Topologies generated by a metric Hi I am new to mathematical proofs/notation and am working through John Lee's Introduction to Topilogical Manifolds. This is the question and my attempt. This is not homework.
2.4
Suppose $M$ is a set and $d$, $d'$ are two different metrics on $M$. Prove that $d$ and $d'$ generate the same topology iff for every $x\in M$ and every $r>0$, there exist positive numbers $r_1$ and $r_2$ such that $B_{r_1}^{(d')}(x) \subseteq B_r^{(d)}(x)$ and $B_{r_2}^{(d)}(x) \subseteq B_r^{(d')}(x)$.
My attempt so far:
Each generated topology is the collection of all the open subsets in the metric space sense. Assuming both generated topologies are the same:
$\tau = \tau^d = \tau^{d'}$ 
$\iff$
$\{ A : A \subseteq M,~and~if~x\in A,\exists~B_r^{(d)}=\{y\in M:d(x,y)<r\}\subseteq A\}=$
$\{ A : A \subseteq M,~and~if~x\in A,\exists~B_r^{(d')}=\{y\in M:d'(x,y)<r\}\subseteq A\}$
$\implies$
if $A\in \tau$,and $x\in A$, $\exists~B_{r}^{(d)}(x)\subseteq A$ and $B_{r}^{(d')}(x)\subseteq A$, for some $r>0$.
This is where I am stuck, I don't know how to compare these two subsets of $A$. I know $x$ has to be a member of both but not sure what to do with that. I am new to this sort of thing and appreciate any help.
 A: the open sets defining a topology $\tau$ on a set $M$ may be defined as a subset $T_{\tau}$ of $\mathfrak{P}(M)$, the power set of $M$ satisfying the open-set axioms. we may write:
$$
\tau_1 \le \tau_2 \Leftarrow \Rightarrow T_{\tau_1} \subseteq T_{\tau_2}
$$
ie $\tau_1$ is weaker (which does not rule out equivalence)  than $\tau_2$ iff every open set of $\tau_1$ is also an open set of $\tau_2$
two topologies are equivalent iff each is weaker than the other. now express this requirement in terms of the open balls. since an arbitrarily small open $\tau_1$ ball around $x$ belongs to $T_{\tau_1}$, then for equivalence this ball must be open also in  $\tau_2$, so it must, in particular contain an open $\tau_2$-ball centered at $x$. and vice versa. this argument shows the necessity of the condition you have stated. can you add a demonstration of its sufficiency?
A: for the first direction lets assume that $\tau_{d'}=\tau_d$
given $x\in{M}$ and $r>0$ we get that $B_r^{(d)}(x)\in \tau_{d'}$ then for every $y\in B_r^{(d)}(x)$ there exicet $r_y>0$ such that  $B_{r_y}^{(d')}(x)\subset B_r^{(d)}(x)$ so lets just take one of them as $r_1$.
and it is symmetrical to find $r_2$  
for the second direction:
we assume the second condition (in the iff)
take $U\in \tau_d$ then for every $x\in U$ theres $r_x>0$ such that $B_{r_x}^{(d)}(x)\subset  U$ so well get $$\cup_{x\in U}B_{r_x}^{(d)}=U$$ 
but for every $r_x$ we have $r_{x_1}$ such that $B_{r_{x_1}}^{(d')}(x)\subset B_{r_x}^{(d)}(x)$ so it is easy to see that: $$\cup_{x\in U}B_{r_{x_1}}^{(d)}=U$$ 
then $U$ is in $\tau_{d'}$
so $\tau_d \subset \tau_{d'}$ and to show that $\tau_{d'} \subset \tau_d$ is symmetrical 
