Is the set of escaping endpoints for $e^z-2$ completely metrizable?

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.

• 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 $G_\delta$ – Pietro Majer Aug 4 at 16:53

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 $$\tilde E(f)$$.