- CONSTRUCTION OF GOOD COVERINGS BY BALLS 161
PROOF. There are two steps to the argument. The first is that the balls
{ B ( xa, 2A. [ra]) : 0 ::::; a ::::; A'} cover for some A'. The second is that we can
find a bound A for A' which is independent of the manifold.
Let A' ( r) ~ max {a : ra ::::; r}. Note that this definition ensures that
ra > r if a> A' (r). We will now show that {B (xa, 2A. [ra]) : 0::::; a::::; A' (r)}
form a cover of B (0, r). Consider p EB (0, r) and let s ~ d (p, 0). If pis
not covered by these balls, then d (p, xa) 2: 2A. [ra] for all a::::; A' (r). Hence
B (p, A. [s]) is disjoint from the balls {B (xa, A. [ra]): 0::::; a::::; A' (s)} since if
q EB (p, A. [s]), then for a::::; A' (s) <A' (r),
d (q, xa) 2: d (p, xa) - d (p, q) 2: 2A. [ra] - A. [s] 2: A. [ra]
since A. is decreasing. This implies that p E sA'(s)+l. This is a contradiction
because rA'(s)+l must be the minimal distance from 0 to a point in 5A'(s)+1,
but since rA'(s)+l > s, the minimum should have beens= d(O,p).
In order to estimate A' ( r) , we shall use the curvature bound to get
volume estimates. The Rauch Comparison Theorem [72] and our injectivity
assumption imply that there is a number c; = c; ( n, Co) depending only on
the dimension and the upper curvature bound of Co such that
Vol (B (xa, A. [ra])) 2: c;,. [raf
for all a ::::; A' (r). By the Bishop Volume Comparison Theorem, there is
a number M = M (n, Co) depending only on the dimension and the lower
curvature bound of -Co such that
Vol ( B ( 0, r)) ::::; M exp [ ( n - 1) VGor J.
Since A. [r] ~A. [ra] for a::::; A' (r), we have
A'(r)
L'.:: Vol (B (xa, A. [ra])) 2: (A' (r) + 1) c;A. [rf.Since the balls B (xa, A. [ra]) are disjoint and are contained in B ( 0, r + A. [OJ),
we also have that
A'(r)
L'.:: Vol (B (xa, A. [ra])) ::::; Vol (B ( 0, r +A. [OJ))
Thus we get the bound
(A' (r) + 1) c;A. [rf::::; M exp [(n -1) VCo (r +A. [o])J
or
1 ( ) M exp [(n - 1) VCo (r +A. [OJ)]
A r ::::; c;,\. [rf - 1.We can take A (r) to be the least integer greater than this number. Since
ra is increases as a increases, we see that if a> A (r), then ra 2: r. D