Confusions about Schwarzchild Geodesic Deviation In schwarzchild metric space gets bigger as you approach the horizon, if you build shells around a black hole infinitesimally small distance apart you can build an infinite number of such shells. However, this effect no longer exists for an observer falling in geodesic, speed cancels the 'space getting bigger' effects. That's why there is no real singularity at the horizon, a falling observer can easily get past it (Gustave-Painleve metric, Eddington Finkelstein metric etc, Ref: Exploring Black holes by Taylor and Wheeler).
So far so good, but here the next problem arises: to a free falling observer the space does not get bigger, so the distance between him and another free falling observer should not accelerate - in other words he will measure a zero geodesic deviation. Of course, in reality he doesn't measure a zero geodesic deviation in r, theta direction - Riemann(theta, r) r has non zero components.
So, what am I missing here? What's wrong with my intuition?
 A: Before answering I need to point out a mistake in the question:
for the Schwarzschild-Droste black hole, the number of one-metre-separated shells between the horizon and any given location is finite not infinite.
To get a visual impression of this, you could look at Flamm's paraboloid: the distance along the surface from the horizon to any given location outside is finite.
The value of this distance is
$$
\int_1^x \frac{1}{\sqrt{1-1/x}} dx = \sqrt{x(x-1)} + \log(\sqrt{x} + \sqrt{x-1})
$$
where $x = r/r_s$ with $r$ the Schwarzschild-Droste radial coordinate and $r_s$ the Schwarzschild radius. (I include the name Droste here for historical reasons; the metric and black hole commonly named after Schwarzschild is more suitably named after both people IMO).
This means the volume between the horizon and a spherical surface of some given area is larger than a Euclidean calculation would suggest, but it is not infinite. However the time for light (or anything else) to propagate out from the horizon is infinite, so a definition of distance based on light travel time would be infinite.
The second paragraph of the question asserts:

to a free falling observer the space does not get bigger, so the
distance between him and another free falling observer should not
accelerate

This is a non-sequitur: the phrase after 'so' does not follow from the phrase before 'so'. To see this, compare with Newtonian gravity, such as the case of two objects falling to Earth, one above the other. The distance between two such objects will get larger as they fall, because the lower one has a larger acceleration. But the spatial geometry is everywhere Euclidean. The lesson is that geodesic deviation is a spacetime idea, and in this example the curvature in spacetime implied by the geodesic deviation of these timelike lines (the two worldlines of the falling objects) does not necessarily imply a curvature in the spatial direction.
