Past exam of the computer science course of the University of Oxford / 2015 / Models of computation / 1 / d

No: regular languages are not closed under countably infinite intersection. Let $\Sigma =\{0,1\}$ and putThe first term in $L_i$ has regular expression $(0\cup 1{)}^i(0\cup 1{)}^{\ast }$, where the fixed power $i$ means $i$ copies. The second term is a finite language, a finite union of single words. Their union is therefore a regular language.

If $w\in K$, then for each $i$ either $|w|\ge i$ or $w$ belongs to the second term, so $w\in L_i$. If $w\notin K$, choose $i>|w|$. It belongs to neither term of $L_i$. ConsequentlyThe formal language $K$ is not a regular language. Indeed, for a proposed pumping length $p\ge 1$, the pumping lemma for regular languages applied to $0^p{1}^p$ must pump a nonempty block consisting only of zeros. Repeating that block increases the number of zeros without increasing the number of ones, leaving $K$. This contradiction proves the counterexample. Finite intersection closure does not extend to this infinite intersection of regular languages.