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Lebesgue's number lemma

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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 such that every subset of having diameter less than 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 is compact and an open cover of is given, then the cover admits some Lebesgue's number .

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

Proof

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Direct Proof

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Let be an open cover of . Since is compact we can extract a finite subcover . If any one of the 's equals then any will serve as a Lebesgue's number. Otherwise for each , let , note that is not empty, and define a function by

Since is continuous on a compact set, it attains a minimum . The key observation is that, since every is contained in some , the extreme value theorem shows . Now we can verify that this is the desired Lebesgue's number. If is a subset of of diameter less than , choose as any point in , then by definition of diameter, , where denotes the ball of radius centered at . Since there must exist at least one such that . But this means that and so, in particular, .

Proof by Contradiction

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Suppose for contradiction that that is sequentially compact, is an open cover of , and the Lebesgue number does not exist. That is: for all , there exists with such that there does not exist with .

This enables us to perform the following construction:


Note that for all , since . It is therefore possible by the axiom of choice to construct a sequence in which for each . Since is sequentially compact, there exists a subsequence (with ) that converges to .

Because is an open cover, there exists some such that . As is open, there exists with . Now we invoke the convergence of the subsequence : there exists such that implies .

Furthermore, there exists such that . Hence for all , we have implies .

Finally, define such that and . For all , notice that:

  • , because .
  • , because entails .

Hence by the triangle inequality, which implies that . This yields the desired contradiction.

References

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  • Munkres, James R. (1974), Topology: A first course, p. 179, ISBN 978-0-13-925495-6