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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.

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    $\begingroup$ I’ve seen the term trace topology before, but for future reference you may find it useful to know that it’s more often called the relative topology or the subspace topology induced on $A$ by $\tau$. $\endgroup$ – Brian M. Scott Apr 30 '13 at 14:52
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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?

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  • $\begingroup$ Thanks for your answer, I have two questions. First, to finish the argument in the first part, I would express $V$ as $U\setminus A$, then $B_d(x,r)\subseteq U\setminus A$ but I do not see how $B_\rho(x,r)\subseteq U$ can be followed. For the second part, I guess $B_d(x,r)\cap A=B_\rho(x,r)$ but why? $\endgroup$ – Alexander May 2 '13 at 9:58
  • $\begingroup$ @Alexander: No, $V$ is not $U\setminus A$: $U\setminus A=\varnothing$, since $U\subseteq A$. Both parts depend on the same observation: for any $x\in A$ and $r>0$, $B_d(x,r)\cap A=B_\rho(x,r)$. That’s just element-chasing: $y\in B_d(x,r)\cap A$ iff $d(x,y)<r$ and $y\in A$ iff $\rho(x,y)<r$ iff $y\in B_\rho(x,r)$. $\endgroup$ – Brian M. Scott May 2 '13 at 11:18
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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 $X$.

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