Notice that . It follows that if the sequence is Cauchy for d2, then the sequence is Cauchy for the usual metric d1 and therefore converges to a finite limit. So must be bounded away from 0.
I'm trying to do a proof on some metric spaces, I think I've almost completed it but I am stuck on the last part of it.
The two metric spaces I am working with are where and
So far I can show that any Cauchy sequence in d2 is also a cauchy sequence in d1, in short because if converges to zero then (y/x) has to converge to one, which means that the difference between x and y has to converge to zero.
I can similarly show that whenever a sequence converges to some limit (notice it cant be zero) in d1, then it converges to the same limit in d2.
Now the only thing that I need to prove next is that if a sequence is Cauchy in both metrics, it cannot converge to zero in d1, or equivalently if a sequence converges to zero in d1, it will not be Cauchy in d2.
I don't really know how to approach this; I see that if x and y are really close to zero, ln(x) and ln(y) will be very large, but I can't manage to turn this into something useful...