I have come to a better understanding of the event horizon, due in part to a comment by Lee Corbin. I also presume the definition that an event horizon is the boundary beyond which no photon can escape to asymptotic infinity. The event horizon is three dimensional. A time slice of space intersects the horizon in a two dimensional surface which is S(2) in the case of a black hole. I call such a 2D intersection an “instant horizon” below. You can imagine an army of photons on the horizon each destined to remain on the horizon, barely escaping being drawn into the interior oblivion. Each such a photon travels perpendicular to the instant horizon and describes a null geodesic in the 3D horizon. These geodesics partition the 3D horizon into fibers, for thru a point (event) on the horizon there is only one direction in which a photon can travel and remain on the horizon; all other directions lead to oblivion. The interval between two points on such a fiber is zero. These fibers are a property of the horizon and not any reference frame choice. As these fibers provide a sense of same place but different time on the horizon, I believe that there is no natural concept of same time, different place, except when one may notice a symmetry.
I claim something above that seems mathematically impossible: that the photon travels perpendicularly to the instant horizon independent of the time slice that produced that instant horizon. This is possible only of a larger yet rare class of sub manifolds that include event horizons. This is connected to the problem of considering a time slice across a region of space-time that includes a black hole. There must be a hole, it seems, in the time slice. The edge of this hole would seem to be an event horizon. Considering that the definition of a horizon seems to require the concept of “asymptotically flat”, this raises questions for non asymptotically flat spaces, such as the one we once thought we lived in.
Consider the one event horizon of a pair of two black holes at some distance. If the holes are far enough apart then the horizon is divided into two components, one about each hole. As photons just outside one component head in a direction nearly toward the other hole, they will be captured. By the Penrose definition we must adjust the slope of the first component to orient it so that no photon emitted perpendicular thereto will be captured by the other hole. It seems clear to me that this requires there to be a point on each component nearest the other component. All of this argument takes place while imagining (falsely) that the holes are not accelerating towards each other. If they spin about each other the argument is more complex but I think still applies.
Here is a little math that helps reason about the event horizon.
The EH is a finite distance away.