Null Hypersurface of Lorentzian Manifold: A hypersurface that admits a null-like or light-like normal vector field($N^a$) to it. i.e. $g_{ab}N^a N^b=0$ (metric signature$(-1,1,1,1,...)$)
In Minkowski spacetime the light cones are null surfaces.If we take the gradient of this surface($f(t,x,y,z)=-t^2+x^2+y^2+z^2=0$) we will find the normal vector on it, and it can be checked that it is a null vector (not normalizable).I also proved that there is no timelike tangent on $f(t,x,y,z)$.
Now, I want to extend this result (a null surface does not admit any timelike tangent to it) to an arbitrary Lorentzian manifold.
I tried this way(i failed,though):
$g_{ab}N^aN^b=0 \\ So,\ N^a\ is\ normal\ to\ itself\ i.e.\ N^a\ is\ also\ a\ tangent\ to\ the\ null\ surface.\\ Now,I\ will\ start\ with\ a\ timelike\ tangent\ T^a, such\ that\ g_{ab}T^aT^b\lt0\ and\ g_{ab}T^aN^b=0.\ Now,\ somehow,\ I\ have\ to\ reach\ to\ the\ conclusion\ that\ g_{ab}T^aT^b\ge0.\\I\ was\ trying\ to\ prove\ that\ g_{ab}T^a(T^b+N^b)\ge0,\ which\ I\ could\ not. $$g_{ab}N^aN^b=0 \\ So,\ N^a\ is\ normal\ to\ itself\ i.e.\ N^a\ is\ also\ a\ tangent\ to\ the\ null\ surface.\\ Now,I\ will\ start\ with\ a\ timelike\ tangent\ T^a, such\ that\ g_{ab}T^aT^b\lt0\ and\ g_{ab}T^aN^b=0.\ Now,\ somehow,\ I\ have\ to\ reach\ to\ the\ conclusion\ that\ g_{ab}T^aT^b\ge,\ to\ contradict\ the\ assumption.\\I\ was\ trying\ to\ prove\ that\ g_{ab}T^a(T^b+N^b)\ge0,\ which\ I\ could\ not. $
Is it possible to prove it, following my approach? I am also curious about other ways to prove it.