Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: mueckenh on 30 Oct 2006 09:05 Virgil schrieb: > > Forget the Turing machines. The diagonal of the list of all finite > > sequences (words) of a finite alphabet is a finite sequence because the > > diagonal cannot have more places than the words in the list. > > > But if the nth word has, say, n+1 places, then one has a list of finite > sequences which are unbounded. > > There is a vital difference between a list of merely finite sequences > and a list of uniformly bounded sequences ( all being bounded by the > same bound) which allow the former to have sequences of arbitrarily > large, but still finite, lengths. Every member of a list of only finite sequences is a finite sequence. Every diagonal of such a list is a finite sequence. That is enough. Regards, WM
From: Sebastian Holzmann on 30 Oct 2006 09:06 mueckenh(a)rz.fh-augsburg.de <mueckenh(a)rz.fh-augsburg.de> wrote: > Every member of a list of only finite sequences is a finite sequence. > Every diagonal of such a list is a finite sequence. That is enough. That is wrong.
From: Bob Kolker on 30 Oct 2006 09:08 mueckenh(a)rz.fh-augsburg.de wrote: > > Correct. And this sequence has to be defined somehow. But there are > only countably many definitions. That is true, if by "definition" you mean a finitary algorithm or recipe. If this constraint does not hold then there are aleph-1 "definitions" Bob Kolker
From: mueckenh on 30 Oct 2006 09:13 Virgil schrieb: > In article <1162190663.531903.51440(a)e64g2000cwd.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Sebastian Holzmann schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de <mueckenh(a)rz.fh-augsburg.de> wrote: > > > > > > > > Sebastian Holzmann schrieb: > > > > > > > >> Virgil <virgil(a)comcast.net> wrote: > > > >> > In article <1161883732.413718.244570(a)k70g2000cwa.googlegroups.com>, > > > >> > mueckenh(a)rz.fh-augsburg.de wrote: > > > >> >> What you propose, namely the infinity of ZF without the axiom INF > > > >> >> would > > > >> >> not be an advance. But meanwhile you may have recognized that your > > > >> >> assertion (ZF even without INF is not finite) is false. > > > >> > It is, however, quite true that ZF without INF need not be finite. > > > >> It is, more than that, quite true that ZF without INF _is_ infinite > > > > Are you really sure? > > > > > > I am. > > > > > So you also must be sure that ZF is yellow. > > And WM is equally sure that what natural numbers which he allows might > exist are all purple with pink polka dots. > > > > >> (the axiom schema of separation alone provides infinitely many axioms). > > > > Are you really sure? > > > > > > I am. > > > > But most of them are very dirty? > > > > > > >> The point is: ZF without INF does not prohibit the existence of infinite > > > >> sets, nor does it force them to exist. > > > > > > > > It prohibits to speak of infinite sets and to recognize such sets. So > > > > one cannot be sure, but you are? > > > > > > Given any model of ZF-INF, one cannot be sure if there are infinite > > > sets. > > > > In particular the sets consisting of infinitely many rabbits would be > > of great interest. > > And like most of WM's notions, such sets would be hare today and gone > tomorrow. Get it right: It is nonsense to talk about infinite sets if there is no axiom of infinity and, therefore, no possible definition of infinity. "Given any model of ZF-INF, one cannot be sure if there are infinite sets" is as silly an argument as talking about sets of rats or rabbits, or sets being clean or dirty. Regards, WM
From: mueckenh on 30 Oct 2006 09:16
Sebastian Holzmann schrieb: > mueckenh(a)rz.fh-augsburg.de <mueckenh(a)rz.fh-augsburg.de> wrote: > > Sebastian Holzmann schrieb: > >> mueckenh(a)rz.fh-augsburg.de <mueckenh(a)rz.fh-augsburg.de> wrote: > >> > Sebastian Holzmann schrieb: > >> >> It is, more than that, quite true that ZF without INF _is_ infinite > >> > Are you really sure? > >> > >> I am. > >> > > So you also must be sure that ZF is yellow. > > > >> >> (the axiom schema of separation alone provides infinitely many axioms). > >> > Are you really sure? > >> > >> I am. > > > > But most of them are very dirty? > > Oh, they are all very similar to each other (but distinct non the less). "dirty" is a property which without the axiom of dirt is as well defined as "infinite" without the axiom of infinity. > > >> >> The point is: ZF without INF does not prohibit the existence of infinite > >> >> sets, nor does it force them to exist. > >> > > >> > It prohibits to speak of infinite sets and to recognize such sets. So > >> > one cannot be sure, but you are? > >> > >> Given any model of ZF-INF, one cannot be sure if there are infinite > >> sets. > > > > In particular the sets consisting of infinitely many rabbits would be > > of great interest. > > Learn about logic, model theory and set theory (preferably in that > order). You might be surprised. "Given any model of ZF-INF, one cannot be sure if there are infinite sets" is as silly an argument as talking about sets of rats or rabbits, or sets being clean or dirty. Regards, WM |