Splitting a Grain of Sand Paradox

Rather than starting with a bucket of sand, let's start with just one grain, which gets split in half, then the two bits are split in half to make 4, and so on. If we call the original grain A1 then after the first split we have A11 and A12, then after the second split A111, A112, A121 and A122, and so forth. As we approach infinity each particle can be represented as a non-terminating series of digits such as A1112211212221112121111111211111112...

Assuming that the collection of such numbers is countable, we can list them. Then we can use the standard diagonalisation procedure to generate a number of the same form, ie 'A' followed by ones and twos, that is not in the list. It will be made to differ from each element of the list in one position, ie it will have a 1 where the element has a 2 or vice versa. This is the same as the standard proof that the set of numbers expressed as non-terminating decimal expansions is uncountable.

So the infinite splitting of the grain of sand produces an uncountable number of particles, hence the paradox does not arise.


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