Let $(X,\mathcal O_X)$ an algebraic variety, so $\mathcal O_X(U)$ is a $k$-subalgebra of $k^U$ for all open set $U\subseteq X$. In the Gortz & Wedorn's book "Algebraic Geometry I", I found the following equation $$\mathcal O_X(U\cup V)=\mathcal O_X(U)\cap\mathcal O_X(V)$$
For any two open sets of $X$.
I don't understand what is the meaning of $O_X(U)\cap\mathcal O_X(V)$; after all functions in $O_X(U)$ and in $O_X(V)$ have different domains!
If $X$ is irreducible then $\mathcal{O}_X(U)$ may be identified with rational functions defined on $U$, hence it naturally sits inside the function field $k(X)$ of $X$. The intersection is as subsets of $k(X)$.