+16 votes
1.9k views

Which of the following statements is false?

1. Every finite subset of a non-regular set is regular
2. Every subset of a regular set is regular
3. Every finite subset of a regular set is regular
4. The intersection of two regular sets is regular

edited | 1.9k views
0
Is non-regular set closed under intersection.

## 3 Answers

+21 votes
Best answer

(b) is False. Any language is a subset of $\Sigma^*$ which is a regular set. So, if we take any non-regular language, it is a subset of a regular language.

(a) and (c) are regular as any finite language is regular.

(d) is regular as regular set is closed under intersection.

by Veteran (434k points)
edited
0
example of option B)a^nb^n is non regular subset of regular (a/b)*.
+2 votes
B is false because if u take a*b* as an example a^n b^n is a subset of a*b* but its not regular
by (31 points)
–2 votes

(c) Every finite subset of a regular set is regular this is false

example:a^n b^n from regular set (a+b)* is not regular

by Boss (14.4k points)
0
No. The set you gave is infinite. Any finite set is trivially regular as there are only finite number of strings. (Finite class of languages is a subset of regular class).

Your example is for (b) and that is the false statement.
+3
sorry for it.it  is a silly mistake i thought it is a very easy question so i just took any option infront of me .it happens with me at times

(b) Every subset of a regular set is regular this is false
Answer:

+15 votes
5 answers
1
+24 votes
7 answers
2
+14 votes
2 answers
3
+12 votes
2 answers
4