Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

3 Answers 3

up vote 1 down vote accepted

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.

share|improve this answer
    
"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 $x \notin K_{\alpha_1} \cap \dots \cap K_{\alpha_n}$.

share|improve this answer

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

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.