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I realize the following simplified claim should suffice for my purpose.

If the intersection of a family of compact sets in $\Bbb R^n$ are empty, then there exist finite many sets in the family with empty intersection.

Previously asked:

Let $S$ be a family of closed sets contained in a compact set $K$ in $\Bbb R^n$, $\bigcap S = \emptyset$ (the intersection of all sets in $S$ is empty), then there exists finite many $s_1,...,s_k\in S,k>0$ such that $\bigcap_{i=1}^k s_i = \emptyset$.

After some investigation, I think this might be related to Standard compactness argument, but I have trouble using the result that "disjoint compact sets have disjoint open neighborhoods" to above desired claim.

River
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1 Answers1

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It is not related to the $T_4$ property.

Let $K$ be a non-empty compact space. Let $F$ be a non-empty family of closed sets and let $F$ have the F.I.P. (Finite Intersection Property). That is, if $G$ is a finite non-empty subset of $F$ then $\cap G\ne \phi.$ Then $\cap F\ne \phi.$

PROOF: Suppose not. Then $C=\{K$ \ $f: f\in F\}$ is an open cover of $K.$ And $K$ is compact. So $C$ has a finite sub-cover $D=\{K$ \ $f: f\in G\}$ where $G$ is a finite subset of $F.$ And $G\ne \phi$ because $K\ne \phi.$ $$\text {Now }\; K=\cup D \;\text { so }\quad \phi=\cap \{K \backslash d: d\in D\}=\cap \{K \backslash (K \backslash f): f\in G\}=\cap G.$$ But since G is finite and non-empty, this contradicts the F.I.P. of $F. $......... QED.

So if $\cap F=\phi$ then $F$ does not have the F.I.P.: There exists non-empty finite $G\subset F$ with $\cap G=\phi.$

Note that this applies to any compact space $K.$

We can use this method to show the converse: If every non-empty family $F$ of closed subsets (of a space $K$ ) that has the F.I.P. also satisfies $\cap F\ne \phi,$ then $K$ is compact.

Every statement about open sets is equivalent to a "dual" statement about their complements (closed sets), and often the dual is useful on its own.

DanielWainfleet
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