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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.

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

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.

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user290895
user290895

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 $\text{Spec}(R)$. It must be either a short argument using closednessA closed point of $Z$ incould map to the wholegeneric point of $U$$\text{Spec}(R)$ without properness of $U\to \text{Spec}(R)$, or there mustso $Z$ could be contained in the generic fiber. I'm just having a counterexamplebit of trouble picturing the situation visually.

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 $\text{Spec}(R)$. It must be either a short argument using closedness of $Z$ in the whole $U$, or there must be a counterexample.

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

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user290895
user290895

Non-empty closed subsets with empty special fiber

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 $\text{Spec}(R)$. It must be either a short argument using closedness of $Z$ in the whole $U$, or there must be a counterexample.