Trace topology with metric topology ident I am just wokring through my topology book and the following question arise:$(M,d)$ is a metric space and $\tau_d$ the topology which is induced by the metric on $M$ and $A\subseteq M$. Now my question: Why is the trace topology from $\tau_d$ on $A$ identical to the topology, which is generated by the restriction of the metric on $A\times A$ as ametric topology on $A$? 
I hope I formulated it in the right way.
The definition I use for the trace topology is the following: Let $(X,\tau)$ a topological space and $A\subseteq X$. $\tau_A:=\{G\cap A | G\in\tau\}$ is the trace topology.
 A: Let $\tau_A$ be the trace topology induced by $\tau_d$, let $\rho$ be the restriction of $d$ to $A\times A$, and let $\tau_\rho$ be the topology on $A$ induced by $\rho$; you want to show that $\tau_A=\tau_\rho$. Doing so is just a matter of showing that each of $\tau_A$ and $\tau_\rho$ is a subset of the other.
To show that $\tau_A\subseteq\tau_\rho$, you need to start with an arbitrary $U\in\tau_A$ and show that $U\in\tau_\rho$. The natural way to try to do this is to let $x\in U$ be arbitrary and show that there is an $r>0$ such that $B_\rho(x,r)\subseteq U$, where $B_\rho(x,r)$ is the open $\rho$-ball of radius $r$ centred at $x$. What do you know about $U$? Since it’s in $\tau_A$, there is a $V\in\tau_d$ such that $U=V\cap A$. Clearly $x\in V$, so there is an $r>0$ such that $B_d(x,r)\subseteq V$. Can you finish it from there?
In the other direction it suffices to show that for any $x\in A$ and $r>0$, $B_\rho(x,r)\in\tau_A$. You know that $B_d(x,r)\in\tau_d$, so $B_d(x,r)\cap A\in\tau_A$. What’s the last step that you need here?
A: The metric topology on $A$, call it $\tau^A_d$ has as a base the balls $B^A_\epsilon(a):=\{x\in A\mid d(x,a)<\epsilon\}$ for positive $\epsilon$ and $a\in A$. But these are just the intersections $B_\epsilon(a)\cap A$. So $\tau^A_d$ is coarser than the subspace topology. To show equality take a ball $B_\epsilon(x)$ for $x\notin A$ and show that for $a\in B_\epsilon(x)\cap A$ there is $\delta>0$ such that $B^A_\delta(a)\subset B_\epsilon(x)\cap A$.
Note that we are using here that a base for the subspace topology consists of intersections of $A$ with the sets in the base of $X$.
