Is the intersection of two quasi-compact open subsets of a scheme quasi-compact? Is there a counterexample?
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This property is known as "quasiseparatedness" and not every scheme has it. Consider $X=\mbox{Spec}~k[x_1,x_2,x_3,\ldots]$ and the maximal ideal $\mathfrak{m} = (x_1,x_2,x_3,\ldots)$. Then $U=X\setminus\{\mathfrak{m}\}$ is an open subset of $X$. Glue together two copies of $X$ at $U$ and call this $Y$. You can think about this example as a generalization of the affine line with doubled origin - it is the infinite-dimensional affine space with doubled origin. This scheme is not quasiseparated. Both copies of $X$ are quasicompact open subsets of $Y$ (in fact they are affine), but their intersection is $U$, and $U$ is not quasicompact. Just take the cover by all $D(x_i)$. |
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No. I believe the following is a counterexample: consider the unit interval $X = [0,1]$ with the following topology: the open subsets of $X$ are all the subsets of $(0,1)$, together with the subsets of $X$ containing 0 or 1, and having finite complement in $X$. Now consider $U = [0,1)$ and $V = (0,1]$; they are open and quasi-compact, but their intersection $U \cap V = (0,1)$ is not. |
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