Two different metrics yielding different uniform structures I have two metrics on $\mathbb R$:
$$d_1(x,y)=|x-y|,$$
$$d_2(x,y)=|x^3-y^3|.$$
And I want to show that the uniform structures $\mathcal{U}_{d_1}$ and $\mathcal{U}_{d_2}$ defined by the respective bases $U_{\epsilon_n}=\{(x,y) : d_n(x,y)<\epsilon_n\}$ for $n=1,2$ are not equivalent.
What I showed is that if we take the same epsilon, then there can be different basis elements, which obviously are in their respective uniform structure but not in the other.
For example, take $\epsilon = 2$, and take the point $(1,2)$, in the basis element with the metric of powers of $3$, this point isn't in it, but in the basis element with the usual metric the point belongs to it, so we have two different basis elements for each uniformity, so we must have one set that is in one uniformity but not in the other.
Is this argument valid?
Thanks.
 A: Recall that the uniform structure given by a metric $d$ on a set $S$ is the collection of the subsets of $S\times S$ which contain a set of the form $\left\{(x,y)\in S\times S, d(x,y)<\delta\right\}$, $\delta>0$.
Here, to show that the uniform structures $\mathcal U_{d_1}$ and $\mathcal U_{d_2}$ are not equivalent, we will use the fact that the map $x\mapsto x^3$ is not uniformly continuous on $\mathbb R$. Put $V_1=\left\{(x,y)\in\mathbb R^2,|x^3-y^3|<1\right\}$, which is an element of the basis of $\mathcal U_{d_2}$. If the two uniform structures were equivalent, we would be able to find a $\delta>0$ such that $U:=\left\{(x,y)\in\mathbb R^2,|x-y|<2\delta\right\}\subset V_1$. For all $x\in\mathbb R$, $\left(x,x+\delta\right)\in U$, but 
$$\left|(x+\delta)^3-x^3\right|=|3\delta^2 x+3\delta x^2+x^3|,$$
which cannot be below bounded by $1$ (for example take $x=\frac 1{\delta^2}$). 
Anyway, you can try to show the following result:

If $f\colon\mathbb R\to\mathbb R$ is bijective and uniformly continuous , then the uniform structures on $\mathbb R$ given by $d_1(x,y):=|x-y|$ and $d_2(x,y)=|f(x)-f(y)|$ are equivalent.   

