# Proving that given metric space is complete: $X := (0,\infty)$ and $d:=|\ln(x)-\ln(y)|$

Given the metric space $(X,d)$ with $X := (0,\infty)$ and $d:=|\ln(x)-\ln(y)|$, how can I show that $(X,d)$ is complete?

I need to prove that any Cauchy sequence converges, so:

If $(x_n)$ is a Cauchy sequence in X, then it follows for all $\epsilon \gt 0$ that there exists an $n_0 \in \mathbb{R} : \forall n,m \leq n_0: d(x_n,x_m) \lt \epsilon$.

I couldn't find a direct way to prove this, I guess an indirect approach might go as follows:

Let $(x_n)$ be a Cauchy sequence in X and assume that it does not converge, then it follows that there exists an $\epsilon > 0$ such that for an arbitrary $x \in X : \forall n \in \mathbb{N}: d(x_n - x) \geq \epsilon$ and this would contradict the fact that $(x_n)$ is a cauchy sequence. I'm assuming that this is incorrect since I didn't even use the given metric and this proof would mean that no cauchy sequence converges.

So how can I prove that this metric space is complete? And in general is there a way how to approach these completeness proofs?

HINT: If $x_n\to x_0$ in the standard metric on $(0,\infty)$, then $\ln x_n \to \ln x_0$.
• Is this only in the direction $x_n \rightarrow x \Rightarrow \ln x_n \rightarrow x$ or is this an iff statement? Jul 23, 2015 at 13:23
• Yes, $(0,\infty)$ and $\bf R$ are homeomorphic (do you know this definition?) by function $\ln$, because it is one-to-one and both $\ln$ and $\exp$ are continuous. Jul 23, 2015 at 13:27
• Yes I know the definition. Ok so I could simply use the fact that any Cauchy sequence in $\mathbb{R}$ with the standard metric converges and the fact that ln is homeomorphic and that would allow me to prove that it also converges with respect to this given metric, right? Jul 23, 2015 at 13:40
Let $(x_n)$ be a Cauchy sequence then $d(x_n, x_m)<\epsilon$ so $\mid \ln(x_n)-\ln(x_m)\mid <\epsilon.$ But $$\mid \ln(x_n)-\ln(x_m)\mid=\mid \ln\frac{x_n}{x_m}\mid<\epsilon$$ So $\frac{x_n}{x_m} \rightarrow 1$ then subsequencec $(x_n)$ and $(x_m)$ have same limit and the sequence is convergent.
Hint If $x_n$ is Cauchy, prove that $e^{x_n}$ is Cauchy with the usual metric of $\mathbb R$. If $y$ is the limit of $e^{x_n}$ then.....