From Local Openness to a Uniform Radius
An open cover supplies a neighborhood around every point, but the size of that neighborhood may depend on the point. A Lebesgue number is a single positive scale that works across the whole space: every sufficiently small ball fits inside some member of the cover. The key fact proved here is that sequential compactness prevents the available scales from shrinking to zero without limit.
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 \(X\). The open ball of radius \(r>0\) centered at \(x\) is \(B(x,r)=\{y\in X:d(x,y)<r\}\). The space is sequentially compact if every sequence in \(X\) has a subsequence converging to a point of \(X\). For the empty space the ball condition below is vacuous, so any positive number works; in the proof assume \(X\ne\varnothing\).
Openness alone gives a radius for each chosen point, not necessarily the same radius for all points. The lemma says that sequential compactness supplies the uniform choice. Its proof is a contradiction argument: if every proposed radius fails, choose a point witnessing failure at each scale, then use a convergent subsequence to contradict openness at its limit.
Proof of the Lebesgue Number Lemma
Proof. The result holds vacuously when \(X=\varnothing\), so suppose \(X\ne\varnothing\). Assume, for contradiction, that no positive Lebesgue number exists. For each \(n\in\mathbb{N}\), the proposed radius \(1/n\) then fails. Thus there is a point \(x_n\in X\) such that \(B(x_n,1/n)\) is not contained in any member of \(\mathcal{U}\). By sequential compactness, \((x_n)\) has a subsequence \((x_{n_k})\) converging to some \(x\in X\), where the indices \(n_k\) increase without bound.
Since \(\mathcal{U}\) covers \(X\), some \(U\in\mathcal{U}\) contains \(x\). Because \(U\) is open in the metric topology, there is \(r>0\) such that \(B(x,r)\subseteq U\). Convergence gives an index after which \(d(x_{n_k},x)<r/2\), and \(n_k\to\infty\) gives an index after which \(1/n_k<r/2\). Choose \(k\) large enough for both inequalities. For any \(y\in B(x_{n_k},1/n_k)\), the triangle inequality gives $$ d(y,x)\le d(y,x_{n_k})+d(x_{n_k},x)<\frac{1}{n_k}+\frac{r}{2}<r. $$ Therefore \(y\in B(x,r)\subseteq U\), so \(B(x_{n_k},1/n_k)\subseteq U\). This contradicts the choice of \(x_{n_k}\), which was chosen so that its ball was contained in no member of \(\mathcal{U}\). The contradiction proves that a positive Lebesgue number exists. \(\square\)
The use of two bounds smaller than \(r/2\) provides a margin: the center \(x_{n_k}\) is close enough to \(x\), and the ball around that center is small enough, for every point in the ball to remain within distance \(r\) of \(x\). The subsequence is essential because the openness radius \(r\) is obtained only at the limit point; convergence brings the selected centers into that one open neighborhood.
Ball and Diameter Formulations
Another common convention defines a Lebesgue number by testing all subsets of small diameter. For a nonempty set \(A\subseteq X\), its diameter is \(\operatorname{diam}(A)=\sup\{d(y,z):y,z\in A\}\). In the following proposition, the diameter condition means that every nonempty \(A\subseteq X\) with \(\operatorname{diam}(A)<\delta\) is contained in some member of \(\mathcal{U}\). Excluding the empty subset avoids a vacuous containment issue if a cover has no members, as can happen when \(X=\varnothing\).
Proof. Suppose first that \(\delta\) is a ball-based Lebesgue number, and let \(A\subseteq X\) be nonempty with \(\operatorname{diam}(A)<\delta\). Choose \(a\in A\). For every \(y\in A\), \(d(y,a)\le\operatorname{diam}(A)<\delta\), so \(A\subseteq B(a,\delta)\). By the ball condition, this ball lies in some \(U\in\mathcal{U}\), and hence \(A\subseteq U\). Conversely, suppose the diameter condition holds, and fix \(x\in X\). If \(y,z\in B(x,\delta/3)\), then the triangle inequality gives \(d(y,z)\le d(y,x)+d(x,z)<2\delta/3\). Thus \(\operatorname{diam}(B(x,\delta/3))\le 2\delta/3<\delta\). The diameter condition places this ball inside some member of \(\mathcal{U}\), proving that \(\delta/3\) is a ball-based Lebesgue number. \(\square\)
The factor \(1/3\) in this direction keeps the diameter strictly less than \(\delta\), even if the supremum of distances in the ball is not attained. The two conventions therefore express the same uniform-neighborhood idea, with a harmless adjustment to the radius.
Worked Example: A Finite Metric Space
Let \(X=\{0,1,3\}\) have the metric inherited from \(\mathbb{R}\), and let \(\mathcal{U}=\{\{0,1\},\{1,3\}\}\). Every subset of \(X\) is open: for each point, the ball of radius \(1\) contains only that point, so any subset is a union of open singletons. The two sets in \(\mathcal{U}\) therefore form an open cover.
Take \(\delta=1\). The distances between distinct points are \(d(0,1)=|0-1|=1\), \(d(1,3)=|1-3|=2\), and \(d(0,3)=|0-3|=3\). Since balls use the strict condition \(d(x,y)<1\), each ball \(B(x,1)\) contains only its center. Specifically, \(B(0,1)=\{0\}\subseteq\{0,1\}\), \(B(1,1)=\{1\}\subseteq\{0,1\}\), and \(B(3,1)=\{3\}\subseteq\{1,3\}\). Hence \(1\) is a Lebesgue number.
This finite space is sequentially compact: in any sequence, at least one of the three points occurs infinitely often, so those occurrences give a constant, convergent subsequence. The computed ball containments also check the conclusion directly for this particular cover.
Worked Example: A Cover of the Closed Interval
Give \(X=[0,1]\) the usual metric and subspace topology, and consider \(U_1=[0,3/4)\) and \(U_2=(1/4,1]\). These sets are open in \(X\), since \(U_1=X\cap(-1,3/4)\) and \(U_2=X\cap(1/4,2)\). They cover \(X\): points at most \(1/4\) lie in \(U_1\), points greater than \(1/4\) lie in \(U_2\), and points in the overlap lie in both. We verify that \(\delta=1/4\) is a Lebesgue number.
If \(x\le 1/2\) and \(y\in B(x,1/4)\), then \(y<x+1/4\le 3/4\), and \(y\in X\) ensures \(y\ge0\). Thus \(y\in U_1\) and \(B(x,1/4)\subseteq U_1\). If \(x>1/2\) and \(y\in B(x,1/4)\), then \(y>x-1/4>1/4\), while \(y\le1\). Thus \(y\in U_2\) and \(B(x,1/4)\subseteq U_2\). The two cases include every \(x\in X\), including the boundary between cases: at \(x=1/2\) the first case applies.
The interval is sequentially compact by the Bolzano–Weierstrass theorem: every sequence in the bounded closed interval \([0,1]\) has a subsequence converging to a point of \([0,1]\). The calculation gives a specific radius for this cover, rather than only the existence supplied by the lemma.
Worked Example: A Cover with No Lebesgue Number
Let \(X=(0,1)\) with its usual metric and subspace topology. For \(n\in\mathbb{N}\), define \(U_n=(1/(n+1),1)\), an open subset of \(X\). These sets cover \(X\): given \(z\in(0,1)\), choose \(n\) large enough that \(1/(n+1)<z\); then \(z\in U_n\).
We show that no \(\delta>0\) is a ball-based Lebesgue number. Fix any \(\delta>0\) and choose \(x=\min(\delta/2,1/2)\). This gives \(x\in(0,1)\) and \(x<\delta\). For each \(n\), set \(y=\tfrac12\min(x,1/(n+1))\). Then \(0<y<x\), \(y<1/(n+1)\), and \(d(x,y)=x-y<x<\delta\). Thus \(y\in B(x,\delta)\), but \(y\notin U_n\). This holds for every \(n\), so \(B(x,\delta)\) is contained in no member of the cover. Since \(\delta\) was arbitrary, no positive Lebesgue number exists.
The failure is consistent with the lemma: \(X\) is not sequentially compact. The sequence \(1/(n+1)\) lies in \(X\) and converges to \(0\) in \(\mathbb{R}\). Every subsequence also converges to \(0\), so none can converge to a point of \(X\). Indeed, if it converged to \(z\in X\), uniqueness of limits in a metric space would force \(z=0\), which is not in \(X\).
A Common Pitfall: Convergence Needs a Margin
A faulty version of the proof might say: “The centers \(x_{n_k}\) eventually lie in \(B(x,r)\), so their balls lie in \(B(x,r)\).” The conclusion does not follow. For example, take \(x=0\), \(r=1\), a center \(x_{n_k}=0.9\), and a ball radius \(0.2\). The center satisfies \(|0.9-0|=0.9<1\), but \(y=1.05\) belongs to \(B(0.9,0.2)\), since \(|1.05-0.9|=0.15<0.2\), and does not belong to \(B(0,1)\), since \(|1.05-0|=1.05>1\).
The correction is to reserve room for both distances. Require \(d(x_{n_k},x)<r/2\) and \(1/n_k<r/2\); then, for every point \(y\) in the small ball, the triangle inequality gives \(d(y,x)<r/2+r/2=r\). A proof about balls must control every point of each ball, not just its center.
The lemma matters because it converts point-by-point openness into one scale that is valid everywhere on a sequentially compact metric space. When checking a proposed radius, keep the quantifiers in view: for each \(x\), the cover member may vary, but the positive radius must be fixed independently of \(x\).
Check Your Understanding
Use the definitions and proof above to answer these questions.
- In the contradiction proof, why does the index \(n_k\) need to become arbitrarily large as well as \(x_{n_k}\) converge?
- For the interval cover, which set contains \(B(1/2,1/4)\), and why does the strict ball inequality matter at its upper endpoint?
- If a ball-based Lebesgue number is \(\delta\), explain why every nonempty set of diameter strictly less than \(\delta\) fits inside a cover member.
- In the cover of \((0,1)\), verify directly that the point \(y=\tfrac12\min(x,1/(n+1))\) is outside \(U_n\).
- Identify the step that fails if one only knows \(d(x_{n_k},x)<r\), with no smaller margin for the radius of the ball.