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In a general topological setting what can be said about an open quasi-compact set? Is it true that a subset of such a set is compact? What if that set is open? I ask because this came up in class today with someones solution to a problem (my class consists mainly of student presentations). I should have asked at the time, but thought about it too much and didn't have an opportunity. So the statement seemed to be that any subset of an quasi-compact open set is quasi-compact- though for the problem we were discussing the subset was also open (though it seemed like this wasn't really used). It feels like it should be easy to prove or disprove but I haven't been able to come up with anything. Thanks. (Note: Hausdorff is not assumed)

Edit: I must apologize I have been away. In all instances where I merely wrote compact I meant quasi-compact. The point being that there is no assumption on the Hausdorff property.

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In your terminology, is there a difference between quasi-compact and compact? (AFAIK they are used for the same thing, some people reserve compact only for $T_2$ spaces.)\\ If $X$ is compact space, then $X$ is open in $X$. If your conjecture would be true, any subset of compact space would be compact. Easy counterexample $(0,1)$ as a subset of the compact space $[0,1]$. – Martin Sleziak Dec 6 '11 at 17:35
@Martin- I must apologize- meant to use quasi-compact throughout, taken to mean it is not necessarily Hausdorff. – MJoszef Dec 7 '11 at 4:18
@Martin again. I'm sorry, I only skimmed what you wrote before. Thank you for answering this. I don't know what the person I heard earlier was going on about then. Thank you though. – MJoszef Dec 7 '11 at 4:25
up vote 3 down vote accepted

As Martin explains, it is not true that an open subset of a quasi-compact space is necessarily quasi-compact. However here is a nice substitute.

A topological space $X$ is said to be noetherian if every decreasing sequence $F_0\supseteq F_0\supseteq F_1\supseteq F_1\supseteq ...$ of closed subsets of $X$ is stationary: there exists $N$ such that $F_N=F_{N+1}=F_{N+2}=...$.
The amazing (but very easy: try to prove it!) result is that every subset, open or not, of a noetherian space is quasi-compact!

You might think that these spaces are rare and useless. This is not the case:
a) Every set $X$ with the cofinite topology (in which the closed subsets are the finite subsets plus $X$ itself) is noetherian.
b) The prime spectrum $Spec(A)$ of any noetherian ring $A$ is a noetherian topological space.
c) More generally the noetherian schemes, those which can be covered by finitely many open affine subschemes as in b), have an underlying noetherian topological space. These schemes are by far the most studied and most useful in algebraic geometry.

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I assume that by "quasi-compact" you mean "every open cover has a finite subcover", and that by "compact" you mean "quasi-compact and Hausdorff."

A closed subset of a quasi-compact (resp. compact) set is again quasi-compact (resp. compact): let $C$ be (quasi-)compact and $F$ a closed subset of $C$. If $\{\mathcal{O}_i\}_{i\in I}$ is an open cover of $F$, then $\{\mathcal{O}_i\}_{i\in I}\cup\{X-F\}$ is an open cover of $C$; since $C$ is (quasi-)compact, there is a finite subcover; we may assume this finite cover includes $X-F$ (add it if necessary), so $C\subseteq \mathcal{O}_{i_1}\cup\cdots\cup \mathcal{O}_{i_n} \cup X-F$. Thus, $$F\subseteq C \subseteq \mathcal{O}_{i_1}\cup\cdots\cup \mathcal{O}_{i_n} \cup X-F.$$ From this, it follows that $F\subseteq \mathcal{O}_{i_1}\cup\cdots\cup \mathcal{O}_{i_n}$; so the cover has a finite subcover.

For the compact case, note that $F$ is Hausdorff in the induced topology.

However, in general it is not true that an open subset of a quasi-compact set is compact or quasi-compact. As Martin Sleziak notes, in $\mathbb{R}$ with the usual topology, $(0,1)$ is not compact, yet it is an open set contained in the compact set $[0,1]$.

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