Büchi automaton with modified acceptance condition

Question

Consider a Büchi automaton $\mathcal{A}$ with the modified acceptance condition, that an $\omega$-word $\mathcal{w}$ is accepted by $\mathcal{A}$ iff every run $\rho$ of $\mathcal{A}$ on $\mathcal{w}$ is accepting (rather than at least one run being accepting). I need to show that this automaton also accepts only $\omega$-regular languages. How do I go about doing that?

Answer 1

Let $S$ and $F$ be the set of all states and all accepting states of $\mathcal{A}$, respectively. For $t \in S \setminus F$, let $\mathcal{A}$ be the automaton obtained by making $t$ the only accepting state. For $t \in S \setminus F$, let $\mathcal{B}_t$ be the automaton obtained by removing all states in $F$, and making $t$ the starting state. We think of $\mathcal{A}_t$ as an NFA, and of $\mathcal{B}_t$ as a Büchi automaton. The following language is $\omega$-regular: $$ L = \overline{\sum_{t \in S \setminus F} L(\mathcal{A}_t) L(\mathcal{B}_t)}. $$ An $\omega$-word $w$ is not in $L$ if and only if on some run of $\mathcal{A}$, no accepting state appears infinitely often. Therefore an $\omega$-word $w$ is in $L$ if and only if on all runs of $\mathcal{A}$, some accepting state appears infinitely often.