What a Lebesgue Number Measures
An open cover may use many sets, and a point of the space may belong to several of them. A Lebesgue number gives a uniform size threshold: any sufficiently small subset of the space fits inside one member of the cover. The threshold is uniform because it does not depend on which subset is chosen.
Let \((X,d)\) be a metric space. An open cover \(\mathcal{U}\) of \(X\) is a family of open subsets of \(X\) whose union is \(X\). For a nonempty subset \(S\subseteq X\), its diameter is \(\operatorname{diam}(S)=\sup\{d(x,y):x,y\in S\}\), where the supremum may be infinite. We use nonempty subsets in the definition below; this avoids any requirement that an empty subset be contained in a member of the cover. If \(X\) itself is empty, the condition is vacuous and every positive number qualifies.
The condition concerns entire subsets, not just pairs of points separately. In practice, one often proves that a candidate \(\delta\) works by assuming a subset is not contained in any cover member and showing its diameter must be at least \(\delta\). To rule out a larger candidate, it is enough to find a subset that is not contained in any member and has diameter strictly less than that candidate.
Two Useful Facts About the Definition
Proof. Suppose \(0<r\leq\delta\), and let \(S\) be nonempty with \(\operatorname{diam}(S)<r\). Then \(\operatorname{diam}(S)<\delta\), so the definition of \(\delta\) puts \(S\) inside some member of \(\mathcal{U}\). Thus \(r\) is a Lebesgue number. Now fix \(x\in X\). The ball \(B(x,\delta/3)\) is nonempty because it contains \(x\). If \(y,z\in B(x,\delta/3)\), then \(d(y,z)\leq d(y,x)+d(x,z)<2\delta/3<\delta\). Consequently its diameter is at most \(2\delta/3\), which is strictly less than \(\delta\). The definition places this ball inside some \(U\in\mathcal{U}\). \(\square\)
Conversely, if for some \(\rho>0\) every \(x\in X\) has a cover member containing \(B(x,\rho)\), then \(\rho\) is a Lebesgue number. Indeed, for any nonempty \(S\) with diameter less than \(\rho\), choose \(x\in S\). For every \(y\in S\), \(d(x,y)\leq\operatorname{diam}(S)<\rho\), so \(S\subseteq B(x,\rho)\), and hence \(S\) lies in that cover member. The two formulations are therefore closely related, though the ball radius obtained directly from a subset-diameter Lebesgue number above is \(\delta/3\), not necessarily \(\delta\).
There need not be a largest Lebesgue number. For example, if a cover member is all of \(X\), every positive number is a Lebesgue number. When we say an example has an exact Lebesgue number, we mean that the number works and no larger number works.
Calculating Numbers for Interval Covers
We use \(I=[0,1]\) with the metric \(d(x,y)=|x-y|\) and its subspace topology from \(\mathbb{R}\). Sets such as \([0,a)\) below are understood as subsets of \(I\); for instance, \([0,a)=I\cap(-1,a)\), so it is open in \(I\). This relative topology matters: an interval containing an endpoint of \(I\) can be open in \(I\) even if it is not open in \(\mathbb{R}\).
Worked Example: A Two-Set Cover of the Unit Interval
Let \(U_1=[0,3/5)\) and \(U_2=(2/5,1]\). These sets are open in \(I\). They cover \(I\): a point at most \(2/5\) lies in \(U_1\), a point at least \(3/5\) lies in \(U_2\), and a point strictly between \(2/5\) and \(3/5\) lies in both.
We claim that \(\delta=1/5\) is an exact Lebesgue number. Let \(S\subseteq I\) be nonempty and suppose it is not contained in \(U_1\) and not contained in \(U_2\). Since \(S\not\subseteq U_1\), it has a point \(x\geq3/5\). Since \(S\not\subseteq U_2\), it has a point \(y\leq2/5\). Therefore \(d(x,y)=x-y\geq3/5-2/5=1/5\), so \(\operatorname{diam}(S)\geq1/5\). It follows that every nonempty \(S\) with diameter less than \(1/5\) must be contained in at least one of \(U_1,U_2\).
To check that no larger number works, take \(S=\{2/5,3/5\}\). Its diameter is \(3/5-2/5=1/5\). The point \(3/5\) is not in \(U_1\), and \(2/5\) is not in \(U_2\), so \(S\) is contained in neither cover member. For any \(\delta>1/5\), this same set has diameter less than \(\delta\) and violates the definition. Thus \(1/5\) is exact.
Check: the distance between the two points of the obstructing set is \(|3/5-2/5|=1/5\), as used above. At the proposed value, the definition tests only sets of diameter strictly less than \(1/5\), so this obstructing set does not contradict that value.
Worked Example: A Three-Set Cover and Its Exact Value
Consider \(V_1=[0,1/2)\), \(V_2=(1/4,3/4)\), and \(V_3=(1/2,1]\), all open in \(I\). They cover \(I\): points below or equal to \(1/4\) lie in \(V_1\); points above or equal to \(3/4\) lie in \(V_3\); and every point strictly between \(1/4\) and \(3/4\), including \(1/2\), lies in \(V_2\). The endpoint \(1\) lies in \(V_3\).
We show that \(1/4\) is a Lebesgue number. Suppose a nonempty \(S\subseteq I\) is contained in none of the three sets. From \(S\not\subseteq V_1\), choose \(x\in S\) with \(x\geq1/2\). From \(S\not\subseteq V_2\), choose \(y\in S\) with either \(y\leq1/4\) or \(y\geq3/4\). From \(S\not\subseteq V_3\), choose \(z\in S\) with \(z\leq1/2\). If \(y\leq1/4\), then \(x-y\geq1/2-1/4=1/4\). If \(y\geq3/4\), then \(y-z\geq3/4-1/2=1/4\). In either case \(\operatorname{diam}(S)\geq1/4\). Thus every nonempty set of diameter less than \(1/4\) is contained in one of the three sets.
For exactness, take \(S=\{1/4,1/2\}\). Its diameter is \(1/2-1/4=1/4\). It is not contained in \(V_1\), because \(1/2\notin V_1\); it is not contained in \(V_2\), because \(1/4\notin V_2\); and it is not contained in \(V_3\), because neither point is greater than \(1/2\). Hence every \(\delta>1/4\) fails, while \(1/4\) works.
Check: the boundary tests use the actual open endpoints: \(1/4\) is excluded from \(V_2\), and \(1/2\) is excluded from both \(V_1\) and \(V_3\). The points are therefore not accidentally covered by a set that includes its endpoint.
An Open Cover Without a Lebesgue Number
Worked Example: A Cover of a Noncompact Interval
Take \(X=(0,1)\) with the usual metric and, for each \(n\in\mathbb{N}\), set \(U_n=(1/(n+1),1)\). Each \(U_n\) is open in \(X\), and the family covers \(X\): given \(x\in(0,1)\), choose \(n\) large enough that \(1/(n+1)<x\); then \(x\in U_n\).
This cover has no Lebesgue number. Let \(\delta>0\) be any proposed number, and choose an integer \(N\geq2\) so large that \(1/N<\delta\). Define the nonempty subset \(S_N=\{1/k:k\in\mathbb{N},\ k\geq N\}\subseteq X\). Its largest point is \(1/N\), and its points approach \(0\), so its diameter is \(1/N\). In particular, \(\operatorname{diam}(S_N)=1/N<\delta\).
But \(S_N\) is not contained in any \(U_n\). For a fixed \(n\), choose \(k\geq N\) with \(k\geq n+1\). Then \(1/k\leq1/(n+1)\), so \(1/k\notin U_n\). This works for every \(n\), and therefore no member of the cover contains \(S_N\). Since this construction works for every \(\delta>0\), the cover has no Lebesgue number.
Check: the strict comparison required by the definition is \(1/N<\delta\), achieved by choosing \(N>1/\delta\). The failure is not caused by a single point: it is the whole small-diameter tail \(S_N\) that fails to fit in every cover member.
A Common Diameter Mistake
Common pitfall. A tempting but incorrect calculation is to treat the radius of a two-point set around its midpoint as its diameter. For the first interval cover, the midpoint of \(\{2/5,3/5\}\) is \(1/2\), and each point is \(1/10\) away from it. The radius is \(1/10\), but the diameter is the distance between the two points: \(|3/5-2/5|=1/5\). So the set rules out every proposed Lebesgue number greater than \(1/5\), not merely every number greater than \(1/10\). A candidate must be tested against diameters because that is the quantity in the definition.
The calculations above illustrate two complementary methods. To prove a lower bound, show that any subset failing to fit in a cover member must contain points separated by at least that bound. To prove an upper bound, exhibit one subset that fails to fit and whose diameter is smaller than any proposed larger number. For compact metric spaces, the general theorem guaranteeing a positive Lebesgue number for every open cover is proved separately; the examples here use only the definition and direct distance estimates.
Check Your Understanding
For each question, attend to whether the diameter inequality is strict and whether interval endpoints belong to the stated sets.
- For the cover \([0,2/3)\), \((1/3,1]\) of \(I\), determine an exact Lebesgue number and justify both that it works and that no larger value works.
- Suppose \(\delta\) is a Lebesgue number. Explain why \(r=\delta/2\) is also a Lebesgue number.
- In the three-set example, why does a subset not contained in \(V_2\) have a point at most \(1/4\) or at least \(3/4\)?
- For the cover \(U_n=(1/(n+1),1)\) of \((0,1)\), verify directly that \(S_N\) is not contained in any one fixed \(U_n\).
- Explain why a set of diameter exactly \(\delta\) does not, by itself, disprove that \(\delta\) is a Lebesgue number.