Here's what I had in mind. Consider a $\Sigma^0_3$-set
$$ T = \bigcup_{n \in \mathbb{N}} \bigcap_{m \in \mathbb{N}} \bigcup_{k \in \mathbb{N}} C_{n,m,k}. $$
where each $C_{n,m,k} \subset 2^\mathbb{N}$ is clopen. We show that there is a continuous map $f : 2^{\mathbb{N}} \to \mathbb{Z}^\omega$ such that $f^{-1}(S) = T$ where
$$ S = \{s \in \mathbb{Z}^\omega \;:\; t^*_{\sigma^k(s)} \rightarrow \infty \text{ as } k \rightarrow \infty \} $$
is the set from the question. This proves that $S$ is not $F_{\sigma \delta}$, since that would imply all $\Sigma^0_3$ sets $T$ in $2^\mathbb{N}$ are $\Pi^0_3$.

You are at the concert. On the stage, there is a conductor and $\omega$ many cellists. The conductor is reading a point $x \in 2^\mathbb{N}$. Whenever she notices $x \in C_{n,m,k}$, the conductor cues the $n$th cellist to play the note $m$, assuming it hasn't been played before, and $n$ has played all the notes before $m$. Only one cellist plays at a time, there is a rest when all the events $x \in C_{n,m,k}$ visible so far are exhausted, and if $m$ cannot be played yet because previous notes have not been played, the conductor makes a note of it and it is played once they have. As you listen to these rising scales, you note that $x \in T$ if and only if one of the cellists plays the entire scale $\mathbb{N}$.

From this continuously revealed information you will construct the continuous function $f : 2^\mathbb{N} \to \mathbb{Z}^\omega$. The construction for $f(t) = s$ is thusly. We go through $\ell = 0, 1, ...$ and by default we just set $s_\ell = 100$ for all $\ell$. Whenever the $n$th cellist plays, we do as follows:

- if one of the cellists $n' < n$ has played between the last time the $n$th cellist played (or the beginning of time if $n$ hasn't played anything) and the present time, then we set $s_{\ell} = 100$. As long as no cellist $n' < n$ plays again we
**ensure** that also $t^*_{\sigma^\ell(s)} = 100$.
- otherwise (if no cellist $n' < n$ has played between), then if the last time $n$ played we set $s_{\ell'} = 100+h$ then we now set such a high value at $s_\ell$ that we have $\lfloor t^*_{\sigma^{\ell'+1}(s)} \rfloor \geq 100+h+1$. Namely, set $s_\ell = \lceil \text{pexp}^{\ell-\ell'-1}(100+h+1) \rceil$ where $\text{pexp}(x) = \exp(x) - 1$. Note that in fact we get precisely $\lfloor t^*_{\sigma^{\ell'+1}(s)} \rfloor = 100+h+1$ because of basic properties of $\text{pexp}$. It also follows, because $\log (103 + h) < 100 + h$ and by induction, that we do not disturb any of the ensured values $t^*_{\sigma^\ell(s)} = 100$ for any $n' \leq n$: those were ensured before setting the value of $s_{\ell'}$.

Now suppose indeed $t \in T$, and some cellist plays infinitely many times. Then if the $n$th cellist is the first cellist that does, then the first item applies only finitely many times for $n$, and after that whenever we set $s_{\ell} = \text{pexp}^{\ell-\ell'-1}(100+h+1)$ we actually set $t^*_{\sigma^{\ell''}(s)} \geq 100+h+1$ for all $\ell'' \in [\ell'+1, \ell]$. So since $n$ plays infinitely many times, actually $t^*_{\sigma^{\ell''}(s)} \rightarrow \infty$ as $\ell'' \rightarrow \infty$.

Suppose then $t \notin T$. If the song is finite, obviously $\lim_\ell t^*_{\sigma^{\ell}(s)} = 100$. Otherwise, whenever $n$ plays for the last time, we have a new ensured value at which $t^*_{\sigma^{\ell}(s)} = 100$, thus $\liminf_\ell t^*_{\sigma^{\ell}(s)} \leq 100$.

An observation:

- As far as I can tell all we are using is that $F$ is monotone, $F(h + 2) < h$ for $h \geq 100$ and that $\lfloor F^n(h) \rfloor \rightarrow \infty$ as $h \rightarrow \infty$ for any $n$, and the values range over $[100, \infty) \cap \mathbb{N}$. And I guess $100$ can be replaced by some other number (probably $1$ or $2$ for your function). Maybe I missed some axioms.

**edit**

Your set is not $G_{\delta \sigma}$ either. Namely, any $\Pi^0_3$ set
$$ T' = \bigcap_{n \in \mathbb{N}} \bigcup_{m \in \mathbb{N}} \bigcap_{k \in \mathbb{N}} D_{n,m,k} $$
clearly continuously reduces to
$$ S' = \{s \in \mathbb{Z}^\omega \;:\; \lim_i s_i = \infty\}. $$
To see this, for each $n$ separately go through $(m,k)$ in lexicographical order, advancing to the next $m$ when you observe the point is not in $D_{n,m,k}$. On step $\ell$, output $n$ if $m$ is updated for $n$, otherwise output $\ell$. This way you construct $g(t) \in \mathbb{Z}^\omega$ for $t \in 2^\mathbb{N}$.

Clearly $t \in T'$ if and only if $m$ is updated for each $n$ only finitely many times. If $m$ is updated infinitely many times for $n$, then the limit of $g(t)$ is at most $n$, while if $m$ is updated finitely many times for each $n = 0, 1, ..., N$ then $g(t)$ stays above $N$ from some point on.

Now it's easy to further reduce to your set, observing that if a sequence satisfies $|s_{i+1} - s_i| \in \{-1,0,1\}$ and $s_i \geq 100$ for all $i$, then $s \in S' \iff s \in S$. Just replace all jumps by arithmetic progressions.

By taking the coordinatewise minimum of this process and the above, I suppose we have

Every set of the form $A \cap B$ for $A \in \Sigma^0_3$ and $B \in \Pi^0_3$ continuously reduces to your set.

But I don't know if your set can be written as $C \cap D$ for $C \in F_{\sigma \delta}$ and $D \in G_{\delta \sigma}$.

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