Let the points \(x_n\) of a sequence \((x_n)_{n\in\mathbb N}\) be contained in the metric space \((X,d)\), i.e. \(x_n\in X\), for all \(n\in\mathbb N\). Let this sequence be convergent and let \(X\) also contain its limit
\[x=\lim_{n\to\infty} x_n,\] i.e. \(x\in X\). Then the subset $U\subset X$ defined by \[U:=\{x_n:~n\in\mathbb N\}\cup \{x\},\] i.e. the subset consisting of all sequence members \(x_n\) and also containing the limit $x$ is compact.
Proofs: 1
Proofs: 1