3 ms·
> It relies on uncountably-infinite division of an object But the theorem claims finite division and not infinite? In R³, given a solid ball B of radius R it
by zokier 4y ago
> It relies on uncountably-infinite division of an object
But the theorem claims finite division and not infinite?
In R³, given a solid ball B of radius R
it is possible to partition B into finitely many
pieces such that those pieces can be reassembled
to form two solid balls B1 and B2 each of
radius R
- moloch-hai 4y agoFinitely many pieces, but on infinitely variable boundaries. It was clever to make the proof allow a finite number in that place. Without, it would have attracted no attention.
- wnkrshm 4y agoIt's still astounding to me that the 'fractal trick' starts working in R3 and not already in R2.
- moloch-hai 4y agoAgreed, each added dimension adds wacky surprises. Packing spheres into a cube leaves you a pretty comprehensible space in the middle. Pack hyperspheres into a hypercube, and all hell breaks loose.
- drdec 4y agoKeep in mind that "pieces" might not mean the same thing you think of it as meaning.