= Compactness from uniformly vanishing sequence tails
Let $a_m\geq0$ and $a_m\to0$. The set $K=\{x=(x_m):x_m\in\mathbb R,\ |x_m|\leq a_m\}$ is a <compact metric space> for the <supremum norm> metric. Every member is a sequence tending to zero.
For any sequence in $K$, successively extract a <convergent subsequence> in each <compact interval> $[-a_m,a_m]$, then take diagonal indices. The coordinate limits define $x\in K$. For any $M$, coordinatewise differences in the tail are bounded by $2\sup_{m>M}a_m$, independently of the sequence index. This tends to zero as $M\to\infty$, while finitely many initial coordinates converge together. Hence the diagonal subsequence converges in the <supremum norm>. This proves <sequential compactness>, equivalent to <compactness> for <metric spaces>.
Back to article page
