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I shall present theorem and its proof by Walter Rudin.

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 $G_\alpha=K_\alpha^c$. Assume that no point of $K_1$ belongs to every $K_\alpha$. Then the sets $G_\alpha$ form an open cover of $K_1$; and since $K_1$ is compact, there are finitely many indices $\alpha_1,…,\alpha_n$ such that $K_1\subset \cup_{i=1...n} G_{\alpha_i}$. But this means that $K_1\cap K_{\alpha_1}\cap\dots \cap K_{\alpha_n}$ is empty, in contradiction to our hypothesis.

My confusion is in the part in bold. Does mean it $K_1\cap K_{\alpha} = \emptyset $ for every $\alpha$? If the answer is "Yes", this not contradict that the intersection of every finite subcollection of $\{K_\alpha\} $ is nonempty?

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  • $\begingroup$ Related/duplicate: math.stackexchange.com/questions/1322540/… $\endgroup$ – egreg Oct 5 '15 at 22:23
  • $\begingroup$ It's a proof by contradiction,He shows that is an untenable assumption because it implies a contradiction to the initial premises. Therefore,the negation of the assumption is true.Therefore some point of $K_1$ belongs to every $K_a$. $\endgroup$ – DanielWainfleet Oct 6 '15 at 2:45
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It means assume that no point of $K_1$ belongs to $\bigcap_{\alpha}K_{\alpha}$. ("Belongs to every $K_{\alpha}$" means "belongs to the intersection of the $K_{\alpha}$".) That means that the complements $G_{\alpha}$ of the $K_{\alpha}$ are open sets that cover the compact set $K_1$ and then $\ldots$

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  • $\begingroup$ makes sense, thanks. $\endgroup$ – Yesid Fonseca V. Oct 5 '15 at 22:59
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No. It means that if you fix $K_1,$ then for each $x\in K_1$ there will exist some $\alpha(x)=\alpha$ (which depends on $x$) such that $x\notin K_\alpha.$ Therefore, for each $x\in K_1$ the open set (a compact set is closed and thus its complement is open) $G_{\alpha(x)}$ contains $x$ and hence $\bigcup\limits_{\alpha(x)}G_{\alpha(x)}\supseteq K_1,$ and since $\bigcup\limits_{\alpha(x)}G_{\alpha(x)}\subseteq\bigcup\limits_{\alpha}G_\alpha $ then $\{G_\alpha\}$ forms an open cover of $K_1$ and all else I think you can understand it

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    $\begingroup$ I think Google's translator failure, I interpreted that "every" is equivalent to "each" in my lenguage. And i like your interpretation, thanks. $\endgroup$ – Yesid Fonseca V. Oct 5 '15 at 23:08

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