Event horizon from the metric Let us suppose we have a metric of this form
$$ds^2=-A(r)dt^2+\frac{dr^2}{B(r)}+r^2d\Omega^2$$
In all documents I can read, I've seen that the event horizon is defined by considering $A(r)=0$
But I don't understand it, because if I consider a surface of equation $r=C^{st}$, a normal vector to this surface is $n_\mu=\nabla_\mu r$ and the norm of this vector is $g^{\mu\nu}\nabla_\mu r \nabla_\nu r$.
So if I want the surface to be null, it gives $g^{\mu\nu}\nabla_\mu r \nabla_\nu r=g^{11}=0$ which means $B(r)=0$.
 A: It is not the definition of an event horizon, and in fact you can choose coordinates that are regular near the event horizon.
A common reason for coordinates that are irregular at the horizon is if the coordinate is primarily made to represent time far away.
In that case, a timelike curve has a negative interval in your convention, so you can have time like curves that don't change r or the angles, so you clearly can change your r to be positive or negative by just having a small change in r go along with your large change in t, so you can be timelike while moving to larger or smaller r.
Inside an event horizon you sometimes can't advance to a larger radius, and that means that r has become a timelike direction the timelike directions don't have enough room (1 out of 4) to turn around to point a different way. So for instance even for a rotating black hole there are regions inside where decreasing r is future timelike and other (different)/regions of spacetime where increasing r is future pointing timelike.
Just because you call it r doesn't mean it isn't timelike, and just because you call it t doesn't mean it isn't spacelike. It is surprising how often a name confuses people, if there is a negative interval that is timelike interval regardless of which letter someone chose to use.
So having the coefficients explode to $\pm\infty$ or to zero is an easy way to transition from it being timelike to spacelike. The point is whether that $r^2$ in front of the $d\Omega^2$ is allowed to get smaller along the continuation of an initially future pointing curve, if not then $\pm r$ is not a spacelike direction.
