Subsets of $\omega$-words which share certain factors and languages accepted by special (prefix-closed) automata

Question

Let $\mathcal A$ be an automaton, then I define the following $\omega$-language accepted by $\mathcal A$: $$ L'(\mathcal A) := \{ \eta \in X^{\omega} : v \sqsubset \eta \mbox{ implies } v \in L(\mathcal A) \} $$ where $v \sqsubset \eta$ means $v$ is a finite prefix of $\eta$ and $L(\mathcal A)$ denotes the finite (regular) language accepted by $\mathcal A$.

For $k > 0$ denote by $PF_k(\eta)$ the set of all $\omega$ words $\xi$ such that the first $k$ letters of $\eta$ and $\xi$ are the same (they share a common prefix of length $k$) and the infixes (or factors) of length $k$ are the same, i.e. if $F_k(\eta)$ denotes the factors of length $k$ then $PF_k(\eta) = \{ \xi : \eta[0...k] = \xi[0...k], F_k(\eta) = F_k(\xi) \}$.

Now I want to show, if $\eta \in L'(\mathcal A)$, then there exists a $k > 0$ such that $PF_k(\eta) \subseteq L'(A)$. First I thought that this does not hold, so that for every $\eta \in L'(A)$ and $k > 0$ you can find some $\xi \notin L'(A)$ such that $\xi \in PF_k(\eta)$, but didn't succeeded in constructing a counter-example, so I guess it holds but I have no idea how to proof it?

Answer 1

The conjecture does not hold:

Let $L$ be the set of prefixes of $(c^*ac^*b)^*$.

Then $L'=(c^*ac^*b)^\omega+ (c^*ac^*b)^*c^\omega+(c^*ac^*b)^*c^*ac^\omega$.

Take the word $\eta=(c^1ac^1b)(c^2ac^2b)(c^3ac^3b)\dots \in L'$.

For all $k\in\mathbb N$, we can show that $PF_k(\eta)\not\subseteq L'$, as witnessed by $$u_k=(c^1ac^1b)(c^2ac^2b)\dots (c^kac^kb)c^kb c^\omega.$$

Indeed we have $u_k\in PF_k(\eta)$ but $u_k\notin L'$.