Consider the following statements. I. If $L_1 \cup L_2$ is regular, then both $L_1$ and $L_2$ must be regular. II. The class of regular languages is closed under infinite union. Which of the above statements is/are TRUE?

• Statement I is FALSE: The union of two non-regular languages can be regular. For example, let $L_1 = \{a^n b^n \mid n \ge 0\}$ (non-regular) and $L_2 = \Sigma^* \setminus L_1$ (non-regular). Then $L_1 \cup L_2 = \Sigma^*$, which is regular, but neither $L_1$ nor $L_2$ is regular.

• Statement II is FALSE: Regular languages are closed under *finite* union, but not *infinite* union. Every non-regular language $L$ is the infinite union of singleton languages containing its strings: $$L = \bigcup_{w \in L} \{w\}$$ Since each singleton set $\{w\}$ is regular, closure under infinite union would imply that every language is regular, which is a contradiction.

Therefore, neither I nor II is true, making (D) the correct answer.