In A. Bondal, M. van den Bergh's paper, **Generators and representability of functors in commutative and noncommutative geometry** , the "reduction principle" of quasi-compact, quasi-separated schemes is shown. It is stated as follows.



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> Assume $X = U\_1 \cup U\_2 $ with $U\_1, U\_2$ open and put $U\_{12} = U\_1 \cap U\_2$.
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> Let $P$ be a property satisfied by some schemes such that
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> (1) $P$ is true for affine schemes.
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> (2) If $P$ holds for $U\_1, U\_2, U\_{12}$ as above, then it holds for $X$.
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> Then $P$ holds for all quasi-compact quasi-separated schemes.
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I understood this statement. However, the next **Remark3.3.2** says



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> It is easy to see that the class of quasi-compact quasi-separated schemes is the biggest class of schemes to which the reduction principle is applicable (for all properties $P$).
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I'm not sure why this characterization of quasi-compact, quasi-separated schemes holds. Is there an obvious property $P$ which satisfies above properties and holds only for quasi-compact, quasi-separated schemes?



The biggest class of schemes which the reduction principle holds
Yes, such a property $P$ exists, namely "being qcqs" (i.e., quasi-compact and quasi-separated). *It is trivial that qcqs answers your question, if it indeed satisfies your conditions on $P$.*


Let's check the conditions:


1. *$P$ is true for affine schemes.* <http://stacks.math.columbia.edu/tag/01S7>
2. *If $P$ holds for $U\_{1}$, $U\_{2}$, $U\_{12}$ as above, then it holds for $X$.* The qc part is (easy) topology. For qs one can use <http://stacks.math.columbia.edu/tag/01KO>, and see that #3 is satisfied for $X$. Indeed, take $S = \mathrm{Spec}(\mathbb{Z})$, and cover it with the trivial covering $\mathrm{Spec}(\mathbb{Z})$. The rest of the lemma says that we have to cover $X$ with affine opens $V\_{j}$: well, we may (and do) choose the $V\_{j}$ to all be in $U\_{1}$ or $U\_{2}$. All we need to check is that $V\_{j} \cap V\_{j'}$ is covered by a finite number of affine open subsets of $X$. Well, take any such cover $W\_{k}$. Suppose $V\_{j} \subset U\_{i}$, and $V\_{j'} \subset U\_{i'}$. We have
$$ V\_{j} \cap V\_{j'} = (V\_{j} \cap U\_{i}) \cap (U\_{i'} \cap V\_{j'}) = V\_{j} \cap (U\_{i} \cap U\_{i'}) \cap V\_{j'}. $$
On the right hand side everything is qcqs (either because affine, or by assumption) **[edit:]** *and* all intersections take place in qcqs schemes (namely $U\_{i}$ and $U\_{i'}$) **[/edit]**. Hence so is the intersection, and therefore $V\_{j} \cap V\_{j'}$ is qcqs (in particular qc). This allows us to take a finite subcover of $W\_{k}$.




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**Edit:** The last paragraph was not very well-phrased. The point is that a priori $V\_{j} \cap V\_{j'}$ need not be qcqs (even though both $V\_{j}$ and $V\_{j'}$ are qcqs). After all, the intersection takes place in $X$, and we do not know that $X$ is qcqs. *(Indeed, it is what we are trying to prove.)* However, we know that $V\_{j} \subset U\_{i}$, and $V\_{j'} \subset U\_{i'}$, and this makes it possible to rewrite the intersection into intersections taking place in qcqs schemes. And of those, we know that the intersection is qcqs: item (6) of <http://stacks.math.columbia.edu/tag/01KU>.


