From Rudin's Principles of Mathematical Analysis:
Theorem 2.36: If {$K_\alpha$} is a collection of compact sets 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 $G_\alpha = K_\alpha^c$. Assume that no point of $K_1$ belongs to every $K_\alpha$. Then the sets $G_\alpha$ form an open cover of $K_1$; and since $K_1$ is compact, there are finitely many indices $\alpha_1, … , \alpha_n$ such that $K_1\subset G_{\alpha_1}\cup … \cup G_{\alpha_n}$.
But this means that $K_1\cap K_{\alpha_1}\cap …\cap K_{\alpha_n}$ is empty, in contradiction to our hypothesis.
My Question: Is it really necessary to use compactness to prove that $\cap K_\alpha$ is nonempty… I mean, we could have just used the fact that since $K_1$ has no elements in common with the other $K_\alpha$'s (from the assumption), then any finite subcollection {$K_\beta$} that would have $K_1$ as one of its elements would then have an empty intersection, and then $\cap K_\alpha$ = $\varnothing$. At least that's what I think… Please tell me what's wrong with my reasoning.
Best Answer
I really don't like the way the proof is presented in Rudin, and I'd go as follows: suppose $\;\bigcap_\alpha K_\alpha=\emptyset\;$ , but then fixing $\;\alpha_0\;$ we get (de Morgan):
$$K_{\alpha_0}=K_{\alpha_0}\setminus\emptyset=K_{\alpha_0}\setminus\bigcap_\alpha K_\alpha=\bigcup_\alpha\left(K_{\alpha_0}\setminus K_\alpha\right)$$
But $\;K_{\alpha_0}\setminus K_\alpha\;$ is open for every $\;\alpha\;$, so by compactness of $\;K_{\alpha_0}\;$ we have that there exists a finite set $\;\{\alpha_1,...,\alpha_n\}\;$ s.t.
$$K_{\alpha_0}=\bigcup_{i=1}^n\left(K_{\alpha_0}\setminus K_{\alpha_i}\right)=K_{\alpha_0}\setminus\bigcap_{i=1}^nK_{\alpha_i}\implies \bigcap_{i=1}^nK_{\alpha_i}=\emptyset$$
and there's our contradiction.
Of course, nothing new is presented above. It is just the way it is presented that changes and that imo makes it easier to grasp.