# Complete metric space

1. Nov 23, 2008

### fk378

1. The problem statement, all variables and given/known data
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.

3. 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.

2. Nov 23, 2008

### morphism

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

3. Nov 23, 2008

### fk378

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

4. Nov 24, 2008

### HallsofIvy

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?

5. Nov 24, 2008

### fk378

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...?

6. Nov 24, 2008

### HallsofIvy

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