Question:
$R_1$ and $R_2$ are regular sets. Which of the following is not correct ?
$R_1$ and $R_2$ are regular sets. Which of the following is not correct ?
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Regular languages are closed under almost all basic set operations. "Needs not be regular" is usually a red flag when discussing Regular Sets.
Updated On: Aug 6, 2026
- $R_1 \cap R_2$ needs not be regular
- $\Sigma^* - R_1$ is regular
- $R_1 \cup R_2$ is regular
- $R_1^*$ is regular
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The Correct Option is A
Solution and Explanation
Concept:
• Regular sets (or Regular Languages) are the simplest class of languages in the Chomsky hierarchy.
• They are defined by the fact that they are accepted by Finite Automata.
• One of their defining characteristics is closure under various algebraic operations.
Step 1: Examine Closure under Union and Kleene Star
By definition, the set of regular languages over an alphabet is closed under the operations used to build regular expressions.
This includes Union (\(R_1 \cup R_2\)) and Kleene Star (\(R_1^*\)).
Therefore, options (C) and (D) represent correct statements about regular sets.
Step 2: Examine Closure under Complementation
Regular languages are closed under complementation.
If a language \(L\) is regular, there exists a Deterministic Finite Automaton (DFA) that accepts it.
By swapping the final and non-final states of that DFA, we obtain a DFA that accepts \(\Sigma^* - L\).
Therefore, option (B) represents a correct statement.
Step 3: Examine Closure under Intersection
Regular languages are closed under intersection.
This can be proven via the "Product Automaton" construction or via De Morgan's Law:
\(R_1 \cap R_2 = \overline{\overline{R_1} \cup \overline{R_2}}\).
Since they are closed under union and complement, they must be closed under intersection.
The statement in (A) says it "needs not be regular", which is false because it is always regular.
• Regular sets (or Regular Languages) are the simplest class of languages in the Chomsky hierarchy.
• They are defined by the fact that they are accepted by Finite Automata.
• One of their defining characteristics is closure under various algebraic operations.
Step 1: Examine Closure under Union and Kleene Star
By definition, the set of regular languages over an alphabet is closed under the operations used to build regular expressions.
This includes Union (\(R_1 \cup R_2\)) and Kleene Star (\(R_1^*\)).
Therefore, options (C) and (D) represent correct statements about regular sets.
Step 2: Examine Closure under Complementation
Regular languages are closed under complementation.
If a language \(L\) is regular, there exists a Deterministic Finite Automaton (DFA) that accepts it.
By swapping the final and non-final states of that DFA, we obtain a DFA that accepts \(\Sigma^* - L\).
Therefore, option (B) represents a correct statement.
Step 3: Examine Closure under Intersection
Regular languages are closed under intersection.
This can be proven via the "Product Automaton" construction or via De Morgan's Law:
\(R_1 \cap R_2 = \overline{\overline{R_1} \cup \overline{R_2}}\).
Since they are closed under union and complement, they must be closed under intersection.
The statement in (A) says it "needs not be regular", which is false because it is always regular.
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