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


Den tisdag 3 oktober 2017 kl. 14:09:14 UTC+2 skrev John Gabriel: > On Tuesday, 3 October 2017 07:44:12 UTC4, WM wrote: > > Am Montag, 2. Oktober 2017 11:30:12 UTC+2 schrieb Alan Smaill: > > > WM <wolfgang.mueckenheim@hsaugsburg.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.hsaugsburg.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 breadthfirst traversal. > > (4) NO, I do *not* agree that you can do a ?left to right? traversal of > the infinitelylong 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 leftneighbor 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 > lefttoright 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/sorememberbackin/
Remember too, you got your arse handed to you.



