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It seems as if every subspace of a compact topological space (equipped with its relative topology) had ought to be compact as well. Is this true in general?

And in particular, I want to use the fact that every quasi-projective variety is compact with the Zariski topology - I remember having proven it for quasi-affine varieties, but the same proof does not work for quasi-projective varieties. Is every quasi-projective variety compact?

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Is the open unit disk, with the relative topology from the closed unit disk, a compact topological space? –  Gerry Myerson Mar 22 '12 at 3:27
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Only closed subsets. $(0,1)$ is not compact, but $[0,1]$ is compact –  Thomas Andrews Mar 22 '12 at 3:28
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5 Answers 5

up vote 6 down vote accepted

A topological space $X$ is called quasi-compact if every open covering of $X$ has a finite subcovering.
(This is the terminology adopted by algebraic geometers to emphasize that Hausdorffness is not required. Compact spaces are then Hausdorff quasi-compact spaces).

A topological space $X$ is said to be noetherian if every non-empty family of closed subspaces has a minimal element.
For example $\mathbb A^n_k$ and $\mathbb P^n_k$ are noetherian: this follows from Hilbert's theorem that the polynomial ring $k[T_i,...,T_n]$ is noetherian .
The following are then equivalent for a topological space $X$:
$\bullet $ $X$ is noetherian.
$\bullet $ Every subset of $X$ is quasi-compact.

Since projective space $\mathbb P^n_k$ over a field is noetherian , any quasi-projective variety is quasi-compact because by definition it is the intersection of an open and a closed subset of some $\mathbb P^n_k$, and any subset of $\mathbb P^n_k$ is quasi-compact by noetherianity of $\mathbb P^n_k$.

To sum-up:
Every quasi-projective variety is quasi-compact.

Edit: as emphasized by Pete, even an arbitrary subset of a quasi-projective variety is quasi-compact, since it too is a subset of $\mathbb P^n_k$.

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+1: I was about to write something very similar. Let me just mention that in $\S 13.5$ of my commutative algebra notes -- math.uga.edu/~pete/integral.pdf -- one can find proofs of these equivalences. –  Pete L. Clark Mar 22 '12 at 5:22
    
(Also, maybe your sum-up should read Every subset of a quasi-projective variety is quasi-compact?) –  Pete L. Clark Mar 22 '12 at 5:23
    
Dear @Pete, thank you for your comments and the link to your very useful notes. Also, I have added an edit to take your suggestion into account. –  Georges Elencwajg Mar 22 '12 at 5:31
    
excellent job answering the question I meant to ask –  smackcrane Mar 22 '12 at 15:41
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A closed subspace $A$ of a compact space $X$ is compact.

Proof (Pardon my terribly sloppy indexing): Take an open cover $\{U_\alpha \cap A\}_{\alpha}$ of $A$ where each $U_\alpha$ is open in $X$. Then $\{U_\alpha\}_{\alpha} \cup \{X \setminus A\}$ is an open cover of $X$, which has a finite subcover. The $U_\alpha$ in this finite subcover necessarily cover $A$. So finitely many $\beta$ can be chosen such that the collection of these $U_\beta \cap A$ cover $A$.

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A space such that all of its subspaces are compact is hereditarily compact. This property turns out to be implied by being Noetherian, which holds in particular for the projective spaces $\mathbb{P}^n$.

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It holds for the projective spaces when they are endowed with a topology one usually never sees :) –  Mariano Suárez-Alvarez Mar 22 '12 at 5:14
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Every closed subspace of a compact Hausdorff space is compact. I'm not actually sure whether that applies to non-Hausdorff spaces; I probably used to know.

But $[0,1]$ is compact, and $(0,1)$ is a non-compact subspace.

If what you propose were true, then one would not speak of "compactifications". A compactification of a topological space $X$ is a space $Y$ such that $X\subseteq Y$ and the closure of $X$ in $Y$ is precisely $Y$. For example, if one adjoins a single point $\infty$ to the real line $\mathbb{R}$ in such a way that an open neighborhood of $\infty$ is any set of the form $$\{\infty\}\cup \{x\in\mathbb{R} : x > a\}\cup \{x\in\mathbb{R} : x < b\},$$ then that is the "one-point compactification" of $\mathbb{R}$. The line also has a two-point compacification. One can show that it has no three-point compactification, but it has some much larger compactifications with infinitely many points.

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Every closed subspace of a compact space is compact. The Hausdorff condition is only necessary for the converse. –  Qiaochu Yuan Mar 22 '12 at 4:48
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You can surely answer this yourself! It is a matter of considering examples.

Let us say you start with the very first topological space you were introduced to: the space of real numbers. It is not compact, but it has lots of compact subspaces, and in fact we can describe them all. Pick one and look at its subspaces now...

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haha, of course it's strange, but the first topological space I was introduced to was affine space under the zariski topology. –  smackcrane Mar 22 '12 at 15:37
    
That may have been the first topological space you were introduced to under that name... –  Mariano Suárez-Alvarez Mar 22 '12 at 15:45
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