Cantor Confusion
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From: mueckenh on 26 Jan 2007 05:07 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > On 25 Jan., 15:57, Franziska Neugebauer > > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > > WM ] > >> | The union of two natural numbers is defined to be the larger one. > >> | This is a set theoretic union. > >> `---- > >> > >> So the union of two natural numbers is _not_ defined to be the larger > >> one but the "larger" of two distint naturals (sets) a and b is > >> defined to be a iff > >> > >> b c a > >> > >> or b iff > >> > >> b c a. > >> > >> So the order _is_ relevant to successfully prove you wrong. > > > > The union of two different natural numbers is defined to be the larger > > one. > > Repeating a falsified claim. "Union" in contemporary set theory is > generally defined (not restricted to natural numbers). This definition > does by no means use a notion of "larger". > > Using e. g. von Neumann ordinals you can define the less than relation > by the subset-property. Not the other way round! Purset nonsense. Using the most basic property of numbers, namely their unary representability, the relation < is defined by c. Everything else is simulating this fact. Regards, WM
From: Franziska Neugebauer on 26 Jan 2007 05:11 mueckenh(a)rz.fh-augsburg.de wrote: > On 25 Jan., 15:57, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > >> > If T(oo) as defined by me is T as defined by me, concerning the set >> > of nodes and edges, > >>There is no room in equality for "concerning" this or that. Either to >> symbols refer to the same entity or they do not. > > That is not so obvious. That *is* obvious. F. N. -- xyz
From: Franziska Neugebauer on 26 Jan 2007 05:17 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > On 25 Jan., 15:57, Franziska Neugebauer >> > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: >> > WM ] >> >> | The union of two natural numbers is defined to be the larger >> >> | one. This is a set theoretic union. >> >> `---- >> >> >> >> So the union of two natural numbers is _not_ defined to be the >> >> larger one but the "larger" of two distint naturals (sets) a and b >> >> is defined to be a iff >> >> >> >> b c a >> >> >> >> or b iff >> >> >> >> b c a. >> >> >> >> So the order _is_ relevant to successfully prove you wrong. >> > >> > The union of two different natural numbers is defined to be the >> > larger one. >> >> Repeating a falsified claim. "Union" in contemporary set theory is >> generally defined (not restricted to natural numbers). This >> definition does by no means use a notion of "larger". >> >> Using e. g. von Neumann ordinals you can define the less than >> relation by the subset-property. Not the other way round! > > Purset nonsense. Using the most basic property of numbers, namely > their unary representability, the relation < is defined by c. You don't need naturals or representations at all to define '<' by 'c'. But since you admit that '<' is defined by 'c' and not the other way round I conclude that you refrain from your claim | The union of two natural numbers is defined to be the larger one. | This is a set theoretic union. F. N. -- xyz
From: mueckenh on 26 Jan 2007 05:19 On 26 Jan., 02:41, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1169740568.082762.116...(a)v33g2000cwv.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > On 25 Jan., 15:57, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: > WM ] > > > | The union of two natural numbers is defined to be the larger one. > > > | This is a set theoretic union. > ... > > The union of two different natural numbers is defined to be the larger > > one. > > In the von Neumann model. But that is a model about ordinal numbers, that > start at 0. There are a few subtle differences. I think you are a bit > confused here. In the von Neumann model we have that an ordinal number > 'k' is the set of all its predecessors ( {0, 1, 2, ..., k-1), if k is not > a limit ordinal). And so the union of two ordinals is the larger one. > In this case, the ordinal 'k', is also the ordinal number of the set of > predecessors. When you shift to '1' base, the latter statement is no > longer true. It has nothing to do with sophisticated models at all. They only have to simulate the basic fact. A natural number in its most basic representation, namely unary or unadic, simply is the superset of all smaller numbers and a subset of all larger numbers. That is the origin which has to be simulated by every correct theory of natural numbers. II is a subset of III. This kowledge as to be taken into account when denoting these numbers by 2 and 3. It has been done by the notation 2 < 3. Everything which Zermelo or v. Neuman or others had to say about this topic is not so new as you want to make me believe. It merely repeats the basics. Regards, WM Regards, WM
From: Dik T. Winter on 26 Jan 2007 07:02
In article <1169804998.026115.299910(a)q2g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 25 Jan., 17:03, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > > > and S is the set of corresponding sequences > > > > > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > > > > > then L cannot be larger than S, because there cannot be more > > > > > > limits than sequences. .... > > Yes? The elements of S as written above are the a_ik for i and k natural > > numbers. > > No. An element of a set S is written as x in S = {x| ...}. Yes, if there is a single x. You can also write: S = { x_i, y_i | i in N } meaning S contains as elements all x_i and y_i for i in N. Your notation is a novelty invented by you. > > So the elements are elements from {0, 1}. S *does* contain > > those paths, but not as elements, but as subsets. S is not the set of > > "corresponding sequences". > > The set of all cars contains every car as an element. But from the > elements of this set we can form the set of all brands. The set of all > brands contains 10 or 20 elements only. Yes, what is the relevance? > > You specifically wrote: > > > However, if L is a set of > > > limits L_k > > > L = { L_k | k in N } > > > and S is the set of corresponding sequences > > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > > then L cannot be larger than S, because there cannot be more limits > > > than sequences. > > this makes no sense if S is *not* a set of sequences. > > S is a set of sequences. You say so, inventing completely new notation. > > Consider the > > following: > > L = { L_k | k in {0, 1, 2}} > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in {0, 1, 2} } > > If S is a set the cardinality of S is at most 2. > > Cardinality is |S| = 3. Its elements are the sequences number k = 0, 1 > , and 2. Wrong. > > If S is an ordered > > *multi-set*, the ordinality is omega. The cardinality of L is 3. > > S is certainly not an ordered set because it contains multiple identical > > elements. > > S is ordered by the numbers k. Wrong. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |