I am just starting to learn about distributions and I have a question about the implications of this definition:
I don't understand why $U_x$ has to be contained in $V$ here. I believe the definition of a distribution vanishing on an open set means that the distribution applied to any test function with compact support contained in the open set is 0.
$x\in V$ means, by definition, that there has to be a neighborhood of $U_x$ $x$ where $u=0$ (if no such neighborhood exists, then $x\in\,\textrm{supp}\,\, u$). But then, being open, every element in this neighborhood has the same property and, therefore, the whole $U_x$ is contained in $V$.