If any finite subcollection of finite sets has non-empty intersection, then the infinite intersection is non-empty as well.

Question

The space is R with usual metric.

Since all the sets are finite, they are compact as well.

Suppose {$A_µ$} is the collection of finite sets and the infinite intersection is empty. Let K belong to {$A_µ$}. So, no element of K is contained in any $A_µ$. Let the complement of $A_µ$ be $G_µ$. So {$G_µ$} forms an open cover for K. Since K is compact, K admits a finite subcover, $G_1$, $G_2$,..., $G_n$ such that K is contained in the union of this finite subcover. This implies that K is not contained in the intersection of $A_1$, $A_2$,..., $A_n$ (Complement of union is intersection of complements). Therefore, intersection of K with {$A_µ$}, 1 ≤ µ ≤ n is empty.

This contradicts the hypothesis that finite subcollection has non empty intersection.

Is my approach correct?

Also, can this be done without using compactness of finite sets? If yes, then please give me a hint.

Thanks in advance :)

Answer 1

Compactness is not needed.

Let $\mathscr{F}$ be a family of finite subsets of $\Bbb R$, and suppose that $\bigcap\mathscr{F}=\varnothing$. Fix $F\in\mathscr{F}$. Then for each $x\in F$ there is an $F_x\in\mathscr{F}$ such that $x\notin F_x$. But then $\{F\}\cup\{F_x:x\in F\}$ is a finite subfamily of $\mathscr{F}$ whose intersection is empty.