We denote $\Sigma:=\{1,\ldots, k\},$ and also denote by $\Sigma_{n}$ the set of words with the length $n$. Let $\Sigma*$ be the set of words over $\Sigma$, i.e., $\Sigma*=\cup_{n}\Sigma_{n}.$
Let $T:\Sigma^{\mathbb{Z}} \to \Sigma^{\mathbb{Z}}$ be a full shift. Assume that $\Sigma' \subset \Sigma^{\mathbb{Z}}$ is a compact and $T-$invariant set and $f:\Sigma' \to \mathbb{R}$ is uniformly continuous.
An $\textit{infinite tree}$ $t$ over a finite alphabet $A$ is a function from $\Sigma∗$ to $A.$ We denote $A^{\mathbb{Z}}$ the set of all infinite trees on $A$.
$$Question:$$
Can we find an alphabet $A$ such that $\Sigma' \subset \Sigma''$ and $f:\Sigma''\to \mathbb{R}$ is uniformly continuous, where $\Sigma'':=A^{\mathbb{Z}}$?In other words, I want to know whether there exists an alphabet $A$ such that every element of $\Sigma'$ can be decomposed into words from $A$ or a similar set.