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

A metric space $X$ is a compact space if every open cover $X={\bigcup }_{\alpha \in I}{U}_{\alpha }$, with all $U_{\alpha }$ open in $X$, has a finite subcover: finitely many indices ${\alpha }_1,\dots ,{\alpha }_r$ 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 $X$. In formulas,A finite subcover is the defining test for compactness; a convergent subsequence is the defining test for sequential compactness.

Let $(x_n)$ be any sequence in the compact metric space $X$. There must be an $x\in X$ such that every open ball about $x$ contains $x_n$ for infinitely many indices $n$. Otherwise, for each $x\in X$ choose a ball $B(x,r_x)$ containing only finitely many terms, counted by indices. These balls are an open cover. A finite subcover would imply that only finitely many indices occur anywhere in $X$, a contradiction.

Now choose successively $n_k>n_{k-1}$ with $x_{n_k}\in B(x,1/k)$; infinitely many available indices make this possible. Then $d(x_{n_k},x)<1/k\to 0$, so $(x_{n_k})$ is a convergent subsequence. ThusThe proof counts indices rather than distinct points, so it also handles a value repeated infinitely often.