Let $(M, d)$ be a metric space. Set (the default bounded metric) $\bar d : M \times M \to R$ by $\bar d(x, y) =$min {$d(x, y), 1$}. Show that:
a) $\bar d$ is a metric in $M$.
b) $d$ and $\bar d$ are equivalent.
c) Every $A \subset M$ is bounded on $(M, \bar d)$, but not necessarily bounded on $(M, d)$.
a) To be a metric we need $\bar d(x,y)\ge 0$; $d(x,x) = 0$ if and only if $x=y$; $d(x,y)=d(y,x)$; $d(x,z) \le d(z,y)+d(y,x)$
$\bar d =$ min {$d(x, y), 1$} and $d$ is metric so $d(x,y) \gt 0$ and min {$d(x, y), 1$} $\ge 0$.
$\bar d (x,x)=$ min{0,1} $= 0$.
$\bar d (x,y)=$ min{$\vert x-y \vert$, 1}=min{$\vert y-x \vert$, 1}=min{$\vert x-y \vert$, 1} = $\bar d (y,x)$.
$\bar d(x,z) \le \bar d(z,y)+\bar d(y,x) = \vert x-z \vert \le \vert z-y \vert + \vert y-x \vert = \vert x-z \vert \le \vert z-x \vert$.
c)$\bar d(x, y) =$min {$d(x, y), 1$} and we know that $ d(x,y) \ge 0$ so $\bar d(x, y) =$min {$0, 1$} therefore $\bar d$ is limited by $0$ and 1.
On the other hand $d(x,y)$ can take any value greater than $0$.
That's what I managed to do, is that correct? But I got stuck on item b).