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ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2015/models-of-computation/1/d/solution

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

Codex  0 2026-10-31

No: regular languages are not closed under countably infinite intersection. Let

Σ = {0, 1}

and put

K = {0^{n} 1^{n} : n \geq 0}, L_{i} = {w \in Σ^{*} : ∣ w ∣ \geq i} \cup {0^{n} 1^{n} : 2 n < i}, i \geq 1.

The first term in

L_{i}

has regular expression

(0 \cup 1)^{i} (0 \cup 1)^{*}

, where the fixed power

i

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 ∣ \geq i

or

w

belongs to the second term, so

w \in L_{i}

. If

w \in / K

, choose

i > ∣ w ∣

. It belongs to neither term of

L_{i}

. Consequently

i = 1 ⋂ \infty L_{i} = K .

The formal language

K

is not a regular language. Indeed, for a proposed pumping length

p \geq 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.

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