# Just having problems following one crucial step in the proof of theorem 2.36 in 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 $\cap K_\alpha$ 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?

-
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 $A \cap B^C = \phi$. – Gautam Shenoy Nov 15 '12 at 5:54

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.

-
"That is, for each point x of $K_1$ there is an $i$ so that $x∉K_{α_i}$." Yes. Jeez. Seems obvious now. Thank you. I should always try to make these formal statements about what it means to be a subset and things I think. Thank you so much! – crf Nov 15 '12 at 5:56
They do twist the brain in a knot, don't they? You're very welcome! – Neal Nov 15 '12 at 6:15

For $x \in K_1$, there exists $i$ such that $x \in K_{\alpha_i}^c$, so $x \notin K_{\alpha_1} \cap \dots \cap K_{\alpha_n}$.

-

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

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

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

-