Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Lester Zick on 7 Dec 2006 16:51 On Thu, 07 Dec 2006 10:43:31 -0800, Mark Nudelman <markn(a)greenwoodsoftware.com> wrote: >On 12/7/2006 3:15 AM, Eckard Blumschein wrote: >> Just try and refute: >> >> Reals according to DA2 are fictitious >> Reals according to DA2 are fictitious >> With fictitious I meant: They must not have a directly approachable >> numerical address. This was the basis for the 2nd DA by Cantor afer an >> idea by Emil du Bois-Raymond. >> >> Good luck >> > >Can you define what you mean by a "directly approachable numerical address"? The definition of "directly approachable numerical address" is "Q" since according to David Marcus definitions are "only abbreviations". ~v~~
From: Lester Zick on 7 Dec 2006 16:51 On 7 Dec 2006 11:40:57 -0800, "MoeBlee" <jazzmobe(a)hotmail.com> wrote: >mueckenh(a)rz.fh-augsburg.de wrote: >> Bob Kolker schrieb: >> > Also the set of natural numbers has a >> > cardinality greater than the cardinality of any set {2*1, 2*2 ..., 2*n} >> > for any integer n. Since the set of integers is equinumerous with the >> > set of even integers (by way of the mapping n<->2*n) >> >> They are not equinumerous by the way of mapping n <--> n. > >AGAIN you show that you don't understand even the basics of this >subject. > >It doesn't even make SENSE to say that two sets are not equinumerous by >a certain mapping. 'x is equinumerous with y' is DEFINED as 'there >exists a bijection between x and y'. But is that true, Moe? ~v~~
From: Lester Zick on 7 Dec 2006 16:58 On 7 Dec 2006 11:49:21 -0800, "MoeBlee" <jazzmobe(a)hotmail.com> wrote: >mueckenh(a)rz.fh-augsburg.de wrote: >> William Hughes schrieb: >> >> > mueckenh(a)rz.fh-augsburg.de wrote: >> > >> > Please clarify a point of nomenclature. >> > Do you consider a potentially infinite >> > set containing only finite elements >> > (e.g. the natural numbers) to be: >> > >> > 1. a set ? >> > 2. a finite set? >> >> It is neither an actually infinite set nor is it a set in the sense of >> set theory. >> I call it a set, because "set" is a handsome word. I call it an infinte >> set, because it is not a finite set. But if I talk to you about that, >> you cannot understand, because you can only think in the notions of set >> theory. > >No, you can't talk about it because you don't HAVE a theory. Do you? Last time I checked no one was particularly anxious to claim what you call a set "theory" was demonstrably exhaustive or particularly true. > If you had >a theory that were not set theory, then that in itself would not >prohibit people who understand set theory from also understanding your >theory. But you dont' have a theory, so the whole matter is nugatory >anyway. Then tell us metaMoe is set "theory" actually true or is it only like metaBob, metaBrian, and metayou really only metatrue? ~v~~
From: Lester Zick on 7 Dec 2006 17:01 On Thu, 07 Dec 2006 14:05:57 -0700, Virgil <virgil(a)comcast.net> wrote: >In article <1165495265.790747.266270(a)f1g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > >> Dik T. Winter schrieb: >> >> > In article <1165492756.322548.255040(a)j72g2000cwa.googlegroups.com> >> > mueckenh(a)rz.fh-augsburg.de writes: >> > > Dik T. Winter schrieb: >> > > > In article <1165421463.339178.48680(a)j44g2000cwa.googlegroups.com> >> > > > mueckenh(a)rz.fh-augsburg.de writes: >> > ... >> > > > The set in the quantifiers you are using is not finite either. The >> > > > quantifier are not over a single line, but over the set of natural >> > > > numbers. >> > > >> > > For finite natural numbers, we have finite lines only. It is not the >> > > question of a single line. Every line is finite. Therefore there is no >> > > line where quantifier reversal could not be applied. >> > >> > But the quantifier reversal is *not* applied to individual lines, it >> > is applied to the set of natural numbers. >> >> No. > >Yes! > >That WM does not know what is involved here does not make his ignorance >correct, nor excuse it. And what makes your ignorance correct and excuses it? ~v~~
From: cbrown on 7 Dec 2006 17:49
mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > So now you return to claiming that your mapping T /is/ a surjection > > from edges onto paths. Perhaps there is some confusion regarding "what > > is the domain/range of T?". > > > > To clarify, let e be any edge; say the first branch to the left in your > > original diagram. According to your above statement, e is in the domain > > of T. Which path is T(e)? > > You know that the two edges mapped on a path consist of shares of many > edges. Why do you put your question? Why should I name a special edge? Because your argument is of the form: "There exists a bijection f between the naturals and the edges. There exists a bijection h between the paths and the reals. There exists a surjection T between edges and paths. Therefore, the composition h o T o f is a surjection of the naturals onto the reals; contradiction." I assume you know what "a surjection T from edges onto paths" is, yes? It doesn't map 1/2 of an edge to 1/4 of a path! It is a function that maps each edge /in the original diagram/ to a path; so that every path is in the image of the function. If you claim to have constructed T, then you claim that you have constructed a function T such that for any edge e, say the first edge to the left, T(e) is a path. So how do you claim to have constructed T? How are we to determine T(e), in principle at least? If you do /not/ claim to have constructed T as a function from edges /in the original diagram/ to paths, then your argument poses no contradiction: there is no composition h o T o f, because the domain of T is not "edges /in the original diagram/", which is the range of f. > I have proved that two edges are collected in form of shares. > No, you have proven that the function g: (paths x edges x N) -> R is defined such that lim n->oo sum (over all edges) g(p, e, n) = 2 for every path p. The range of the function g is not "the set edges /in the original diagram/", nor is it "the set of full edges"; it is R. The real number "2" is not some specific edge /in the original diagram/; nor is it "2 specific edges /in the original diagram/". It is simply the real number, 2. You have yet to define what you mean by saying that, therefore, we can use g to biject "shares of edges /in the original diagram/" to "shares of full edges" to produce "full edges", which can then be bijected/injected to "edges /in the original diagram/". Until you have you defined these ideas and constructed these mappings, you have not proved that therefore there exists a surjection T (edges /in the original diagram/ -> paths); you at best have a function mapping something called "full edges" to paths. You need to actually /demonstrate/ that g can be used to /construct/ T as required; instead of merely asserting it by waving your hands and claiming that it is obvious. > > > I have proved it by rational relation and by a random mapping. > > > > You haven't proved it until you provide a surjective function T : > > (edges -> paths). (And I am only interested in your "rational relation" > > argument). > > It has been proved. See above. I do not see the necessity to do more > than to show that there are enough shares. Since you refuse to describe what you mean by "shares" in a mathematical sense, I can't assign any particular meaning to the assertion "there are enough shares". How do you define this phrase so that we can construct a surjection T (edges /in the original diagram/ -> paths)? > > > > > > > > > > > I add an appendix to one of my papers, where this is underlined I (here > > > > > the arguing is based on nodes instead of edges, but that doesn't matter > > > > 4> much): > > > > > > > > > > > > > Your appendix fails to address the key question: what is the domain of > > > > the function T? If e is an edge, what is the set of "shares of the > > > > divided edge e"? > > > > > > What is your problem? The complete set of shares of one edge is the > > > full edge. > > > > So you have a bijection S : (edges -> complete sets of shares of an > > edge). But given an edge e, what are the /elements/ of the set of > > shares, S(e)? How many elements does S(e) have? 1? 32? A countable > > number? An uncountable number? > > Everything in this tree is countable. Could you instead answer my questions regarding S(e)? What is the set of shares associated with edge e, where e is an edge in the original diagram? You claim it is a set; so one assumes it has members. Is "e/32" a member of this set of shares? Is "1/sqrt(2)" a member of this set of shares? Is "0" a member of S(e)? Cheers - Chas |