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Is infinite union of non-regular languages $L_i$ that form a chain such that $L_i\subseteq L_{i+1}$ always non-regular?

Or is there a possibility that it be ever regular?

Is there an easy way to see this?

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  • $\begingroup$ I'm certain we have a duplicate of this. $\endgroup$ – Raphael Nov 3 '15 at 19:04
  • $\begingroup$ @Raphael The question has changed. $\endgroup$ – Yuval Filmus Nov 3 '15 at 20:24
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    $\begingroup$ Consider $L_i=\{w\in\{a,b\}^*\ |\ |w|_a-|w|_b\le i\}$. $\endgroup$ – Klaus Draeger Nov 3 '15 at 20:54
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Let $L$ be a non-regular language over $\Sigma$, and let $w_1,w_2,\ldots$ be an enumeration of $\Sigma^*\setminus L$. Define $L_i = L \cup \{ w_j : j \geq i \}$. Each $L_i$ is non-regular (why?), $L_i \subset L_{i+1}$, and $\bigcup_i L_i = \Sigma^*$.

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  • $\begingroup$ So in general kleene star of a non-regular language is regular? $\endgroup$ – T.... Nov 3 '15 at 20:04
  • $\begingroup$ No, for example $\{0^n1^n:n\geq0\}^*$ is not regular (why?). $\endgroup$ – Yuval Filmus Nov 3 '15 at 20:05
  • $\begingroup$ Since it contains $0^n1^n0^m1^m$ $\endgroup$ – T.... Nov 3 '15 at 20:08
  • $\begingroup$ Since it contains $\mathsf{only}$ strings of form $0^n1^n0^m1^m\dots$? $\endgroup$ – T.... Nov 3 '15 at 20:18
  • $\begingroup$ Pumping lemma? Take $xyz = 0^n1^n$ with $y=0$ and pump $y$? $\endgroup$ – T.... Nov 3 '15 at 20:26

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