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Let $L$ be an infinite regular language, then does there exists a strictly locally testable infinite language $P$ such that $P \subseteq L$?

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    $\begingroup$ 0 down vote If $L$ is the empty language you need $P$ empty too. Luckily it is star-free. And works for other $L$. Or is your question different from what I understood? $\endgroup$ – phs Oct 16 '13 at 11:04
  • $\begingroup$ yes, so it is quite trivial, I mean for every infinite $L$ if there exists an infinite $P$ which is locally testable. $\endgroup$ – StefanH Oct 16 '13 at 11:21
  • $\begingroup$ Now for $L=(ab)^*$, any regular $L'\subseteq L$ is some $\{(ab)^i~|~ i\in U\}$ for an ultimately periodic $U\subseteq\mathbb{N}$. It seems that this language is not star-free when $L'$, i.e. $U$, is infinite. $\endgroup$ – phs Oct 16 '13 at 11:41
  • $\begingroup$ @phs $(ab)^+$ is strictly locally testable. $\endgroup$ – J.-E. Pin Oct 16 '13 at 15:44
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The answer is no. The argument was suggested by phs, but you have to start with a different language. Take $L = (aa)^*$. Then any regular language $K$ contained in $L$ is of the form $$ K = \{(aa)^n \mid n \in U \} $$ for some ultimately periodic subset $U$ of $\mathbb{N}$. If $K$ is infinite, then the period $p$ of $U$ is not $0$ and thus $K$ is not star-free (hence not locally testable and not strictly locally testable).

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  • $\begingroup$ thx for your answer. one question related, what about prefix-closed languages, do you have there a counter-example too? $\endgroup$ – StefanH Oct 18 '13 at 14:36

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