Complete metric space on the interval $[0,\infty)$ I want to check if metric $d(x,y)=|x-y|+|\frac{1}{x}-\frac{1}{y}|$ is complete on the interval $[0,\infty)$.
I know that $d(x,y)=|\frac{1}{x}-\frac{1}{y}|$ is not complete (counterexample: $x_n=n$)  but what happens when we add $|x-y|$ to it?
 A: I assume that, the set is $(0,\infty)$ instead $[0,\infty)$.
Now, if $(x_n)$ is a Cauchy sequence in this metric.
Then, $(x_n),(\frac{1}{x_n})$ both are Cauchy sequence in usual metric defined on $\bf{R}$.
So $(x_n)$ converges to some $c\ne 0$ in usual metric on $\bf{R}$ and $(\frac{1}{x_n})$ also converges to $(\frac{1}{c})$
Thus, in this metric $d$ $(x_n)$ converges to  $c\ne 0$.
So, $((0,\infty),d)$ is complete.
A: Apparently the metric is defined as
$$
 d(0, 0) = 0 \, ,\\
 d(0, x) = d(x, 0) = x + \frac 1x \text{ for } x > 0 \, ,\\
 d(x, y) = |x-y| + \left|\frac 1x -\frac 1y \right | \text{ for } x, y > 0 \, .
$$
Now let $(x_n)$ be a Cauchy sequence with respect to $d$. Since $d(0, x) \ge 2$ there are only two possible cases:


*

*All but finitely many $x_n$ are zero. In that case $d(x_n, 0) \to 0$ and we are done.

*All but finitely many $x_n$ are non-zero. Without loss of generality we can assume that all $x_n \ne 0$.


In the second case, both $(x_n)$ and $(1/x_n)$ are Cauchy sequences with respect to the Euclidean metric, and therefore convergent in that metric. If the respective limit values are $a$ and $b$ then $ab=1$, so that both are non-zero.
Then
$$
 d(x_n, a) = |x_n - a | + \left|\frac 1{x_n} - b\right| \to 0
$$
so that $(x_n)$ is convergent with respect to the metric $d$.
A: $d'(x,y)=|x-y|$ is already a complete metric on $[0,\infty)$, being a closed subset of the complete $(\Bbb R, d')$. Why do you want another ill-defined (at $0$) one?
A: If at all you want to bring $0$ into the picture you should define the metric like the this:
$$d(0, 0) = 0 \, ,\\
 d(0, x) = d(x, 0) = x + \frac 1x \text{ for } x > 0 \, ,\\
 d(x, y) = |x-y| + \left|\frac 1x -\frac 1y \right | \text{ for } x, y > 0 \, .$$
Before we proceed just verify it gives you a metric on $[0,\infty)$ .


*

*Now observe that $0$ is an isolated point in the given metric space $([0,\infty),d)$.


(Why?)
Since $d(x,0)\ge 2, \forall x \in (0,\infty)$ .


*

*Also observe that there can't be any Cauchy sequence which has infinitely many Zero and infinitely many non-zero terms in it.


(Why?)
Again the same reason $d(x,0)\ge 2, \forall x \in (0,\infty)$ .
If ${x_n}$ is a Cauchy-sequence then it(our discussion) leaves us with two possibilities:


*

*The Cauchy sequence all but finitely many terms zero.
$\implies$ the sequence converges to $0$ (trivial)

*The Cauchy sequence all but finitely many terms non-zero.
As you already know you should be bothered about the tail of the sequence, so you can ignore the finitely many zeros of the sequence. Then just observe just observe that 
$d(x_n,x_m)<\epsilon$$\implies |x_n-x_m|<\epsilon \text{ and } \left|\frac{1}{x_n} -\frac {1}{x_m} \right | <\epsilon $ Then you can say that ${x_n}$ converges in $([0,\infty),d)$ .


Hence you can say that $([0,\infty),d)$ is complete!
If you have read this far Thanks!
