Complete metric space

Homework Statement

We say that two metrics d, d' on a space S are equivalent if each "dominates" the other in the following sense: there exist constants M, M'>0 such that
d'(x,y)<=M' d(x,y) and d(x,y)<=M d'(x,y) for all x,y in S.

If metrics d, d' are equivalent, prove that (S,d) is complete<==>(S,d') is a complete metric space.

The Attempt at a Solution

If (S,d) is complete then every Cauchy sequence in S converges to a limit in S. I want to go somewhere along the lines of saying that multiplying by a constant will not change its convergence.....am I going along the right lines here? I don't really know how else to go about this.

morphism
Homework Helper
Let's do the => direction first. Take a cauchy sequence {x_n} in (S,d'). Is {x_n} cauchy in (S,d)?

I'm guessing it is yes. Because we have M' that we can multiply the sequence by?

HallsofIvy
Homework Helper
And, of course, a sequence convertes if and only if "given $\epsilon> 0$, there exist N such that ...

Given $\epsilon> 0$ in d', what does that tell you about $\epsilon/M'$ in d?

And, of course, a sequence convertes if and only if "given $\epsilon> 0$, there exist N such that ...

Given $\epsilon> 0$ in d', what does that tell you about $\epsilon/M'$ in d?

To finish the first part of your response, "there exists N such that..."
is the end of that "n larger than N will be equal to some epsilon>0"?

For the second part, epsilon/M' will be some small value in d...?

HallsofIvy
If $d(x_n, L)< \epsilon[/tex] for all n> N, then [itex]d'(x_n,L)= M'\epsilon$ and conversely.