= Solution
Let $(x_n)$ be any <sequence> in a complete, totally bounded metric space $X$. Cover $X$ by finitely many balls of radius $2^{-1}$; one contains infinitely many terms. From those terms choose an infinite subsequence lying in one ball of radius $2^{-2}$, and continue. The <diagonal argument> produces a subsequence $(x_{n_j})$ whose tail from its $j$th term onward lies in a ball of radius $2^{-j}$. Hence
$$
d(x_{n_k},x_{n_\ell})<2^{1-j}
$$
whenever $k,\ell\geq j$, so the subsequence is Cauchy. Completeness makes it converge in $X$.
Every sequence in $X$ therefore has a convergent subsequence, so $X$ is <sequentially compact space>[sequentially compact]. Sequential compactness is equivalent to compactness for metric spaces, and thus $X$ is compact. This proves that a <complete totally bounded metric space is compact>.
Back to article page
