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Definition 1. Two metrics $d_{1}$, $d_{2}$ on a set $X$ are equivalent, if there exist two positive numbers $c$, $c'\in\mathbb{R}_{> 0}$, such that if $x$, $y\in X$ then$$ c d_{2}(x, y)\leq d_{1}(x, y)\leq c'd_{2}(x, y). $$

Question. What does the equivalence of metrics mean? I think there are some types of equivalence:

  1. $d_{1} = d_{2}$ i.e. $d_{1}(x, y) = d_{2}(x, y)$ for all $x$, $y\in X$.
  2. The induced topological structure with respect to $d_{1}$ is equal to that of $d_{2}$.
  3. The induced uniform structure with respect to $d_{1}$ is equal to that of $d_{2}$.
  4. The induced "structure of metric space" with respect to $d_{1}$ is equal to that of $d_{2}$, although I don't know what it means.

I tried to show 2, but it seems to be false. As for 3, I don't know enough about uniform spaces. If 4 is true, does the definition 1 defines the equivalence of "structure of metric space", or is there a clearer definition ?

Didier
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K. Y.
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  • Why do you think 2 is false? Did you find any counterexample where such equivalent metrics give different open sets? However if you meant whether the definition is equivalent to 2, then indeed 2 should be weaker. – Tesla Daybreak Feb 10 '21 at 01:44
  • Oh is it true? I have no idea to show 2. implies Def. 1. – K. Y. Feb 10 '21 at 01:48
  • I see, so you are trying to establish equivalence rather than implication. See wikipedia for some explanations: Def. 1 is the 'strong equivalence,' while 2 is the 'topological equivalence.' – Tesla Daybreak Feb 10 '21 at 01:50
  • @K.Y.: (2) does not imply Def’n. 1: it is weaker than Def’n. 1. – Brian M. Scott Feb 10 '21 at 01:50
  • Thanks. Then, is the following true? : In theory of metric spaces, a metric space is a equivalent class of a pair of set and its metric, with respect to def 1. Also, all definitions about metric spaces (for example, totally boundedness) should be well-defined up to the equivalence. – K. Y. Feb 10 '21 at 01:56
  • What I stated above is false, because it won't work in defining isometry. Is there standard name for the equivalence stated in Def. 1? – K. Y. Feb 10 '21 at 02:02
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    What you have is usually referred to as "Lipschitz equivalence" (in honour of Rudolf Lipschitz), because the identity $(X,d_1)\to (X,d_2)$ (and back) are Lipschitz continuous functions. Note that it is stronger than preservation of uniform structure. – user10354138 Feb 10 '21 at 02:05

1 Answers1

2

2 is certainly true, as I explain in this answer, but more can be said: having this "equivalence via constants", it's quite clear that the uniformities induced by $d_1$ and $d_2$ are the same: for every entourage of the form $\{(x,y) \mid d_1(x,y) < r\}$ we can find another entourage of the form $\{(x,y)\mid d_2(x,y) < r'\}$ that is a subset of it and vice versa, and as these kinds form a (filter) base for the respective uniformities from $d_1$ resp. $d_2$, we get that the uniformities are the same.

So $d_1$ and $d_2$ induce the same uniform properties: uniform continuity, total boundedness and Cauchy-ness and completeness means exactly the same under either.

3 alwyas implies 2 as well, but 3 is stronger and so probably the intended answer, as 4 is false:

If a map is an isometry wrt $d_1$ it need not be an isometry w.r.t. $d_2$ or vice versa. So really metric stucture (where the actual values of the metric matter!) can be different. We can alraedy see this by the "shapes" of the balls under the different equivalent metrics on $\Bbb R^2$. Ths shows that point 4 is false.

So the most informative is 3: the uniform structure is the same (and hence the topologies too).

Henno Brandsma
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