Let $d$ and $\rho$ be metrics on $X$. If there exist positive numbers $a$ and $b$ such that $ad \le \rho \le bd$, then the two metrics give rise to the same topology on $X$.
Let $\tau_p$ and $\tau_d$ denote the topologies induced by $\rho$ and $d$, respectively. I am having trouble showing either inclusion. Here is what I have tried in attempting to show $\tau_d \subseteq \tau_\rho$. Let $B_d(x,\epsilon)$ be some basis element of $\tau_d$, and let $y \in B_d(x, \epsilon)$. I want to find a $\delta > 0$ such that $B_\rho(y,\delta) \subseteq B_d(x,\epsilon)$. Let's try $\delta = b(\epsilon – d(x,y))$. Then $\rho(y,z) < b(\epsilon – d(x,y)$ or $\frac{\rho(y,z)}{b} + d(x,y) \le \epsilon$ or $\frac{\rho(y,z)}{b} + \frac{\rho(x,y)}{b} \le \frac{\rho(y,z)}{b} + d(x,y) \le \epsilon$ or $\frac{\rho(x,z)}{b} \le \frac{\rho(y,z)}{b} + \frac{\rho(x,y)}{b} < \epsilon$…Nope not going to work…Doing these same manipulations, I also tried setting $\delta$ equal to $\frac{\epsilon – d(x,y)}{b}$, and then tried $a(\epsilon – d(x,y))$, and then tried $\epsilon – \frac{\rho(x,y)}{b}$, and then I gave up. I could use a hint…
Best Answer
Observation. If $d$ and $\delta$ are two metrics on $X$ such that $\delta<k d$ then the identity $j:(X,d)\longrightarrow (X,\delta)$ is $k$-lipschitz (in particular continuous). Apply twice the observation to conclude that $j$ is an omeomorphism between $(X,\tau_d)\stackrel{j}{\cong}(X,\tau_\delta)$ (so $\tau_d=\tau_\delta$)