Let $A$ be a regular set. Consider the two sets below.

\begin{align*}
L_1 &= \{ x \mid \exists{n}\geq 0 , \exists{y} \in A : y =x^{n} \}, \\
L_2 &= \{ x \mid \exists{n}\geq 0 , \exists{y} \in A : x =y^{n} \}.
\end{align*}

Which of the following is true?

 1. $L_1$ and $L_2$ are regular
 2. $L_1$ is regular but $L_2$ is not
 3. $L_2$ is regular but $L_1$ is not
 4. Both are not regular

**What I knew**

$Y \in A $ only but now the relation between $y$ and $x$ is:  $Y  =  X^{n}$  and given the clause there exist associated with the value of $n$, so we can assign any value of $n$ which is $\geq 0$. So if $n = 1$, then $Y = X$ and hence $X$ is obviously regular set.

Similarly on setting $n = 2$, we get $Y = X^{2}$  meaning $Y$ can be partitioned on exactly 2 halves and hence we can say $X$ is $\mathrm{half}(Y)$. And we know given a language or a set $L$ is regular, then $\mathrm{half}(L)$ is also regular.

With this understanding I came to a conclusion that $L_1$ is regular. 

But, **How to check the regularity of $L_2$?** I also wanted to confirm my approach to $L_1$ being regular is correct.