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Quick question: let $V$ be a closed subset in the Zariski topology on $k^n$ ($k$ algebraically closed), and let $U$ be an open subset of $V$. Is it possible to find another open subset $U_1$ of $V$ such that $U_1$ is contained in $U$, and the subringed space structure on $U_1$ is affine (that is, $U_1$ is isomorphic in the category of ringed spaces to an affine variety)? Here I'm not requiring varieties just to be irreducible.

I think the answer should be no, for example if you take $k$ in the Zariski topology, and any open set $U$ is just $k$ with finitely many points missing, I don't think you can find an open subset of $U$ that's affine (that is, I doubt any cofinite subset of $k$ besides $k$ itself is affine).

D_S
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    Affine open subsets form a base for the topology. Remember that if you have an affine variety $X$ and a regular function $f$ on $X$ then the open set of points of $X$ where $f$ does not vanish is also affine. The trick is to add a new variable $y$ and the equation $fy - 1$. This should be in any algebraic geometry book! – Hoot Sep 18 '15 at 21:56

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The answer, if I am not misreading your question, is yes. I do not understand your suggestion of $k$ (usually denoted by $\mathbb{A}^1_k$), which is affine and any open subset of it is also affine.

Mohan
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