1 Let Md be a metric space and let abc be any three points o

1. Let (M,d) be a metric space and let a,b,c be any three points of M . Prove that |d(a, c) d(b, c)| d(a, b).

If (M, d) is a metric space then

1. d (x, y) 0 (non-negativity), for any x, y in M

2. d (x, y) = 0 if and only if x = y

3.d(x, y) = d(y, x) (symmetry), for any x, y in M

4. d(x, y) = d(x, z) + d(y, z) for any x, y, z in M (triangular inequality)

In the triangular inequality if we take x=a, y=c, z=b then we have

d(a, c) d(a, b) + d(b, c)

this implies that |d(a, c) – d(b, c)| d(a, b) which is the required result.