## Expert Answer

- The concatenation of two non regular languages can be regular , It can be proved as , Find the two non regular languages , for example L
_{0}= {a^{x}b^{y}| x≠y } and L_{1}= {b^{q}a^{p}| q≠p } ,

The concatenation of two languages can be wriiten as, L_{0}L_{1} = { a^{x}b^{y+q}a^{p} | x≠y ∧ q≠p}

The concatenation can be proved by following conditions ,

when x=0,p=0 : y+q≥2

when x=0,p≠0 : y+q≥1

when x≠0,p=0: y+q≥1

when x=1,y+q≠1: p=1

Or else: y+q≥0

Hence the union of two regular languages is closed and concatenated L_{0}L_{1} is regular.

- To prove that there are a countably infinite number of examples , Consider 5 set of languages P1 to P5 , It is obvious to say P1 to P5 are countably infinite , which can be proved by a set P for which exists , P={L|L=L
_{1}∪L_{2}∪L_{3}∪L_{4}∪L_{5},L_{1}∈P_{1}∧L_{2}∈P_{2}∧L_{3}∈P_{3}∧L_{4}∈P_{4}∧L_{5}∈P_{5}} , The cross product of P_{1}to P_{5}is the cardinality set of P , and it is countably infinite set , Thus amount of such examples of countably infinite exists . - Consider n and m be the set of P
_{1}to P_{5}are numbers in natural numbers set , therefore it consists of infinite number of countably infinite sets , and the cross product of these countable sets is uncountably infinite , Hence uncountable infinite amount of such exmaples exists.