Let $\mathcal{U}$ be an [[Cover of a Set|open cover]] of a [[Metric Space|metric space]] $(X,d)$. A number $\delta>0$ is called a Lebesgue number for $\mathcal{U}$ if for any $x\in X$, there exists $U \in \mathcal{U}$ such that $B(x,\delta)\in U$. > [!NOTE] The existence of Lebesgue's numbers for compact metric spaces > Every open cover $\mathcal{U}$ of a [[Compactness|compact]] metric space $(X, d)$ has a Lebesgue number. **Proof**: For every $x\in X$ there exists $r(x)>0$ such that $B(x, r(x)) \subset U_{x}$ for some $U_{x} \in \mathcal{U}$. The collection $\{ B(x, r(x)/2): x\in X \}$ forms an open cover of $X$, so has a finite subcover $\{ B(x_{j}, r(x_{j})/2): j=1, 2, \dots, n \}.$ Set $\delta = \min_{j} r(x_{j})/2$ (which exists by [[Well-Ordering Principle|well-ordering]]), we claim that $\delta$ is a Lebesgue number for $\mathcal{U}$. Given $x\in X$, we must have $x \in B(x_{j}, r(x_{j})/2)$ for some $1 \leq j\leq n$. Then, since $\delta \leq r(x_{j})/2$ for all $j$, the triangle inequality yields $B(x, \delta) \subset B \left( x_{j}, \frac{r(x_{j})}{2} + \delta \right) \subset B(x_{j}, r(x_{j})) \subset U_{x_{j}}$which completes the proof. $\square$