Find languages A and B such that $A^* \cup B^* \neq (A \cup B)^*$.
Is this even possible?
I tried:
$A:\{$ $\epsilon $ $\}$ and B:$\{$ $1$ $\}$
$A^*= \{\epsilon \}$ and $B^*=\{\epsilon,1,11,111,....\}$
$A \cup B = \{ \epsilon , 1\}$
$A^* \cup B^*= \{\epsilon, 1, 11, 11,111,.....\}$
$(A \cup B)^*=\{\epsilon, 1, 11, 11,111,.....\}$
because lets say $C=(A \cup B)$ and $C=\{ \epsilon,1 \}$, so $C^*=\{\epsilon, 1, 11, 11,111,.....\} $ which is exactly not what I want.
Problem I have is whatever $A \cup B$ outputs, they will get the kleene star outside of the parenthesis which will basically make it the same as the union of kleene stars. Also kleene star will add $\epsilon$ to the resulting languages so I can't trick it by using that. Is my logic flawed?
If $A=\{a\}, B=\{b\}$ then $ab \in (A \cup B)^\ast$ but $ab \notin A^\ast \cup B^\ast$.