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I want to show that $\mathbb{A}^n$ is quasi-compact. I'm kind of stuck, I really don't know where to go with my proof, so I'll show what I have

Proof: So suppose that $\cup U_i$ was an open cover for $\mathbb{A}^n$, then we look at $\mathbb{A}^n - (U_1 \cup \dots \cup U_i)$ which closed.

I'm stuck here, I wanna use the fact that the Zariski topology has the Noetherian property but I can't really see how to do it in this case.

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  • $\begingroup$ What’s your definition of noetherian? Ascending sequences of open sets terminate? $\endgroup$
    – k.stm
    Jan 21, 2015 at 21:40
  • $\begingroup$ Yes, all subsets of affine space, including $\mathbb{A}^n$ itself, are quasi-compact,see the discussion here. $\endgroup$
    – Dietrich Burde
    Jan 21, 2015 at 21:42
  • $\begingroup$ Descending chain of closed subsets which stabilize. $\endgroup$
    – Alex
    Jan 21, 2015 at 21:44

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A space is noetherian if and only if every ascending chain of open subspaces stabilize.¹

For any open cover $(U_i)_{i∈I}$ of $\mathbb A^n$, look at the collection of finite unions of its cover members $$\bigcup_{j ∈ J} U_j;~\text{$J ⊂ I$ is finite}.$$ Use the noetherian property to show that every chain in this collection has an upper bound in this collection. Apply Zorn’s Lemma and examine what you got. Then you’re done.

This works for all noetherian spaces. I’m not sure that you have to use Zorn’s Lemma, though.

¹This is because the closed subsets of a space are dual to the open subsets of the space by taking complements. So the descending chain condition on closed sets directly translates to the ascending chain condition on open sets.

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  • $\begingroup$ You don't need Zorn's lemma. Every chain has an upper bound, and any such upper bound must be $\mathbb{A}^n$, or you could add one more open set to have a bigger finite union. $\endgroup$
    – Kevin Carlson
    Jan 21, 2015 at 21:58
  • $\begingroup$ @KevinCarlson But $U_1 ⊂ U_1 ⊂ U_1 ⊂ …$ might be a chain with an upper bound which is not all of $\mathbb A^n$? Don’t you need something guaranteeing the existence of a chain running through suffiently many open sets? Oh, one can try to construct strictly increasing chains and fail to do so, right? $\endgroup$
    – k.stm
    Jan 21, 2015 at 22:03
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    $\begingroup$ You often hear it said that you don't need the axiom of choice to get maximal ideals in Noetherian rings... $\endgroup$
    – Hoot
    Jan 21, 2015 at 22:27
  • $\begingroup$ Right, it's only strictly increasing chains that we're interested in bounding here. $\endgroup$
    – Kevin Carlson
    Jan 21, 2015 at 23:41
  • $\begingroup$ Ok, I’m confused now: Does or doesn’t one need the axiom of choice? @KevinCarlson $\endgroup$
    – k.stm
    Jan 22, 2015 at 7:51

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