Is there a prefix-continuous bijection between finite words and eventually zero words?

Question

Let $$ X = \{x \in \{0,1\}^{\omega} \;|\; \exists m: \forall i \geq m: x_i = 0\} $$ (one-way infinite eventually zero words). Let $\{0,1\}^*$ denote the finite (not necessarily nonempty) words over $\{0,1\}$, and write $\{0,1\}^{\leq k} = \{w \in \{0,1\}^* \;|\; |w| \leq k\}$ where $|w|$ denotes length.

Is there a bijection $\phi : X \to \{0,1\}^*$ such that $$ \exists n \in \mathbb{N}: \forall a \in \{0,1\}: \forall x \in \{0,1\}^{\mathbb{N}}: \exists b, c \in \{0,1\}^{\leq n}: \exists y \in \{0,1\}^*: \phi(x) = b \cdot y \wedge \phi(a \cdot x) = c \cdot y $$ holds, where $\cdot$ is concatenation?

This is a kind of coarse uniformity / bornologousness assumption: $\phi$ needs to be bornologous between the two sets, seen as metric spaces with the path metric of the graph structure where $x$ and $y$ are adjacent if $y = ax$ or $x = ay$ for some $a \in \{0,1\}$. This seems vaguely familiar to me but I don't know from where, and I'm not seeing how to construct $\phi$. The straightforward idea of cutting out the zero tail doesn't work because it's not surjective, and I run into trouble trying to fix that. But I also didn't manage to prove impossibility because there's a lot of freedom.

The question arises in some (leisurely) research, so asking here instead of math.SE even if it might be safer to start there with this one. Geometric group theory tag because this is related to Thompson's $V$, even if I didn't elaborate and I doubt it's useful (every countable group acts freely on $\{0,1\}^*$).

Answer 1

I believe the following works, but I might be missing something.

If $x$ has at least two 1s, then $\phi(x)$ is the sequence cut just before the last 1: $$\phi(0011101000\cdots) = 001110 $$

If $x$ has at most one 1, then $\phi(x)$ is the sequence cut before the one, with an additional zero: $$ \phi(000\cdots) = \varnothing,\qquad \phi(001000\cdots) = 000 $$

The inverse bijection is $\psi(\varnothing)=000\cdots$, $\psi(y)=y\cdot1000\cdots$ when $y$ has at least one 1, and $\psi(y\cdot0)=y\cdot 1$ for $y$ consisting only of zeros (possibly $y=\varnothing$).

Then $\phi(a\cdot x)=a\cdot\phi(x)$ if $x$ has at least two 1s, and the case of $000\cdots$ is unimportant, up to taking a large $n$ (although $n=1$ works so far). Now if $x$ consists only of zeros, $$\phi(0\cdot y\cdot 1000\cdots)=0 \cdot y\cdot 0=00\cdot y\qquad\text{ and }\qquad\phi(1\cdot y\cdot 1000\cdots)=1\cdot y$$ while $\phi(y\cdot1000\cdots)=y\cdot0=0\cdot y$.

So the hypotheses are satisfied with $n=1$.