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 $\bar{B}_r (x) \subseteq B_{r'} (x)$
Best Answer
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 $(X,\overline{d})$.