Let $X$ be a Noetherian scheme, $U\subset X$ be an affine open dense subscheme. Is the underlying space of $U$ necessarily homeomorphic to the underlying space of $X$?
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$\begingroup$ A variant was asked here: mathoverflow.net/questions/331673/… but Francesco's answer applies to both. $\endgroup$– BortCommented May 16, 2019 at 12:16
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$\begingroup$ OK I see, I lost the assumption "affine" from the linked question. $\endgroup$– BortCommented May 17, 2019 at 12:07
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2 Answers
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Let $k$ be an algebraically closed field, ad take $X=\mathbb{P}^2_k$, $U=\mathbb{A}^2_k$.
Then $X$ and $U$ are not homeomorphic, since $U$ contains two disjoint, Zariski-closed, irreducible subsets made of more than one point (think of two parallel lines), but this is not possible in $X$ because of Bézout theorem.
answered May 16, 2019 at 9:09
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Let $X=\operatorname{Spec}R$, where $R$ is a discrete valuation ring. It consists if two points $x,y$ where $x$ is the generic point and $y$ is a closed point. Then $\{x\}$ is an open, dense, affine subset which is not homeomorphic to $X$, since it has fewer points.
