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I am working on the exercise where the hypothesis is : Let $X$ be a scheme such that there exists affine open subsets $U_i \ (1 \leq i \leq n)$ such that $X = \cup U_i$. Further any two of the $U_i$'s have intersection which can be covered by a finite number of affine open subsets.

Suppose I take $U_i \cap U_j$. Does this mean $U_i \cap U_j \subseteq V_1 \cup ... \cup V_m$ or $U_i \cap U_j = V_1 \cup ... \cup V_m$, where each $V_j$ is an affine open subset? (I was also wondering if they were equivalent statemetns?) Thanks!

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It means that $U_i \cap U_j = V_1 \cup \dotsc \cup V_m$ for affine opens $V_i$. (By the way, the condition means exactly that $X$ is quasi-compact and quasi-separated. This is a very natural finiteness condition for schemes. "noetherian" is often too strong and is no good relative notion.)

It is not enough to assume that $U_i \cap U_j$ is contained in such a finite union, for we always have $U_i \cap U_j \subseteq U_i$ and not every quasi-compact scheme is quasi-separated. Take for instance an affine scheme $X$ which has an open subscheme $U$ which is not quasi-compact. Then $X \cup_U X$ is quasi-compact, but not quasi-separated.

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