494 I. ASYMPTOTIC CONES AND SHARAFUTDINOV RETRACTION
By (i) we have
. ds(p s) ( ai n s (p, s) '/3i n s (p, s))
hm ' =0;
s--+oo S
hence (ii) follows.
(iii) From (ii) and Lemma I.52, it suffices to prove the existence of
. ds(p s) (a ( s) , /3 ( s))
hm '
s--+oo S
for any a, f3 E Ray M (p). It follows from parts ( 1) and ( 3) of Presumed
Theorem I.50 that
8
r---+ ds(p,s) (a (s), /3 (s))
s
is monotonically nonincreasing for s 2:: so (p); hence the limit exists and the
lemma is proved. D
6.2.3. Points at infinity. Based on the above discussion, one expects
that the following is true.
PRESUMED THEOREM l.54 (Points at infinity). Let (Mn,g) be a com-
plete noncompact Riemannian manifold with nonnegative sectional curva-
ture. If Presumed Theorem I.50 is true, then
(I.65)
is a compact metric space and diam ( M ( oo)) -< oo. We call the metric space
M ( oo) in (I.65) the space of points at infinity.
PROOF. To see that M ( oo) is a metric space, note that by Lemma
I.53(i) we have dR ([a], [/3]) = 0 if and only if [a] = [/3]. On the other hand,
that dR ([a], [/3]) = dR ([/3], [a]) is trivial, whereas the triangle inequality,
i.e.,
dR ([a] 'b]) ~ dR ([a] ' [/3]) + dR ([/3], [1])
~r any [a], [/3], ['y] E Ray Mf rv, follows from the triangle inequality for
ds(p,s)·
From the proof of Lemma I. 53 (iii), the fact that diam ( M ( oo)) < oo
follows from the monotonicity of ds(p,s)(aipnS(~,s),a^2 pnS(p,s)) and the property
that diam(S(p,so(p)),ds(p,so(P))) < oo.
Next we prove that M ( oo) is compact. By Lemma I.52 each element a
in M ( oo) can be represented by ap E Ray M (p). But elements in Ray M (p)
are in one-to-one correspondence to a closed subset n of the unit tangent
sphere Sp ( 1) c TpM. Since M ( oo) is a quotient space of Ray M (p) and n
is compact, M ( oo) is compact. D