Cantor Confusion
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From: Gregor H. on 1 Feb 2007 09:01 On 1 Feb 2007 03:49:09 -0800, mueckenh(a)rz.fh-augsburg.de wrote: > > Here is the definition of the set N > > 1 > 2 > 3 > ... > > Here is the definition of the sum of all elements of N > > 1 > 23 > 456 > ... > Hahahahaha... -- E-mail: info<at>simple-line<Punkt>de
From: Gregor H. on 1 Feb 2007 09:03 On 1 Feb 2007 04:17:25 -0800, mueckenh(a)rz.fh-augsburg.de wrote: >> >> A listing of all natural numbers is not the same as the summing of all >> natural numbers, and WM has, at most, produced a listing. >> > The sum is a listing: II + III = IIIII = 1,2,3,4,5. > > The sum of all natural numbers can be obtained (if all natural nunbers > exist) by the following "diaogonalization" > > 1 > 23 > 654 > ... > It is countable. > Hahahahahaha... :-))) -- E-mail: info<at>simple-line<Punkt>de
From: Dik T. Winter on 1 Feb 2007 09:15 In article <1170337444.507110.282460(a)k78g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 1 Feb., 13:48, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1170327430.425220.185...(a)a34g2000cwb.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > On 31 Jan., 03:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > The union of trees implies the union of outmost > > > > > left-hand side paths for example. This union is a path. > > > > > > > > Indeed, but it is *not* in the union of the sets of paths of finite trees. > > > > In the above example, that set does contain *all* the paths {0}, {0, 1}, > > > > {0, 1, 2}, {0, 1, 2, 3}, etc., but *not* the path {0, 1, 2, 3, ...}, > > > > because the last path is in *none* of the constituent sets. > > > > > > This is the same with the infinite tree. Why do you accept the > > > existence of the infinite tree? > > > > Why is it the same? The infinite tree *does* contain the path > > {0, 1, 2, 3, ...}. But that path is *not* in the union of the sets > > of finite paths. > > And that path is ot contained in the union of sets of finite trees. Indeed. But it is in the union of finite trees. There is a difference. > > > The union of all finite paths of type p(r) establishes the infinite > > > path r. > > > The union of all finite paths of all finite trees establishes all > > > infinite paths. > > > > Yes. I do not contradict *that*. The "set of the union of paths" is > > *not* the "union of the sets of paths". > > No, it is only a subset. That is false because the "set of union of paths" does contain infinite paths while the "union of the sets of paths" does *not* contain infinite paths. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 1 Feb 2007 09:18 In article <1170337797.588981.135470(a)q2g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 1 Feb., 14:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1170328870.612961.224...(a)v33g2000cwv.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: .... > > > > Well, it is certainly not finite, so it must be infinite (translation: > > > > "not finite"). > > > > > > Not finite is correct. But you translate it as actually existing set, > > > i.e., finished infinity. That is incorrect. > > > > Not finite is identical to infinite. But the axiom of infinity states that > > it exists. > > Not finite is identical to infinite. The axiom of infinity, however, > finishes infinity. Eh? > > > Because there is nothing else in the union but natural numbers. > > > You cannot get an apple if you collect nuts. > > > > The union of all finite natural numbers is not a natural number. Why > > do you think it is? > > The union of paths is a path. The projection of a finite number on a > pathlength remains a finite number. The projection of all finite > numbers on a pathlength remains a finite number. Yes, so what? > > > Nevertheless, if you assume that the length of the union of all paths > > > (not the cardinal number !) is infinite, then you advocate infinite > > > number sizes are required for an infinite union of numbers. > > > > That is quite confusing as stated. > > It is only unfamiliar to think about that. > > > But indeed, the infinite union of > > all natural numbers (considered as sets) is not a natural number. > > Maybe. But the natural numbers consideed as paths remain natural > numbers in any case. When you do so, yes. On the other hand their union is not a natural number but can also be considered a path. > > > On the other hand many people assert that there is an infinite union > > > of finite pathlengths without containing or establishing an infinite > > > pathlength. > > > > Eh? Who? > > Everybody who says that there is "an infinite set of finite > pathlengths". What is the difference between that and asserting that there is an infinite set of finite numbers? But in that statement I do not see a "union" expressed. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 1 Feb 2007 11:45
On 1 Feb., 14:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170330236.064219.62...(a)a75g2000cwd.googlegroups.com> mueck....(a)rz.fh-augsburg.de writes: > >  > If you have a set and you add something to this set (in unary >  > representation), then you have a super set of the former. The former >  > is a subset of the latter. > > This is not defining what I asked for.  What are the elements of the sets > you are talking about? The type of element is not restricted other than by the requirement that it be part of the real world. > >  > >  > > Bizarre.  So 3 is the set of all existing (what does that mean) sets >  > >  > > with 3 elements, >  > >  > >  > >  > The number 3 is the set of all sets of 3 elements (unless they were >  > >  > existing, they would not have 3 elements). But 3 is also here III. It >  > >  > is every set of 3 elements. >  > > >  > > You failed to respond to what followed.  If you *now* define it as every >  > > set of 3 elements, that is a circular definition.  "3 is any set of 3 >  > > elements" looks quite circular to me. >  > >  > It is circular. But please do not forget that I do not *define* a >  > natural number! I leave that to Peano etc. I look for its *existence*. > > So it was not a definition at all.  Note also that the Peano axioms to > *not* talk about representation. Therefore we have to investigate whether such a representation exists. Note that I use the definition of a number by Peano in order to look for the existence of that number. > >  > >  > > I do not ask what possible answers there are, I only ask how you would >  > >  > > like to propose it.  And, pray extend to 111111111. >  > >  > >  > >  > This is a set of 9 distinguishable elements, it is a set representing >  > >  > the number 9. And: It is the nunmber 9. Every set of 9 elements and >  > >  > avery set of sets of 9 elements is the number 9. >  > > >  > > So you are now going to equivalence classes of sets with the same number >  > > of elements?  Or what?  As you write it your definition is still circular. >  > >  > Of course it is. You must start at some point and eventually you will >  > come back. Circular reasoning cannot be avoided. But that is not a >  > problem when existence is concerned. The definition of a number is >  > given on p. 3 and p 130 of my book. > > A definition that uses circular reasoning is not a definition.  If a write: >   a foo is a foo The definition of a number is given as non-circular as possible: p. 3: 1) 1 ist eine natürliche Zahl. 2) Jede Zahl a in N hat einen bestimmten Nachfolger a' in N. 3) Es gibt keine Zahl mit dem Nachfolger 1. 4) Aus a' = b' folgt a = b. 5) Jede Menge M von natürlichen Zahlen, welche die Zahl 1 und zu jeder Zahl a ï M auch den Nachfolger a' enthält, enthält alle natürlichen Zahlen. p. 130: 1) 1 in M. 2) If n in M then n + 1 in M. 3) IN ist Durchschnitt aller Mengen M, die (1) und (2) erfüllen. To investigate the *existence* of number 3 we can use the definition of 3. Regards, WM |