Showing $\rho (x,y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric

That post has a more generalized form of a metric I occasionally see $d(x,y) = \frac{|x-y|}{1+|x-y|}$. When would using this metric be useful exactly? It occasionally comes up when I study analysis, but I don't know why, I don't know what people use it for or what benefit it could ever bring over the standard metric. I've merely only seen it as an example of a metric in books or sites, but if so many sources mention it, then it's very unlikely that it's useless.

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    $\begingroup$ Since this bounded metric $\rho$ defines the same open sets and closed sets and same notions of convergence and continuity as the original metric $d$, this shows that the properties of a topology from a metric defined in terms of boundedness are sensitive to the choice of metric. For example, a subset of $\mathbf R^n$ is compact if and only if it is closed and bounded, but that is not a characterization of compactness for general metric spaces. For that, see math.stackexchange.com/questions/774111/… $\endgroup$ – KCd Sep 29 at 18:24

It's a metric that is bounded above by $1$, while maintaining the same topology. This means that bounded metrics are just as powerful as general metrics (which is arguably interesting in itself).

More concretely, there's a commonly used construction for turning a countable product of metric spaces into a metric space itself. Specifically, if we have spaces $(X_n, d_n)$ where $n \in \Bbb{N}$ and $d_n$ is bounded uniformly (e.g. $d_n \le 1$ for all $n$), then $\prod_n X_n$ is a metric space with the metric $$d(x, y) = \sum_{n=1}^\infty \frac{d_n(x_n, y_n)}{2^n}.$$ Boundedness is important to guarantee convergence. This function is a metric, and it proves that a countable product of metrisable spaces are metrisable. This, in turn, is used to prove a bunch of interesting metrisability theorems. Coming from a functional analysis background, one consequence I'm partial to is the metrisability of the weak topology of a separable normed linear space when restricted to the unit ball. From this, we get the handy Eberlein-Smulian theorem.

Of course, this is just one field's use of this metric!

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    $\begingroup$ If you want to bound the metric above by $1$ while maintaining the same topology, is there an advantage to using $d(x,y)/(1 + d(x,y))$ rather than $\min\{d(x,y), 1\}$? $\endgroup$ – John Gowers Sep 30 at 9:33
  • $\begingroup$ @JohnGowers I don't think so, no. $\endgroup$ – Theo Bendit Sep 30 at 9:35

One advantage of $\rho$ is that $\rho\le1$ regardless of the size of $d$. Suppose you're fitting a model to data, penalising the model for each data point's distance from the model's predictions, with the aim of parameter estimation. If $d$ is unbounded, a sum of $d$ penalisations is very sensitive to outliers, especially if large $d$ values aren't all that improbable (they're not always Gaussian). By contrast, $\rho$ gives at most a penalty of $1$ to any one data point, so the sensitivity to outliers is reduced.


That function is a metric because $f(x)=\frac{x}{1+x}$ is a monotone increasing function on $(0,\infty)$ so if you consider point $x,y,z$ if your space, then

$d(x,y)\leq d(x,z)+d(z,y)$


$f(d(x,y))\leq f(d(x,z)+d(z,y))= \frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)}= $

$\frac{d(x,z)}{1+d(x,z)+d(z,y)}+ \frac{d(z,y)}{1+d(x,z)+d(z,y)}\leq $

$\leq \frac{d(x,z)}{1+d(x,z)}+ \frac{d(z,y)}{1+d(z,y)}$

So $d’(x,y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric

Why it is useful use this metric? Because this metric is always limited, in fact

$d’(x,y)< 1$

This is useful to prove, for example, that a countable product of metric space is also a metric space.

In fact you can observe that $d$ and $d’$ induce the same Topology on the space, so if you consider a countable family of metric spaces $\{(X_n,d_n)\}_n$ then

$D(x,y):=\sum_{n=1}^\infty \frac{d_n’(x_n,y_n)}{2^n}$ is a metric of $\prod_{n}X_n$ that induces the product Topology on this space.

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    $\begingroup$ That's not what the question is about. $\endgroup$ – J.G. Sep 29 at 9:48
  • $\begingroup$ @J.G. One moment please $\endgroup$ – Federico Fallucca Sep 29 at 9:50
  • $\begingroup$ Is there a name for any specific theorems where this is useful? What is the name of the theorem that proves a countable product of metric spaces is also a metric space and relies on that metric in its proof? $\endgroup$ – stackexchangequestions2 Sep 29 at 9:51
  • $\begingroup$ @stackexchangequestions2 i don’t know if this result it’s a name, but it is remarkable fact $\endgroup$ – Federico Fallucca Sep 29 at 9:57
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    $\begingroup$ That is interesting then, I'll have to read up on it. $\endgroup$ – stackexchangequestions2 Sep 29 at 10:06

I know this thing: if $(X, d)$ supports a measure $\mu$, then convergence in measure, for a sequence of functions, is the same as convergence with respect to the (integral of the) metric you cite. That is: on $\mathrm{Meas}(X)$, define the metric $$ d_\mu(f, g) = \int \frac{d(f(x), g(x))} {1+d(f(x), g(x))} d\mu(x) $$ Then a sequence of measurable functions converges in measure iff it converges with respect to the metric $d_\mu$.

This is stated in Tao's book on measure theory, if I recall correctly.


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