The Fivefold Covering Lemma in R^n

Why a Geometric Selection Is Useful

In covering arguments, a family of balls may overlap too much to estimate its union directly. A useful remedy is to retain a disjoint subfamily and enlarge its members by a fixed factor. The fivefold covering lemma provides exactly this: the selected balls do not overlap, while their fivefold dilations cover every ball in the original family. The key is to select in order of radius scale, so that a ball discarded because it intersects a selected ball cannot be much larger than that selected ball.

Throughout, \(n\ge1\), and \(B(c,r)=\{x\in\mathbb{R}^n:|x-c|<r\}\) denotes the open Euclidean ball with center \(c\in\mathbb{R}^n\) and radius \(r>0\). Its fivefold dilation is \(5B(c,r)=B(c,5r)\), with the same center. We assume the family has uniformly bounded radii: there is a finite \(R>0\) such that \(r\le R\) for every ball in the family. No bound on the centers is required. The balls are Borel sets in \(\mathbb{R}^n\); when measure is used below, it is Lebesgue measure \(m\) on \((\mathbb{R}^n,\mathcal{L},m)\), a complete, sigma-finite measure space.

Selection by Radius Scales

Divide the radii into the classes \((2^{-(k+1)}R,2^{-k}R]\), for integers \(k\ge0\). Process these classes in order from largest to smallest. In each class, discard any ball that intersects a ball selected in an earlier class. From the balls remaining in the current class, choose an inclusion-maximal pairwise disjoint subfamily. “Maximal” here means that no additional remaining ball can be added while preserving pairwise disjointness; it does not mean that the subfamily has the largest possible cardinality.

Such a maximal subfamily exists: order the pairwise disjoint subfamilies by inclusion. The union of any chain of them is again pairwise disjoint, so Zorn’s lemma gives a maximal element. This selection also works when a class contains uncountably many balls. The maximal subfamily is countable: each nonempty open ball contains a point of \(\mathbb{Q}^n\), and distinct disjoint balls contain distinct rational points. More explicitly, enumerate \(\mathbb{Q}^n\) and assign each ball the first point in the enumeration that it contains; disjoint balls cannot receive the same point. The union of the selected subfamilies over the countably many classes is therefore countable.

For the radius comparison, suppose first that \(B(c,r)\) and \(B(c',r')\) are in the same class \((2^{-(k+1)}R,2^{-k}R]\). Then \(r\le2^{-k}R\) and \(r'>2^{-(k+1)}R\), so \(r<2r'\). If \(B(c,r)\) is in a later class than \(B(c',r')\), the earlier class has index \(j<k\), and \(r\le2^{-k}R\le2^{-(j+1)}R<r'\). Thus \(r\le2r'\) in either case. Each original ball is either selected itself or was discarded because it met a selected ball in its own or an earlier class.

Proof. Let \(B(c,r)\) be an original ball that is not selected, and let \(B(c',r')\) be a selected ball that it meets, as guaranteed by the selection procedure. Choose \(q\in B(c,r)\cap B(c',r')\). The triangle inequality gives \(|c-c'|\le |c-q|+|q-c'|<r+r'\). For every \(x\in B(c,r)\), another application of the triangle inequality gives $$ |x-c'|\le |x-c|+|c-q|+|q-c'|<2r+r'\le5r'. $$ The last inequality follows from \(r\le2r'\). Hence \(B(c,r)\subseteq B(c',5r')\). A selected ball is also contained in its own fivefold dilation, so every original ball is covered. The selected balls are pairwise disjoint by construction, and their family is countable by the rational-point argument. If the original family is empty, the asserted containment holds for the empty selection. \(\square\)

Worked Applications

Finite Families and a Common Pitfall

Proof. A finite family has bounded radii, so the fivefold covering lemma supplies a finite disjoint subfamily \(\mathcal{G}\) whose fivefold dilations cover the original union. The original union is a finite union of open, hence Lebesgue measurable, sets. Subadditivity gives \(m(\bigcup_{B\in\mathcal{F}}B)\le\sum_{B\in\mathcal{G}}m(5B)\). For a ball \(B=B(c,r)\), the scaling formula for Lebesgue measure gives \(m(5B)=5^n m(B)\). The selected balls belong to \(\mathcal{F}\), so their measures sum to no more than the sum over \(\mathcal{F}\), since all terms are nonnegative. Combining these inequalities proves both bounds. If \(\mathcal{F}\) is empty, both sums and the measure of its union are zero. \(\square\)

A common mistake is to infer fivefold coverage merely from intersection. Intersection alone does not ensure that the ball being covered is small enough. For instance, \(B(1,100)\) intersects \(B(0,1)\), because the centers are distance \(1<101\) apart. But \(100\in B(1,100)\), while \(100\notin B(0,5)\). Thus \(B(1,100)\) is not contained in the fivefold dilation of \(B(0,1)\). The selection argument avoids this failure: a ball discarded because it intersects a selected ball satisfies \(r\le2r'\). This radius control, not intersection by itself, is what makes the triangle-inequality estimate work.

The lemma is useful whenever a later estimate is easier for disjoint balls than for overlapping ones. In the finite measure bound, disjointness keeps the selected balls’ measures additive, while dilation transfers their coverage to the original family at the cost of the fixed factor \(5^n\). The proof depends on Euclidean geometry and the scale ordering; it does not claim that arbitrary overlapping balls can be replaced by disjoint balls without enlargement.