Two Different Metrics Induce the Same Topology 
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...
 A: 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 $\tau_d=\tau_\delta$)
A: First, choose $\delta' > 0$ such that $B_d(y,\delta')\subset B_d(x,\varepsilon)$. This is clearly possible. Now, choose $\delta < a\delta'$. Then for $z\in B_\rho(y,\delta)$ we have
$$
d(z,y)\le\frac 1 a\rho(z,y)\le\frac \delta a < \delta'.
$$
Thus, $z\in B_d(y,\delta')\subset B_d(x,\varepsilon)$, which establishes $B_\rho(y,\delta)\subset B_d(x,\varepsilon)$.
A: You want to show that $B_d (x,\varepsilon) \in \tau_\rho$. 
Let $y\in B_d (x,\varepsilon)$. We need to show that there exists some $\delta>0$ such that $B_\rho (y,\delta)\subseteq B_d (x,\varepsilon)$. Let $z \in B_\rho (y,\delta)$, and choose $\delta=a(\varepsilon -d(x,y))$.
Since $ad\le\rho$, $d(z,y)<\frac{\delta}{a}$.
We have that $d(x,z)\le d(x,y)+d(y,z)<d(x,y)+\varepsilon-d(x,y)=\varepsilon$, so $z \in B_d (x,\varepsilon)$.
A similar choice of $\delta$, making use of the fact that $\rho\le bd$ will prove that $\tau_\rho \subseteq \tau_d$.
