Do non-equivalent metrics induce the same uniformity? I found the following statement in "Introduction to General Topology , K.D. Joshi" Page:354
"It may of course happen two distinct metrics induce the same uniformity."
Question. Let $d, \rho$ do not have same topology. Do $d, \rho$ induce the same uniformity?
Recall a uniformity for a set $X$ is a non-empty $\mathcal{U}$ of subsets of $X\times X$ such that:


*

*If $U\in\mathcal{U}$, then $U^{-1}\in\mathcal{U}$;

*Each member of $\mathcal{U}$ contains diagonal of $X$;

*If $U\in \mathcal{U}$, then $V\circ V\subseteq U$ for some $V\in\mathcal{U}$;

*If $U$ and $V$ are elements of $\mathcal{U}$, then $U\cap V\in\mathcal{U}$;

*If $U\in\mathcal{U}$ and $U\subseteq V\subseteq X\times X$, then $V\in \mathcal{U}$.


The pair $(X, \mathcal{U})$ is a uniform space.
 A: The uniformity determines the topology, by the definition that a set $O$ is open if for all $x\in O$ there exists $V\in \mathcal U$ such that $V[x]=\{y\mid (x,y)\in V\}\subseteq O.$ And this matches the usual definition for a metric space. So if the topologies are different, the uniformities are different.
A: A metric induces a unique uniformity; a uniformity induces a unique topology. 
The composition is just the usual way that metrics give rise to a topology (and in a first course the step via uniformities is usually skipped).
So different topologies means different uniformities. Also, if the metrics are uniformly equivalent (A sufficient condition for this is if there are constants $A,B > 0$ such that $Ad(x,y) \le \rho(X,y) \le Bd(x,y)$ for all $x,y$), they give the same uniformities and thus the same topology.
We can have that two different metrics have different uniformities but that these uniformities induce the same topology. So then continuity means the same for these metrics, but uniform continuity or Cauchy-ness does not. So the uniformity can give more information, in that sense. 
