On an openess property

Question

Let $S=Spec(A)$ where $A$ is a noetherian integral domain. Let $f:X\rightarrow S$ be a flat, proper morphism of schemes. Let $U\subset X$ be an open and $V=f(U)$ (in particular $V$ is open by flatness). Assume that the fibers $X_s\subset U$ for $s$ in a dense set of points of finite type of $S$. Is it true that I can find a non empty open $V_1\in V$ such that $U\times_S V_1=X\times_S V_1$?

Answer 1

It is enough to suppose $X_s\subseteq U$ for one point $s\in S$ and $f$ proper (flatness is useless).

Indeed, $X\setminus U$ is closed, so $f(X\setminus U)$ is closed and is different from $S$ because it does not contain $s$. Now take $V_1=S\setminus f(X\setminus U)$. It is dense in $S$ because the latter is irreducible. By construction, $f^{-1}(V_1)\subseteq U$, hence $$X\times_S V_1=f^{-1}(V_1)=U\cap f^{-1}(V_1)=U\times_S V_1.$$