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?

It's a real vector space with an indefinite bilinear form.
