Show that a sequentially compact space is totally bounded.
Exercise
Solution
1. Let \((X,d)\) be sequentially compact. Suppose, toward a contradiction, that \(X\) is not totally bounded.
Then there exists \(\varepsilon>0\) such that no finite family of open \(\varepsilon\)-balls covers \(X\).
2. Choose inductively \(x_1,x_2,\dots\) so that\[d(x_m,x_n)\ge\varepsilon\qquad(m\ne n).\]
After choosing \(x_1,\dots,x_n\), their \(\varepsilon\)-balls do not cover \(X\), so choose \(x_{n+1}\) outside their union.
3. This sequence has no convergent subsequence.
Every convergent sequence in a metric space is Cauchy, but any two distinct terms of any subsequence remain at distance at least \(\varepsilon\).
4. This contradicts sequential compactness. Therefore \(X\) is totally bounded.
Sequential compactness requires every sequence to have a convergent subsequence.