188 5. Exact Groups and Related Topicsparticular, countable. We leave it as an exercise to check that every x EK
can be connected to every z E fJK via a geodesic path.
Although the topology on k can be defined like that of a tree, we give
a different description. Fix a base point o E K. For z E fJK and R > 0, we
set
U(z, R) = { x E k : 3 geodesic paths a, /3 with a+ = x, /3+ = z
such that liminf (a(m),/3(n)) 0 > R},
m,n-+oo
where in the case x EK, we choose the "geodesic" a(m) = x for all m. We
also define
U' (z, R) = { x E k : \::/ geodesic paths a, /3 with a+ = x, /3+ = z,
we have m,n-+oo liminf(a(m),/3(n)) 0 > R}.It turns out these sets satisfy the axioms for a neighborhood basis. The
resulting topology on k is as expected: k is compact and K is a dense
open discrete subset. Let's prove this.
It is clear that U'(z,R) c U(z,R). On the other hand, we have
Lemma 5.3.11. There exists C = C(K) > 0 with the following property:
If a, a' and /3, /3^1 are geodesic paths such that a+ =a'+ and /3+ = f3'+, thenm,n-+oo liminf (a'(m),/3'(n))^0 2': m,n-+oo liminf(a(m),/3(n))^0 - C.
In particular, U'(z,R)::) U(z,R+ C) for every z E fJK and R > 0.Proof. This follows from Lemma 5.3.8 and the inequality
(a' ( m'), /3^1 (n') ) 0 2': (a(m ), /3(n) ) 0 - ( d( a' (m'), a(m)) + d(/3^1 (n'), /3( n)) ).
D
Lemma 5.3.12. For any R > 0, there exists S > 0 with the following
property: For any y,z E fJK with y E U(z,S), we have U(y,S) c U(z,R).Proof. Choose some 5 > 0 such that every geodesic triangle is 5-thin. By
Lemma 5.3.11, it suffices to show that if y E U'(z, N) for y, z E fJK and NE
N, then U'(y, N) C U(z, N-5). Let x E U'(y, N) and take geodesic paths a,
/3 and/ connecting o to x, y and z, respectively. Since liminf(t(n), /3(m)) 0 >
N, we have d(t(N),/3(N)) < 5. Similarly, d(/3(N), a(N)) < 5 and hence
d(t(N),a(N)) < 25. It follows that for every m,n 2': N,
2(a(m), 1(n)) 0 = m + n - d(a(m), 1(n))
2': m + n - (m - N + d(a(N), 1(N)) + n - N)
2': 2N - 25.
This shows x E U(z, N - 5). D