Subset of a scheme where a section vanishes

Question

Suppose $(X,\mathscr{O})$ is a scheme, $s\in \mathscr{O}(U)$. I am trying to prove that $\{x\in U: s_x=0\in \mathscr{O}_x(\text{the stalk})\}$ is open in $U$ but not necessarily closed. I can see why it's open, but why is it not necessarily closed? Is there any example?

Answer 1

Let $A=k[x,y]/(y^2,xy)$ for a field $k$, and set $X=\operatorname{Spec} A$. Then $y\in\mathcal{O}_X(X)=A$ is a section which is zero in the stalk at every point in $D(x)$, but is not zero in the stalk at the origin. As $D(x)$ is open but not closed, this does the trick.