Lebesgue's number lemma explained

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

l U be an open cover of X . Since X is compact we can extract a finite subcover \{A _{1,} ... , A _{n\}} \subseteq l 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 , ... , n\} , let C _{i} : = X \smallsetminus A _{i} , note that C _{i} is not empty, and define a function f : X → R byf(x) : = (1)/(n) \sum ^{n} _{i =1} 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 , \exists A _{1} \subset X where \operatorname{diam} (A _{1} ) < \delta _{1} and \neg\exists \beta (A _{1} \subset U _{\beta} ) \delta _{2} = (1)/(2) , \exists A _{2} \subset X where \operatorname{diam} (A _{2} ) < \delta _{2} and \neg\exists \beta (A _{2} \subset U _{\beta} ) \vdots \delta _{k} = (1)/(k) , \exists A _{k} \subset X where \operatorname{diam} (A _{k} ) < \delta _{k} and \neg\exists \beta (A _{k} \subset U _{\beta} ) \vdots

Note that

A _{n} ≠ \emptyset for all n \in 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 Z _{>} ) 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 Z ^{+} such thatL \le k implies x _{n _{k}} \in B _{r/2} (x _{0} ) .

Furthermore, there exists

M \in Z _{>} such that \delta _{M} = \tfrac{ 1 }{M} < \tfrac{r}{ 2 } . Hence for all z \in Z _{>} , we have M \le z implies \operatorname{diam} (A _{M} ) < \tfrac{r}{ 2 } .

Finally, define

q \in 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 \operatorname{diam} (A _{n _{q}} )< (r)/(2) , because n _{q} \geq M .d(x _{n _{q}} ,x _{0} )< (r)/(2) , because q \geq L entails x _{n _{q}} \in B _{r/2} \left(x _{0} \right) .

Hence

d(x _{0} ,x')<r by the triangle inequality, which implies that A _{n _{q}} \subset U _{\alpha _{0}} . This yields the desired contradiction