Omega-regular language

Remarks

Example: over $\set{a, b}$, the set of infinite words with infinitely many $a$'s is omega-regular, as $(b^* a)^\omega$ with $n = 1$, $U_1 = \set{\varepsilon}$ and $V_1 = b^* a$. Non-example: $L = \set{a^k b^k c^\omega : k \ge 0}$ over $\set{a, b, c}$ is not omega-regular. For an omega-regular $L$, the set $\set{u \in \Sigma^* : u c^\omega \in L}$ is regular, and here it would be $\set{a^k b^k : k \ge 0} \cup \set{a^k b^k c^m : k \ge 0,\ m \ge 1}$, whose intersection with $a^* b^*$ is not regular. By a theorem of Büchi, a language $L \subseteq \Sigma^\omega$ is omega-regular if and only if it is accepted by a nondeterministic Büchi automaton. The condition $V_i \subseteq \Sigma^+$ excludes the empty word, for which an infinite repetition would not give an infinite word.