So, if I understand this correctly, essentially you're saying that any event horizon (no matter when it was created or its exact size) experiences infinite time dilation and as a result the object will end up asymptotically approaching the new horizon instead (or whichever horizon is most readily accessible in a general case) rather than the old one. It's sort of like the new horizon hides the old one, so that the object can no longer try to approach the one that's behind the barrier of the new horizon.

Yes (subject to someone managing to find a horizon type that I haven't thought of that doesn't give a divergent integral :-).

What about if we then consider the second version of my syllogism, wherein for instance one object starts falling into a black hole and after a long time gets very close to the horizon and then another object starts also falling into the black hole? In that case the total radius would increase again and it wouldn't envelop the second object, but is it possible that it could increase enough to envelop the first?

@FigrothFelanor regardless of what happens it is always true that the time dilation at the horizon must be infinite at the spacetime point where the first object crosses it. So my argument still applies.

@TimRias OK :-)

@John Rennie I get the argument that an object can never reach an event horizon (when viewed from outside), however what happens if an object reaches a radius where there is no event horizon at the time but where one will form in the future due to the influence of other objects? The object does not then need to approach the event horizon, because it has already passed it at the time of its formation (I imagine that this is more or less how black holes would form in the first place, because otherwise a collapsing star would never reach the growing event horizon either).

@FigrothFelanor no, the horizon cannot move discontinuously so it has to grow outwards smoothly. That means the first object must cross the horizon - the horizon cannot jump from inside the object to outside the object. Incidentally a collapsing star never crosses the horizon, so for an observer far from the black hole the horizon never forms. That is, for us there are no black holes in the universe, only close approximations to black holes.

See Why does Stephen Hawking say black holes don't exist?

I think I see. The problem with my thinking is probably that in the first place the second object would have to be inside the new SR it created, for it to become an event horizon. This is never the case, so it wouldn't affect the first object either. I'm not sure I get how an event horizon could grow smoothly honestly (though it doesn't necessarily matter). The way I see it, it's just that each object approaches a different event horizon, but that event horizon has always been there (more or less), since the total SR calculated for the combined mass does not require the object to be close.

Does that mean that the collapsing star also never actually reaches the event horizon either? It's just that by the time we discover these black holes, the light from the star has been so red-shifted we can't detect it any more? Or is there a mechanism that allows the collapsing star to collect in the region that the black hole will occupy, because time dilation is not extreme yet?

@FigrothFelanor, Re, "in the first place the second object would have to be inside the new SR it created, for it to become an event horizon." An event horizon is not a physical boundary. It's more like a "point of no return." If you were falling in to a super massive black hole, you would not notice anything happening when you passed that point of no return because literally, nothing happens there. Also, it's not true that you must either be completely inside or completely outside of that boundary. Each individual particle of your body crosses the boundary in its own time.

@SolomonSlow Oh I get that, but the discussion here is exclusively about what an outsider observer sees, not about the perspective of the infalling object. Of course in that perspective, the horizon is crossed.

@FigrothFelanor, Right, but that observer cannot see anything enter a black hole, so if the discussion really was just about that, then it would be over and done. You seemed surprised when John Rennie said an event horizon cannot expand in discontinuous jumps. That suggested to me that maybe you think there is some qualitative physical difference between a black hole with a freely falling object just outside of its event horizon and the same black hole with the same falling object just inside of its event horizon. I am trying to say that there is no discontinuity between those two examples.

@SolomonSlow My point in regards to that was that the SR as calculated for a BH + an object A inside the EH, for a BH + A just outside and for a BH + A very far away is always the same (assuming no interference from other objects B getting closer to the black hole than the object we're speaking of). Thus, the event horizon approached by A does not grow smoothly or discontinuously, it's always been that size. I guess, though, in the case where B comes closer to the BH than A, then the EH would gradually grow, as its volume would gradually pass A, which might be the point you were making.

@SolomonSlow This might be better suited for a new question, but why does an event horizon expand continuously if mass isn't a continuous quantity? I'm by no means a physicist, but it feels strange to me that an in-falling fundamental particle would contribute partial mass as its position becomes more and more likely to be within the event horizon, since there's still the possibility for interactions that could deflect it away when it's only "partially" within the event horizon. Or would that be getting into an area that'd require quantum gravity to answer?

@Idran, I don't know how to calculate the locus of a black hole's event horizon, but if it changes because of the smooth, continuous motion of an additional mass toward the singularity, then I would expect the change to be smooth and continuous. That's the only thing I've been trying to say, and after this, I'm out of ideas for new ways to say it. Have fun y'all!

@TimRias “Your argument fails exactly when t is chosen to be a hyperboloidal time coordinate” - John’s argument stands. Hyperboloidal time is based on the raindrop proper time with some conformal tweaking, not on the coordinate time of an external observer.

@Idran I'm no expert, but I think GR doesn't assume a discrete form of matter. The Quantum description of matter is not included in the theory and a smoothly continuously distributed mass is probably a good enough approximation for macroscopic object. It makes sense to want to apply GR on actual particle matter, of course, rather than hypothetical smooth matter, but it sounds like it wouldn't be immediately obvious how a particle obeying QFT would interact with an event horizon. You probably need QG for a real answer.

This answer assumes a point-sized test object. In reality all objects are extended, and we know that a smaller black hole can merge into a larger black hole in a finite time as measured by an observer at infinity.

@Prof.Legolasov merging black holes never actually merge. (Actually they weren't black holes in the first place.(

@Idran The radius of the horizon is defined by the mass just outside the horizon. Nothing inside the horizon has any effect on the outside. If matter could fall trough the horizon in our view, the black hole would lose its gravity and disappear. In reality, when a particle $m$ gradually falls to a black hole $M$, the radius of the horizon gradually increases from $r_s=2M$ to $r_s=2(M+m)$ at which point the particle meets the horizon. Plus indeed GR is classical with no particles, but a smooth distribution of mass, yet according to the above this is not important and makes no difference here.