# Show that the entire collection has non-empty intersection and its counter example. (non-empty intersection property)

Let $$(X, d)$$ be a compact metric space. Suppose that $$(K_\alpha)_{\alpha \in I}$$ is a collection of closed sets in $$X$$ with the property that any finite subcollection of these sets necessarily has non-empty intersection, thus $$\bigcap_{\alpha \in F} K_\alpha \not= \emptyset$$ for all finite $$F \subseteq I$$. (This property is known as the finite intersection property.) Show that the entire collection has non-empty intersection, thus $$\bigcap_{\alpha \in I}K_\alpha \not= \emptyset$$. Show by counterexample that this statement fails if $$X$$ is not compact.

Suppose that $$\bigcap_{\alpha \in I}K_\alpha = \emptyset$$. This implies that $$\bigcup_{\alpha \in I}K_\alpha^c = X$$. $$X$$ is compact and $$\bigcup_{\alpha \in I}K_\alpha^c$$ is an open cover of $$X$$. Thus, there exist a finite subcover. This implies that $$I$$ must be finite. But, this is contradiction due to the finite intersection property.

I am struggling to come up with a counter example. I appreciate if you give some hint.

Let $$X = (0,1)$$ and $$K_n = (0,\frac{1}{n}]$$. Can you finish?