Past exam of the topology course of the University of Oxford / 2017 / 1 / a

A subset $K$ is compact if every open cover has a finite subcover. Explicitly, if $U_i$ are open sets of $X$ and $K\subseteq {\bigcup }_{i\in I}{U}_i$, there are finitely many indices $i_1,\dots ,i_m$ such thatEquivalently, the subspace topology makes $K$ a compact space. An open set in this subspace topology is $K\cap U$ for an ambient open set $U$, which proves equivalence of the two open cover formulations. The empty set is compact: the empty selection is a finite subcover.

Assume $K$ is compact, and let the closed subsets $F_i$ of $K$ have the finite intersection property. If their total intersection were empty, their complements $K\setminus F_i$ would form an open cover of $K$. A finite subcover would givecontradicting the finite intersection property. Therefore the total intersection is nonempty.

Conversely, assume the stated nonempty-intersection property for every family of closed subsets of $K$. Given an open cover $\{V_i:i\in I\}$ of $K$ in its subspace topology, set $F_i=K\setminus V_i$. Their total intersection is empty. By the contrapositive of the assumed property, this family fails the finite intersection property; hence some finite $J\subseteq I$ satisfiesThis is a finite subcover, proving compactness. Both directions are instances of De Morgan's laws. Thus compactness is equivalent to the finite intersection characterization of compactness.

For completeness, the intersection of the empty family of subsets of $K$ is $K$. If $K$ is empty, no family has the finite intersection property when the empty finite subfamily is included, so the implication remains valid.

Let $F$ be a closed subset of the compact space $X$. Express an arbitrary open cover of $F$ in its subspace topology as $\{F\cap U_i:i\in I\}$, where the $U_i$ are open sets of $X$. Since $X\setminus F$ is open, the familyis an open cover of $X$. Take a finite subcover and discard $X\setminus F$ if it occurs. The remaining finitely many $U_i$ still cover $F$, since the discarded set contains no point of $F$. Their traces on $F$ give the required finite subcover. Hence every closed subset of a compact space is compact, with no Hausdorff assumption.