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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 $1$.

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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.

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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 Aug 4 at 16:53
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It seems to be that $\tilde E(f)$ is first category in itself. So the answer to both questions is NO.

Let $$F_n=\{z\in \tilde E(f):|f^k(z)|\geq 2\text{ for all }k\geq n\}. $$ Apparently $\tilde E(f)=\bigcup \{F_n:n<\omega\}$ and each $F_n$ is a relatively closed subset of $\tilde E(f)$.

Fix $n<\omega$ and let $U$ be any non-empty open subset of $\tilde E(f)$. Let $z\in U$. By Montel's Theorem the backward orbit of $z$ is dense in the Julia set $J(f)$. Also, $\{z\in J(f):|z|<2\}$ is a non-empty open subset of $J(f)$. So there exists $m<\omega$ such that $|f^{-m}(z)|<2$. Again by density of the backward orbit there exists $k\geq n$ such that $$z':=f^{-m-k}(z)\in U.$$ But $|f^k(z')|<2$, so $z'\notin F_n$. This argument shows $F_n$ is nowhere dense in $\tilde E(f)$.

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