Proving that the metric space $((0,\infty),d)$ is complete, with $d(x,y)=|\ln x-\ln y|$ 
Let $X$ denote $(0,\infty)\subseteq \mathbb{R}$, and let $d:X\times X\to \mathbb{R}$ be defined as $d(x,y)=|\ln x- \ln y|$. Show that $(X,d)$ is a complete metric space.

I am taking for granted here that $d$ actually defines a metric on $X$, since I have already showed this. However, I am quite stuck on how to show that this space is complete. My initial thought was to take an arbitrary Cauchy sequence in $X$, and show that it converges and that the limit is in $X$. 
Now if we let $\{x_n\}_{n=1}^\infty$ be some such Cauchy sequence in $X$, we know that given any $\varepsilon>0$ we can find a corresponding $N\in\mathbb{N}$ such that $d(x_n,x_m)=|\ln x_n - \ln x_m|=|\ln(x_n/x_m)|<\varepsilon$ whenever $n,m>N$.
Now, since $\ln(x_n/x_m)$ should tend to zero, this would mean that $x_n$ and $x_m$ become arbitrarily close to each other as $N$ grows large. This in turn would imply that a limit exists, however I am not sure how to make these arguments rigorous and also how to show that such a limit lies in $X$.
Any hints on how to proceed would be greatly appreciated!
 A: Here is another way of looking into this problem: If you know about isometry then the following will help.
Let $Y=\Bbb R$ with the standard abbslute value metric $|\cdot|.$     As notation in your problem, the map given by $f:(X,d)\to (Y,|\cdot|)$ given by $f(x)=\log x$ is a surjective isometry  (as $\forall x,y \in X, \, \left | f(x)-f(y) \right|= |\log x-\log y|=d(x,y)).$ So the metric spaces $X$ and $Y$ are isometric. Since isometry preserves completeness (more precisely Cauchy sequences) and since $Y=\Bbb R$ is complete, it follows that $X$ is also complete with respect to the metric $d$. 
A: Hint: First show that a Cauchy sequence in this metric is contained in an interval of the form $[\epsilon, M]$, because for a sufficiently large $N$ we have $|\ln(x_m) - \ln(x_N)| < 1$ whenever $m > N$.
Now since the derivative of $\ln$ is bounded in such an interval, the function is Lipschitz, i.e., there exists $K > 0$ such that $|\ln(x) - \ln(y)| \leq K|x - y|$ for $x,y \in [\epsilon,M]$: this shows that the sequence is also Cauchy in the usual metric.
Hope this gets you started.
