On Thursday, 27 June 2013 01:17:19 UTC+2, Virgil wrote:
> Suppose that x and y are members of |N [...] with no maximum member and f(x,y) means x < y then "for all x, there is a y such that x < y" is true but "there is a y such that for all x, x < y" is false.
>> In a linear set there is no chance to apply the quantifier trick.
> Then I have created the impossible!!!
No, you have just done sober mathematics, probably unconsciously. You cannot have more than all naturals that have a larger successor. You cannot have more than all initial segments (d_1, ..., d_n) of the diagonal.
Set theory-based mathematics assumes that there is a diagonal d with more than any natural number of digits d_n, but that is nonsense, as you just have argued. There is no cardinal y = aleph_0 such that for all cardinals x < y.
All (d_1, ..., d_n) is all that you can have of d (although for each one there is a larger one). And that is what I have in my rationals-complete Cantor-list.