Suppose $A = \{0^{n}1^{n} \mid n \ge0\}$, which is not regular, and let $B$ be a finite subset of $\Sigma^* \setminus A$. Is $A \cup B$ regular?
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Note the closure properties of regular languages. What is $(A \cup B) \setminus B$? What does that imply about $(A \cup B)$ since we know $B$ is regular (due to being finite)?
