Is it possible to have languages A and B where A*⊄B* and B*⊄A* such that (A∪B)* = A*∪B*? (Specific cases)
Below is the closest I have to being correct.
A = {ε} and B = {1} - This doesn't work because B*={ε, 1, 11, 111, ...} and A*={ε,εε,εεε}={ε} and A*⊂B*.
I am assuming that strict subset is a hint here, but I can't seem to figure it out. Any help or insight?
Let $X = \{0,1\}$ and $A = \{ 1^m01^n \mid n, m > 0 \}, B = \{ 1 \}$. Then $$ A^{\ast} = \{ 1^{n_1}01^{n_2}0 \cdots 01^{n_k} \mid k > 1, n_1, n_k > 0 \mbox{ and } n_2, \ldots, n_{k-1} > 1 \} $$ and $$ B^{\ast} = \{\varepsilon, 1, 11, 111, \ldots \}. $$ For this languages all your requirements are fulfilled and we have $$ (A \cup B)^{\ast} \subseteq A^{\ast} \cup B^{\ast} $$ as could be easily checked.