Theorem: If $(X,d)$ is a metric space and $d' = \min (d(x,y), 1)$ is the standard bounded metric then $d$ and $d'$ induce the same topology.
Equivalently, for all $x_0$ there are $a,b$ such that for all $y$: $a d'(x_0,y) \le d(x_0,y) \le b d'(x_0, y)$. Clearly, $a=1$. But: how to determine $b$? The statement appears to be false: $d'$ is bounded while $d$ is not. Yet, see here on page 3. Thank you.