A problem about topologically equivalent metrics I tried to solve this problem:
Let $(X,d)$ a metric space. Show that $d$ and $\bar{d} =\min({d(x,y),1})$ are topologically equivalent metrics. 
I proved that $\bar{d}$ is a distance, then I tried to show that every open ball on $(X,d)$ is contained on an open ball on $(X, \bar{d})$ and vice versa.
if $r<1$, and $\forall x \in X$ $B_r (x)=\bar{B}_r (x)$.
But if $r \ge1$, $\bar{B}_r (x)=X$, then I can't find a ball on $(X,d)$ such that $\bar{B}_r (x) \subseteq B_{r'} (x)$
 A: To show that the topologies of $(X,d)$ and $(X,\overline{d})$ coincide, it suffices 
to show that every open set in $(X,d)$ is a $\textit{union}$ of open balls in $(X,\overline{d})$ and vice versa.
As you already showed, the open balls in $(X,\overline{d})$ are open in $(X,d)$.
On the other hand every open set $U$ in $(X,d)$ can be written as $U = \bigcup_{x \in U} B_{r_x}(x)$ for suitable $r_x > 0$. We can assume $r_x < 1$ without loss of generality. But then we have $B_{r_x}(x) = \overline{B}_{r_x}(x)$, hence $U = \bigcup_{x \in U} \overline{B}_{r_x}(x)$ which shows that $U$ is open in $(X,\overline{d})$.
A: If $r<1$ then $\bar B_r(x)=B(r)$ and we are done.
If $r\ge 1$ then $\bar B_r(x)=X$. This is not an open ball, but the full space is always open, and that is enough. 
A: It is enough to show that any $\bar d$-neighborhood $\bar U$ of any $x\in X$ contains a $d$-neighborhood $U$ of $x$, and vice versa. 
One way is trivial: Given an $x$ and an $\epsilon>0$ the condition $d(x,y)<\epsilon$ implies $\bar d(x,y)\leq d(x,y)<\epsilon$. It follows that the $d$-neighborhood $U$ of radius $\epsilon$ is contained in the given $\bar d$-neighborhood of radius $\epsilon$.
Conversely: Given an $x$ and an $\epsilon>0$ the condition $\bar d(x,y)<\epsilon':=\min\bigl\{\epsilon,{1\over2}\bigr\}$ implies $d(x,y)\leq{1\over2}$ and therefore $d(x,y)=\bar d(x,y)<\epsilon$. It follows that the $\bar d$-neighborhood $\bar U$ of radius $\epsilon'$ is contained in the given $d$-neighborhood $U$ of radius $\epsilon$.
A: Let $i$ be the identity function on $X$. That is, $i(x)=x$ for every $x\in X$. 
If $x_n$ is a sequence converging in $(X,d)$, then it converges in  $(X,\overline{d})$. 
If $x_n$ is a sequence converging in $(X,\overline{d})$, then it converges in  $(X,d)$.
Therefore, $i$ is a homeomorphism. This means that as topological spaces
$(X,\overline{d})$ and  $(X,d)$ are equivalent. 
