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  1. If all finite subsets of LL are regular, then LL is regular.
  2. If a proper subset of LL is not regular, then LL is not regular.
  3. Subsets of finite sets are always regular.
  4. Subsets of finite sets are always regular.
  5. Every subset of language is regular than L is regular.

1 Answer

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If a proper subset of L is not regular, then L is not regular.

 ∑* is a universal language and it is RL.

Any language is subset to ∑* .

take L = {a$^n$.b$^n$ | n > 0 } ==> it is non-Regular, but it is subset to Regular

∴ Given statement is False.

 

Subsets of finite sets are always regular.

We know that every Finite set is regular ===> given statement is true.

 

If all finite subsets of L are regular, then L is regular.

False., finite subsets are always regular but we can't judge L depend upon this.

 

Every subset of language is regular than L is regular.

Let take L is Non-Regular language, then it must be an infinite language. ==> it have infinite number of subsets.

In those subsets, atleast one subset is Non-Regular to make L is Non-regular, otherwise How L is NON-Regular ?

So, if L is NON-REGULAR, there exist atleast one subset which is Non-Regular.

if all subsets are regular, then it must be not a non-regular language ==> it must be a Regular language.

 

One can think like, for some languages, there are infinite subsets.

if all these are RL's, then we make it as, union of all these subsets.

Then it is representation the statement " Infinite union of Regular languages " --- which is not closed under Regular languages.

So, given statement is false. But note that, Not closed doesn't mean always gives Non-RL.

i mean to say, CFL are not closed under intersection but there are some CFL's which intersection also lead to a CFL.

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