13.13. BESICOVITCH COVERING THEOREM 377

ball except that its measure does not get smaller. Thus the thing required is a coveringtheorem which does not depend on enlarging balls.

This all works in a normed linear space (X ,∥·∥) which has dimension p. Here is asequence of balls from F in the case that the set of centers of these balls is bounded. I willdenote by r (Bk) the radius of a ball Bk.

A construction of a sequence of balls

Lemma 13.13.1 Let F be a nonempty set of nonempty balls in X with

sup{diam(B) : B ∈F}= D < ∞

and let A denote the set of centers of these balls. Suppose A is bounded. Define a sequenceof balls from F ,

{B j}J

j=1 where J ≤ ∞ such that

r (B1)>34

sup{r (B) : B ∈F} (13.13.31)

and ifAm ≡ A\ (∪m

i=1Bi) ̸= /0, (13.13.32)

then Bm+1 ∈F is chosen with center in Am such that

r (Bm)> r (Bm+1)>34

sup{r : B(a,r) ∈F , a ∈ Am} . (13.13.33)

Then letting B j = B(a j,r j) , this sequence satisfies{

B(a j,r j/3)}J

j=1 are disjoint,.

A⊆ ∪Ji=1Bi. (13.13.34)

Proof: First note that Bm+1 can be chosen as in 13.13.33. This is because the Am aredecreasing and so

34

sup{r : B(a,r) ∈F , a ∈ Am}

≤ 34

sup{r : B(a,r) ∈F , a ∈ Am−1}< r (Bm)

Thus the r (Bk) are strictly decreasing and so no Bk contains a center of any other B j.If x∈ B(a j,r j/3)∩B(ai,ri/3) where these balls are two which are chosen by the above

scheme such that j > i, then from what was just shown∥∥a j−ai∥∥≤ ∥∥a j−x

∥∥+∥x−ai∥ ≤r j

3+

ri

3≤(

13+

13

)ri =

23

ri < ri

and this contradicts the construction because a j is not covered by B(ai,ri).Finally consider the claim that A ⊆ ∪J

i=1Bi. Pick B1 satisfying 13.13.31. If B1, · · · ,Bmhave been chosen, and Am is given in 13.13.32, then if Am = /0, it follows A⊆ ∪m

i=1Bi. SetJ = m.