Properness is not a compactness claim on rational points

Properness is a property of a scheme morphism: it asks for separatedness, finite type, and universal closedness, as in Proper morphisms and Universally closed morphisms. It does not put a topology on the set $X(k)$ of $k$-valued points for an arbitrary field $k$, and it therefore makes no blanket compactness assertion about that set. The scheme-theoretic content that replaces compactness is universal closedness, together with the valuative criterion on this page: every solid valuative diagram has a unique filler.

On a page about $k$-varieties the word complete is used only as a name for properness of the structure morphism $X\to \mathrm{Spec}k$ (Complete varieties); it is not a claim that $X(k)$ is compact in any topology. For $k=\mathbb{C}$ there is a classical analytic statement, not proved here, that a proper $\mathbb{C}$-scheme has compact $\mathbb{C}$-points in the analytic topology; that statement needs the analytic topology as extra structure and is a theorem about it, not part of the definition of properness. The empty scheme is proper over every base and its set of points is empty, so a compactness reading would also have to handle that boundary case separately.

Care is needed because an empty set of rational points does not imply properness. For example, let $X={\mathbb{A}}_{\mathbb{C}}^1=\mathrm{Spec}\mathbb{C}[x]$, viewed as an $\mathbb{R}$-scheme. It has no $\mathbb{R}$-points, since such a point would give an $\mathbb{R}$-algebra map $\mathbb{C}[x]\to \mathbb{R}$, and in particular a unital $\mathbb{R}$-algebra map $\mathbb{C}\to \mathbb{R}$, which does not exist. This morphism is not proper: after base change to $\mathbb{C}$, $\mathbb{C}{\otimes }_{\mathbb{R}}\mathbb{C}\cong \mathbb{C}\times \mathbb{C}$, so the base change is two copies of ${\mathbb{A}}_{\mathbb{C}}^1$ over $\mathrm{Spec}\mathbb{C}$. After a further base change along ${\mathbb{A}}_{\mathbb{C}}^1\to \mathrm{Spec}\mathbb{C}$, the closed subset $V(tx-1)\subseteq \mathrm{Spec}\mathbb{C}[t,x]$ in one component has image $D(t)\subseteq \mathrm{Spec}\mathbb{C}[t]$, which is not closed. Thus universal closedness fails. The point of this remark is that properness must be tested with the morphism definition and the valuative criterion, never by inspecting $X(k)$ alone.