For $x_1,x_2 \in X = (1, 3) \subset \mathbb R$ it is defined a metric $ d(x_1, x_2) = |\frac{1}{x_1}-\frac{1}{x_2}|$ I must determine if $(X; d)$ is a complete metric space
The solution I was given:
Because $ \frac{1}{9}|x_1-x_2| \leq d(x_1,x_2)=\frac{|x_1-x_2|}{|x_1 x_2|} \leq |x_1-x_2|$, $x_n$ is a Cauchy sequence if and only if it is a Cauchy sequence for $d_e$(the usual euclidean metric). The sequence $x_n=\frac{n}{n+1}$, is a Cauchy sequence for $d_e$ and therefore also for $d$, but it does not converge in $X$. Then $(X,d)$ is not complete
I have two questions:
1) It looks to me that there is a mistake here. The given sequence does not belong to the given interval $X$, did they meant $x_n=\frac{n+1}{n}$ or is it fine and the defined metric does the trick of inverting it?
2) Before seeing the solution I found the space was complete and can't figure out what is wrong with my solution, which is as follows:
Once proved $x_n$ is a Cauchy sequence for $d_e$, I did:
Let $x_0$ be the limit of the sequence with respect to $d_e$ since it converges with respect to $d_e$: $$\lim_{n\to\infty} d_e(x_n-x_0)= \lim_{n\to\infty} |x_n-x_0|= 0.$$Then I used it to prove: $$\lim_{n\to\infty} d(x_n-x_0)=\lim_{n\to\infty}\frac{|x_n-x_0|}{|x_n x_0|}=0,$$ concluding that the sequence converged with respect to $d$, and that therefore the space was complete. Being done for a generic sequence, I thought it was right, what was wrong with it? Why is it not considering the counterexample?
