Answer to Question #157918 in Real Analysis for Surada

Question

Let (X d) be metric space and X is unbounded then d_(1) defined as d_(1)=(d(x y))/(1+d(x y)) x y in X is

a)metric and unbounded

b)metric but may be bounded or unbounded c)metric and bounded

Answer 1

First of all, let's prove that this define a distance :

1. This distance is obviously positive and $d_1(x,y)=0 \leftrightarrow x=y$ trivially, as $d$ is a distance.
2. This distance is symmetric, as $d_1(y,x) = \frac{d(y,x)}{1+d(y,x)}=d_1(x,y)$ as $d$ is symmetric.
3. It satisfies the triangle inequality, as the function $f(x)=\frac{x}{1+x}$ is subadditive and increasing for $x\geq 0$ ($f(x+y)\leq f(x)+f(y)$ ), which can be proven by direct calculation and as $d$ satisfies triangle inequality, we have, by composition, $d_1(x,z)=f(d(x,z))\leq f(d(x,y)+d(y,z))\leq f(d(x,y))+f(d(y,z))=d_1(x,y)+d_1(y,z)$

Now we just need to study whether $d_1$ is bounded, but it trivially is : $d_1(x,y)=1-\frac{1}{1+d(x,y)}\leq 1$.

Therefore the answer is C : it is metric and bounded.