I have seen the propity of $L_1\cup L_2$ is a regular language if $L_1$ & $L_2$ are regular languages.
But it works backwards? If i have a regular language L, always exist two others regular languages that together makes L?
Can i have a demostration?
Corrected and expanded 15 January 2024.
Provided that $|L|>1$, we can always find disjoint, non-empty languages $L_0$ and $L_1$ such that $L=L_0\cup L_1$. Every finite language is regular, and regular languages are closed under taking relative complements, so we may take $L_0$ to be any non-empty proper subset of $L$ and let $L_1=L\setminus L_0$.
However, if $L$ is infinite it is actually possible to choose $L_0$ and $L_1$ both to be infinite as well. Let $L$ be an infinite regular language, let $p$ be its pumping length, and fix $w\in L$ such that $|w|\ge p$. Then $w$ can be decomposed as $w=xyz$ in such a way that $|xy|\le p$, $|y|\ge 1$, and $xy^kz\in L$ for each $k\in\Bbb N$. (Note that my $\Bbb N$ includes $0$.)
Let $L_0=\left\{xy^{2k}z:k\in\Bbb N\right\}$; $L_0$ is described by the regular expression $x(yy)^*z$, so it is regular, and clearly it is infinite. Again let $L_1=L\setminus L_0$; clearly $\left\{xy^{2k+1}z:k\in\Bbb N\right\}\subseteq L_1$, so $L_1$ is infinite, and $L=L_0\cup L_1$ is the desired decomposition of $L$.