# Show that a metric space is complete

## Homework Statement

Given (R+, d), R-Real #
d= | ln(x/y) |

Show that this metric space is complete

## The Attempt at a Solution

Firstly, I know that to show it is complete I need to have that all Cauchy sequences in that space converge...

So I'm not 100% sure, but if I know I have to generalize so that it works for every Cauchy sequence, so can I find subsequences that converge, and then say that each sequence converges??? :S if so, how do I start this without picking specific cases, or can I pick a specific sequence in that space?

Hello.

Trying to find a convergent subsequence of a cauchy sequence is indeed enough, but in a lot of cases you don't need to specifically focus on subsequences. I will tell you that in this problem you won't need to bother with subsequences, it'll be more direct to directly proof that a cauchy sequence has a limit.

Specific cases are a good way to get a feel for the problem. Problems where specific cases are really useful are those where the question is more open-ended: "Take this metric space ... Is it complete? Yes/no and prove this" but here you already know that it will be complete, so you can jump in:
$$\textrm{Take a cauchy sequence (x_n)_n in the metric space M := (\mathbb R^+,d) with d(x,y) = |\ln(\frac{x}{y})|. We will now prove that there is an x \in M such that x_n \to x in M:}$$
And then it's up to you ;)

Of course, first we need to get an idea of what that limit would look like, before we can prove that it exists. As a hint, rewrite |ln(x/y)| as |ln(x) - ln(y)|. Hint #2: literally write down what it means for x_n to be a cauchy sequence, using the rewritten form of the metric in hint #1. Let this inspire you.

Hm, ok.. I noticed the rewrite you mentioned, and attempted to do something with it to somehow get a lovely conclusion but I'm not sure what it is.

I have | ln(xj) - ln(xi) |, what i need is something greater than or equal to that.. i first thought of the triangle inequality but i don't know what i would use...

Maybe there is another trick I am not seeing.