You can't prove it's not on the list , but you can't prove it is on the list either (if you could prove it , you would match the number 0.111.... to one of your finite definitions (eg. 0.1111 ) on the list ).
> Therefore there are no actually infinite sequences at all - and there > is no Cantor-list, other than finitely defined lists.
False . Either there are only finite-sequence lists with with a finite number of terms , or there are infinite lists with infinite terms .
0 0 1 0 1 1 0 1 1 1 0 1 1 1 1 ............
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?
is not valid? What's possible for the columns of a list must be possible for the rows . The same constraints must apply both to the indexes of the real numbers in your list (vertical indices) , and to the indexes of your digits (horizontal indices) .
More specifically : you have a list of (FINITE sequences of digits) . Your list an INFINITE number of (horizontal FINITE sequences of digits) :
The same indexing method for members of a list is used for digits of a member. What works for the columns must work for the rows . What works for the vertical must work for the horizontal . If you have infinite members you must allow infinite digits in a member . If you have infinite digits in any member, you must allow infinite members .