Defining a new metric d' wrt to a metric d such that d' is bounded and d and d' induce the same topology This question was asked in my real analysis exam and I have a trouble proving it .

Let $(X,d)$ be a metric space . Then prove that there exists a metric $d'$ on $X$ such that $d$ and $d'$ define the same topology .

I chose the metric $d'=\min\{1,d(x,y)\}$ and proved $B_d(x,r)\subseteq B_{d'}(x,r)$ but in converse case if $ y \in B_{d'}(x,r) $ and $\min\{d(x,y),1\} = 1$ then I am not able to prove $B_{d'}(x,r)\subseteq B_{d}(x,r)$ as $d(x,y)$ could be greater than $d'(x,y)$ .
The case when $\min \{1,d(x,y) \}= d'(x,y)$ has been solved .
Kindly tell how should I approach that case .
Thanks!!
 A: Your choice for the metric $d'$ is already a very good idea!
But you do not actually need to show that $B_{d'}(x,r)\subseteq B_{d}(x,r)$ holds.
If we say that $d$ and $d'$ define the same topology,
we do not mean that they have the same
balls with radius $r$, it just means that they define the same open sets.
Suppose a set $A\subseteq X$ is open with respect to $d'$.
Then, for every $x$ there exists an $r>0$ such that $B_{d'}(x,r)\subseteq A$.
Since you already showed that $B_d(x,r)\subseteq B_{d'}(x,r)$, this implies
$B_d(x,r)\subseteq A$, and therefore $A$ is open with respect to $d$.
Converseley, suppose a set $B\subseteq X$ is open with respect to $d$.
Then, for every $x$ there exists an $r>0$ such that $B_{d}(x,r)\subseteq B$.
But to show that $B$ is open with respect to $d'$, it suffices to find
an $r'>0$ (that can differ from $r$) such that
$B_{d'}(x,r)\subseteq B$.
Therefore, it should suffice to show that for each $r>0$ there exists an $r'>0$
such that $B_{d'}(x,r')\subseteq B_{d}(x,r)$ holds.
This can be done by choosing $r':=\min(\frac12,r)$.
Then for $y\in B_{d'}(x,r')$ we have
$d'(x,y)\leq r'\leq \frac12$, and therefore $d'(x,y)= d(x,y)$
by your definition.
Since we also have $r'\leq r$, it follows
that $y\in B_{d}(x,r)$.
A: Hint:  Define $d':=\dfrac d{1+d}$.
It is well-known that this is a bounded metric which induces the same topology as $d$.
Symmetry and positive definiteness are obvious.  The triangle inequality is a little less obvious, but not too difficult. Use that $t\mapsto \dfrac t{1+t}$ is monotone.   Thus $d'$ is a metric.
It turns out they induce the same topology;  that is, the identity function between $(X,d)$ and $(X,d')$ is a homeomorphism.
