# Show that the two topological spaces arising from these metrics are homeomorphic.

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)

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)

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

• 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. Apr 15, 2020 at 18:42
• I'm gonna put my proof and could you please tell if its ok? Apr 15, 2020 at 18:53
• Do please post your proof. The two topologies are actually not just homeomorphic: they are the same. Apr 15, 2020 at 19:46
• @RobArthan Hold on, im on it Apr 15, 2020 at 19:47
• My edit was for a $y'$ that shoulsd have been $y.$ Apr 16, 2020 at 0:51

$$id_Y$$ 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 $$(Y,e).$$
• 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? Apr 16, 2020 at 0:46
• 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. Apr 16, 2020 at 1:01
• 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. Apr 16, 2020 at 1:13