This question concerns the closure properties of regular languages (sets). We need to identify the statement that is incorrect regarding operations on regular sets.

Analyzing Regular Set Properties

Regular languages possess specific closure properties, meaning that if you perform certain operations on regular languages, the result is also a regular language. Let's examine each option:

Option 1: Intersection - $R_1 \cap R_2$

The statement claims "$R_1 \cap R_2$ needs not be regular". This is incorrect. A fundamental property of regular languages is that they are closed under intersection. If $R_1$ and $R_2$ are both regular sets, their intersection, $R_1 \cap R_2$, is guaranteed to be regular. This can be proven using state minimization techniques on Finite Automata or using closure proofs involving other operations.

Option 2: Complementation - $\Sigma^* - R_1$

The statement claims "$\Sigma^* - R_1$ is regular". This is correct. The set difference $\Sigma^* - R_1$ represents the complement of the regular set $R_1$ (denoted as $\overline{R_1}$). Regular languages are closed under complementation. If $R_1$ is a regular language over alphabet $\Sigma$, its complement $\overline{R_1}$ is also regular.

Option 3: Union - $R_1 \cup R_2$

The statement claims "$R_1 \cup R_2$ is regular". This is correct. Regular languages are closed under the union operation. If $R_1$ and $R_2$ are regular sets, their union $R_1 \cup R_2$ is also a regular set.

Option 4: Kleene Star - $R_1^*$

The statement claims "$R_1^*$ is regular". This is correct. Regular languages are closed under the Kleene star operation. If $R_1$ is a regular set, then the set of all possible concatenations of strings from $R_1$, denoted by $R_1^*$, is also a regular set.

Conclusion

Based on the closure properties, the statement that is not correct is that the intersection of two regular sets ($R_1 \cap R_2$) might not be regular. It is always regular.