The standard bounded metric (see slide 3 of 8) is defined as $\overline d(x,y)=\min\{1,d(x,y)\}$. For balls of this metric $B_{r,\overline d}(p) = \{x \in M|\overline d(x,p)<r\}$, they're the same as the balls of the original metric $B_{r,d}(p) = \{x \in M|d(x,p)<r\}$ for $r<1$.
For $r \ge 1$, are all the balls in the standard bounded metric $B_{r,\overline d}(p)$ equal to the whole of $M$?
For another metric $d_q(x,y)=\min\{q,d(x,y)\}$, are the balls in this metric $B_{r,d_q}(p) = \{x \in M|\overline d_q(x,p)<r\}$ also equal to the whole of $M$ for $r\ge q$?
My thinking is that if a point's $x's$ distance $d$ from some central point $p$ is greater than the bound $q$ then $\overline d$ or $d_q$ will force the distance to be $1$ or $q$, respectively, which will be less than $r$, which we choose to be at least $1$ or $q$, respectively.
Actually, the balls in the standard bounded metric are the same as the balls of the original metric when $r \le 1$, and when $r > 1$ we have $B_{r,\overline d}(p) = M$.
Indeed, consider a discrete metric space $M = \{x,y\}$. Then $d = \overline{d}$ so $B_{1,\overline d}(x) = \{x\} \ne M$.
The situation is exactly the same for the metric $d_q$, we have
$$B_{r,d_q}(p) = \begin{cases} B_{r,d}(p), &\text{ if } r \le q\\ M, &\text{ if } r > q \end{cases}$$