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Consider the projective space ${\mathbb P}^{n}_{k}$ with field $k$. We can naturally give this the Zariski topology.

Question: What are the (proper) compact sets in this space?

Motivation: I wanted nice examples of spaces and their corresponding compact sets; usually my spaces are Hausdorff and my go-to topology for non-Hausdorff-ness is the Zariski topology. I wasn't really able to find any proper compact sets which makes me think I'm doing something wrong here.

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$\mathbb{P}^n_k$ is a Noetherian topological space, so every subset is (quasi)compact. Moreover every closed subset is complete (the analogue of compactness in algebraic geometry). – Zhen Lin Nov 18 '11 at 17:04
Everything Zhen writes in his great comment is absolutely correct and relevant:+1. Readers not used to algebraic geometry (this excludes Zhen and many others!) should beware, however, that completeness is not a purely topological notion, but depends on the scheme-theoretic structure. For example $\mathbb A^1_k$ is not complete but is homeomorphic to $\mathbb P^1_k$ which is complete. – Georges Elencwajg Nov 18 '11 at 17:37
up vote 5 down vote accepted

You are in for a big surprise, james: every subset of $\mathbb P^n_k$ is quasi-compact.
This is true more generally for any noetherian space, a space in which every decreasing sequence of closed sets is stationary.
However: the compact subsets of $\mathbb P^n_k$ are the finite sets of points such that no point is in the closure of another.

A topological space $X$ is quasi-compact if from every open cover of $X$ a finite cover can be extracted. A compact space is a Hausdorff quasi-compact space.

Bibliography Bourbaki, Commutative Algebra, Chapter II, §4,2.

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Ah, this is nice. I kept seeing "quasi-compact" but I mistook the definition! You note then that there are compact subsets though --- what is the difference between the compact subsets and the quasi-compact subsets? – james Nov 19 '11 at 1:55

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