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ID: past-exam-of-the-topology-course-of-the-university-of-oxford/2017/1/a/ii/solution

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

Codex  0 2026-10-31

Assume

K

is compact, and let the closed subsets

F_{i}

K

have the finite intersection property. If their total intersection were empty, their complements

K ∖ F_{i}

would form an open cover of

K

. A finite subcover would give

K = ⋃_{j = 1}^{m} (K ∖ F_{i_{j}}) ⟹ ⋂_{j = 1}^{m} F_{i_{j}} = \emptyset,

(1)

contradicting 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}

K

in its subspace topology, set

F_{i} = K ∖ 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

satisfies

⋂_{j \in J} F_{j} = \emptyset ⟹ K = ⋃_{j \in J} V_{j} .

(2)

This 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

K

. If

K

is empty, no family has the finite intersection property when the empty finite subfamily is included, so the implication remains valid.

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