In article <46cd79fe-ca18-4b2c-bcc4-5d2032fb3769@w31g2000yqb.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 23 Jun., 14:51, Sylvia Else <syl...@not.here.invalid> wrote: > > > Cantor doesn't rely on being able to identify a last digit. He's just > > saying that no matter how far down the list you look, you'll find that > > the element at that point doesn't match the anti-diagonal. > > > That is wrong. Cantor uses the alleged "fact", that infinity can be > completed, i.e., that that the infinite list can be finished. > If he would only assume what you say, then the anti-diagonal could > remain in the unknown part of the list. You know and appreciate that > after *any* line number n there are infinitely many more lines? > > > But you can't > > even begin to formulate his proof if you can't identify the first > > element of the list (and hence first digit of the anti-diagonal) either. > > Isn't it enough, also in my case, to know that every antidiagonal has > a first digit?
No. You have to be able to tell how that, and any other digit of that "antidiagonal" was determined.
In fact it has. Every antidiagonal is constructed from > a list with a first line (that is the previous antidiagonal) and the > remaining list. Every line has a finite number n. > > > > First and last are interchangeable, of course, but with your > > construction above, you can't specify either the first or the last. > > As I told you, my notation is only an abbreviation for the following > definition: > 1) Take a list L0 of all rational numbers. > 2) Construct its antidiagonal A0. > 3) Add it at position 0 to get (A0,L0) > 4) Construct the antidiagonal A1. > 5) and so on. > > There occurs never a problem, because we know Hilberts hotel, don't > we?
There is also no problem in finding AN antidiagonal for the doubly open ended list (...,a2,a1,a0, LO.1, L0.2, L0.2, ...) once a surjection from N to it has been chosen.
