Proposition: Let $\{K_\alpha\}$ be a collection of compact subsets of metric space $X$. If any finite subcollection of $\{K_\alpha\}$ has a nonempty intersection, then $\bigcap_\alpha K_\alpha \neq \emptyset$.

Attempt(by contradiction):

Let $\{K_\alpha\}$ be a collection of compact subsets of metric space $X$. Assuming that the intersection of any finite subcollection of $\{K_\alpha\}$ is not empty; $\bigcap_{i=1}^{n}K_{{\alpha}_{i}}\neq\emptyset$, we want to show that $\bigcap_\alpha K_\alpha\neq\emptyset$. If $\bigcap_\alpha K_\alpha=\emptyset$, then $\forall x\in X, \exists \alpha$ such that $x\notin K_\alpha$. Since $\bigcap_{i=1}^{n}K_{{\alpha}_{i}}\subseteq X$, then for all $k \in \bigcap_{i=1}^{n}K_{{\alpha}_{i}}$ there exists $\gamma$ such that $k \notin K_\gamma$. This implies $(\bigcap_{i=1}^n K_{\alpha_i})\bigcap K_\gamma=\emptyset$ and $\{K_{\alpha_i}\}_{i=1}^n\bigcup\{K_\gamma\}$ is a finite subcollection with an empty intersection - a contradiction.

  • $\begingroup$ Where did you use compactness? Also, please state what you are trying to prove. One immediate problem is that $\bigcap_{i=1}^{\infty} K_{\alpha_i}$ doesn't look like a finite intersection to me, unless you are making some assumption about the $\alpha_i$ values. $\endgroup$ – Bungo Aug 8 '18 at 5:41
  • $\begingroup$ I made a typo and yes I did not use compactness. But, do we really need to use it? $\endgroup$ – TheLast Cipher Aug 8 '18 at 5:47
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    $\begingroup$ Yes, the result is fale if they are not compact : there are exemples in $\mathbb{R}$ $\endgroup$ – Max Aug 8 '18 at 5:53
  • $\begingroup$ I wonder where my argument fails though. $\endgroup$ – TheLast Cipher Aug 8 '18 at 5:55
  • $\begingroup$ You picked a specific $k$ and showed that there was some finite intersection in which it wasn't; you didn't show that the finite intersection in question was empty. Try to look for a counterexample without compactness to see what happens $\endgroup$ – Max Aug 8 '18 at 6:00

For a $k \in \bigcap_{1}^{n} K_{\alpha_{i}}$ there is $\gamma$ such that $k \not\in K_{\gamma}$, but may exist $l \in \left(\bigcap_{1}^{n} K_{\alpha_{i}}\right)\cap K_{\gamma}$. The problem is: you cannot find a $\gamma$ that works for every $k$.

Couterexample. Consider the family (in $\mathbb{R}$) $\{[n, \infty)\}_{n \in \mathbb{N}}$. So, given any finite subcollection has a nonempty intersection. But, if $\alpha \in \bigcap_{n \in \mathbb{N}}\{[n, \infty)\}$, then $\alpha \in [n,\infty)$ for every $n \in \mathbb{N}$, that is, $\alpha \geq n$ for every $n \in \mathbb{N}$, an absurd!

Here's a proof using your idea:

Proof. Suppose that $\bigcap_{\alpha}K_{\alpha} = \emptyset$ and fix $\alpha_{0}$. Note that if $x \in K_{\alpha_{0}}$, since $\bigcap_{\alpha}K_{\alpha} = \emptyset$, then $$x \in \left(\bigcap_{\alpha}K_{\alpha}\right)^{c} = \bigcup_{\alpha}K_{\alpha}^{c}.$$ But, $K_{\alpha_{0}}$ is compact, there is $\alpha_{1},..., \alpha_{m}$ such that $$K_{\alpha_{0}} \subset \bigcup_{i=1}^{m}K_{\alpha_{i}}^{c} = \left( \bigcap_{i=1}^{m}K_{\alpha_{i}}\right)^{c}.$$ Therefore, $K_{\alpha_{0}} \cap K_{\alpha_{1}} \cap ... \cap K_{\alpha_{m}} = \emptyset$, a contradiction.


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