Compactness · Definition

Lebesgue number

topologycompactnessmetric spaces

Before you start you need to know

It gives a uniform scale that turns an open cover into a tool for passing from the local to the global.

Definition

A Lebesgue number of an open cover \(\mathcal{U}\) of a metric space is a \(\delta>0\) such that every subset \(A\) of diameter less than \(\delta\) fits inside some member of \(\mathcal{U}\).

\[\operatorname{diam}A<\delta\ \Rightarrow\ \exists\,U\in\mathcal{U}:\ A\subseteq U\]

The diameter of a set \(A\) is the supremum of the distances between its points, \(\operatorname{diam}A=\sup\{d(a,b):a,b\in A\}\). A Lebesgue number \(\delta\) says that, at that scale, the cover no longer «cuts» any small set: everything smaller than \(\delta\) fits entirely in one piece. If \(\delta\) works, so does every smaller positive number, so existence matters more than the optimal value.

Without compactness there is no guarantee. The cover \(U_n=(1/n,1)\) of \((0,1)\) has no Lebesgue number, since \((0,\delta/2)\) has diameter less than \(\delta\) and fits in no \(U_n\). With compactness there is one: this is the Lebesgue lemma. If \(X\) is a compact metric space, the function \(f(x)=\max_U d(x,X\setminus U)\) is continuous and positive, and its minimum \(\delta>0\) is a Lebesgue number, because a set of diameter less than \(\delta\) containing \(a\) lies inside \(B(a,\delta)\), and that ball fits in some member.

A computable example: the cover \([0,0.6)\), \((0.4,1]\) of \([0,1]\) admits exactly the \(\delta\le0.2\), because the set \(\{0.4,0.6\}\) has diameter \(0.2\) and fits in no member. For intervals it is enough to look at pairs of points, since a set contained in an interval lies in a member if and only if its minimum and maximum do.

The lemma is the tool that passes from the local to the global: it lets you split \([a,b]\) into subintervals of length less than \(\delta\), each inside a member of the cover, and from there one proves that every continuous function on a compact set is uniformly continuous. It also enters the equivalence between compactness and sequential compactness in metric spaces.

Think about it

Does the cover of \(\mathbb{R}\) by the intervals \((n-1,n+1)\), \(n\in\mathbb{Z}\), have a Lebesgue number?

Yes: \(\delta=1\) works, even though \(\mathbb{R}\) is not compact. Compactness guarantees a Lebesgue number for every cover, but a particular cover can have one without it.

If \(\delta\) is a Lebesgue number, why is \(\delta/2\) one too?

Because a set of diameter less than \(\delta/2\) has diameter less than \(\delta\), and by hypothesis it fits in a member.

Exercises 10

References

  • James R. Munkres, Topology, 2nd edition, Pearson, 2000. §27 (Lema 27.5)
  • Bert Mendelson, Introduction to Topology, 3rd edition, Dover Publications, 1990. cap. 2