If $d$ is a metric then $\frac{d}{1+d}$ is a metric. What are the applications of this? Where does it come up?

Question

This is a question that if $d(x,y)$ is a metric, then $\frac{d(x,y)}{d(x,y)+1}$ is a metric.

Is this true for other functions of $d$ and what are the applications of this property?

If $d$ is the usual Euclidean metric, then this new metric has all points contained in the unit ball: If $x = 0$, then the limit of the new metric, as $y$ gets further from the origin, is $1$. — J126, Oct 01 '22 at 14:19
Sometimes we need a metric that is bounded by a constant we can control. Particularly when we (when dealing with series and such) need to add infinitely many metrics together. — Jyrki Lahtonen, Oct 01 '22 at 14:35
[1] Defining a metric on a countable Cartesian product of metric spaces in a way that is topologically equivalent to the usual product topology on the product of the corresponding topological spaces -- see Show that the countable product of metric spaces is metrizable. [2] Defining a metric for complex-analytic functions on an open set in a way to give uniform convergence on compact sets -- see Proving that a metric on space of analytic functions is equivalent to compact convergence — Dave L. Renfro, Oct 01 '22 at 14:36
here are many other functions of $d$ that also provide metrics, and more importantly, as here, metrics that are equivalent to $d$. That is, a set will be open under the new metric if and only if it is open under $d$. Another example of how you can replace an unbounded metric with an equivalent bounded metric is just $d′=\min{d,1}$. — Paul Sinclair, Oct 02 '22 at 18:09

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