Find languages L1 and L2, neither of which contains the other, such that (L1* ∪ L2*) = (L1 ∪ L2)*.

Question

I'm trying solve this question in several ways, but only textbook has not helped me alot.

Answer 1

With a one-symbol alphabet $\Sigma = \{a\}$, let

$$ \begin{align} L_1 &= \{a^{2n+1} \mid n \in \mathbb{N}\} \\ L_2 &= \{a^{2n} \mid n \in \mathbb{N}\} \text{.} \\ \end{align} $$

Then $L_1 \nsubseteq L_2$ and $L_1 \nsubseteq L_1$, in fact $L_1 \cap L_2 = \emptyset$.

We also have $L_1^{*} = \Sigma^{*}$, and $L_2^{*} = L_2$.

So: $$ L_1^{*} \cup L_2^{*} = \Sigma^{*} \text{.} $$ Clearly, $L_1 \cup L_2 = \Sigma^{*}$, so: $$ (L_1 \cup L_2)^{*} = \Sigma^{*} \text{.} $$