points. It is easy to show $I(f)$ is an $F_{\sigma\delta}$-subset of the plane using only continuity of $f$.

**Question.** Is $I(f)$ completely metrizable?

**EDIT:** Note that completeness of $I(f)$ would imply $E(f)\setminus\tilde E(f)=J(f)\setminus I(f)$ is $F_\sigma$ in the plane and therefore $\sigma$-compact. Clearly $E(f)$ is totally disconnected, and so this implies $E(f)\setminus\tilde E(f)$ is zero-dimensional. So the one-point extension $(E(f)\setminus\tilde E(f))\cup\{\infty\}$ is also zero-dimensional. It was only recently proved that $(E(f)\setminus\tilde E(f))\cup\{\infty\}$ is totally separated; see Theorem 1.2 in the second reference.

*Alhabib, Nada; Rempe-Gillen, Lasse*, **Escaping endpoints explode**, Comput. Methods Funct. Theory 17, No. 1, 65-100 (2017). ZBL1381.37051. for further information.

*Evdoridou, Vasiliki; Rempe-Gillen, Lasse*, **Non-escaping endpoints do not explode**, Bull. Lond. Math. Soc. 50, No. 5, 916-932 (2018). ZBL1411.37046.