If (X,d) is a metric space (not defined), prove that (X,p) is a metric space defined by p(x,y) = d(x,y)/(1+d(x,y)).
I'm having trouble proving that p has the triangle equality property.
1st attempt:
I wrote d(x,z)/(1+d(x,z)) <= d(x,y)/(1+d(x,y)) + d(y,z)/(1+d(y,z)), multiplied the whole thing by (1+d(x,z))(1+d(x,y))(1+d(y,z)), then cancelled out as many terms as possible, but I only got d(x,z) <= d(x,y) + d(y,z) + 2d(x,y)d(y,z) + d(x,y)d(x,z)d(y,z).
I think it will be simpler to solve this problem by solving an analogous problem, that x <= y + z implies that x/(1+x) <= y/(1+y) + z/(1+z).
You know that and you want to show that .
Start with what you are given, and get to what you want:
Add in the extra terms to each side of the equation so that you can get:
Then the final step is to divide both sides by .
Then you have
To figure it out, you start with the conclusion and simplify. To prove the conclusion, you start with what you are given and work your way backwards through whatever work you did to simplify the expression. In other words, do the simplifications of the conclusion on scratch paper. It should not be part of the proof.
Thanks SlipEternal. Looking back at the way I originally did it, I think I pretty much had it. All I had to do was show that if d was a metric, then d(x,z) <= d(x,y) + d(y,z) + 2d(x,y)d(y,z) + d(x,y)d(x,z)d(y,z) holds since d(x,y) + d(y,z) was inbetween, and this long ugly inequality is equivalent to p(x,z) <= p(x,y) + p(y,z) so it all works out.