# What is a practical use for this metric?

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.

• 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/… – 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!

• 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 $\min\{d(x,y), 1\}$? – John Gowers Sep 30 at 9:33
• @JohnGowers I don't think so, no. – 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

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

Thus

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

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

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

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.

• That's not what the question is about. – J.G. Sep 29 at 9:48
• @J.G. One moment please – Federico Fallucca Sep 29 at 9:50
• 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? – stackexchangequestions2 Sep 29 at 9:51
• @stackexchangequestions2 i don’t know if this result it’s a name, but it is remarkable fact – Federico Fallucca Sep 29 at 9:57
• That is interesting then, I'll have to read up on it. – 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.