It's easy to verify this for yourself in a simple example in something like python. Draw from a multivariate normal distribution (or almost anything) in a 1000-dimensional space and look at the norm of your vector. If you do this a lot of times you'll see that the answer is close to the same each time. This even holds true if you replace the vector norm by the distance to an arbitrary fixed point.
edit: here's some code:
import numpy import matplotlib.pyplot norms = [] dists = [] for i in range(5000): vec = numpy.random.normal(size=4096) norms.append(numpy.linalg.norm(vec)) dists.append(numpy.linalg.norm(vec-numpy.ones(4096))) matplotlib.pyplot.hist(norms) matplotlib.pyplot.show() matplotlib.pyplot.hist(dists) matplotlib.pyplot.show()
(didn't render right, sorry)
For instance, consider a unit cube in n dimensions. The volume of this cube (1 unit on each side) remains constant, but the distance from the center to a corner increases with the square root of the number of dimensions. This implies that in high dimensions, most of the volume of a hypercube is located at its corners. This is bizarre to comprehend because it's hard to stop imagining a "cube" in the 3D sense.
Algorithms that rely on distance measures (like k-nearest neighbors or clustering) may perform poorly in high-dimensional spaces without dimensionality reduction techniques like PCA (Principal Component Analysis) or t-SNE (t-Distributed Stochastic Neighbor Embedding). And even with these techniques, you can get very misleading results.
This is incorrect. If you put a little box of small side length x at each corner, then each box has volume x^n and there are 2^n many of them, so the total volume they take up is 2^nx^n=(2x)^n, which is very small. Maybe what you meant is that most of the volume of a high-dimensional hypersphere is near its surface.
Surface area is 2-volume.