Lebesgue's number lemma

Short description: Given a cover of a compact metric space, all small subsets are subset of some cover set

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 [math]\displaystyle{ (X, d) }[/math] is compact and an open cover of [math]\displaystyle{ X }[/math] is given, then there exists a number [math]\displaystyle{ \delta \gt 0 }[/math] such that every subset of [math]\displaystyle{ X }[/math] having diameter less than [math]\displaystyle{ \delta }[/math] is contained in some member of the cover.

Such a number [math]\displaystyle{ \delta }[/math] is called a Lebesgue number of this cover. The notion of a Lebesgue number itself is useful in other applications as well.

Proof

Direct Proof

Let [math]\displaystyle{ \mathcal U }[/math] be an open cover of [math]\displaystyle{ X }[/math]. Since [math]\displaystyle{ X }[/math] is compact we can extract a finite subcover [math]\displaystyle{ \{A_1, \dots, A_n\} \subseteq \mathcal U }[/math]. If any one of the [math]\displaystyle{ A_i }[/math]'s equals [math]\displaystyle{ X }[/math] then any [math]\displaystyle{ \delta \gt 0 }[/math] will serve as a Lebesgue number. Otherwise for each [math]\displaystyle{ i \in \{1, \dots, n\} }[/math], let [math]\displaystyle{ C_i := X \smallsetminus A_i }[/math], note that [math]\displaystyle{ C_i }[/math] is not empty, and define a function [math]\displaystyle{ f : X \rightarrow \mathbb R }[/math] by

[math]\displaystyle{ f(x) := \frac{1}{n} \sum_{i=1}^n d(x,C_i). }[/math]

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

Proof by Contradiction

Assume [math]\displaystyle{ X }[/math] is sequentially compact, [math]\displaystyle{ \mathcal{A} = \{U_{\alpha} | \alpha \in J\} }[/math] is an open covering of [math]\displaystyle{ X }[/math] and the Lebesgue number [math]\displaystyle{ \delta }[/math] does not exist. So, [math]\displaystyle{ \forall \delta \gt 0 }[/math], [math]\displaystyle{ \exists A \subset X }[/math] with [math]\displaystyle{ diam (A) \lt \delta }[/math] such that [math]\displaystyle{ \neg\exists\beta \in J }[/math] where [math]\displaystyle{ A\subset U_{\beta} }[/math].

This allows us to make the following construction:

[math]\displaystyle{ \delta_{1}=1 }[/math], [math]\displaystyle{ \exists A_{1} \subset X }[/math] where [math]\displaystyle{ (diam (A_{1})\lt \delta_{1}) }[/math] and [math]\displaystyle{ \neg\exists \beta (A_{1} \subset U_{\beta}) }[/math]

[math]\displaystyle{ \delta_{2}=\frac{1}{2} }[/math], [math]\displaystyle{ \exists A_{2} \subset X }[/math] where [math]\displaystyle{ (diam (A_{2})\lt \delta_{2}) }[/math] and [math]\displaystyle{ \neg\exists \beta (A_{2} \subset U_{\beta}) }[/math]

⋮

[math]\displaystyle{ \delta_{k}=\frac{1}{k} }[/math], [math]\displaystyle{ \exists A_{k} \subset X }[/math] where [math]\displaystyle{ (diam (A_{k})\lt \delta_{k}) }[/math] and [math]\displaystyle{ \neg\exists \beta (A_{k} \subset U_{\beta}) }[/math]

⋮

For all [math]\displaystyle{ n \in \mathbb{Z}^{+} }[/math], [math]\displaystyle{ A_{n} \neq \emptyset }[/math] since [math]\displaystyle{ A_{n} \not\subset U_{\beta} }[/math].

It is therefore possible to generate a sequence [math]\displaystyle{ \{x_{n}\} }[/math] where [math]\displaystyle{ x_{n} \in A_{n} }[/math] by axiom of choice. By sequential compactness, there exists a subsequence [math]\displaystyle{ \{x_{n_{k}}\}, k \in \mathbb{Z}^{+} }[/math] that converges to [math]\displaystyle{ x_{0} \in X }[/math].

Using the fact that [math]\displaystyle{ \mathcal{A} }[/math] is an open covering, [math]\displaystyle{ \exists \alpha_{0} \in J }[/math] where [math]\displaystyle{ x_{0} \in U_{\alpha_{0}} }[/math]. As [math]\displaystyle{ U_{\alpha_{0}} }[/math] is open, [math]\displaystyle{ \exists r \gt 0 }[/math] such that [math]\displaystyle{ B_{d}(x_{0},r) \subset U_{\alpha_{0}} }[/math]. By definition of convergence, [math]\displaystyle{ \exists L \in \mathbb{Z}^{+} }[/math] such that [math]\displaystyle{ x_{n_{p}} \in B_{d} \left(x_{0},\frac{r}{2}\right) }[/math] for all [math]\displaystyle{ p \geq L }[/math].

Furthermore, [math]\displaystyle{ \exists M \in \mathbb{Z}^{+} }[/math] where [math]\displaystyle{ \delta_{M}=\frac{1}{K}\lt \frac{r}{2} }[/math]. So, [math]\displaystyle{ \forall z \in \mathbb{Z}^{+} z \geq M \Rightarrow diam (A_{M})\lt \frac{r}{2} }[/math].

Finally, let [math]\displaystyle{ q \in \mathbb{Z}^{+} }[/math] such that [math]\displaystyle{ n_{q} \geq M }[/math] and [math]\displaystyle{ q \geq L }[/math]. For all [math]\displaystyle{ x' \in A_{n_{q}} }[/math], notice that:

• [math]\displaystyle{ d(x_{n_{q}},x')\leq diam (A_{n_{q}})\lt \frac{r}{2} }[/math] because [math]\displaystyle{ n_{q} \geq M }[/math].
• [math]\displaystyle{ d(x_{n_{q}},x_{0})\lt \frac{r}{2} }[/math] because [math]\displaystyle{ q \geq L }[/math] which means [math]\displaystyle{ x_{n_{q}} \in B_{d}(x_{0},\frac{r}{2}) }[/math].

By the triangle inequality, [math]\displaystyle{ d(x_{0},x')\lt r }[/math], implying that [math]\displaystyle{ A_{n_{q}} \subset U_{\alpha_{0}} }[/math] which is a contradiction.

References

• Munkres, James R. (1974), Topology: A first course, p. 179, ISBN 978-0-13-925495-6, https://archive.org/details/topologyfirstcou00munk_0/page/179