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Let $X$ be a topological space. Prove $X$ is quasi-compact iff for any family $(V_i)_{i\in I}$ of closed subset of $X$ with $\bigcap_{j\in J}V_j\neq \emptyset$ for any $J\subseteq I$ with $J$ finite it follows that $\bigcap_{i\in I}V_i \neq \emptyset$


This confuses me, since I have closed sets here. Shouldn't $\bigcap_{i\in I}V_i =\emptyset$?

Because if $X$ is quasi-compact, there exist for every open sets $X=\bigcup\limits_{i\in I}U_i$ a finite $J\subset I$ such that $X=\bigcup\limits_{i\in J}U_i$. This means $$\emptyset =(\bigcup\limits_{i\in J}U_i)^c=\bigcap\limits_{i\in J}U_i^c=\bigcap\limits_{i\in J}V_i$$ And because $\bigcap\limits_{i\in I}V_i=\bigcap\limits_{i\in J}V_i \cap \bigcap\limits_{i\in I}V_i$ we have that $\bigcap\limits_{i\in I}V_i=\emptyset$

This confuses me and I don't know how to solve the problem.

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If you assume that $\bigcap_{i\in I}V_i=\varnothing$, you get a contradiction. Specifically, for each $i\in I$ let $U_i=X\setminus V_i$. Then each $U_i$ is open, and

$$\bigcup_{i\in I}U_i=\bigcup_{i\in I}(X\setminus V_i)=X\setminus\bigcap_{i\in I}V_i=X\setminus\varnothing=D\;,$$

so $\{U_i:i\in I\}$ is an open cover of $X$. Therefore there is a finite $J\subseteq I$ such that $\bigcup_{i\in J}U_i=X$; can you finish it by showing that this implies that $\bigcap_{i\in J}V_i=\varnothing$ and thereby getting a contradiction?

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  • $\begingroup$ Not really. I don't see what the contradiction should be. It's clear that $\bigcap_{j\in J}V_i=\emptyset$ when $\bigcap_{j\in I}V_i=\emptyset$ since $\bigcap_{j\in J}V_i\subset \bigcap_{j\in I}V_i$. $\endgroup$ – MarcE Sep 25 '16 at 17:26
  • $\begingroup$ @MarcE: No, it certainly is not: you have the inclusion backwards. The correct inclusion is $\bigcap_{i\in I}V_i\subseteq\bigcap_{i\in J}V_i$. The intersection gets smaller when you intersect more things, not when you intersect fewer. $\endgroup$ – Brian M. Scott Sep 25 '16 at 17:28
  • $\begingroup$ Since $X$ is compact we have $$X=\bigcup\limits_{j=1}^n U_j=\bigcup\limits_{j=1}^n(X\setminus V_j)=X\setminus\bigcap\limits_{j=1}^n V_j$$ therefore $\bigcap\limits_{j=1}^n V_j=\emptyset$ Sadly, I still don't see what this contradicts. $\endgroup$ – MarcE Sep 25 '16 at 18:06
  • $\begingroup$ @Marc: Look at your hypotheses: what do they say about intersections of finite subfamilies of $\{V_i:i\in I\}$? $\endgroup$ – Brian M. Scott Sep 25 '16 at 18:08
  • $\begingroup$ I say if the intersection of the finite subfamilies $\{V_j:j\in J\}$ is non-empty, then the intersection of infinite families $\{V_i:j\in I\}$ is non-empty. But now I showed that if $\bigcap\limits_{i\in I} V_i=\emptyset$ then $\bigcap\limits_{j=1}^n V_j=\emptyset$. Where is the contradiction? Or does this last statement mean if $\bigcap\limits_{i\in I} V_i=\emptyset\implies \bigcap\limits_{i\in I} V_i=\emptyset$ which implies $\bigcap\limits_{i\in I} V_i\neq\emptyset\implies\bigcap\limits_{j=1}^n V_j\neq\emptyset$? $\endgroup$ – MarcE Sep 25 '16 at 18:28

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