The sequence $(x_n) = 1/n$ is not convergent in $(\mathbb{R} \backslash \{0\}, | · |)$ (the usual distance).
Knowing that a sequence is Cauchy if there exists an N such that n,m>N we have |x(n)-x(m)| < ε.
Is the reason the sequence is divergent in $(\mathbb{R} \backslash \{0\}, | · |)$ is because the limit point $(x=0)$ is not in the set? As in it is not closed (by sequence characterization)?
Thank you!
The sequence $\left(\frac1n\right)_{n\in\Bbb N}$ is divergent in $\Bbb R\setminus\{0\}$ because there is no $x\in\Bbb R\setminus\{0\}$ such that $\lim_{n\to\infty}\frac1n=x$. The fact that $\Bbb R\setminus\{0\}$ is not a closed subset of $\Bbb R$ is relevant, but not essential. After all $(-2,2)$ is also a subset of $\Bbb R$ whay is not closed, but the sequence $\left(\frac1n\right)_{n\in\Bbb N}$ converges in it.