Open properties of quasi-compact schemes I am following Ravi Vakil's Math 216: Foundations of Algebraic geometry notes, and there is a remark following an exercise that I don't understand at all, and if anyone could enlighten me then that would be brilliant.
The exercise asks one to show that if $X$ is a quasicompact scheme, then every point has a closed point in its closure, which is clear from the preceding exercise asking to show that $X$ is quasicompact if and only if it can be written as a finite union of affine schemes. These I am fine, as well as the following implication that every nonempty closed subset of $X$ contains a closed point. 
However, then the notes then go on to state that this will be used in the following way: If a property $P$ is open (that is, if some point $x$ has $P$, then there exists an open neighbourhood $U$ of $x$ such that all points in $U$ have $P$), then to check that all points of a quasicompact scheme have $P$, then it suffices to check only the closed points.
I do not seem to be able to see how this follows at all. It seems to me that everything in the exercises is regarding closed points being in closures, and to show the remark, I want to show that other points are in (all) open neighbourhood(s) if closed points. These seem relatively distinct to me - is this wrong?
These comments/exercises are on pages 139-140 of the notes.
 A: Let $\eta$ be a generic point with closure $Y = \overline{\{ \eta \}}$, and let $P$ be an open property. Then, the set of all points of $X$ with property $P$ is open, so its complement is closed; in particular, if $\eta$ does not have property $P$, then no point of $Y$ has property $P$. (This is just point set topology.)
Now, suppose $X$ is a quasicompact scheme. Then, $Y$ must have a closed point; so if all closed points of $X$ have property $P$, then $\eta$ must have property $P$ as well.
A: First: Are you sure that your argument about existence of closed points is correct?  The reason I ask is that you say that it's clear from writing $X$ is a finite union of open affines, but (while I agree that $X$ is quasi-compact if and only if this is possible) I don't myself see how it follows immediately from this. (The arguments I know, e.g. this one, use the topological property of quasi-compactness in an explicit manner.)
Secondly: Suppose $U$ is an open subset of $X$ that contains all the closed points of $X$.  The complement of $U$ is a closed subset of $X$.  Can it contain any closed point of $X$?  If not, then taking into account the facts you state in your question, can it contain any points at all?
A: This is a purely topological statement.  Suppose that an open property $P$ holds for all closed points.  Let $\eta$ be a non-closed point, and let $x \in \bar{\eta}$ be a closed point in the closure (which exists by the exercise!).  Then by assumption $x$ has $P$, and since $P$ is open there exists an open neighborhood $U$ of $x$ so that all points of $U$ have $P$.
It is clear that any open neighborhood $U$ of $x$ must contain $\eta$.  If $\eta \notin U$, then $U^c:=X-U$ is a closed subset of $X$ containing $\eta$.  But $\eta \in U^c$ implies $\bar{\eta} \subset U^c$ which then means that $x \in \bar{\eta} \subset U^c$, a contradiction since $x \in U$.
Since $\eta \in U$, then $\eta$ has property $P$ also.  Hopefully this gives you some intuition about the strangeness of non-closed points.
