stalk-local implies affine-local I am looking for some clarification regarding the following statements from Vakil's  Algebraic geometry notes:
Any property P that is stalk-local(i.e. A scheme has property P iff its stalks have property Q) is necessarily affine-local(i.e. A scheme has property P iff every open affine has property R).
I have some intuitive idea of why it might be true.If i'm not mistaken, a scheme has a property P means that by definition every open set have property P'. Hence, the affine opens have property P'. Now P' may lead to some other property R' of affine opens. The statement therefore says that in case P stalk-local,  then in the set of such R' properties, there should exist some property R such that P would become affine-local. How do i prove the existence of such a property? 
Thanks in advance!
 A: Let $R$ be the property that the open affines have property $P$. 
You seem to be a bit confused about things. A scheme having a property $P$ doesn't imply every open set necessarily has property $P'$ for some property $P'$, also what is $P'$ supposed to be anyway? A good example of such properties that don't come from local properties are things that are inherently global such as properness over a field (since properness is target local, see here) or again (and somewhat relatedly) for a scheme over a base field, that the global sections of the scheme are just the constant functions to the base field.
Anyway, returning to the particular question at hand. Suppose $P$ is stalk-local, then the following are equivalent for a scheme $X$.


*

*$X$ has property $P$ 

*Every open subscheme has property $P$

*Every open affine subscheme has property $P$ (this equivalence gives us that $P$ is affine local)

*$X$ has an open cover by affine subschemes with property $P$

*$X$ has a cover by open subschemes with property  $P$
Clearly 2 $\implies$ 3 $\implies$ 4 $\implies$ 5. We just need to show 5 $\implies$ 1 and 1 $\implies$ 2.
1 $\implies$ 2: Let $U$ be an open subscheme of $X$. For every $x\in U$, the stalk of $U$ at $x$ is the same as the stalk of $X$ at $x$, and every stalk of $X$ has property $Q$ since $P$ is stalk-local, so every stalk of $U$ has property $Q$. Thus $U$ has property $P$.
5 $\implies$ 1: Let $x\in X$. Then there is some $U$ in the open cover with $x\in U$. Once again, the stalk of $U$ at $x$ is the same as the stalk of $X$ at $x$, so since $U$ has $P$, the stalk of $U$ at $x$ has property $Q$. Since $x$ was arbitrary, every stalk of $X$ has property $Q$, so $X$ has property $P$.
