there is no such thing as an inertial frame or near-inertial frame that crosses an event horizon.
Sure there is. Consider an object falling into the black hole; you can set up a local inertial frame around the event of that object crossing the horizon and it will work just fine even though it straddles the horizon.
What you can't have is a static inertial frame that crosses the horizon; i.e., you can't have one that stays at the same radius, even for an instant. The inertial frame I described above is infalling--i.e., an observer at rest in the frame is falling into the hole.
But there is no near-inertial frame that, from within the frame itself, appears to cross the Rindler horizon.
Yes, there is. This is even easier than the black hole case because you can have a Rindler horizon in flat spacetime, where all inertial frames are global--they cover the entire spacetime, including the region behind the Rindler horizon.
I should note that I do not mean to imply that the "finbot" article linked to is correct; it isn't. As I described above, inertial frames that straddle the horizon can be constructed, and they are perfectly consistent with the equivalence principle.
I tentatively don't agree with your Rindler horizon analysis. Inside an inertial frame, you don't see the Rindler horizon, because only the accelerating observer can see it. If you are in the accelerating frame of the observer, the apparent position of the Rindler horizon will vary, depending on your position, so you can't actually cross it; but in any case you are not in an inertial frame. Physical laws that hold in any inertial frame do not necessarily have to hold when observing from a non-inertial frame (for example ability to signal between any two points).
At this point, my understanding of physics gets hazy so I'm not sure how much this extrapolates to the black hole case.
Inside an inertial frame, you don't see the Rindler horizon, because only the accelerating observer can see it.
Actually, the accelerating observer can't "see" the Rindler horizon; light emitted at the Rindler horizon never reaches the accelerating observer (that's the definition of the Rindler horizon). The inertial observer is the one who can actually "see" the horizon, because he passes it.
Physical laws that hold in any inertial frame do not necessarily have to hold when observing from a non-inertial frame (for example ability to signal between any two points).
Any two points within the "accelerating frame" of the accelerating observer can signal to each other. The Rindler horizon is not, strictly speaking, "within" that frame; Rindler coordinates become singular at the Rindler horizon.
That said, I don't see what the properties of the accelerating frame have to do with the statement I made that any inertial frame in flat spacetime will include the Rindler horizon, since all inertial frames in flat spacetime are global.
Sure there is. Consider an object falling into the black hole; you can set up a local inertial frame around the event of that object crossing the horizon and it will work just fine even though it straddles the horizon.
What you can't have is a static inertial frame that crosses the horizon; i.e., you can't have one that stays at the same radius, even for an instant. The inertial frame I described above is infalling--i.e., an observer at rest in the frame is falling into the hole.
But there is no near-inertial frame that, from within the frame itself, appears to cross the Rindler horizon.
Yes, there is. This is even easier than the black hole case because you can have a Rindler horizon in flat spacetime, where all inertial frames are global--they cover the entire spacetime, including the region behind the Rindler horizon.
I should note that I do not mean to imply that the "finbot" article linked to is correct; it isn't. As I described above, inertial frames that straddle the horizon can be constructed, and they are perfectly consistent with the equivalence principle.