164 4. PROOF OF THE COMPACTNESS THEOREM
such that K (ae, f3e) E Ke is a monotone function of£. If k 2: K (ae, f3e),
then k E Kg. Hence either B~£ n B~£ is empty for all k 2: K ( ae, f3e) or it is
nonempty for all k 2: K ( ae, f3e). D
DEFINITION 4.30. Let K (r) ~ max (K' (r), { K (a, /3) : a, f3 _:S A (r)}).By definition, we have that if a, f3 :SA (r), then K (a, /3) :SK (r). Note
also that K (r) is increasing. This proves statement (6) of Lemma 4.18.
3.4. Estimates on inclusions of intersecting balls. Now to com-
plete the proof of Lemma 4.18 (only statements (5) and (7) remain to be
proved), we recaUDefinition 4.26 and prove the following proposition.
PROPOSITION 4.31. If Bk n B~ i= 0' where k 2: K (max { rk' r~})' then
-13 - ->(3.we have Bk C Bk and Bk C Bk.
PROOF. Recall estimate (4.9), which givesrZ :S rk + 10>. [OJ.
Now, we can estimate >./3 by
./3 2: ~>. [r~] 2: ~>. [rk + 10>. [OJ].
Recall that the definition of >. is
>. [rJ = ce-^0 r.
So >. [OJ = c and >. [rk + 10>. [OJ] = e-lOcC >. [rk]. Hence
>./3 > ~e-lOcC>.[raJ > ~e-lOcCAa
- 2 k - 4
or
). a :S 4 e10cC >./3.
Choose a y E Bk n BZ. Then for each x E Bk we get
d ( x, x~) :S d (x, x/:) + d (xk', y) + d (y, x~)
< 5>.0i + 5>.0i + 5>,/3
:S 20e^10 cC >./3 + 20elOcC >./3 + 5>./3
:S 45 elOcC >,/3.Similarly, if x E Bk, then
d (x,x%) :S d(x,xk') + d(xk',y) + d (y,x%)
< 45e10cC A a + 5>. a + 5>,/3
:S 45e10cC ( 4e10cC >,/3) + 5 ( 4elOcC >,/3) + 5>,/3
:S 205e20cC >./3.D