Sequentially compact space

A space is sequentially compact when every sequence has a convergent subsequence. A metric space is compact exactly when it is sequentially compact.

Every continuous map $T:X\to X$ on a sequentially compact space with a metric satisfying $d(Tx,Ty)\ge d(x,y)$ is a surjective isometry.

A convergent subsequence of the paired orbit $(T^{k}x,T^{k}y)$ has two late terms arbitrarily close. Removing their common number of iterates by the distance inequality shows that $(x,y)$ returns arbitrarily close to itself after a common positive time $n$. Thus $d(Tx,Ty)\le d(T^{n}x,T^{n}y)\le d(x,y)+2ϵ$, and letting $ϵ\downarrow 0$ proves the isometry claim. For each $x$, choose return times $n_k\ge 1$ with $T^{n_k}x\to x$. A convergent subsequence of $T^{n_k-1}x$ has limit $z$, and continuity gives $Tz=x$. This proves surjectivity.

Let $a_m\ge 0$ and $a_m\to 0$. The set $K=\{x=(x_m):x_m\in \mathbb{R},|x_m|\le a_m\}$ 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.