# Scheme of finite Krull dimension, with a closed point, whose closed subsets are all comparable

Let $$X$$ be a scheme which contains a closed point and also assume that for every two closed subsets $$Y_1$$ and $$Y_2$$ of $$X$$, we have either $$Y_1 \subseteq Y_2$$ or $$Y_2 \subseteq Y_1$$. Also assume that $$X$$ has finite Krull dimension. Then, is it true that $$X$$ is affine ?

My thoughts: Since the closed subsets of $$X$$ are comparable, so $$X$$ has exactly one closed point, say $$x \in X$$ and also every non-empty closed subset contains $$x$$ . Hence for any abelian sheaf $$\mathcal F$$ on $$X$$, we have $$\mathcal F_x =\mathcal F(X)$$. Since taking stalks of sheaves is exact, we get $$H^j(X, \mathcal F)=0, \forall j>0$$ for any abelian sheaf $$\mathcal F$$ on $$X$$. However , since I'm not assuming the scheme to be quasi-compact , I cannot quite apply Serre's criteria now.

The assumption of finite Krull dimension is unnecessary. Let $$x\in X$$ be a closed point and note that for any $$y\in X$$, we must have $$\{x\}\subseteq \overline{\{y\}}$$. This means that any open set containing $$x$$ must contain $$y$$, so the only open set containing $$x$$ is all of $$X$$. Since there is some affine open set containing $$x$$, this means $$X$$ itself is affine.