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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.

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    I don't understand your argument, the bases doesn't need to be the same (you can take two different bases for the same uniformity). Here, you have to use the fact that the map $f\colon\mathbb R\to\mathbb R$, $f(x)=x^3$ is not uniformly continuous. For all $k$, $(k+1/(2k),k)$ is in $U_{1,k}=\{(x,y),|x-y|<1/k\}$, but it's not contained in $V_{1,1}$.2011-11-19
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    I don't understand where do you use here the fact that $x^3$ isn't uniformly continuous? and how do you define $V_{1,1}$? Is it with the second metric?2011-11-19
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    $V_{1,1}$ is the set $\{(x,y),|x^3-y^3|\}$ (I wanted to denote it by $V_{2,1}$, but I maid a typo). You can show that if $f$ is bijective and uniformly continuous on $\mathbb R$, then the uniform structures associated to the metrics $d(x,y)=|x-y|$ and $d(x,y)=|f(x)-f(y)|$ are equivalent. Here if the uniform structure were equivalent, $U_{1,k}$ would be contained in $V_{2,1}$ for some $k$.2011-11-19
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    Isn't this what I argued in my OP? I said take $U_{1}=\{(x,y): |x-y|<2\}$ and $U_2=\{ (x,y): |x^3-y^3|<2 \}$, then the point (1,2) is in $U_{1}$, but not in $U_2$.2011-11-19
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    It's not exactly that, since you use $\varepsilon_1=\varepsilon_2=2$. It doesn't prove that if you choose $\varepsilon_1$ very small you won't have the inclusion.2011-11-19

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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.