Answer to Question #157918 in Real Analysis for Surada

Question #157918

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


1
Expert's answer
2021-01-26T04:08:18-0500

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

  1. This distance is obviously positive and d1(x,y)=0x=yd_1(x,y)=0 \leftrightarrow x=y trivially, as dd is a distance.
  2. This distance is symmetric, as d1(y,x)=d(y,x)1+d(y,x)=d1(x,y)d_1(y,x) = \frac{d(y,x)}{1+d(y,x)}=d_1(x,y) as dd is symmetric.
  3. It satisfies the triangle inequality, as the function f(x)=x1+xf(x)=\frac{x}{1+x} is subadditive and increasing for x0x\geq 0 (f(x+y)f(x)+f(y)f(x+y)\leq f(x)+f(y) ), which can be proven by direct calculation and as dd satisfies triangle inequality, we have, by composition, d1(x,z)=f(d(x,z))f(d(x,y)+d(y,z))f(d(x,y))+f(d(y,z))=d1(x,y)+d1(y,z)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 d1d_1 is bounded, but it trivially is : d1(x,y)=111+d(x,y)1d_1(x,y)=1-\frac{1}{1+d(x,y)}\leq 1.

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


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