Which metric is not complete? Problem
Which one of the following metric is not complete on $\mathbb R$:


*

*$|x-y|$

*$|arctan(x) - arctan(y)|$

*$|x^3 - y^3|$

*$|x^{1/3} - y^{1/3}|$
1 is complete. I am not sure about the 2, 3 and 4.
For 2, I do not know the definition of $arc$ function
For $3. 4$, they can be viewed as the special case of $|f(x) - f(y)|$, note that
$$
|f(x) - f(y)| = |f'(\theta)||x-y|
$$
where $x< \theta < y$.
it seems that 3&4 are complete with respect to the intervals, but what about $\mathbb R$ 
 A: You know 1 is complete. Denote this distance by $d$.
For 2 (presumably $\arctan$), the sequence $x_n = n$ is Cauchy but
clearly does not converge.
For 3 & 4, suppose $f$ is odd, continuous, strictly increasing, unbounded
and $f(0) = 0$ (the latter follows since $f$ is odd). Then I claim that $\rho(x,y) = |f(x)-f(y)|$ is a complete
metric.
Note that $f$ has an inverse $g$ such that $g(f(x)) = x$.
It is straightforward to show that $g$ is continuous.
Suppose $x_n$ is $\rho$-Cauchy. It follows that $f(x_n)$ is bounded. Since $g$ is uniformly continuous on bounded sets, it follows that
for any $\epsilon>0$, there is some $\delta>0$ such that if
$\rho(x_n,x_m)=|f(x_n)-f(x_m)| < \delta$, then $|g(f(x_n))-g(f(x_m))| = |x_n-x_m| < \epsilon$. It follows that $x_n$ is Cauchy with respect to the
standard distance, and hence there is some $x$ such that $x_n \to x$ with respect to $d$. It follows from continuity that $\rho(x_n,x) \to 0$
as well and hence $\rho$ is a complete metric.
Now check that $x \mapsto x^3, x \mapsto \sqrt[3]{x}$ satisfy the
conditions above.
