Cantor Confusion
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From: Dik T. Winter on 7 Dec 2006 19:54 In article <45781DC1.1000804(a)et.uni-magdeburg.de> Eckard Blumschein <blumschein(a)et.uni-magdeburg.de> writes: > On 12/7/2006 1:54 AM, Dik T. Winter wrote: .... > > Oh, well. In Bourbaki's mathematics R+ and R- both contain 0. So you > > are a follower of Bourbaki after all? But of course the 0's are the > > same. If they were different you would have quite a few problems with > > limits and continuity. > > According to my reasoning, any really real number is not unique but must > rather be void because even the tiniest interval is thought to contain > indefinitely not just many rational numbers but indefinitely much of > real numbers. But what you state here for real numbers is also valid for rational numbers. So you make not clear why in one case they are "void" (whatever that may mean) and in the other case they are not void. > Therefore, unreachable the very nil on top of the nested > intervals has not any significance at all. I can not parse this statement, so I have no idea what it means. > It cannot even be > distinguished from numbers 0- and 0+ left and right from it, > respectively, because the diffence is zero. What "It"? > So I agree with the > Bourbakis perhaps for the first time: 0+ and 0- are indiscriminable in > IR. However among the rationals, the nil is the first negative number > according to my reasoning and my old encyclopedia. Oh. Well in English nil is not a number, I think you mean zero. But pray quote your old encyclopedia where it is stated that it is the first negative number, and I think we can outline where you are wrong. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 7 Dec 2006 22:03 In article <MPG.1fe275887bc3ca5a9899fa(a)news.rcn.com>, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > Virgil wrote: > > WM should read some Korzybski to help him get his head straightened out. > > I fear that reading books will not help WM. Beating him over the head with them is not a viable option in NGs.
From: Virgil on 7 Dec 2006 22:05 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.
From: Virgil on 7 Dec 2006 22:12 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.
From: cbrown on 7 Dec 2006 23:29
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? 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}? We all know that the sequence of rational numbers (1, 2, 5/2, 8/3, ..., sum(0..n) 1/n!, ...) approaches the real number e. Could you please prove that lim 2^n = 2^N? :) Cheers - Chas |