Question

Why is $a^{n}b^{n} \cup a^{*}b^{*}$ regular ?

Does this imply that a subset of non regular language can be regular ?

Answer 1

It is necessary to understand  the set of strings that are accepted by both the languages.

a^n b^n = {∈, ab,aabb, aaabbb, aaaabbbb, aaaaabbbbb,...........}

a*b*= {∈, a, b, aa, ab, aab, aabb, abbb, aabbbb,aaabbb,aaabb, aaaabbbb,........}

Clearly former one is the subset of later one.

So when union is done on the two languages, the resultant language is the later one i.e. a*b*, which is a regular language. Hence the union of the two is a regular language.

The empty language is a regular subset of any language at all.

More generally, every finite subset of any language is regular.

e.g Let L be the language {a^n|n is prime}={aa,aaa,aaaaa,aaaaaaa,…}.

Let R be the subset of L consisting just of {aa, aaaaa,aaaaaaaaaaa}.

Then L is not regular but R is.

Comments

here you are saying union is a regular language,and it is known that union of two regular  languages is always regular .so if the answer is regular then the two languages should be regular but the first language a^nb^n is not a regular language ,it is a context free language.

Answer 2

a*b* U a^nb^n = a*b*. That's why it is regular.

No it does not imply that a subset of non regular language can be regular.

It implies that subset of regular language can be non regular !

In fact every non regular language is subset of regular Language Σ*  !

Comments

i think finite subset of any language whether it is regular or not regular is always regular