I have a brief question about Theorem 2.36 in Baby Rudin.
The theorem is as follows:
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.
I actually follow Rudin's proof, but the whole theorem seems a bit counterintuitive for me. After all, it is quite easy to draw, say, three sets, $A$, $B$ and $C$ such that $A\cap B$ is nonempty, $A\cap C$ is nonempty, $B\cap C$ is nonempty, but where $A\cap B \cap C$ is empty. Imagine for instance you draw to circle-sets, $A$ and $B$, on top of each other, where the top of the lower circle intersect the bottom of the upper circle. Then you draw one more set, $C$, sorta formed like a snake, where the "tail" of the snake intersect the right side of the lower circle, extends a while out, curves $180$ degrees (sort of forming a donut-shape), and then its "head" intersects the top circle on the right side (wish I could upload a picture – but hope you get the point). Then $A\cap B$ is nonempty, $A\cap C$ is nonempty, $B\cap C$ is nonempty, but $A\cap B \cap C$ is empty. So if $A, B, C$ are compact, wouldn't the theorem then be invalid?
If someone can explain to me what is wrong with my reasoning above, then I would be very grateful! Is it perhaps that we here assumes that $K_\alpha$ is supposed to be a collection of inifinitely many sets? And in my reasoning above I only use $3$ sets?
Best Answer
Perhaps it would help to think of an analogy with the open cover definition of compactness. A space is compact if every open cover has a finite subcover. However, you can easily come up with examples of compact sets that have a covering with 3 open sets, but no subcover with 2 open sets. (Of course, trivially you can have covers with 1 set and no subcover with 0 sets.)
Compactness allows you to reduce an open cover to a finite open cover, or to check that an intersection of closed sets is nonempty by only checking that every finite intersection is nonempty (and here, it is that last every that is key, as Qiaochu and David pointed out in the comments). In both cases, it is a condition that reduces infinite to finite, roughly speaking. But this says nothing about reducing finite to smaller finite.