Compactness implies total boundedness

Compactness implies total boundedness: If $(X,d)$ is a compact metric space , then for every $\varepsilon>0$ there exist points $x_1,\dots,x_N\in X$ such that $X\subseteq \bigcup_{j=1}^N B(x_j,\varepsilon).$ Equivalently, $X$ is totally bounded .

Total boundedness strengthens boundedness by requiring finitely many $\varepsilon$-balls to cover the set. Together with completeness , it characterizes compactness in metric spaces.

Proof sketch (optional): The family $\{B(x,\varepsilon):x\in X\}$ is an open cover of $X$. Compactness yields a finite subcover.