Is $(0,1]$ equipped with the metric $d(x,y) = |1/x-1/y|$ complete? Is $(0,1]$ equipped with the metric $d(x,y) = |\frac{1}{x}-\frac{1}{y}|$ complete?
How would you dis/prove that this is complete?
 A: Let $\{x_n\}_{n=1}^{\infty}$ be a Cauchy sequence in $(0,1]$ with respect to this metric. Since $d(x_n,x_m) \to 0$ as $n,m\to\infty$, we have
$$
d(x_n,x_m) = \frac{|x_n-x_m|}{|x_nx_m|} \ge \underbrace{|x_n-x_m|}_{\to 0,\text{as}\ n,m\to\infty}
$$
where the inequality is due to the fact that $|x_n|\leq 1$ for every $n$. Now, as $[0,1]$ with the standard metric is complete, the sequence $\{x_n\}$, due to the inequality above, either converges to $0$ or converges to a non-zero element, which is in $(0,1]$. In the latter case, if $x_n\to c$ with respect to standard metric with $c\neq 0$, then from continuity of $f:x\mapsto \frac{1}{x}$, we get $f(x_n)\to f(c)$ with respect to standard metric, namely 
$$\left|\frac{1}{x_n}-\frac{1}{c}\right|\rightarrow 0.
$$
Finally, to eliminate the former case, suppose $x_n\to 0$. Then, $\frac{1}{x_n}\to \infty$ (both with respect to standard metric). But under this, $x_n$ can not be a Cauchy with respect to $d(\cdot,\cdot)$, since, otherwise for a fixed $\epsilon>0$, an integer $N$, and $n,m \geq N$ we would have
$$
\left|\frac{1}{x_n}-\frac{1}{x_m}\right|<\epsilon,
$$
which is false after sending $n\to\infty$.
A: Draw the graph. You can see that as $x$ goes from $1$ down to $0$, then $1/x$ goes from $1$ up to $\infty$. So you're taking about the ordinary distance between points in the space $[1,\infty)$. That is a closed subset of the set of real numbers, so within that space, every Cauchy sequence converges. If we had had $(1,\infty)$ instead of $[1,\infty)$ then it would not have been complete because Cauchy sequences approaching $1$ do not converge to a limit within that set. If we had had $(0,1]$ with the more usual metric, then it would not have been complete because some Cauchy sequences within that space converge to $0$. However, with that altered metric, those cease to be Cauchy sequences. For example, the distance between $1/n$ and $1/m$ for integers $n$ and $m$ is $1/n-1/m,$ and that does not go to $0$ as $n,m\to\infty.$
