Past exam of the options II course of the University of Oxford / 2013 / E4 / a

A topological space is compact if every open cover of it has a finite subcover. That is, if $X={\bigcup }_{i\in I}{U}_i$ with all $U_i$ open, finitely many of those same open sets already cover $X$.

If the intersection were empty, the open complements $X\setminus F_i$ would cover the compact space. A finite subcover would give an empty finite intersection of the corresponding $F_i$. But the sets are nested, so any finite intersection is the set with largest index, which is nonempty. This contradiction provesThis is the nested-closed-set form of the finite intersection property for compact spaces.

Suppose no $F_N$ is contained in $U$. Then $K_i=F_i\cap (X\setminus U)$ are nested nonempty closed subsets of the compact space. Part (ii) gives a point in their intersection. Such a point lies both in ${\bigcap }_i{F}_i$ and outside $U$, contradicting the hypothesis. Thus some $F_N\subseteq U$, and nesting givesCompactness upgrades containment of the limiting intersection to eventual containment of the whole nested sets.