Let $\mathcal A = (X, Q, \delta, q_0, F)$ be a deterministic finite automata with the following acceptance condition on infinite words:

The automata accepts $\xi \in X^{\omega}$ with respect to $F$ iff $$ \forall i : \delta(q_0, \xi[0...i]) \in F. $$ Meaning that every prefix $\xi[0...i]$ of $\xi$ goes to an acceptance state. With $L'(\mathcal A)$ I denote the set of all accepted infinite words.

Now such automata have a special form, a word is accepted iff it just passes through certain allowed states in $F$, otherwise it is rejected if it ever enters a state from $Q\setminus F$. This by the way means that if $\mathcal A$ is reduced and complete, it has exactly one non-accepting trapping state $s$, because every word which enters $s$ needs to ultimately stay there, because the word gets never accepted, regardless of what comes there after $s$ was entered.

Now I define $F_n(\xi) := \{ w \in X^* : w \in \mbox{infix}(\xi) \cap X^n \}$, the set of all factors (or infixes) of $\xi$ of length $n$. Then $$K_n(\xi) := \xi[0...n] \cdot X^{\omega} \cap \{ \eta \in X^{\omega} : F_n(\eta) = F_n(\xi) \}$$ is the set of all infinite words which share with $\xi$ a common prefix of length $n$ and which have the same set of factors of length $n$.

Now I conjecture: If $\xi$ is accepted by $\mathcal A$ according to the above mentioned acceptance condition, then there exists a $n > 0$ such that $$ K_n(\xi) \subseteq L'(\mathcal A) $$ meaning that if $\xi$ is accepted there exists a $n > 0$ such that every word which share with $\xi$ a common prefix of length $n$ and all factors of length $n$ (and up to $n$) is also accepted.

Intuitively I guess this is right, because if $\xi$ is accepted, and because $\mathcal A$ is a finite automata, then as $\xi$ goes through the states it eventually ends in some cycle, which I guesss bound the possible factors, and the prefix condition could be used to ensure that another word would end in the same cycle.

If $\xi = uv^{\omega}$ for finite $u, v$, i.e. if $\xi$ is ultimately periodic I guess a construction would be to determine the smallest $k$ such that $F_k(\xi) = F_{k+1}(\xi)$, so the number of factors of length $k' > k$ is the same as $|F_k(\xi)|$, and set $n := k + 1$. Then I conjecture $$ K_{n}(\xi) \subseteq L'(\mathcal A). $$ (I have no proof) Maybe this works also for non-ultimately periodic words, but I could not proof it. So any ideas? Or maybe an idea how to proof my conjecture?