 Solution

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

K

is compact if every open cover has a finite subcover. Explicitly, if

U_{i}

X

and

K \subseteq ⋃_{i \in I} U_{i}

, there are finitely many indices

i_{1}, \dots, i_{m}

such that

K \subseteq U_{i_{1}} \cup \dots \cup U_{i_{m}} .

Equivalently, the subspace topology makes

K

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.

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