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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?

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3 Answers 3

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HINT: If $x_n\to x_0$ in the standard metric on $(0,\infty)$, then $\ln x_n \to \ln x_0$.

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  • $\begingroup$ Is this only in the direction $x_n \rightarrow x \Rightarrow \ln x_n \rightarrow x$ or is this an iff statement? $\endgroup$ Commented Jul 23, 2015 at 13:23
  • $\begingroup$ 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. $\endgroup$ Commented Jul 23, 2015 at 13:27
  • $\begingroup$ 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? $\endgroup$ Commented Jul 23, 2015 at 13:40
  • $\begingroup$ Yes. I hope that this not an exercise, in which this method is prohibited. $\endgroup$ Commented Jul 23, 2015 at 14:15
  • $\begingroup$ Thanks for your help. I'd accept your answer if you have any general tips on how to prove that a metric space is complete, what are the things I should look out for or pay attention to when looking at a specific metric? $\endgroup$ Commented Jul 23, 2015 at 14:47
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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.

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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.....

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