On 4 Mai, 22:48, Dan <dan.ms.ch...@gmail.com> wrote:
> You can have a formula for the list , f(a,b) , that gives the b'th > digit of the a'th number of the list . > You're saying b must be bounded (no infinite sequences of digits) , > but a can be infinite (an infinite amount of finite sequences) . > This is inconsistent . Why should f(a,b) be a valid list but f(b,a) be > an invalid list?
A very clear and correct argument! Of course there cannot be any asymmetry. This can be proven by the obvious bijection between all FIS of the first column and of the lines:
1 2, 1 3, 2, 1 ...
But we know from mathematics that a strictly incresing sequence cannot contain its limit and also that all natural numbers are finite. Therefore all lines are finite. They are potentially infinite, i.e., they are all finite, but you cannot find any threshold, let alone a supremum.
Vice versa this implies that the columns, including the first column, do not contain their limit. There is no actual infinity, i.e. more than every natural number of numbers in the first column.