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Posts: 1,176
Registered: 9/18/17

Re: When has countability been separted from listability?
Posted: Oct 5, 2017 6:52 AM

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Den tisdag 3 oktober 2017 kl. 14:09:14 UTC+2 skrev John Gabriel:
> On Tuesday, 3 October 2017 07:44:12 UTC-4, WM wrote:
> > Am Montag, 2. Oktober 2017 11:30:12 UTC+2 schrieb Alan Smaill:
> > > WM <wolfgang.mueckenheim@hs-augsburg.de> writes:
> > >
> > > > Cantor has shown that the rational numbers are countable by
> > > > constructing a sequence or list where all rational numbers
> > > > appear. Dedekind has shown that the algebraic numbers are countable by
> > > > constructing a sequence or list where all algebraic numbers
> > > > appear. There was consens that countability and listability are
> > > > synonymous. This can also be seen from Cantor's collected works
> > > > (p. 154) and his correspondence with Dedekind (1882).
> > > >
> > > > Meanwhile it has turned out that the set of all constructible real
> > > > numbers is countable but not listable because then the diagonalization
> > > > would produce another constructible but not listed real number.
> > >
> > > Wrong;
> >
> > In my opinion correct, but not invented by me. "The constructable reals are countable but an enumeration can not be constructed (otherwise the diagonal argument would lead to a real that has been constructed)." [Dik T. Winter in "Cantor's diagonalization", sci.math (7 Apr 1997)] More and links are given in https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf on p. 275 of
> >
> > Regards, WM
>
> I remember in my debate with mark chu carroll how he harped on representation and enumeration. What he failed to realise is that representation is nine tenths of enumeration.
>
> Here was his response which came after a lot of sturm und drang:
>
> (1) Yes, I?ll agree that all real numbers are *representable* using
> infinite decimal notation.
>
> (2) I?ll also agree that taken to infinity, your tree contains all real
> numbers between zero and 1.
>
> (3) Yes, I?ll agree that all finitely representable numbers can be
> enumerated from your tree using a breadth-first traversal.
>
> (4) NO, I do *not* agree that you can do a ?left to right? traversal of
> the infinitely-long representations. This is the problem with your whole
> damned argument: you?re mixing together notions from finite representations
> with infinite representations. You cannot do an ordered traversal of the
> leaves of an infinite tree. It?s *meaningless*. What?s the left-neighbor of
> 1/3 in your tree? You cannot specify it ? it doesn?t really exist: there
> simply is no real number which is ?closest? to 1/3 without being 1/3. But a
> left-to-right traversal supposes that there *is*. And that?s the problem.
> You?re trying to get a result using a property of a finite tree, when that
> property doesn?t exist on a tree extended to infinity.
>
> Now, notice that in (4) Crank Carroll has no problem with 1/3 being represented as an infinite string of decimals but it cannot exist as an "infinite branch" on my tree. Chuckle.
>
> Actually I had won the debate after he responded with YES to point (1).
>
> http://scienceblogs.com/goodmath/2010/02/04/so-remember-back-in/

Remember too, you got your arse handed to you.