Does a black hole have size if it's collapsed to a singularity?

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If a black hole is a stellar structure that has collapsed on itself to a singularity does that mean it has no size? But is so defined by its mass and schwarzschild radius?
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The black hole has a size defined by the event horizon. The actual volume and such is a bit more complicated to determine due to spacetime curvature I believe. A singularity, IF it exists at all, would have no size.
 
Current physical theory is unable to accurately describe what is going on inside the event horizon of a black hole. Quantum theory and general relativity don't mesh, and they both come into play.
 
The thing to realize is that the term 'black hole' doesn't mean: 'the bit at the center where all of the mass is concentrated, that may or may not be a singularity'.

'black hole' refers to the whole volume inside the event horizon, which clearly can have a radius, surface area and volume.

my understanding is that non-rotating black holes are perfect spheres with a radii equal to their Schwarzschild radius, and that rotating black holes are distorted into oblate spheroids, as are most rotating stellar objects.
 
Drakkith said:
The black hole has a size defined by the event horizon. The actual volume and such is a bit more complicated to determine due to spacetime curvature I believe. A singularity, IF it exists at all, would have no size.

A Schwarzschild black hole singularity is a mathematical point, yes, and so is a Nordström black hole singularity. But Kerr and Newman black hole singularities are one dimensional - they have no thickness, but they have circumference, radius etc.
 
How can it be one dimensional with a circumference and a radius? does that make it 2 dimensional?