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I know that there's another question very similar to this one, but I couldn't understand the answers since they work with another definition of compact set. So, here's the problem: Let $\{A_i\}$ be an infinite family of compact sets in a normed vector space $E$ such that $\cap_{i \in \Bbb N} A_i=\emptyset$, show that there exists a finite number of elements of the family $A_{i_1},...,A_{i_n}$ such that $\cap _{j=1}^nA_{i_j}= \emptyset$

So, I'm working with the definition by sequences of a compact set (every sequence in the set has a convergence subsequence to an element of the set), I can see why this can be true for a "well behaved" family, but I don't know what propierty of a compact set can I use in order to prove this.

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    are you working in a metric space? – 2017-01-22
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    @JorgeFernándezHidalgo no, I'm working on vector spaces – 2017-01-22
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    in a normed vector space? – 2017-01-22
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    @JorgeFernándezHidalgo yep – 2017-01-22

2 Answers 2

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We prove it by contradiction, suppose no finite intersection is empty.

Define the set $B_n=\bigcap\limits_{m=1}^nA_m$. We have that each $B_n$ is non-empty, and compact.

Pick a sequence $x_1,x_2,\dots,$ such that $x_k\in B_k$.

Notice that this sequence has a converging subsequence because it is inside the compact set $A_1$. We can show that the limit of that sequence is contained in every $A_n$, because the tail of the sequence is contained in $A_n$, and $A_n$ is closed. This is a contradiction because we proved $\bigcap\limits_{n=1}^\infty A_n$ contains a point.

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    $B_n$ is the intersection of the whole family? Or the intersection of the first $n$ elements? – 2017-01-22
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    just the first $n$, my bad. – 2017-01-22
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Hint

Define recursively $B_1=A_1$, $B_{n+1}=B_n\cup A_{n+1}$. Each $B_i$ is compact and $B_1\subset\bigcup_i B_i^c$.

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    How come $B_1\subset\bigcup_i B_i^c$? Since $A_1 \subset B_i$, for all i? – 2017-01-22