Let $d_1(x,y)$, $d_2(x,y) $, equivalent metrics on a arbitrary set $Y$ with $x,y \in Y$. Show that the two topological spaces arising from these metrics are homeomorphic.

$\alpha d_1(x,y)\leq d_2(x,y)\leq \beta d_1(x,y)$.

I've been trying to figure out by implying that homeomorphic to itself imply the same topology, also i thought that considering the identity mapping on Y may help but a smart guy told me it wouldn't be that general. I still think that the correct approach is by the identity mapping,but any extra ideas on this or any help?

Proof verification

I'm gonna show that the identity mapping $g:(Y,d_1)\mapsto (Y,d_2)$ is uniformly continuous, let $\epsilon>0$ we can define $\delta =k_{2}^{-1}\epsilon$ then $\forall$ $x,t \in Y$ if $d_1(x,t)<\epsilon$ follows:

$d_2(g(x),g(y))=d_2(x,y)\leq k_2 d_1(x,y)<k_2 \delta=\epsilon$

As desired.

Similarly, the mapping $f:(Y,d_2) \mapsto (Y,d_1)$ is uniformly continuous, and let $\epsilon>0$, we can define $\delta=k_1 \epsilon$ and then $\forall$ $x,t \in Y$ if $d_2(x,t)<\epsilon$ then:

$d_1(f(x),f(y))=d_1(x,y)\leq k_{1}^{-1}d_1(x,y)<k_{1}^{-1}\delta=\epsilon$

Since the identity functions in both directions are continuous and inverse to each other they are homeomorphisms

  • $\begingroup$ Yes, you want to show that the map $(Y,d_1) \to (Y,d_2)$ which is the identity on the set $Y$ is a continuous mapping of metric spaces. Which amounts to showing that open balls in the $d_2$ metric are open sets in the $d_1$ metric. $\endgroup$ Apr 15, 2020 at 18:42
  • $\begingroup$ I'm gonna put my proof and could you please tell if its ok? $\endgroup$ Apr 15, 2020 at 18:53
  • 1
    $\begingroup$ Do please post your proof. The two topologies are actually not just homeomorphic: they are the same. $\endgroup$
    – Rob Arthan
    Apr 15, 2020 at 19:46
  • $\begingroup$ @RobArthan Hold on, im on it $\endgroup$ Apr 15, 2020 at 19:47
  • 1
    $\begingroup$ My edit was for a $y'$ that shoulsd have been $y.$ $\endgroup$ Apr 16, 2020 at 0:51

1 Answer 1


$id_Y$ is a Lipschitz-continuous bijection from $(Y,d_1)$ to $(Y,d_2)$ with Lipschitz constant $\beta,$ and its inverse (which, of course, is also $id_Y$) is a Lipschitz-continuous bijection from $(Y,d_2)$ to $(Y,d_1)$ with Lipschitz constant $1/\alpha.$ So $id_Y$ is a homeomorphism from $(Y,d_1)$ to $(Y,d_2).$

Remark. Suppose $d,e$ are metrics on $Y.$ Suppose $\gamma>0$ and $f:Y\to Y$ such that $\forall x,y \in Y\,\,(\,e(f(x),f(y)) \le \gamma \cdot d(x,y)\,)....$

.... Now if $(y_n)_n$ is a sequence in $Y$ and $y\in Y$ with $0=\lim_{n\to \infty}d_1(y,y_n)$ then we have $$\lim_{n\to \infty}\sup_{m\ge n}e(f(y),f(y_m))\le \lim_{n\to \infty}\sup_{m\ge n}\gamma \cdot d(y,y_m)=0$$ so $\lim_{n\to \infty}e(f(y),f(y_n))=0.$ Therefore $f$ is continuous from $(Y,d)$ to $(Y,e).$

  • $\begingroup$ Nice and different way to prove it, you use the other definition of equivalent norm which uses the convergence of a succession ${x_n}$ no? $\endgroup$ Apr 16, 2020 at 0:46
  • $\begingroup$ Yes. In general topology there are many equivalent definitions of the continuity of a function $f:A\to B. $ One is that $f[Cl_A(C)]\subset Cl_B(f[C])$ for all $C\subseteq A,$ which for metric spaces can be refined to the condition I used in my answer. $\endgroup$ Apr 16, 2020 at 1:01
  • $\begingroup$ Metrics on $Y$ that generate the same topology are called equivalent. When $\alpha, \beta$ exist as in your Q then $d_1,d_2$ are called uniformly equivalent. $\endgroup$ Apr 16, 2020 at 1:13

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