Here, we show that any [[Compact spaces|compact]] metric space $(X,d)$ is [[Complete metric spaces|complete]]. Recall that for metric space, [[Equivalence of compactness and sequential compactness for metric spaces|sequential compactness is equivalent to compactness]]. Hence any Cauchy sequence $(x_{n})$ has a convergent subsequence $x_{n_{j}}$ whose limit we denote by $x$. Now we can show that $x_{n}\to x$.