3 ms·
I don't think I agree, but this is difficult to communicate in words. Cantor's diagonalization argument doesn't rely on the contents of an infinite list of num
by epgui 3y ago
I don't think I agree, but this is difficult to communicate in words.
Cantor's diagonalization argument doesn't rely on the contents of an infinite list of numbers, or on numbers themselves. You don't need to know anything about what "..." represents and you don't need to be "at" anywhere. There is no need to compute anything with concrete numbers.
The only thing the argument relies upon is whether the fundamental structure of numbers allows for defining a bijection. The proof/computation is an abstract relation, not a concrete one with actual number objects.
- cubefox 3y ago> The only thing the argument relies upon is whether the fundamental structure of numbers allows for defining a bijection. That there is no bijection must be the conclusion, not an assumption, otherwise the argument would simply be question begging. If you can't define a new diagonal number in the first place, you don't have an argument why you can't have a bijection. With finitism it doesn't seem like you can define a diagonal number, because then Cantor's infinite table is merely potentially infinite in width and height. For the diagonal to be finished, the table has to be at least as wide as it is tall. But if you list numbers by their lexicographic order, the list grows much faster in height than in width. For base 10, for n columns (digits), you get 10^n rows (numbers). Then you can't ever finish a diagonal that isn't on the list. The number of digits doesn't grow fast enough to cover the entire list. You don't get the diagonal in the limit. Now if you assume infinitism, you can simply declare that the height and width of the list is the same (or even that the width is larger), and therefore the diagonal can be constructed. But that's not a convincing assumption for a finitist (or ultrafinitist).