A Topological Space $K$ is sequentially compact if for every sequence $( x_{n} )_{n\in \mathbb{N}}\subseteq K$, one can find a convergent subsequence in $K$ i.e. $$\forall(x_{n})_{n\in \mathbb{N}}\subseteq K , \exists (n_{k})_{k\in \mathbb{N}}:\lim_{ k \to \infty } x_{n_{k}}=x\in K$$