From Small Balls to a Finite Subcover
A finite net gives a finite collection of small balls covering a metric space. A Lebesgue number for an open cover says that sufficiently small sets fit inside individual cover members. The useful combination is to make each net ball small enough for that second property: then one cover member can be chosen for each ball, and the chosen members cover the space. The factor of two in the radius condition comes from the diameter of a ball, so it is important not to omit it.
Throughout, \((X,d)\) is a metric space with its metric topology, whose open sets are unions of open balls. An open cover \(\mathcal{U}\) of \(X\) is a collection of open subsets of \(X\) whose union is \(X\). A finite subcover is a finite subcollection of \(\mathcal{U}\) whose union is still \(X\). For a nonempty subset \(E\subseteq X\), its diameter is \(\operatorname{diam}(E)=\sup\{d(x,y):x,y\in E\}\).
We use the diameter convention for a Lebesgue number. Some statements instead call \(\lambda>0\) a Lebesgue number when, for each \(x\in X\), the ball \(B(x,\lambda)\) lies in some cover member. Under that convention, if \(F\) is an \(r\)-net with \(r\leq\lambda\), then \(B(c,r)\subseteq B(c,\lambda)\) for every \(c\in F\), so the same finite-selection argument applies. The factor of two below belongs to the diameter convention: a ball of radius \(r\) has diameter at most \(2r\).
The Finite-Net Theorem
Proof. If \(X=\varnothing\), the empty subcollection is a finite subcover, so assume \(X\ne\varnothing\). For each \(c\in F\), consider \(B(c,r)\). If \(x,y\in B(c,r)\), then \(d(x,c)<r\) and \(d(y,c)<r\). The triangle inequality gives \(d(x,y)\leq d(x,c)+d(c,y)<2r\). Thus every distance between two points of \(B(c,r)\) is at most \(2r\), so \(\operatorname{diam}(B(c,r))\leq 2r<\delta\). In particular, this ball is a nonempty subset of diameter less than \(\delta\). By the Lebesgue-number property, there is some \(U_c\in\mathcal{U}\) such that \(B(c,r)\subseteq U_c\). There are only finitely many centers in \(F\), so the collection of chosen sets \(\{U_c:c\in F\}\) is finite. Since \(F\) is an \(r\)-net, every \(x\in X\) lies in \(B(c,r)\) for some \(c\in F\), and hence lies in \(U_c\). Therefore the chosen collection covers \(X\) and is a finite subcover. \(\square\)
The proof uses two separate ingredients: the net covers \(X\), and the Lebesgue number places each net ball inside a cover member. The strict inequality \(2r<\delta\) ensures that even the supremum of the distances within a ball is less than \(\delta\). It is not necessary for the ball’s diameter to equal \(2r\); the upper bound is enough.
Worked Example: A Two-Set Cover of the Unit Interval
Give a finite subcover using a finite net for \(I=[0,1]\) with its usual metric, where \(U_1=(-0.1,0.7)\cap I\) and \(U_2=(0.4,1.1)\cap I\).
First, \(\{U_1,U_2\}\) is an open cover of \(I\): the sets are open in the subspace topology, and the intervals overlap on \((0.4,0.7)\). We can take \(\delta=0.3\) as a Lebesgue number. Indeed, suppose a nonempty set \(E\subseteq I\) with diameter less than \(0.3\) were contained in neither member. Not being contained in \(U_1\) means that \(E\) has a point \(x\geq0.7\); not being contained in \(U_2\) means that it has a point \(y\leq0.4\). These points satisfy \(d(x,y)=x-y\geq0.7-0.4=0.3\), contradicting \(\operatorname{diam}(E)<0.3\).
Take \(F=\{0,0.25,0.5,0.75,1\}\) and \(r=0.14\). Consecutive centers have gaps \(0.25,0.25,0.25,0.25\); each midpoint is \(0.125\) from its two neighboring centers, and \(0.125<0.14\). The endpoints are themselves centers. Therefore every point of \(I\) is at distance less than \(0.14\) from a center, so \(F\) is an \(r\)-net. The required inequality is \(2r=2(0.14)=0.28<0.3=\delta\).
For the centers \(0\) and \(0.25\), the balls of radius \(0.14\), taken in \(I\), lie in \(U_1\): their points are respectively in \([0,0.14)\) and \((0.11,0.39)\). The ball around \(0.5\) is \((0.36,0.64)\), which lies in both \(U_1\) and \(U_2\). The ball around \(0.75\) is \((0.61,0.89)\), and the ball around \(1\) is \((0.86,1]\); both lie in \(U_2\). Thus choosing \(U_1\) for the first two centers and \(U_2\) for the last three gives a finite subcover.
Check: the two selected sets do cover \(I\): \(U_1\) covers all points below \(0.7\), and \(U_2\) covers all points above \(0.4\), with overlap between them. All the balls used above have radius \(0.14\), and the comparison needed for the theorem is \(0.28<0.3\), not merely \(0.14<0.3\).
Worked Example: An Infinite Cover with an Explicit Lebesgue Number
On \(I=[0,1]\) with the usual metric, consider the cover consisting of \(U_n=(-1/n,1-1/n)\cap I\) for \(n\geq2\), together with \(V=(0.8,1.1)\cap I\). Verify a Lebesgue number and use a finite net to select a finite subcover.
Each \(U_n\) and \(V\) is open in \(I\). Every \(x<1\) belongs to some \(U_n\): choose an integer \(n\geq2\) with \(1/n<1-x\), which gives \(x<1-1/n\); also \(x\geq0>-1/n\). The remaining point \(1\) belongs to \(V\). Thus this is an open cover.
We claim \(\delta=0.2\) is a Lebesgue number. Let \(E\subseteq I\) be nonempty with \(\operatorname{diam}(E)<0.2\). If every point of \(E\) is greater than \(0.8\), then \(E\subseteq V\). Otherwise choose \(x\in E\) with \(x\leq0.8\). For every \(y\in E\), \(y\leq x+\operatorname{diam}(E)<0.8+0.2=1\). Hence all points of \(E\) have a common upper bound strictly less than \(1\): specifically \(y\leq x+\operatorname{diam}(E)<1\). Choose \(n\geq2\) large enough that \(1-1/n\) is greater than this upper bound. Since every \(y\in E\) also satisfies \(y\geq0>-1/n\), we obtain \(E\subseteq U_n\). This proves the claim.
Use the finite set of centers \(F=\{0,0.15,0.30,0.45,0.60,0.75,0.90,1\}\) with \(r=0.09\). The gaps between consecutive centers are \(0.15\), except the final gap \(1-0.90=0.10\). Half of each gap is at most \(0.075\), which is less than \(0.09\); the endpoints are centers. Thus every point of \(I\) lies in a radius-\(0.09\) ball around a center. Also \(2r=2(0.09)=0.18<0.2=\delta\).
For centers \(0,0.15,0.30\), their radius-\(0.09\) balls lie in \(U_2=[0,0.5)\). The ball around \(0.45\) lies in \(U_3=[0,2/3)\), because its points are below \(0.54<2/3\). The ball around \(0.60\) lies in \(U_4=[0,0.75)\), because its points are below \(0.69<0.75\). The ball around \(0.75\) lies in \(U_7=[0,6/7)\): its points are below \(0.84=21/25<6/7\), since \(147<150\). Finally, the balls around \(0.90\) and \(1\) lie in \(V\): their points are respectively greater than \(0.81>0.8\) and greater than \(0.91>0.8\). These choices select the finite subcover \(\{U_2,U_3,U_4,U_7,V\}\).
Check: the selected cover members contain all the net balls, and those balls cover \(I\). The cover has infinitely many members, but the argument selects only five of them.
Worked Example: A Four-Point Metric Space
Let \(X=\{a,b,c,d\}\) have the metric with \(d(a,b)=d(c,d)=1/5\), every distance between a point of \(\{a,b\}\) and a point of \(\{c,d\}\) equal to \(1\), and distance \(0\) from each point to itself. The distances are symmetric. They satisfy the triangle inequality: for three distinct points, the side lengths are either \(1,1,1\) or \(1,1,1/5\), and the largest side is no greater than the sum of the other two. Let \(U=\{a,b\}\) and \(V=\{c,d\}\).
The sets \(U\) and \(V\) are open in the metric topology. In fact, \(B(a,1)=\{a,b\}=U\) and \(B(b,1)=\{a,b\}=U\), since the within-pair distance \(1/5\) is less than \(1\), while both cross-pair distances equal \(1\) and are excluded from the open ball. Likewise, \(B(c,1)=B(d,1)=V\). Thus \(\mathcal{U}=\{U,V\}\) is an open cover.
The number \(\delta=1\) is a Lebesgue number for this cover. If a subset \(E\subseteq X\) met both \(U\) and \(V\), it would contain a cross-pair at distance \(1\), so its diameter would be at least \(1\). Therefore every nonempty subset with diameter less than \(1\) lies entirely in \(U\) or entirely in \(V\). Now take \(F=\{a,c\}\) and \(r=1/3\). Since \(1/5<1/3\), \(B(a,1/3)=U\) and \(B(c,1/3)=V\); these balls cover \(X\). The inequality \(2r=2/3<1=\delta\) holds, and selecting \(U\) and \(V\) gives the finite subcover. In this example the balls \(B(a,1/2)\) and \(B(c,1/2)\) are also the two-point sets \(U\) and \(V\), not singletons.
Check: the radius-\(1/3\) balls include the paired points because \(1/5<1/3\), and exclude the cross-pair points because \(1>1/3\). The open balls of radius \(1\) used to verify openness exclude points at distance exactly \(1\).
Why the Radius Inequality Matters
A common mistake is to infer that every net ball fits in a cover member from \(r<\delta\). That conclusion does not follow under the diameter definition: points near opposite sides of a ball can be almost \(2r\) apart.
For a concrete calculation, take \(X=\mathbb{R}\) with its usual metric and the open cover \(U=(-\infty,0.6)\), \(V=(0.4,\infty)\). The number \(\delta=0.2\) is a Lebesgue number: if a set of diameter less than \(0.2\) were contained in neither member, it would contain a point \(x\geq0.6\) and a point \(y\leq0.4\), giving \(d(x,y)\geq0.6-0.4=0.2\), a contradiction. The tempting choice \(r=0.15\) satisfies \(r<\delta\), but \(B(0.5,0.15)=(0.35,0.65)\) is contained in neither \(U\) nor \(V\). It extends above \(0.6\), so it is not contained in \(U\), and below \(0.4\), so it is not contained in \(V\). The correction is to require \(2r<\delta\); for example \(r=0.09\) gives \(2r=0.18<0.2\). This illustrates why the proof compares the diameter bound \(2r\) with the Lebesgue number, rather than comparing \(r\) alone.
The finite-net argument is useful because it turns a local cover condition into a finite selection: one cover member is chosen for each of finitely many balls. In metric compactness arguments, finite nets and Lebesgue numbers therefore work together as a concrete mechanism for obtaining finite subcovers. The exact numerical condition depends on the stated convention for “Lebesgue number,” so recording whether the condition concerns diameters or balls prevents a factor-of-two error.
Check Your Understanding
For each question, state which Lebesgue-number convention you are using and keep strict inequalities explicit.
- If \(\delta=0.6\) is a diameter Lebesgue number, does an \(r=0.3\) net satisfy the theorem’s hypothesis? What if \(r=0.29\)?
- Explain why the diameter of an open ball \(B(c,r)\) is at most \(2r\), even when it does not equal \(2r\).
- In the four-point metric example, what changes if the balls are defined using the closed-ball condition \(d(x,c)\leq r\) instead of the open-ball condition?
- For the cover of \(I\) by the sets \(U_n\) and \(V\), why does a set \(E\) of diameter less than \(0.2\) that has a point at most \(0.8\) have all its points bounded above by one fixed number less than \(1\)?
- If a Lebesgue number is defined by requiring \(B(x,\lambda)\) to lie in a cover member, why does an \(r\)-net with \(r\leq\lambda\) allow the same finite-selection argument?