Let A = {0,1} and L = A*. Let R = {0ⁿ1ⁿ, n > 0}. The languages L∪R and R are respectively

Since $R = \{0^n 1^n \mid n > 0\}$ is a subset of $L = \{0, 1\}^*$, their union is $L \cup R = L = \{0, 1\}^*$, which is a Regular language. However, $R = \{0^n 1^n \mid n > 0\}$ itself requires unbounded counting and is NOT regular.