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I am learning algebraic geometry and I came across the folowing question. Prove that the complement of a point is compact in $\mathbb{A}^n$. Does anyone know how to do this?

Alex Youcis
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dunkindonuts
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  • Open sets are complements of solutions of polynomial equations. Covering $\mathbb{A}^n\setminus{a}$ with open sets, is the same as having closet sets with intersection ${a}$. So we have infinitely many equations with solution set ${a}$. By Hilbert's there are finitely many of them that define have the same solution. Take the open sets they are associated with. – OR. Aug 07 '13 at 20:26
  • @QiL'8 In most Topology textbooks compact is just having the finite intersection property, but in Commutative Algebra like in Atiyah Macdonald quasi-compact means a compact non-hausdorff space, so in Atiyah notation he meant quasi-compact – ADR Aug 08 '13 at 03:03

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Hint: A space is Noetherian if and only if every subset of the space is compact.

Alex Youcis
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