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.
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.
