The set of all recursively enumerable languages is

Final answer: the set of all recursively enumerable languages is closed under intersection.

Construction and reasoning:

• Let M1 and M2 be Turing machines that recognise L1 and L2 (they accept strings in the language and may loop on others).

• Build a machine M that on input w simulates M1 and M2 in parallel (dovetailing their steps). If both simulations have accepted, M accepts.

• Correctness: if w ∈ L1 ∩ L2 then both M1 and M2 will eventually accept, so M will accept. If w ∉ L1 ∩ L2 then at least one simulation never accepts, so M may run forever — which is allowed for a recogniser.

Brief note: This closure under intersection does not imply closure under complementation. Many r.e. languages (for example, the halting set) have complements that are not r.e., so complementation fails and the class is not a subset of recursive languages; also the set of r.e. languages is countable.