Algebraic Geometry – Non-Empty Closed Subsets with Empty Special Fiber

Question

Let $R$ be a dvr and $U\to \text{Spec}(R)$ an affine smooth $R$-scheme with non-empty special fiber $U_0$.

Let $Z\subset U$ be a closed subset. Assume the intersection of $Z$ with $U_0$ is empty.

Is $Z$ empty?

If $U\to \text{Spec}(R)$ was proper then the answer would be yes because the image of $Z$ could only be the closed point in $\text{Spec}(R)$.

A closed point of $Z$ could map to the generic point of $\text{Spec}(R)$ without properness of $U\to \text{Spec}(R)$, so $Z$ could be contained in the generic fiber, so the answer should be no.

I'm just having a bit of trouble picturing the situation visually.

Answer 1

$R=\mathbb{Z}_p, U=\mathrm{Spec}\:\mathbb{Z}_p\times \mathbb{Q}_p, Z=\mathrm{Spec}\:\mathbb{Q}_p\neq \emptyset$