
Re: A BLATENT FLAW in Cantor's diag proof
Posted:
Jun 8, 2010 3:33 PM


On Jun 8, 2:42 pm, Marshall <marshall.spi...@gmail.com> wrote: > On Jun 8, 9:48 am, jbriggs444 <jbriggs...@gmail.com> wrote: > > > > > > > On Jun 8, 10:03 am, Marshall <marshall.spi...@gmail.com> wrote: > > > On Jun 7, 8:40 pm, "ercules" <radgray...@yahoo.com> wrote: > > > > "Marshall" <marshall.spi...@gmail.com> wrote: > > > > > On Jun 7, 7:51 pm, "ercules" <radgray...@yahoo.com> wrote: > > > > > >> I can compute the list of all computable reals. There's just some numbers that show > > > > >> up blank. > > > > > >> It's trivial to compute a list that covers every digit sequence to all (infinite) finite lengths. > > > > > > How are you going to do that? > > > > > > Write a program that first prints out an infinite sequence of zeroes, > > > > > and then... > > > > > > Oops! Already a problem! There is no "then" that comes after > > > > > writing the zeroes, because the process of writing the > > > > > zeroes will never finish. > > > > > > Please show us this trivial program that computes every infinite digit > > > > > sequence. > > > > > Here you go: > > > > > 1 000000 > > > > 2 31415 > > > > 3 2818 > > > > 4 141 > > > > 5 22 > > > > 6 7 > > > > > It's not finished yet! Next digit is the 7th 0 on the first number. > > > > At no point will this process ever produce even a single > > > infinite string of digits. > > > > Also I see no reason to think it's going to be anything vaguely > > > comprehensive in the vertical direction either. > > > With a serpentine traversal, for every position on the grid > > there will be (for an algorithm that doesn't end up looping > > quietly or halting) a time by which the algorithm will have > > provided a digit for that position. > > > In this sense, the algorithm specifies every digit on every > > line. > > Sure. However, at no point will this process ever produce > even a single infinite line of digits.
I certainly agree that if we set this thing running and let it print on a hypothetical infinite roll of infinitely wide paper then we can never ever point to a line on the paper and say "there, look at it, the completed decimal expansion of line number 1".
You are absolutely correct. In that sense, the algorithm never "produces" a single number.
> > When we talk about enumerating the natural numbers, > for example: > > f(x) = { emit x; f(x+1); } > f(0); > > then at no point will we be finished with all of them. > However, for every natural number n, there is > a point at which n will be emitted.
Agreed.
> At no point, ever, in the serpentine traversal, will > we have even a single infinitely long line of digits. > We never ever produce even one infinite digit > sequence this way; we certainly cannot say that > we are producing all of them.
I think you are stretching your point when you use the phraseology "are producing". At best the relevant point would be that we cannot say that we "have produced" all of them.
However...
One can evaluate a limit without computing the values of all of its terms. And one can characterize a completely populated infinite square grid of numbers in terms of a generating algorithm without actually having to run that algorithm and without requiring that the algorithm ever arrive at a halt state.
But perhaps we are in vigorous agreement.

