Lebesgue's number lemma

In topology, Lebesgue's number lemma 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" of diameter < &delta; is contained in some member of the cover. The number &delta; is called the Lebesgue Number of this cover for "X".

Proof

Let $mathcal\; A$ be an open cover of "X". If $X\; in\; mathcal\; A$, then any &delta; will work, so assume $X\; otin\; mathcal\; A$. Choose $\{A\_1,\; dots,\; A\_n\}\; subseteq\; mathcal\; A$ that covers "X". For each "i", set $C\_i\; =\; X\; -\; A\_i$ and define $f:X\; ightarrowmathbb\; R$ by letting $f(x)\; =\; frac\{1\}\{n\}sum\_\{i=1\}^n\; d(x,\; C\_i)$. Given "x" in "X", choose "i" so that $x\; in\; A\_i$. Then choose $epsilon\; ge\; 0$; so that the &epsilon;-neighborhood of "x" lies in $A\_i$. Then $d(x,\; C\_i)\; ge\; epsilon$, so $f(x)\; ge\; frac\{epsilon\}\{n\}$. Thus $f(x)\; ge\; 0$ for all "x".

Since "f" is continuous, it has a minimum value, &delta;. Since $f(x)\; ge\; 0$ for all "x", $delta$ is not 0. We will show that &delta; is the Lebesgue number. Let "B" be a subset of "X" of diameter less than &delta;. Choose $x\_0\; in\; B$, thus "B" lies in the &delta;-neighborhood of $x\_0$. Now $delta\; le\; f(x\_0)\; le\; d(x\_0,\; C\_m)$ where $d(x\_0,\; C\_m)$ is the largest of the number $d(x\_0,\; C\_i)$. So the &delta;-neighborhood of $x\_0$, and thus "B", is contained in the element $A\_m\; =\; X\; -\; C\_m$ of the covering $mathcal\; A$

Applications

The Lebesgue number lemma is useful in the study of compact metric spaces and functional metric spaces, since it can often be used to obtain approximations of distances when the space is compact.