TITLE: Every subspace of a compact space is compact?
QUESTION [5 upvotes]: 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?

REPLY [8 votes]: 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  $\mathbb P^n_k$.