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Opoitsev states in 'A Converse to the Principle of contracting Maps', 1976:

Equivalent metrics are also topologically equivalent. The converse is false, but the following easily checked statement is true: If two metrics d1 and d2 are topologically equivalent and the spaces (X, d1) and (X, d2) are both complete, then d1 and d2 are equivalent.

He says it is easy to check, but I am not sure. Is that statement true and if so, how can it be proved? Or are there other necessary conditions for topological equivalence to imply equivalence?

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    $\begingroup$ Here's a possible hint (I'm still thinking about this myself). Metric equivalence requires that we can write $ad_1 \leq d_2 \leq bd_2$ for constants $a,b$ that work everywhere, whereas topological equivalence lets the constants change point from to point. Perhaps try taking a Cauchy sequence of points for which the constants get bigger and bigger (if this can be done), and then looking at what happens at the limit point. $\endgroup$ – Elchanan Solomon Sep 28 '13 at 15:08
  • $\begingroup$ Thank you for your hint. Do the constants in topological equivalence depend on one of the points, or both, such as a=a(x), b=b(x) or a=a(x,y), b=b(x,y)? $\endgroup$ – john matthew Sep 28 '13 at 15:34
  • $\begingroup$ It should only depend on one of the points, I think. At each $x \in X$, you can nest a small ball in $d_1$ with balls from $d_2$, but the ratio of the the $d_1$ ball's radius, and the radii of the $d_2$ balls might change. $\endgroup$ – Elchanan Solomon Sep 28 '13 at 15:38
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    $\begingroup$ Can you give the definitions of "equivalent" and "topologically equivalent" that you are using? $\endgroup$ – Nate Eldredge Sep 28 '13 at 16:05
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If "equivalent" means $ad_1\leq d_2\leq bd_1$, this is wrong.

Take $X=\mathbb N$, with $d_1(u,v)=\vert u-v\vert$ and $d_2(u,v)=1$ if $u\neq v$. Then both metrics induce the discrete topology and $\mathbb N$ is complete with respect to both. But certainly you cannot find a constant $a>0$ such that $d_2\geq a d_1$ since $d_1(0,n)=1$ for all $n\geq 1$ and $d_2(0,n)=n$.

If "equivalent" means "uniformly equivalent", as suggested by Daniel, this is wrong as well.

Take $X=\mathbb N\cup\{ n+3^{-n};\; n\geq 1\}$, with the same distances as above. Then $d_1$ and $d_2$ again induce the discrete topology on $X$, and $X$ is of course complete wrt $d_2$. Moreover, $X$ is also complete wrt to $d_1$ since it is closed in $\mathbb R$ and $\mathbb R$ is complete wrt $d_1$. Finally, $d_1(n, n+3^{-n})\to 0$ as $n\to\infty$ but $d_2(n,n+3^{-n})=1$ for all $n\geq 1$; so $d_1$ and $d_2$ are not uniformly equivalent.

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    $\begingroup$ I'm pretty sure "equivalent" means uniformly equivalent, and not Lipschitz-equivalent here. So $\bigl(\forall \varepsilon > 0\bigr)\bigl(\exists \delta > 0\bigr)\bigl(d_1(x,y) < \delta \Rightarrow d_2(x,y) < \varepsilon\bigr)$, and of course with the roles of $d_1, d_2$ interchanged. $\endgroup$ – Daniel Fischer Sep 28 '13 at 18:18
  • $\begingroup$ I think you're right... $\endgroup$ – Etienne Sep 28 '13 at 18:21
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    $\begingroup$ My example was going to be $\mathbb{N}\times\mathbb{N}$ with $d_2(x,y) = \delta_{xy}$, and $d_1((k,\ell),(m,n)) = \frac1k$ if $k=m$ but $\ell \neq n$, $1$ if $k\neq m$, and of course $0$ if $(k,\ell) = (m,n)$. $\endgroup$ – Daniel Fischer Sep 28 '13 at 18:44
  • $\begingroup$ It works fine as well; in fact it is the same. $\endgroup$ – Etienne Sep 28 '13 at 19:21
  • $\begingroup$ Equivalence of metrics means that both spaces have the same Cauchy sequences. $\endgroup$ – Andrés E. Caicedo Sep 28 '13 at 19:24

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