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Let $\mathcal{U}$ be an oper cover of a topological space $A \subseteq \mathbb{R}^n$. The Lebesgue number of $\mathcal{U}$ is defined as the least upper bound for all numbers $\delta \geq 0$ such that any subset $B \subseteq A$ of diameter less than $\delta$ is contained in some element of the cover.

Is this definition correct? How is the existence of the supreme guaranteed? . In most texts they define the number of Lebesgue as a number that satisfies the aforementioned condition, but they omit the supreme one to give a unique definition.

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In fact there are two definitions of a Lebesgue number of an open cover $\mathcal{U}$ of a metric space $(X,d)$. As José Carlos Santos pointed out, the first one is the most common version:

(1) A Lebesgue number for $\mathcal{U}$ is a number $\delta > 0$ such that any open ball $B(x;\delta)$ with radius $\delta$ is contained in some $U \in \mathcal{U}$.

(2) A Lebesgue number for $\mathcal{U}$ is a number $\delta > 0$ such that any subset $M \subset X$ having diameter $< \delta$ is contained in some $U \in \mathcal{U}$.

These concepts are equivalent.

If $\delta$ is a Lebesgue number in the sense of (1), then it is also a Lebesgue number in the sense of (2): Let $M \subset X$ have diameter $< \delta$. Then for any $x \in M$ we have $M \subset B(x;\delta)$.

If $\delta$ is a Lebesgue number in the sense of (2), then any $\delta' < \delta/2$ is a Lebesgue number in the sense of (1): The diameter of $B(x;\delta')$ is $\le 2 \delta' < \delta$.

The definition of the Lebesgue number of $\mathcal{U}$ as the supremum $\lambda$ of all Lebesgue numbers for $\mathcal{U}$ is not really common. Note that $\lambda = \infty$ is possible.

Let us show that based on definition (2) $\lambda$ is the biggest Lebesgue number for $\mathcal{U}$. So let $M \subset X$ have diameter $< \lambda$. Hence there exists a Lebesgue number $\delta$ such that $M$ has diameter $< \delta$. We conclude that $M$ is contained in some $U \in \mathcal{U}$.

Based on definition (1) it is not guaranteed that $\lambda$ is a Lebesgue number for $\mathcal{U}$. Certainly each $B(x;\lambda)$ is the union of all $B(x;\delta)$ such that $\delta$ is a Lebesgue number for $\mathcal{U}$. Each of these sets is contained in some $U_{x,\delta} \in \mathcal{U}$, but it is not guaranteed that all these $U_{x,\delta}$ are contained in a single $U \in \mathcal{U}$.

The existence of Lebesgue numbers depends on $\mathcal{U}$. If $X$ is compact, then each $\mathcal{U}$ admits a Lebesgue number.

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For a compact metric space $X$ and an open cover $\mathcal{U}$ of it, there is at least one number $\delta$ that obeys the property $$l(\delta, \mathcal{U}): \forall B \subseteq X: (\operatorname{diam}(B) < \delta) \implies (\exists U \in \mathcal{U}: B \subseteq U)$$

This is a well-known fact, as you mention. The set of all numbers $\delta$ that can obey $l(\delta,\mathcal{U})$ is downward closed (a smaller number than one that works also works) and there are numbers like any $\delta > \operatorname{diam}(X)$ that cannot work, unless the cover is $\{X\}$. So it makes sense to define the $\lambda(\mathcal{U}):=\sup \{\delta: l(\delta, \mathcal{U})\}$ to define a number only dependent on the cover.

It's unusual, but why not? In the compact case we can also check that $ \delta=\lambda(\mathcal{U})$ also obeys the defining property $l(\delta, \mathcal{U})$.

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First of all, the concept of Lebesgue number is defined only for metric spaces, not for topological spaces in geberal.

I believe that the most common notion of Lebesgue number of an open cover is this: it is a number $\delta\in(0,\infty)$ such that any open ball with radius $\delta$ is a subset of some element of the cover. In general, not every open cover has a Lebesgue number. On the other hand, the existence of the supremum of such numbers is not guaranteed. Of course, if the metric is unbounded and if one of the elements of the open cover is the whose space, the supremum doesn't exist, but there are other cases in which that happens. For instance, take a metric space with an unbounded metric and let $\mathcal U$ be the set of all open balls.

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  • $\begingroup$ Why if an open cover has a Lebesgue number, then the set of all Lebesgue numbers of that cover is bounded ? $\endgroup$ – Juan Daniel Valdivia Fuentes Dec 5 '18 at 3:02
  • $\begingroup$ For no reason. I was wrong and I've edited my answer. Sorry about that. $\endgroup$ – José Carlos Santos Dec 5 '18 at 6:57

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