Answer to Question #217016 in Differential Geometry | Topology for Prathibha Rose

Question

Given a metric d on a set X prove that there exists an equivalent bounded metric d' on X

Answer 1

Solution;

Let d(x,y)=0

Let a,b and c be non negative numbers such that,

1+b+c$\leq$ (1+b)(1+c)

$\frac{2+b+c}{(1+b)(1+c)}$ $\leq$ $\frac{2+b+c}{1+b+c}$

That is;

$\frac{1}{1+b}$ +$\frac{1}{1+c}$ $\leq$ 1+$\frac{1}{1+b+c}$ $\leq$ 1+$\frac{1}{1+a}$

If 0$\leq{a}\leq{b+c}$

Add 1 to both sides of the above equation and rearrange;

$\frac{a}{1+a}$ $\leq$ $\frac{b}{1+b}$ +$\frac{c}{1+c}$ .......(i)

For any non negative numbers such that a$\leq$b+c

For x ,y ,z$\epsilon$ X the none negative numbers a=d(x,y),b=d(x,z) and c=d(z,y) satisfy a$\leq$ b+c by the triangle of inequality of metric d.

Equation (i) gives the triangle of inequality of d'(x,y)=$\frac{d(x,y)}{1+d(x,y)}$ in which d' is metric on X.