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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 $1$.

enter image description here

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?

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

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    $\begingroup$ A subset of a complete metric space is completely metrizable if and only if it is a $G_\delta$, so the questions are equivalent to ask whether these sets are $G_\delta$ $\endgroup$
    – Pietro Majer
    Commented Aug 4, 2019 at 16:53

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$\tilde E(f)$ and $I(f)$ are first category, so the answer to each question is NO. https://doi.org/10.1017/etds.2019.111

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