Krull dimension of complement of an open subset containing all generic points

Let $X$ be a Noetherian scheme of krull dimension $1$. Let $U$ be an open subset of $X$ such that $U$ contains the generic points of irreducible components of $X$.

Then, is it true that $X\setminus U$ has krull dimension $0$ ?

More generally, without the restriction of $\dim (X)=1$, do we always have $\dim (X\setminus U)\le \dim(X)-1$ , when $U$ is an open subset of $X$ such that $U$ contains the generic points of irreducible components of $X$ ?

Solution 1:

Yes.

Note that since every irreducible closed subset has a unique generic point, there is a bijection between irreducible closed sets and points of $X$ given by $p \rightarrow \overline{p}$ and $Z \rightarrow \eta_Z$. This also preserves inclusion.

Let $n = \dim(X)$. Suppose for a contradiction that $\dim(X \backslash U) \geq \dim(X)$. There is a chain $Z_0 \subsetneq \dots \subsetneq Z_{n}$ of irreducible closed subsets of $X - U$ (with $n$ inclusions). Let $\eta_i$ be the generic points of $Z_i$. Then, we have $\eta_i \in \overline{\eta_{i+1}}$ for $i = 0, \dots, n-1$.

Now, let $\eta_{n+1}$ be the generic point of the irreducible component that $\eta_n$ is in. Since $\eta_{n+1} \in U, \eta_{n} \notin U$, $\eta_{n+1} \neq \eta_n$. Then, we have a chain $\overline{\eta_0} \subsetneq ... \subsetneq \overline{\eta_{n+1}}$ in $X$, contradicting $n = \dim(X)$.