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?