# Is the absolute value function a metric?

I was trying to find out more information about absolute value, and I came upon the fact that AV satisfies a whole set of properties that usually defines a distance function or metric. But in the Wikipedia article on metrics, there's no mention of the AV function, so I'm a bit confused now. Is it some sort of metric subspace instead?

P.S.: I'm not at all well-versed on metrics, so I'd appreciate simple answers :)

• If you mean that $(x, y) \mapsto |x - y|$ is a metric, then yes, you're correct.
– AJY
Aug 15 '16 at 18:31
• If $x \in \mathbb R$, then $$|x| = \|x\|_1 = \|x\|_2 = \|x\|_{\infty}$$ Aug 15 '16 at 18:31
• More generally, absolute value is a norm, which is a function $\| x \|$ on a vector space $X$ that assigns to every vector a non-negative real number that satisfies certain properties. Moreover, every norm induces a metric by $d(x, y) = \| x - y \|$.
– AJY
Aug 15 '16 at 19:14
• Thanks, so it's basically a norm which can also take the form of a metric. Right? Aug 15 '16 at 19:53

The absolute value $x\mapsto |x|$ is not a metric but a norm on $\mathbb R$ (or $\mathbb C$), viewed as a one-dimensional vector space. However, from any norm you can derive a metric in a standard way.

In the case of the absolute value, this gives the well-known metric $d(x,y)=|x-y|$.

• Thanks. So what's the difference between a norm and a metric? Aug 15 '16 at 19:05
• @Matt24: A norm takes a single vector and tells you its length. A metric takes two points and tells you the distance between them. Norms work only for vector spaces, whereas metric spaces do not need to have a vector space structure. Aug 15 '16 at 19:15

Yes, for $\Bbb R$ (or $\Bbb C$ for that matter), the map $d(x,y):=|x-y|$ is a metric.

No,the absolute value function $|\cdot|:\mathbb R\to[0,\infty)$ is not a metric.

A metric on a set $X$ is a function $d:X\times X\to[0,\infty)$ defined on the cartesian square of the set $X$ satisfying certain properties, and the absolute value function is not of that form.

What it is true is that out of the absolute value you can construct a metric: indeed, the function $$(x,y)\in\mathbb R\times\mathbb R\longmapsto|x-y|\in[0,\infty)$$ is a metric.

• Oh, hold on, so the AV function isn't a metric in itself because it have any variables? Aug 15 '16 at 19:08