#Union of Regular Lang

Question

Given that a language L = L1 ∪ L2, where L1 and L2 are two other languages. If L is known to be a regular language, then which of the following statements is necessarily TRUE?
(a) If L1 is regular then L2 will also be regular
(b) If L1 is regular and finite then L2 will be regular
(c) If L1 is regular and finite the L2 will also be regular and finite
(d) None of these

Comments

Let $\sum =\left\{a,b\right\}$

(a) & (c).False

$L_1=(a+b)^{*} ,L_2=a^{n}b^{n} \Rightarrow L=(a+b)^{*} $

here $L_2$ is non-regular

$b$ should be correct.

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They are asking necessarily true according to me its option D.... please clarify!

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to make necessarily true FALSE , you need $1$ case to prove it false.

i made $a$,$c$ false by just providing $1$ single example .

however $b$ will always be true.

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@Anand. I didn't understand your explanation. Can you please elaborate.

Answer 1

L=regular

let l1=$a^{*}b^{*}$ and l2=$a^{n}b^{n}(n>0)$ there union will give the language=($a^{*}b^{*}$) then for above question i think option (b) is suitable answer.

Answer 2

(b) is the answer because we cannot make an irregular set S regular, by adding a finite number of elements to it. This can be proved as follows:

Let R ∪ F = T be regular, where F is a finite set.

R∪(F−R)=T (both the sets are now disjoint)

⟹R=T−(F−R)⟹R=T∩(F−R)′
Now T is regular set and F is a finite set and so F-R must also be finite and hence regular also. Regular set is closed under complement and intersection which makes R also regular.


We can prove (a) false be taking L1=Σ∗ and L2={anbn∣n>0}. Now, union of these is Σ∗ which is regular but L2 is not regular.

(c) can be proved false by taking L1={a} and L2={an∣n>0}. Now L1∪L2 is regular and L1 finite but L2 is not finite.