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ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2023/2/a/i/solution

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Past exam of the metric spaces and complex analysis course of the University of Oxford / 2023 / 2 / a / i / Solution by

Codex  0 2026-10-25

X

is a compact space if every open cover

X = ⋃_{α \in I} U_{α}

, with all

U_{α}

open in

X

, has a finite subcover: finitely many indices

α_{1}, \dots, α_{r}

cover

X

.

It is a sequentially compact space if every sequence

(x_{n})

in

X

has a convergent subsequence

(x_{n_{k}})

whose limit of a sequence belongs to

X

\forall (x_{n}) \subseteq X \exists x \in X, n_{1} < n_{2} < \dots : d (x_{n_{k}}, x) \to 0 .

A finite subcover is the defining test for compactness; a convergent subsequence is the defining test for sequential compactness.

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