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Theorem. If $\{K_\alpha\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_\alpha\}$ is nonempty, then $\cap K_\alpha$ is nonempty.

Proof. Fix a member $K_1$ of $\{K_\alpha\}$ and put $G_\alpha=K_\alpha^c$ [this denotes the complement of $K_\alpha $]. Assume that no point of $K_1$ belongs to every $K_\alpha$. Then the sets $G_\alpha$ form an open cover of $K_1$ [this took me a bit but that's by the last assumption]; and since $K_1$ is compact, there are finitely many indices $\alpha_1,\dots,\alpha_n$ such that $K_1\subset \bigcup_{i=1}^nG_{\alpha_i}$ [so far so good]. But this means that $K_1\cap K_{\alpha_1}\cap\dots K_{\alpha_n} $is empty, in contradiction to our hypothesis.

I just don't see how the very last part is implied. Can someone help me see it?

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    It's a contradiction argument. You start by assuming $\cap K_{\alpha}$ is empty. Then that would imply no point of $K_1 $ belongs to every alpha. So when $K_1 \subset \cup_{i=1}^n G_{\alpha_i}$, it means $K_1 \subset \cup_{i=1}^n K_{\alpha_i}^C$. Now use that if $A \subset B$ then $A \cap B^C = \phi$.2012-11-15

4 Answers 4

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Since $K_1\subset \cup_i G_{\alpha_i} = \cup_i K_{\alpha_i}^c$, each point $x$ of $K_1$ is contained in the complement of some $K_{\alpha_i}$. That is, for each point $x$ of $K_1$ there is an $i$ so that $x\notin K_{\alpha_i}$. This means the intersection of all of the sets $K_1\cap K_{\alpha_1}\cap\cdots\cap K_{\alpha_n}$ is empty.

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    They do twist the brain in a knot, don't they? You're very welcome!2012-11-15
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For $x \in K_1$, there exists $i$ such that $x \in K_{\alpha_i}^c$, so $x \notin K_{\alpha_1} \cap \dots \cap K_{\alpha_n}$.

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It follows from the following identity. Let $\{E_\alpha\}_{\alpha \in A}$ be a collection of subsets of $X$. Then,

$\bigcup_{\alpha \in A} X - E_\alpha = X - \bigcap_{\alpha \in A} E_\alpha$

$\bigcap_{\alpha \in A} X - E_\alpha = X - \bigcup_{\alpha \in A} E_\alpha$

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This is a contrapositive argument. We supposed that the second part of the argument is false. That is, we constructed a collection $\{K_{\alpha}\}$ for which $\bigcap K_{\alpha}$ is empty, and then derived that the first part would be false in this case.

The first part says that the intersection of an arbitrary finite subcollection would not be empty, but the intersection of a subcollection of our specifically constructed collection is empty contrary to that.

The choice of words in the book ("in contradiction to") is confusing.