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No. The intersection of an infinite set of regular languages is not necessarily even computable. The closure of regular languages under infinite intersection is, in fact, all languages. The language of “all strings except s” is trivially regular.
Regular Languages are closed under intersection, i.e., if L1 and L2 are regular then L1 ∩ L2 is also regular. L1 and L2 are regular • L1 ∪ L2 is regular • Hence, L1 ∩ L2 = L1 ∪ L2 is regular.
Regular languages are closed under union, concatenation, star, and complementation.
Let L be a regular language, and M be an NFA that accepts it. Here, δR is δ with the direction of all the arcs reversed. Thus, it is proved that L is closed under reversal.
Intersection is the easiest example to show directly. Finite-state automata are closed under intersection because we can always create a pairwise state representing the operation of both of the original automata, and accept a string only if both automata accept. This effectively runs both automata in parallel.
Closure properties on regular languages are defined as certain operations on regular language that are guaranteed to produce regular language. Closure refers to some operation on a language, resulting in a new language that is of the same “type” as originally operated on i.e., regular.
The closure properties of a regular language include union, concatenation, intersection, Kleene, complement , reverse and many more operations.
Closure under Union For any regular languages L and M, then L ∪ M is regular. Proof: Since L and M are regular, they have regular expressions, say: Let L = L(E) and M = L(F). Then L ∪ M = L(E + F) by the definition of the + operator.
Closure Properties of Regular Languages Given a set, a closure property of the set is an operation that when applied to members of the set always returns as its answer a member of that set. For example, the set of integers is closed under addition.