TITLE: Equivalent bounded metric: Why should one prefer $\frac{d}{1+d}$ over $\min\{d,1\}$?
QUESTION [9 upvotes]: This is my first question and I hope it is not considered too argumentative.
It is often useful to change the metric on a space to an equivalent bounded metric.
Traditionally, people use
$$
\delta(x,y) = \frac{d(x,y)}{1+d(x,y)}.
$$
For example for the metric on product spaces: one equips
$$
\prod_{i=1}^\infty X_i
$$
with the metric $d(x,y) = \sum_{i=1}^\infty \frac{1}{2^{i+1}} \delta_i(x,y) = \sum_{i=1}^\infty \frac{1}{2^{i+1}} \frac{d_i(x,y)}{1+d_i(x,y)}$.
Similarly, when equipping the unit ball of the dual space of a separable normed space with a metric inducing the weak$^\ast$-topology, people tend to use a  construction using $\delta$. Or when metrizing convergence in measure on a probability space.
Is there any real advantage of this metric over, say,
$$
d'(x,y) = \min\{d(x,y),1\}\: ?
$$
It is much easier to check that $d'$ is an equivalent metric on $(X,d)$ than proving that $\delta$ is an equivalent metric. So why do people use $\delta$?
The only reason I can imagine (apart from tradition) is that knowing $\delta$ one can recover $d$. But is this reason enough to use this more complicated construction?

REPLY [8 votes]: Each has some advantages. The truncated metric $d\,'$ is very easy to define and very easy to show equivalent to $d$, and it is locally identical to $d$. On the other hand, $\delta$ is also easy to define and not much harder to prove equivalent to $d$ if you approach it properly, and it has the advantage of preserving more of the distant metric structure: for all $x,y,z\in X$, $\delta(x,y)=\delta(x,z)$ iff $d(x,y)=d(x,z)$, something that isn’t true if you replace $\delta$ by $d\,'$.
If you’re working in a setting in which differentiability is relevant, the fact that $f(x)=\frac{x}{1+x}$ is a nice, differentiable function of $x$ on $[0,\to)$ may be useful; the truncation function $f(x)=\min\{x,1\}$ isn’t nearly so nice from that point of view.