I'm facing the problem which defines $\ll$ as on a set X with metrics $d_1,d_2$ $\forall \varepsilon \exists\delta>0 \forall x,y\in X s.t. d_2(x,y)<\delta \Longrightarrow d_1(x,y)<\varepsilon $ and $d_1$ and $d_2$ are uniformly equivalent if $d_1 \ll d_2$ and $d_2\ll d_1$. It's easy to show that if $d_1$ and $d_2$ are uniformly equivalent then theny generate the same topology. However now the question is to show that if they generate the same topology, they are not necessarily to be uniformly equivalent.

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    $\begingroup$ The topology you generate only depends on the small $\epsilon$-balls, while uniform equivalence basically says that all the open balls are "more or less" equivalent in size. So take the standard metric on $\mathbb{R^2}$ and envision a second metric which expands open balls proportional to their size. $\endgroup$
    – Neal
    May 13, 2014 at 23:24

1 Answer 1


$X=\mathbb{R}$, $d_1(x,y)=|x-y|$, $d_2(x,y)=|x^3-y^3|$ - because $x\mapsto x^3$ isn't uniformly continuous on $\mathbb{R}$, these aren't uniformly equivalent; take $x=n$, $y=n+\frac{1}{n}$, then $d_1(x,y)=\frac{1}{n}$, $d_2(x,y)\geq 3$, so for $\epsilon = 3$, every $\delta$ fails by taking $\frac{1}{n}<\delta$. However, essentially because $x\mapsto x^3$ is continuous, they do generate the same topology.

  • $\begingroup$ Nice, good example. $\endgroup$
    – Rustyn
    May 13, 2014 at 23:29

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