First of all, let's prove that this define a distance :
- This distance is obviously positive and d1(x,y)=0↔x=y trivially, as d is a distance.
- This distance is symmetric, as d1(y,x)=1+d(y,x)d(y,x)=d1(x,y) as d is symmetric.
- It satisfies the triangle inequality, as the function f(x)=1+xx is subadditive and increasing for x≥0 (f(x+y)≤f(x)+f(y) ), which can be proven by direct calculation and as d 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)
Now we just need to study whether d1 is bounded, but it trivially is : d1(x,y)=1−1+d(x,y)1≤1.
Therefore the answer is C : it is metric and bounded.
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