A language $L \subseteq X^*$ is non-counting of order $n > 0$ iff for all $u,v, w \in X^*$$$ uv^nw \in L \Leftrightarrow uv^{n+1} w \in L.$$A $\omega$-language (set of infinite sequences) $L \subseteq X^{\omega}$ is non-counting of order $n > 0 $ iff for all $u,v \in X^*, \eta \in X^{\omega}$$$ uv^{n}\eta \in L \Leftrightarrow uv^{n+1} \eta \in L.$$When talking about languages over finite words, i.e. subsets $L$ of $X^*$ every regular, noncounting languages of order $n > 0$ fullfills the following property: Let $u,v \in X^*$ with$$ P_n(u) = P_n(v), I_n(u) = I_n(v), S_n(u) = S_n(v)$$and $u \in L$ (where $P_n, S_n, I_n$ denote the prefix, suffix and infixes of length $n$) then also $v \in L$, meaning if two words coincide in their prefix, suffix and infixes of fixed length $n > 0$ then they are either both in $L$ or both not in $L$.Now when considering subsets of infinite words (and so just prefix and infixes), i.e. $L \subseteq X^{\mathbb N}$, then this does not hold any longer. For consider $L = X^* 0^{\omega}$, then $L$ is regular and non-counting for each $n > 0$. Fix some $n > 0$ and consider$$ \xi = 0^n1^n0^{\omega} \quad \mbox{ and } \quad \eta = 0^n1^n0^n1^{\omega}$$then $P_n(\xi) = P_n(\eta)$ and $I_n(\xi) = I_n(\eta)$ but $\xi \in L$ and $\eta\notin L$.The space $X^{\omega}$ is naturally topologized as the Cantor space, i.e. the space with the common-prefix-metric, or equivalently generated by the basis $v\cdot X^{\omega}$, $v \in X^*$. Now the set $L = X^* 0^{\omega}$ is not closed in Cantor space, for example $1^n0^{\omega}$ is in $L$, but its limit $1^{\omega}$ is not in $L$. So I conjecture that if $L$ is closed then the property from finite sets also hold for infinite sequences, i.e. if $L$ is regular and non-counting, then if two words coinside in their prefix and infixes of some specified length, then they are either both in $L$ or not both. A language $L$ is closed in Cantor space iff if it has the property that if $\xi$ is such that every prefix of $\xi$ is the prefix of some words from $L$, then $\xi \in L$.I conjecture that for closed sets $L$, that if $L$ is regular and non-counting, then membership could be decided by comparing prefix and infixes, but I am stuck with the proof, so any hints or suggestions for me?EDIT: There is a certain problem with the way it is stated, instead of locally testable use star-free, and instead of just presuppose a set to be closed I will presuppose it to be open or closed, so that the proposition is stable under complementation (which are the star-free and non-countable languages), see the comments.
Transfering properties from subsets of $X^*$ to subsets of $X^{\omega}$ by using the topology induces by Cantor space
fl.formal languages;automata theory;regular language;topology