Here we are concerned with the space $X^{\omega}$ of infinite sequences. Denote by $F\_n(\xi)$
the set of factors (consecutive finite subsequences) of length $n$ and consider the set
$$
 K\_n(\xi) = \xi[1\ldots n] \cdot X^{\omega} \cap \{ \eta \in X^{\omega} : F\_n(\eta) = F\_n(\xi) \}
$$
of all words which share with $\xi$ the first $n$ symbols and all factors of length $n$. Then I want to show that there does not exists a finite number of sets $X\_1, X\_2, \ldots, X\_m$ which represent these sets in the sense that for every $\xi$ and $n$ every set $K\_n(\xi)$ could be written in the form $w \cdot X\_i$
for some $i$ and some $w\in X^\*$, i.e. for each $K\_n(\xi)$ there is some $i$ and a finite sequence $w$ with
$$
 K\_n(\xi) = w \cdot X\_i.
$$
I tried to prove that a finite number is not enough, but run out of ideas. Any ideas or suggestions? (A note which might be helpful, the $K\_n(\xi)$ are $\omega$-regular sets, meaning there exists a finite automaton accepting them, and so contain an ultimately periodic infinite sequence, and two such sets are equal if they coincide in all there ultimately periodic sequences)



Representability of sets of infinite sequences sharing common prefixes and factors (i.e. infixes)
If I get your question right, you are given an infinite word $\xi$ of $X^\omega$ and a fixed integer $n$. Let $p$ be the prefix of length $n$ of $\xi$ and let $F$ be the set of all infixes of length $n$ of $\xi$. Let 
$$
 K = \{ \eta \in X^\omega \mid F\_n(\eta) = F\}
$$
Then $K$ is $\omega$-regular and 
$$
K\_n(\xi) = pX^\omega \cap K = p(p^{-1}K)
$$
where $p^{-1}K$ is the $\omega$-regular set $\{\eta \mid p\eta \in K \}$.
Now, for a fixed $n$ and a finite alphabet $X$, there are only finitely many possible values for $p$ and for $F$ (and hence for $K$ and $p^{-1}K$).


