Q: Show that $\sum^* = L_1\sum^*L_2$
if $e\in L_1 \subseteq \sum^*$ and $e\in L_2 \subseteq \sum^*$ for any alphabet $\sum$.
In these equivalence proofs, one way is usually easy. I guess it is RHS to LHS in this case.
Suppose $w \in L_1\sum^*L_2$. Then $w = xyz$ for any three strings belonging to $L_1$, $\sum^*$, $L_2$. But since $L_1, L_2 \subseteq \sum^*$, we can say that $w = xyz$ belongs to $\sum^*\sum^*\sum^* = \sum^*$.
So $L_1\sum^*L_2 \subseteq \sum^*$. This completes the one way.
Now we want to show $\sum^* \subseteq L_1\sum^*L_2$. Let's do induction on the string length.
BASIS STEP: $|w| = 0$ corresponds to the empty string, which is of course a member of $L_1\sum^*L_2$. This completes the basis step.
INDUCTIVE STEP: Suppose $\sum^* \subseteq L_1\sum^*L_2$ for strings of length $\leq n$. Now form the string $w = ua$ where $u$ is the string of length $n$ and $a \in \sum$. Since both strings have lengths $\leq n$, by induction hypothesis we have $w \in L_1\sum^*L_2$. This completes the inductive step.
I feel like I am doing something wrong in the proof by induction, because it was way too easy. I didn't even use any other fact than the induction hypothesis.
I'd be glad to learn the correct solution to this problem, and also learn what I've done wrong.
Induction is overkill.
Take $x \in \Sigma^\ast$. Then $x = exe$. And $exe \in L_1 \,\Sigma^\ast \,L_2$, as $e \in L_1$ and $e \in L_2$. This proves the other inclusion.
On the induction proof itself: in the base step with the empty string you do use the assumption that $e \in L_1, e \in L_2$, albeit implicitly (make it explicit!). The induction step itself is unclear to me: we have $w = ua$, so by assumption $w \in L_1 \Sigma^\ast L_2$ and $u \in L_1 \Sigma^\ast L_2$. But how then does it follow that $ua$ is in this concatenation as well? Say: $u = l_1 s l_2, a = l'_1 s' l'_2$ where $l_1, l'_1 \in L_1$, $l_2, l'_2 \in L_2, s,s' \in \Sigma^\ast$, then $w = ua = l_1 s l_2 l'_1 s' l'_2$, which is in $L_1 \Sigma^\ast L_2$ how? Not quite clear. Even choosing $l'_1 = e, l'_2 = e, s' = a$, as we can, does not quite solve that, because then you're left with $w = l_1 s l_2 a$, and how to get rid of the $a$? So I don't think induction can really work here. Induction is a nice tool, but if all you have is a hammer...