Question

Let d: R XR + R be defined to be d(x, y) = |arctan(x) – arctan(y)]. Show that d is a metric on R.

Answer 1

Let X be a non empty set. A metric on X is any non negative function d:X×X<-> R^+ such that the following properties hold ,. 1)d(x,y)>=0. for all x,y$\epsilon$X 2)d(x,y)=d(y,x) for all x,y$\epsilon$X 3)d(x,z)<=d(x,y)+d(y,z) for all x,y,z$\epsilon$X 4)d(x,y)=0 iff x=y for all x,y$\epsilon$X Clearly d(x,y)=| arctanx -arctany|>=0 for all x,y$\epsilon$X Now d(x,y)=|arctanx-arctany| =|arctany-arctanx| =d(y,x) for all x,y$\epsilon$X The function arctant is strictly increasing on R. hence d(x,y)=0 if and only if x=y. Now to prove triangle inequality d(x,z)=|arctanx-arctanz| =|arctanx-arctany+arctany-arctanz| <=|arctanx-arctany|+|arctany-arctanz| =d(x,y)+d(y,z). So d(x,z)<=d(x,y)+d(y,z). Thus d(x,y) satisfies all the metric property on R.So d is a metric on R.