On Jun 23, 6:42 am, Sylvia Else <syl...@not.here.invalid> wrote: > On 23/06/2010 11:10 PM, WM 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? > > Yet it's clear that if you look at any later line m, you will still find > that it doesn't match the anti-diagonal. > > > > >> 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? 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. > > I agree that every anti-diagonal you added is well defined. But for your > proof to work you also have to look at the anti-diagonal of the list > after you've added all of the (infinitely many) constructed > anti-diagonals. To construct the first digit of that anti-diagonal you > have to look at the first element in the list. But it has no first > element - any element you might claim is the first is in fact preceded > by infinitely many other elements. So you have no way of deciding what > the first digit of the anti-diagonal should be, let alone any of the > other digits. > > Your proof falls apart if you cannot construct the anti-diagonal which > you claim should have been in the list.
Why does WM have to construct the anti-diagonal and Cantor does not? > > As I observed earlier, this problem can be obviated by changing how the > list is constructed. I can't see why you're opposing that, unless you > consider it more than a cosmetic change for reasons that escape me. > > > > >> 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. > > With a resulting 'list' which is infinite at both ends. > > > There occurs never a problem, because we know Hilberts hotel, don't > > we? > > I'd have thought the Grand Hotel Cigar Mystery was the one to watch. > > Sylvia.