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[edit]

Direct Proof[edit]

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 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 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, $Y\subseteq A_{i}$ .

Proof by Contradiction[edit]

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

This allows us to make the following construction:

\delta _{1}=1

,

\exists A_{1}\subset X

where

(diam(A_{1})<\delta _{1})

and

\neg \exists \beta (A_{1}\subset U_{\beta })

\delta _{2}={\frac {1}{2}}

,

\exists A_{2}\subset X

where

(diam(A_{2})<\delta _{2})

and

\neg \exists \beta (A_{2}\subset U_{\beta })

⋮

\delta _{k}={\frac {1}{k}}

,

\exists A_{k}\subset X

where

(diam(A_{k})<\delta _{k})

and

\neg \exists \beta (A_{k}\subset U_{\beta })

⋮

For all $n\in \mathbb {Z} ^{+}$ , $A_{n}\neq \emptyset$ since $A_{n}\not \subset U_{\beta }$ .

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

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

Furthermore, $\exists M\in \mathbb {Z} ^{+}$ where $\delta _{M}={\frac {1}{K}}<{\frac {r}{2}}$ . So, $\forall z\in \mathbb {Z} ^{+}z\geq M\Rightarrow diam(A_{M})<{\frac {r}{2}}$ .

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

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