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Suppose that for every $n\in\mathbb{N}$ the $V_n$ is a non-empty closed subset of a compact space $X$ and suppose that $V_{n+1}\subset V_n$,I have to show that $\bigcap_{n=1}^{\infty}V_n\neq \emptyset$

I'm not sure about my solution as follows:

So suppose that the intersection is empty, $\bigcap_{n=1}^{\infty}V_n = \emptyset$.

Now let $U_n=X\setminus V_n$ then the $U_n$ would form a covering of $X$ as $$\bigcup U_n=\bigcup(X\setminus C_n)=X\setminus(\cap C_n)=X$$

Then we have a finite subcovering $\{U_1,U_2,.....U_n\}$

Am I on the right track with this?

Thanks for any help

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I would say that yes, you are on the right track. Now the obvious question is, what are you going to do with your finite subcovering? –  MJD Oct 30 '12 at 21:45
    
Yes, you're on the way to the solution. –  Beni Bogosel Oct 30 '12 at 21:45
    
Note also that your finite subcovering is not $\{U_1, U_2,\ldots, U_n\}$ but rather $\{U_{a_1}, U_{a_2},\ldots, U_{a_n}\}$. –  MJD Oct 30 '12 at 21:46
    
The subsets $U_i$ are ordered and there is one containing all the rest. –  i. m. soloveichik Oct 30 '12 at 21:48
    
This is a special case of a more general fact: "If $X$ is compact and $F_i, \ i\in I$ a family of closed subsets of $X$ with the property that every finite intersection of them is nonempty then $\cap_{i \in I}F_i \neq \emptyset$ ". The proof is the same as the suggested one. –  P.. Oct 30 '12 at 21:52
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2 Answers

up vote 2 down vote accepted

Yes you are on the right track.

You have a finite covering, $U_1,\ldots,U_n$, which in turn means that $$X=(X\setminus V_1)\cup(X\setminus V_2)\cup\ldots\cup(X\setminus V_n)=X\setminus(V_1\cap\ldots\cap V_n)$$

Therefore $V_1\cap\ldots\cap V_n=\varnothing$. However since $V_{k+1}\subseteq V_k$ we have that $V_n=\varnothing$, in contradiction to the assumption that $V_k\neq\varnothing$ for all $k$.

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Since there are only finitely many subsets $U_i$ and they are ordered then there is one containing all the rest. So one of the $U_i$ is $X$ and hence $V_i=\emptyset$, a contradiction.

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