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Call a topological space $X$ Artinian if every nested sequence of closed sets

$$C_1 \supset C_2 \supset C_3 \supset \cdots$$

is eventually constant.

Prove that if $X$ is Artinian then it is also compact.

I don't have a concrete strategy for attacking this. I am assuming (maybe wrongly so) that appealing to the finite intersection property might be of some help. Is this the case? Should I employ a different strategy?

Thank you very much in advance!

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  • $\begingroup$ There exists some relation between noetherian and artinian topological spaces as in the case of rings and modules? Sorry for this question here. $\endgroup$ – Diego Silvera Mar 16 '13 at 1:44
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    $\begingroup$ This is weird terminology. Don't most people call this property of a topological space being Noetherian? $\endgroup$ – Matt Mar 16 '13 at 2:47
  • $\begingroup$ Indeed, this space is Noetherian. $\endgroup$ – Wuestenfux Sep 26 '18 at 7:55
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Take an open cover $\{U_i\mid i\in I\}$. Using the axiom of choice we can assume that $I$ is well-ordered and so we can write $V_i=\bigcup_{j<i}U_i$, then $\{V_i\mid i\in I\}$ is an open cover as well.

Use the usual trick of complements and the Artinian property to conclude there is a finite subcover, or else.

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  • $\begingroup$ ...or else what? ;) $\endgroup$ – Dan Rust Mar 16 '13 at 1:47
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    $\begingroup$ @Daniel: Or else $0=1$, that's what! :-) $\endgroup$ – Asaf Karagila Mar 16 '13 at 1:48

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