A black hole itself (the singularity) is 1 dimensional - a single infinitesimal point. The event horizon around it is roughly a sphere. The diagrams that you are talking about are a visual metaphor that represent 3D space as a 2D plane and the 3rd dimension standing in for the influence of gravity. IRL, spacetime has 3 spatial dimensions, not 2, and gravity is not a dimension but a force. It's hard to visually represent gravitational distortion of 3 dimensional space without a 4th spatial dimension to do it with, so textbook diagrams use a 2D plane.
The singularity is likely a mathematical artifact of the fact that GR is insufficient to describe black holes. In reality (quantum gravity) they probably do not exist.
Could you elaborate on your definition of volume here (are you talking about spatial volume or spacetime volume?) and how the curvature going to infinity at the singularity should imply its infiniteness?
My thought process here is the following: The inside of an (eternal) black hole carries four (Schwarzschild) coordinates t, r, theta, phi – r now being timelike and confined to the interval (0, 2M) and t now being spacelike and being any real number. That is, depending on when (at what time t) you cross the event horizon, you end up at a different point in space. The singularity at r=0 is then a point in your future which, like your own death, you cannot actually see but which you will nevertheless hit in finite proper time.
So in this sense I'd say the volume is very finite (if we disregard the (trivially unbounded) spacelike coordinate t which, as mentioned before, simply corresponds to the time of entering the BH).
1. In the interior of the BH, the determinant of the metric is bounded since the Schwarzschild factors in the metric cancel out. So the volume measure doesn't do anything crazy as one gets closer to the singularity and boundedness of coordinates implies boundedness of the volume. Again, I'm disregarding the spacelike t coordinate because to me the relevant fact is that all matter reaches the singularity in finite proper time, so while we could theoretically stack lots of (actually, an infinite amount of) (massless) cubes inside a black hole, they would soon all get crushed.
2. Of course the situation is slightly different if we're talking about a growing black hole whose mass (and, therefore, radius) increases as we throw matter into it.
Unfortunately, I am not strong on mathematical part of GR, so feel free to disregard the rest of this idea, if math's the sole thing you are looking for.
A thought experiment: imagine we have a clock, falling into Schwarzschild black hole. Obviously, any real clock would have some non-zero size in all space dimensions. Here we will be concerned with just two: r and any orthogonal one. So for simplicity let the clock be a simple rubber-like oscillating ring with a fixed k and infinite resistance to tearing. (You could also take infinite k, but I'd argue that would not be physically meaningful in this setup)
As the ring is closing to the r = 0, its oscillations will slow down and come to a halt due to physical stretching along the r dimension. What I am trying to say is that maybe these oscillations make more sense as the measure of time for the ring people, than what a numerical value of proper time tells us. In a similar way the time singularity at the horizon is nothing special for a freely falling observer.
I am unsure how to interpret the fact, that the number of oscillations per proper time unit is going down though. Seems to be quite the opposite of my original note about the volume, yet something is ringing.
Unfortunately can't reply to the original question anymore. But here's another hypothesis I just read elsewhere (RU: https://don-beaver.livejournal.com/212422.html): the matter-energy falling towards "center" inevitably has some non-zero momentum, orthogonal to the direction along r. That means its fall is going to be (sort of) spiral, causing it to produce gravitational waves. The argument here is that the whole mass will be converted into gravitational waves before reaching the central point. Now what happens to those waves is a question still.
A point is 0-dimensional, but in a space-time diagram the time dimension is added, which makes it 1-dimensional. Penrose diagrams for black holes assume spherical symmetry, so all of space is represented by a single radial coordinate, which makes it possible to display such diagrams in 1+1=2 dimensions.
A singularity doesn't have a dimension. It is a portion of spacetime that is missing, not a point or set of points. We can't define its dimensionality, either.(×) What we can say is that the singularity in Schwarzschild black holes is spacelike.
×) Counterexample: Consider the manifold M := R³\B, where B is the closed unit ball, equipped with the standard Euclidean metric. This manifold is certainly not Cauchy-complete and we can reach the singularity at r=1 in finite time. Now, if we had to define the dimension of the singularity, what dimension n should it have? n=2 (a sphere)? Maybe. At least we could extend M by the unit sphere to make it complete. But could the singularity also be a point (i.e. n=1)? Yes, certainly. By diffeomorphism invariance, we could simply find new coordinates and map R³\B to R³\{0}, so the singularity would suddenly become a point. So, as you can see, interpreting the singularity as a point or set of points that have a topological dimension doesn't work.
Not sure why you're getting downvoted, this is a legitimate question. However, in this particular case, I think the final sentence of the accepted answer on Physics.SE, namely that
> In this diagram the singularity is a line in spacetime i.e. a one dimensional object in spacetime.
is wrong or at least very misleading – the answer does (correctly) say that asking for the "dimensionality of a singularity […] is a meaningless question because the spacetime geometry is undefined at a singularity".
