Cantor Confusion
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From: Virgil on 8 Dec 2006 00:50 In article <1165552164.399063.145330(a)f1g2000cwa.googlegroups.com>, cbrown(a)cbrownsystems.com wrote: > Virgil wrote: > > In article <MPG.1fe27d2a5a4ae60a989a01(a)news.rcn.com>, > > David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > I do nothing. The tree cares that even in the limit the number of paths > > > > cannot become uncountable. 2^n remains the cardinal number of a > > > > countale set, even in the limit n --> oo. That's why I devised the > > > > tree! > > > > > > The number of paths in a tree of height n is 2^n. Are you saying that > > > the number of paths in the infinite tree is lim_{n -> oo} 2^n? > > > > Actually, if one interprets 2^n as the number of functions from > > {0,1,2,...,n-1} to {0,1} in NBG, or. equivalently. as the number of > > functions from {1,2,3,...,n} to {1,2}, then 2^n -> 2^N as n increases > > without limit and 2^N, interpreted that same way, is an uncountable set. > > What do you mean by "the number" of functions? I mean the cardinality of the set of all such functions, of course. > In what sense does the > sequence of natural numbers (1,2,4,8, ..., 2^n, ...), approach "the > number" of the set of functions N->{0,1}? For each finite n in NBG, n = {0,1,2,...,n-1}, so the set 2^n consists of all functions from n to {0,1} = 2, and is of cardinality equal to the number of such functions. So as n --> N, 2^n --> 2^N.
From: Eckard Blumschein on 8 Dec 2006 01:27 On 12/8/2006 4:05 AM, Virgil wrote: > In article <MPG.1fe27757af15f1f89899fd(a)news.rcn.com>, > David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > >> Eckard Blumschein wrote: >> > On 12/7/2006 4:38 AM, Virgil wrote: >> > > In article <4576DF19.7070005(a)et.uni-magdeburg.de>, >> > > Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> wrote: >> > >> > >> The abstract concept of numbers >> > >> must not be misused as to declare rationals and embeded rationals >> > >> likewise existent. >> > > >> > > The "abstract concept of number" can be used in any way that >> > > mathematicians choose to use it, >> > >> > If there was really general agreement among mathematicians, then there >> > would be an acceptable printed definition. Since Cantor's definition of >> > set has been declared untennable without substitute, I do not expect a >> > clean definition of number either. >> >> What definition of set is this and why is it untenable? > > And who declared it so? If it was TO or WM or EB, forget it. Cantor's pandemic set "theory" originated from Germany. Therefore I recommend to the doctors who intend to cure mathematics: Learn German. Then you can read and understand Cantor's own papers and also what Adolf Fraenkel wrote: Einleitung in die Mengenlehre, Berlin: Springer 1923, p. 3: "Cantor hat den Begriff der Menge folgenderma�en definiert: Eine Menge ist eine Zusammenfassung bestimmter wohlunterschiedener Objekte unserer Anschauung oder unseres Denkens - welche die Elemente der Menge genannt werden - zu einem Ganzen." p. 185: " Von einer Definition des Begriffs 'Menge' unmd der Bezeichnung 'm ist ein Element der Menge M' wird �berhaupt abgesehen; die durch die Paradoxien als unhaltbar erwiesene Definition Cantors (S. 3 und 10f.) - d. h. letzten Endes das Verfahren, einem beliebigen logischen Begriff eine Menge zuzuordnen - wird also aufgegeben und keine andere Mengendefinition an ihre Stelle gesetzt."
From: cbrown on 8 Dec 2006 02:52 Virgil wrote: > In article <1165552164.399063.145330(a)f1g2000cwa.googlegroups.com>, > cbrown(a)cbrownsystems.com wrote: > > > Virgil wrote: > > > In article <MPG.1fe27d2a5a4ae60a989a01(a)news.rcn.com>, > > > David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > I do nothing. The tree cares that even in the limit the number of paths > > > > > cannot become uncountable. 2^n remains the cardinal number of a > > > > > countale set, even in the limit n --> oo. That's why I devised the > > > > > tree! > > > > > > > > The number of paths in a tree of height n is 2^n. Are you saying that > > > > the number of paths in the infinite tree is lim_{n -> oo} 2^n? > > > > > > Actually, if one interprets 2^n as the number of functions from > > > {0,1,2,...,n-1} to {0,1} in NBG, or. equivalently. as the number of > > > functions from {1,2,3,...,n} to {1,2}, then 2^n -> 2^N as n increases > > > without limit and 2^N, interpreted that same way, is an uncountable set. > > > > What do you mean by "the number" of functions? > > I mean the cardinality of the set of all such functions, of course. > > > > In what sense does the > > sequence of natural numbers (1,2,4,8, ..., 2^n, ...), approach "the > > number" of the set of functions N->{0,1}? > > For each finite n in NBG, n = {0,1,2,...,n-1}, so the set 2^n consists > of all functions from n to {0,1} = 2, and is of cardinality equal to the > number of such functions. > > So as n --> N, 2^n --> 2^N. Ah! So, because (1,2,4, ..., 2^n, ...) is a subsequence of (1,2,3, ..., n, ...), then as n->N, n->2^N. Cheers - Chas
From: Han de Bruijn on 8 Dec 2006 03:06 Franziska Neugebauer wrote: > Han de Bruijn wrote: > >>Franziska Neugebauer wrote: >> >>>I don't know any mathematician who "pretends" an ability of >>>"knowing" (what does this mean?) every integer. Who "pretends" so? >> >>Yeah, sure .. Everybody on earth knows what "knowing" means in common >>conversation, except mathematicians. > > WM: "Those who pretend to be able of knowing every integer [...]" > are those who "overestimate their capabilities so grossly." > > The questions is still unanswered: > > 1. Who pretends to "know" every integer? If you know "the set of all integers" as a completed entity, does that also mean that you "know" each and every integer in that set? Do _you_ know the set of all integers? And if you don't, how can it nevertheless "exist"? > 2. Does "to know" mean the same as in the sentence "I don't know every > person"? Han de Bruijn
From: Han de Bruijn on 8 Dec 2006 03:08
Bob Kolker wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > >> You cannot imagine the integer [pi*10^10^100]. > > That is not an integer, dummkopf. It is an irrational real number. Dummkopf? Who? Doesn't "[ .. ]" stand for the "floor" function? Han de Bruijn |