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Using the definition of compactness alone, how can one show that if $A_i~i\in I$ are compact sets in $\mathbb{R}$ such that any finite subcollection of them has a non-empty intersection, then $\exists$ an $x$ which belongs to all $A_i$?

Edit: The definition I have is the following: $F$ is called compact if from every open cover of $F$, we can select a finite subcover.

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    There are many equivalent definitions of "compactness", so you are going to have to tell us which is the one that you use as "the" definition of compactness. Is it "every open cover has a finite subcover"? Is it "every collection of closed sets that has the finite intersection property has nonempty intersection"? Something else?2011-11-22
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    @Arturo, Hopefully it's not the latter definition that you gave!2011-11-22
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    @The Chaz: Well, if it is the latter, then it's a trivial problem (-:. It could also be, for this context, "closed and bounded".2011-11-22
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    I have added the definition. Sorry, I didn't add it before.2011-11-22
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    @AKM: Do you know that compact sets are necessarily closed, or do you not know that?2011-11-22
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    @AKM: This might be helpful: http://en.wikipedia.org/wiki/Nested_intervals2011-11-22
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    @Arturo: I know that a set is compact if and only if it is closed and bounded.2011-11-22
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    Indeed, since the real numbers are complete, we have that "compact" <=> "closed and bounded" (edit... 15 seconds too late!)2011-11-22

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We prove it by contrapositive. (I assume you know that compact implies closed in general; if you don't want to invoke this fact, then see ahead).

For each $j\in I$, let $\mathscr{O}_j = \mathbb{R}-A_j$. Since $A_j$ is closed, $\mathscr{O}_j$ is open.

If $\bigcap\limits_{i\in I}A_i=\emptyset$, then $$\mathbb{R} = \mathbb{R}-\bigcap_{j\in I}A_j = \bigcup_{j\in I}(\mathbb{R}-A_j)= \bigcup_{j\in I}\mathscr{O}_j.$$

In particular, for any $i\in I$, $A_i\subseteq \bigcup_{j\in I}\mathscr{O}_j$. Fix $i_0\in I$. Since $A_{i_0}$ is compact, this open cover has a finite subcover, $\mathscr{O}_{j_1}\cup\cdots\cup\mathscr{O}_{j_n}$. So $$A_{i_0}\subseteq \mathscr{O}_{j_1}\cup\cdots\cup\mathscr{O}_{j_n}.$$ That means that $$\emptyset = A_{i_0}\cap\left(\mathbb{R}-\bigcup_{k=1}^n\mathscr{O}_{j_k}\right) = A_{i_0}\cap\left(\mathbb{R}-\bigcup_{k=1}^n(\mathbb{R}-A_{j_k})\right) = A_{i_0}\cap\left(\bigcap_{k=1}^n A_{j_k}\right).$$ Thus, there is a finite subcollection that has empty intersection.

We have shown that if the entire family has empty intersection, then there is a finite subcollection that has empty intersection. By contrapositive, this gives the desired result.

Added. If you want to show that compact implies closed (at least in $\mathbb{R}$; in fact, it is true in any Hausdorff space) directly using the definition, let $C$ be compact in $\mathbb{R}$, and let $x\notin C$. Then for every $c\in C$ there is an open set $\mathscr{O}_c$ that contains $c$ and an open set $\mathscr{U}_c$ that contains $x$ with $\mathscr{U}_c\cap \mathscr{O}_c=\emptyset$. Since $\cup \mathscr{O}_c$ contains $C$, which is compact, there is a finite subcover $C\subseteq \mathscr{O}_{c_1}\cup\cdots\cup \mathscr{O}_{c_n}$. Let $\mathscr{U}=\mathscr{U}_{c_1}\cap\cdots \cap \mathscr{U}_{c_n}$. Then $\mathscr{U}$ is an open set that constains $x$, and $$\mathscr{U}\cap C\subseteq \mathscr{U}\cap(\mathscr{O}_{c_1}\cup\cdots \cup \mathscr{O}_{c_n}) = \emptyset,$$ so $x\notin \overline{C}$. Thus, $\overline{C}\subseteq C$, which gives $C=\overline{C}$, so $C$ is indeed closed.

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    Thanks very much for your answer. I understand almost everything except this part: $$\emptyset = A_{i_0}\cap\left(\mathbb{R}-\bigcup_{k=1}^n\mathscr{O}_{j_k}\right)$$ I'm failing to see why it is true.2011-11-22
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    @AKM: If $A$ is contained in $\cup \mathscr{O}_{j_k}$, then it cannot intersect its complement.2011-11-22
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    huh...thanks${}{}{}{}{}$2011-11-22
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    @AKM: So, it's clear now?2011-11-22
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    Very very clear. Thanks once again.2011-11-22