# Is Minkowski space not a metric space?

I've just started reading a book on functional analysis, and first definition given there is for a metric and metric space:

Let $\mathfrak{M}$ be an arbitrary set. A function $\rho\colon \mathfrak M\times\mathfrak M\to[0,\infty)$ is called metric if it has the following properties:

1) $\rho(x,y)=0 \iff x=y$ (axiom of identity)

2) $\rho(y,x)=\rho(x,y)\;\forall x,y\in\mathfrak M$ (axiom of symmetry)

3) $\rho(x,y)\le\rho(x,z)+\rho(z,y)\;\forall x,y,z\in\mathfrak M$ (triangle inequality)

The pair $(\mathfrak M,\rho)$ is called metric space.

First and second identities don't make any surprise, I understand them. But what about image of $\rho$ and third inequality? They don't seem to hold in Minkowski space, where if we check interval as candidate for metric, we get $$s^2=t^2-x^2-y^2-z^2,$$ which can be negative and violate triangle inequality (and if take square root, it becomes complex and inequality makes no sense in this case).

So this clearly isn't a metric. But is Minkowski space then not a metric space? What is it then?

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I've wondered this for so long. Favourited. – goblin Oct 14 '13 at 18:55
from wikipedia (search metric): "In differential geometry, one considers metric tensors, which can be thought of as "infinitesimal" metric functions. They are defined as inner products on the tangent space with an appropriate differentiability requirement. While these are not metric functions as defined in this article, they induce metric functions by integration." – Dubious Oct 14 '13 at 18:59

The topology is the standard one of ${\mathbb R}^4$. Which metric (in the "metric space" sense) you use for that is not particularly relevant. – Robert Israel Oct 15 '13 at 6:52