Lebesgue's number lemma

In topology, Lebesgue's number lemma, named after Henri Lebesgue, is a useful tool in the study of compact metric spaces. It states:

:If the metric space $(X, d)$ is compact and an open cover of $X$ is given, then there exists a number $\delta > 0$ such that every subset of $X$ having diameter less than $\delta$ is contained in some member of the cover.

Such a number $\delta$ is called a Lebesgue number of this cover. The notion of a Lebesgue number itself is useful in other applications as well.

Proof

Let $\mathcal U$ be an open cover of $X$. Since $X$ is compact we can extract a finite subcover $\ \subseteq \mathcal U$.
If any one of the $A_i$'s equals $X$ then any $\delta > 0$ will serve as a Lebesgue number.
Otherwise for each $i \in \$, let $C_i := X \smallsetminus A_i$, note that $C_i$ is not empty, and define a function $f : X \rightarrow \mathbb R$ by

: $f(x) := \frac \sum_^n d(x,C_i).$

Since $f$ is continuous on a compact set, it attains a minimum $\delta$.
The key observation is that, since every $x$ is contained in some $A_i$, the extreme value theorem shows $\delta > 0$. Now we can verify that this $\delta$ is the desired Lebesgue number.
If $Y$ is a subset of $X$ of diameter less than $\delta$, then there exists $x_0\in X$ such that $Y\subseteq B_\delta(x_0)$, where $B_\delta(x_0)$ denotes the ball of radius $\delta$ centered at $x_0$ (namely, one can choose $x_0$ as any point in $Y$). Since $f(x_0)\geq \delta$ there must exist at least one $i$ such that $d(x_0,C_i)\geq \delta$. But this means that $B_\delta(x_0)\subseteq A_i$ and so, in particular, $Y\subseteq A_i$.

References

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Theorems in topology
Lemmas

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