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A quasi-affine variety is an open subset of an affine variety.

Open under Zariski topology? How does this make sense?

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    $\begingroup$ Meaning "relatively open". So a quasi-affine is the intersection of an affine variety with an open subset of the ambient affine space. The usual example is $\mathbf{A}^2 - \{(0, 0)\}$. $\endgroup$ – Hoot Dec 31 '14 at 17:08
  • $\begingroup$ @Hoot- Would $x+y=0\cap [(0,1)\times(2,3)]$ be another example? $\endgroup$ – algebraically_speaking Dec 31 '14 at 17:10
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    $\begingroup$ @algebraically_speaking: No that set would not be quasi-affine: you have to intersect with a Zariski-open set, not a Euclidean-open set. The set $(0,1)\times (2,3)$ is not Zariski-open because its complement isn't the solution set of a polynomial. Hoot: He's referring to a piece of the line $y=-x$, intersected with an open rectangle. $\endgroup$ – pre-kidney Dec 31 '14 at 17:50
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Yes, open under the Zariski topology. Here is how it makes sense:

An affine variety is the solution set of polynomials.

A quasi-affine variety is a solution set of polynomials minus another solution set.

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