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Is the intersection of two quasi-compact open subsets of a scheme quasi-compact? Is there a counterexample?

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The property you ask about, quasi-separatedness (also known as quasi-separatedness of $\mathrm{Spec}(\mathbf{Z})$) is equivalent to the diagonal $X\rightarrow X\times_{\mathrm{Spec}(\mathbf{Z})}X$ being quasi-compact. –  Keenan Kidwell Dec 11 '12 at 12:10

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up vote 3 down vote accepted

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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Interesting, but $X$ is not a scheme... –  Makoto Kato Dec 11 '12 at 11:24
Sorry, I missed that in your question! –  Adeel Dec 11 '12 at 11:34

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