# Understanding this proof about the intersection of compact subsets

The following proof is theorem 2.36 from Rudin's Principles of Mathematical Analysis:

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 $\bigcap K_\alpha$ is nonempty.

Proof: Fix a member $K_1$ of $\{K_\alpha\}$ and put $G_\alpha = K^c_\alpha$. Assume that no point of $K_1$ belongs to every $K$. Then the sets $G_\alpha$ form an open cover of $K_1$; and since $K_1$ is compact, there are finitely many indices $\alpha_1,\dots,\alpha_n$ such that $K_1\subset G_{\alpha_{1}}\cup\dots\cup\,G_{\alpha_{n}}$. But this means that $$K_1\cap\,K_{\alpha_{1}}\cap\dots\cap\,K_{\alpha_{n}}$$ is empty, in contradiction to our hypothesis.

I can make neither heads nor tails of any part of this proof and am especially lost about the last sentence. He doesn't seem to have made any hypothesis. In fact, this seems to prove the opposite of the theorem. What's going on here?

• It's a proof by contrapositive. To prove $A\implies B$, one proves $(\lnot B) \implies (\lnot A)$. – Daniel Fischer Sep 3 '15 at 16:09
• The hypothesis he refers to is that no point in $K_1$ belongs to every $K$. – Alex S Sep 3 '15 at 16:13
• But if no point in $K_1$ belongs to every K then i don't understand why the conclusion is a contradiction. – Zachary F Sep 3 '15 at 16:27
• "Assume that … $\bigcap K_\alpha = \varnothing$". Then he shows that already some finite subfamily has empty intersection. – Daniel Fischer Sep 3 '15 at 16:28
• Could I ask you to outline that in more detail? I still don't understand. – Zachary F Sep 3 '15 at 16:31

## 1 Answer

Since no point of $K_1$ belongs to every $K_\alpha$ $\implies$ $K_1$ belongs to $\cup G_\alpha$. Note: $G_\alpha$'s are open and they form an open cover of $K_1$

Since $K_1$ is compact, there will be a finite subcover.

Let $G_{\alpha_1},G_{\alpha_2},...,G_{\alpha_n}$ be the finite subcover.

$\implies K_1 \subset G_{\alpha_1} \cup G_{\alpha_2} \cup ... \cup G_{\alpha_n}$

Since $K_1$ is in this finite subcover, $\implies$ $K_1 \cap (G_{\alpha_1} \cup G_{\alpha_2} \cup ... \cup G_{\alpha_n})^c = \phi$

$\implies K_1 \cap (G_{\alpha_1})^c \cap (G_{\alpha_2})^c \cap ... \cap (G_{\alpha_n})^c = \phi$

$\implies K_1 \cap K_{\alpha_1} \cap K_{\alpha_2} \cap ... \cap K_{\alpha_n} = \phi$

This contradicts the finite intersection property.