Regularity of a language

Question

$$ L = \{w \mid w \text{ does not contain } 000\} $$
$$ L_2 = \{w \mid xwy \in L \text{ for some } x,y \in (0+1)^*\} $$
Is $L_2$ regular? I am thinking regular language is closed under concatenation, but it seems that this language should be irregular. And I have no idea how to prove this, how do I apply pumping lemma or is there a counter example?

Answer 1

$L_2$ is regular, because $L_2 = L_1$ (and $L_1$ is regular).

($L_2 \subseteq L_1$) If $w\in L_2$, then there are $x,y$ such that $xwy\in L_1$, so $xwy$ does not contain $000$. But then $w$ can't contain $000$, so $w\in L_1$.

($L_1 \subseteq L_2$) If $w\in L_1$, then does not contain $000$, so if $x=y=\varepsilon$, then $xwy = w \in L_1$, thus $w\in L_2$.

Answer 2

Actually, you have $L = L_2$.

Let $w\in L$. Then $w\in L_2$, as $\varepsilon w \varepsilon \in L$.

Let $w\in L_2$. Then, let $x$ and $y$ be such that $xwy\in L$, i.e. $xwy$ does not contain $000$. But if $w$ contained $000$, then $xwy$ would too. Thus $w\in L$.