Compactness implies completeness

Compactness implies completeness: If $(X,d)$ is a compact metric space , then $X$ is complete .

This shows compactness is a strong finiteness condition: it forces not only boundedness but also the existence of limits for all Cauchy sequences .

Proof sketch (optional): Let $(x_n)$ be Cauchy in $X$. By compactness (sequential compactness ), it has a convergent subsequence $x_{n_k}\to x$. Cauchy-ness forces the full sequence to converge to the same $x$, hence $(x_n)\to x\in X$.