# Are deterministic Büchi automata omega-closed?

As in, given a regular language $$V$$, does there exist a deterministic Büchi automaton $$\mathcal{A}$$, or equivalently a regular language $$W$$ such that $$\mathcal{L}(\mathcal{A})=\vec{W}=V^\omega$$?

For clarity, if $$L$$ is a finite-word language, then $$L^\omega$$ and $$\vec{L}$$ are resp. the $$\omega$$-closure and the limit of $$L$$; see here.

Edit: My guess is that $$V^\omega=\vec{V^\ast}$$, but I'm unable to (dis)prove it. see comments.

• Regarding your conjecture: it probably is not simple as that. In a paper by Choueka, Theories of automata on $\omega$-tapes, we find the following Lemma 5.2: For every regular $V \subseteq X^*$ one can effectively find some regular $\widetilde V \subseteq X^*$ such that $V^\omega = V^* ( \lim \widetilde V)$. Commented Apr 28, 2022 at 16:34
• Quoting from the same paper, just above the lemma: Certainly one cannot say, in general, that $V^\omega = \lim V$ or $V^\omega = \lim V^*$: the set $V = \{1\}\{0\}^*$ is a counterexample to both assertions, since $10^\omega$ is not in $V^\omega$ although it is in $\lim V$ and in $\lim V^*$. Commented Apr 28, 2022 at 16:39