The nonemptiness of the intersection of compact sets such that all finite intersections are nonempty

Question

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.

Answer 1

$K_1$ might have lots of points in common with other $K_\alpha$s, indeed it must since every finite sub-collection (such as every intersection of two sets) has nonempty intersection. The point is that we assume that no point of $K_1$ is in every $K_\alpha$ (by way of contradiction). I suppose you could assume (again by way of contradiction) that there was no elements in common with $K_1$ at all, but then you aren't satisfying the assumption of the theorem in the first place.

Answer 2

For each $a\in\mathbb R$, let $K_a=[a,\infty)$. Note that $K_a$ is a family of closed subsets of $\mathbb R$ in which every finite subcollection has non-empty intersection, however $\bigcap K_a=\varnothing$. So you need compactness at some point.

Answer 3

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.