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