Sequentially compact set

Let $(X,d)$ be a metric space and let $K\subseteq X$. The set $K$ is sequentially compact if for every sequence $(x_n)$ in $K$ there exist:

• a subsequence $(x_{n_k})$, and
• a point $x\in K$ such that $x_{n_k}\to x \quad\text{as }k\to\infty$ (see convergence ).

In metric spaces, sequential compactness is equivalent to compactness , but the sequential formulation is often easier to use in analysis.

Examples:

• In $\mathbb{R}^k$, any closed and bounded set is sequentially compact (Bolzano–Weierstrass + closedness).
• The set $\{1/n:n\in\mathbb{N}\}\subset\mathbb{R}$ is not sequentially compact as a subset of itself (the sequence $1/n$ converges to $0\notin\{1/n\}$).
• Any finite subset of a metric space is sequentially compact (every sequence has an eventually constant subsequence).