I've proved that if $X$ is a quasi-compact scheme than every point has a closed point in its closure and so every closed subset of $X$ contains a closed point of $X$. Why this statement implies the following statement: if there is a property $P$ of points of a scheme that is "open" then to check if all points of a quasicompact scheme have $P$ it suffices to check only the closed points?
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What do you mean by an open property? – Bruno Joyal Sep 28 '13 at 17:13
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A property that if holds in a point, holds in $U_p$ (an open subset of $X$ containing $p$) – ArthurStuart Sep 28 '13 at 17:18
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Open sets are stable under generization. This means that if $U$ is open, $x\in U$ and $x \in \overline{\{y\}}$ (in which case we say that $y$ is a "generization" of $x$, or that $x$ is a "specialization" of $y$), then $y\in U$. Indeed, suppose otherwise that $y$ belongs to the closed set $X-U$. Then so does its closure $\overline{\{y\}}$, which contradicts that $x\in U$.
In your situation, this implies that any open set containing all closed points must be all of $X$, since you have proved that every point of $X$ is a generization of some closed point.
Now if $P$ is an open property which holds for all closed points, then it holds for all points in an open set containing all closed points, and therefore holds on all of $X$.
Bruno Joyal
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