X dx be a metric space define dx y min dxx y 1 Show that d

(X, d_x) be a metric space, define d(x, y) = min {dx(x, y), 1}. Show that d is a metric

Solution

clearly

d(x,y) = min{d_x(x,y) ,1} >=0

d(x,y) = 0 => min{d_x(x,y) ,1} =0 => d_x(x,y) = 0 = > x=y (since d_x is a metric)

d(x,y) = min{d_x(x,y) ,1} = min{d_x(y,x) ,1} = d(y,x)

d(x,z) = min{d_x(x,z) ,1} <= min{d_x(x,y) + d_x(y,z) ,1} <= min{d_x(x,y) ,1} + min{d_x(y,z) ,1} = d(x,y) + d(y,z)

satisfies all conditions

d is a metrix