Proof. Suppose no such \(\delta \) exists. Then for each \(n\) the number \(\frac 1n\) fails, so there is a set \(A_n\subset X\) with \(\diam (A_n)<\frac 1n\) lying inside no member of \(\xi \). Choose \(x_n\in A_n\).

By sequential compactness some subsequence \(x_{n_k}\rightarrow x\). Now \(x\) lies in some \(G\in \xi \), and \(G\) is open, so \(B(x;\varepsilon )\subset G\) for some \(\varepsilon >0\). Choose \(k\) so large that \[d(x_{n_k},x)<\frac {\varepsilon }{2}\qquad \text {and}\qquad \frac {1}{n_k}<\frac {\varepsilon }{2}.\] Every point of \(A_{n_k}\) is within \(\frac {1}{n_k}\) of \(x_{n_k}\), hence within \(\varepsilon \) of \(x\), so \[A_{n_k}\subset B(x;\varepsilon )\subset G ,\] contradicting the choice of \(A_{n_k}\) as lying inside no member of \(\xi \). □

Proof.

(i) \(\implies \) (ii)

Let \((x_n)\) be a sequence in \(X\) with no convergent subsequence. Then no point \(x\) of \(X\) is a limit of a subsequence, so each \(x\) has a ball \(B(x;r_x)\) containing \(x_n\) for only finitely many indices \(n\). These balls form an open cover; a finite subcover \(B(y_1;r_1),\dots ,B(y_m;r_m)\) would then contain \(x_n\) for only finitely many \(n\) in total, yet it must contain every term of the sequence. That is impossible, so some subsequence converges.

(ii) \(\implies \) (iii)

Total boundedness is the theorem of the previous subsection: every sequence in a sequentially compact space has a convergent, hence Cauchy, subsequence, and that is equivalent to total boundedness. For completeness, let \((x_n)\) be Cauchy. By hypothesis it has a subsequence converging to some \(x\). A Cauchy sequence with a convergent subsequence converges to the same limit, since given \(\varepsilon >0\) one may choose \(N\) with \(d(x_m,x_n)<\frac {\varepsilon }{2}\) for \(m,n\geq N\) and then a term \(x_{n_k}\) of the subsequence with \(n_k\geq N\) and \(d(x_{n_k},x)<\frac {\varepsilon }{2}\); the triangle inequality gives \(d(x_n,x)<\varepsilon \) for all \(n\geq N\).

(iii) \(\implies \) (ii)

Let \((x_n)\) be a sequence. Total boundedness gives a Cauchy subsequence, and completeness makes it converge.

(ii) \(\implies \) (i)

Let \(\xi \) be an open cover of \(X\) and let \(\delta >0\) be a Lebesgue number for it. By the implication already proved, \(X\) is totally bounded, so it can be covered by finitely many sets \(E_1,\dots ,E_n\) each of diameter less than \(\delta \). By the choice of \(\delta \) each \(E_k\) lies in some \(G_k\in \xi \), and then \[X = \bigcup _{k=1}^{n}E_k \subset \bigcup _{k=1}^{n}G_k .\] So \(\{G_1,\dots ,G_n\}\) is a finite subcover. □

Solution. Let \(F\) be closed in the compact space \(X\), and let \((x_n)\) be a sequence in \(F\). Regarded as a sequence in \(X\) it has a subsequence converging to some \(x\in X\), by sequential compactness. Since \(F\) is closed and every term of the subsequence lies in \(F\), the limit \(x\) lies in \(F\). So every sequence in \(F\) has a subsequence converging in \(F\), and \(F\) is sequentially compact, hence compact.

The cover argument is just as short, and either is acceptable; the point of having both descriptions is that one of them is usually the easier.