Suppose $D$ is a non-empty set and $\{ R\_i : i \in \mathbb{N} \}$ is a family of binary relations on sequences over $D$ so that $R\_i \subseteq D^i \times D^i$. Let $R\_\omega \subseteq D^\omega \times D^\omega$ be a relation that holds between two denumerable sequences $x$ and $y$ over $D$ iff for all $n\in \mathbb{N}$, $x\_1,\ldots,x\_n R\_n y\_1,\ldots,y\_n$. 


Now for every ordinal $n \leq \omega$, there is a Kripke frame $F\_n = (D^n, R\_n)$. I would like to show that any formula of (standard propositional) modal logic is valid in $F\_\omega$ iff it is valid in all $F\_n$ with finite $n$.


For example, clearly $R\_\omega$ is reflexive iff all $R\_n$ with finite $n$ are reflexive, so $\Box p \to p$ is valid in $F\_\omega$ iff it is valid in all $F\_n$ with finite $n$. But is this true for all formulas? If so, how would I go about showing it? 



Validity in Kripke frames whose points are finite or infinite sequences
It is an attractive idea, but unfortunately, it seems not to be true.


The reason is that we can have that every $R\_n$ is nontrivial, in the sense that the relation sometimes holds between different two different sequences, but there is no path through these relations so that $R^\omega$ never holds between two different sequences. For example, let $D=\{0,1\}$, and when $n$ is even, let $R\_n$ be the reflexive lexical order on binary sequences, but when $n$ is odd, let it be the reverse lexical order. Thus, $R\_\omega$ will never hold except reflexively, since the initial segments of a two infinite sequence can't be related lexically in both directions unless they are equal.


In this case, the formula $\Box p\leftrightarrow p$ will be valid in $R\_\omega$, but not in any $R\_n$. 


Another simple counterexample would occur where one $R\_n$, say $R\_{17}$, never holds, but all the other $R\_n$'s always hold. In this case, $R\_\omega$ will never hold, and so it's validities will agree with the validities of $R\_{17}$, but not with any other $R\_n$. 


