Lebesgue's number lemma

In topology, the Lebesgue covering lemma is a useful tool in the study of compact metric spaces.

Given an open cover of a compact metric space, a Lebesgue's number of the cover is a number $\delta > 0$ such that every subset of $X$ having diameter less than $\delta$ is contained in some member of the cover.

The existence of Lebesgue's numbers for compact metric spaces is given by the Lebesgue's covering lemma:

:If the metric space $(X, d)$ is compact and an open cover of $X$ is given, then the cover admits some Lebesgue's number $\delta > 0$.

The notion of Lebesgue's numbers itself is useful in other applications as well.

Proof

Direct 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's 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's number.
If $Y$ is a subset of $X$ of diameter less than $\delta$, choose $x_0$ as any point in $Y$, then by definition of diameter, $Y\subseteq B_\delta(x_0)$, where $B_\delta(x_0)$ denotes the ball of radius $\delta$ centered at $x_0$. 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, $Y\subseteq A_i$.

Proof by contradiction

Suppose for contradiction that $X$ is sequentially compact, $\$ is an open cover of $X$, and the Lebesgue number $\delta$ does not exist. That is: for all $\delta > 0$, there exists $A \subset X$ with $\operatorname (A) < \delta$ such that there does not exist $\beta \in J$ with $A \subset U_$.

This enables us to perform the following construction:

$\delta_ = 1, \quad \exists A_ \subset X \quad \text \quad \operatorname (A_) < \delta_ \quad \text \quad \neg\exists \beta (A_ \subset U_)$

$\delta_ = \frac, \quad \exists A_ \subset X \quad \text \quad \operatorname (A_) < \delta_ \quad \text \quad \neg\exists \beta (A_ \subset U_)$

$\vdots$

$\delta_=\frac, \quad \exists A_ \subset X \quad \text \quad \operatorname (A_) < \delta_ \quad \text \quad \neg\exists \beta (A_ \subset U_)$

$\vdots$

Note that $A_ \neq \emptyset$ for all $n \in \mathbb^$, since $A_ \not\subset U_$. It is therefore possible by the axiom of choice to construct a sequence $(x_)$ in which $x_ \in A_$ for each $i$. Since $X$ is sequentially compact, there exists a subsequence $\$ (with $k \in \mathbb_$) that converges to $x_$.

Because $\$ is an open cover, there exists some $\alpha_ \in J$ such that $x_ \in U_$. As $U_$ is open, there exists $r > 0$ with $B_(x_) \subset U_$. Now we invoke the convergence of the subsequence $\$: there exists $L \in \mathbb^$ such that
$L \le k$ implies $x_ \in B_ (x_)$.

Furthermore, there exists $M \in \mathbb_$ such that $\delta_= \tfrac < \tfrac$. Hence for all $z \in \mathbb_$, we have $M \le z$ implies $\operatorname (A_) < \tfrac$.

Finally, define $q \in \mathbb_$ such that $n_ \geq M$ and $q \geq L$. For all $x' \in A_$, notice that:
*$d(x_,x') \leq \operatorname (A_)<\frac$, because $n_ \geq M$.
*$d(x_,x_)<\frac$, because $q \geq L$ entails $x_ \in B_\left(x_\right)$.

Hence d(x_,x') by the triangle inequality, which implies that $A_ \subset U_$. This yields the desired contradiction.

References

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{{DEFAULTSORT:Lebesgue's Number Lemma
Lemmas
Theorems in topology