On 29 Jan., 10:18, William Hughes <wpihug...@gmail.com> wrote: > On Jan 29, 10:09 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > On 29 Jan., 09:54, William Hughes <wpihug...@gmail.com> wrote: > > > > On Jan 29, 9:33 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > "All" and "every" in impredicative statements about infinite sets. > > > > > Consider the following statements: > > > > > A) For every natural number n, P(n) is true. > > > > B) There does not exist a natural number n such that P(n) is false. > > > > C) For all natural numbers P is true. > > > > > A implies B but A does not imply C. > > > > Which is the point. Even though A > > > does not imply C we still have > > > A implies B. > > > > Let L be a list > > > d the antidiagonal of L > > > P(n), d does not equal the nth line of L > > > > We have (A) > > > > For every natural number n, P(n) is true. > > > > This implies (B) > > > > There does not exist a natural number n > > > such that P(n) is false. > > > > In other words, there is no line of L that > > > is equal to d. > > > And how can C be correct nevertheless? Because "For all" is > > contradictory. > > B: There is no line of L that is equal to d > > does not imply > > C: For all n, line n is not equal to d. > > B correct does not mean "C correct nevertheless"-
But we know of cases where C is correct nevertheless. I quoted four of them in the § 203. Or do you disagree to one of them?
In case you have forgotten the old discussion concerning the configurations of the Binary Tree construction, here it is repeated:
The complete infinite binary tree is the limit of the sequence of its initial segments B_k: ____________________ B_0 =
The structure of the Binary Tree excludes that there are any two initial segments, B_k and B_(k+1), such that B_(k+1) contains two complete infinite paths both of which are not contained in B_k. Nevertheless the limit of all B_k is the complete binary tree including all (uncountably many) infinite paths. Contradiction. There cannot exist more than countably many infinite paths.
Alternative consideration: Obviously every B_k is finite. None does contain any infinite path. The infinite paths come into the play only after all B_k with k in N (by some unknown mechanism). If that is possible, however, this mechanism can also act in Cantor's diagonal proof such that the anti-diagonal enters the list only after all lines at finite positions.