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 existance 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 $\{A_{1},\dots ,A_{n}\}\subseteq {\mathcal {U}}$ . If any one of the $A_{i}$ 's equals $X$ then any $\delta >0$ will serve as a Lebesgue's number. Otherwise for each $i\in \{1,\dots ,n\}$ , 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 {1}{n}}\sum _{i=1}^{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 that $X$ is sequentially compact, $\{U_{\alpha }\mid \alpha \in J\}$ 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 $\operatorname {diam} (A)<\delta$ such that there does not exist $\beta \in J$ with $A\subset U_{\beta }$ .

This enables us to perform the following construction:

\delta _{1}=1,\quad \exists A_{1}\subset X\quad {\text{where}}\quad \operatorname {diam} (A_{1})<\delta _{1}\quad {\text{and}}\quad \neg \exists \beta (A_{1}\subset U_{\beta })

\delta _{2}={\frac {1}{2}},\quad \exists A_{2}\subset X\quad {\text{where}}\quad \operatorname {diam} (A_{2})<\delta _{2}\quad {\text{and}}\quad \neg \exists \beta (A_{2}\subset U_{\beta })

\vdots

\delta _{k}={\frac {1}{k}},\quad \exists A_{k}\subset X\quad {\text{where}}\quad \operatorname {diam} (A_{k})<\delta _{k}\quad {\text{and}}\quad \neg \exists \beta (A_{k}\subset U_{\beta })

\vdots

Note that $A_{n}\neq \emptyset$ for all $n\in \mathbb {Z} ^{+}$ , since $A_{n}\not \subset U_{\beta }$ . It is therefore possible by the axiom of choice to construct a sequence $(x_{n})$ in which $x_{i}\in A_{i}$ for each $i$ . Since $X$ is sequentially compact, there exists a subsequence $\{x_{n_{k}}\}$ (with $k\in \mathbb {Z} _{>0}$ ) that converges to $x_{0}$ .

Because $\{U_{\alpha }\}$ is an open cover, there exists some $\alpha _{0}\in J$ such that $x_{0}\in U_{\alpha _{0}}$ . As $U_{\alpha _{0}}$ is open, there exists $r>0$ with $B_{d}(x_{0},r)\subset U_{\alpha _{0}}$ . Now we invoke the convergence of the subsequence $\{x_{n_{k}}\}$ : there exists $L\in \mathbb {Z} ^{+}$ such that $L\leq k$ implies $x_{n_{k}}\in B_{r/2}(x_{0})$ .

Furthermore, there exists $M\in \mathbb {Z} _{>0}$ such that $\delta _{M}={\tfrac {1}{M}}<{\tfrac {r}{2}}$ . Hence for all $z\in \mathbb {Z} _{>0}$ , we have $M\leq z$ implies $\operatorname {diam} (A_{M})<{\tfrac {r}{2}}$ .

Finally, define $q\in \mathbb {Z} _{>0}$ such that $n_{q}\geq M$ and $q\geq L$ . For all $x'\in A_{n_{q}}$ , notice that:

$d(x_{n_{q}},x')\leq \operatorname {diam} (A_{n_{q}})<{\frac {r}{2}}$ , because $n_{q}\geq M$ .
$d(x_{n_{q}},x_{0})<{\frac {r}{2}}$ , because $q\geq L$ entails $x_{n_{q}}\in B_{r/2}\left(x_{0}\right)$ .