Cantor Confusion
in [Math]
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From: Virgil on 12 Feb 2007 15:22 In article <1171283208.696664.25590(a)a75g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 12 Feb., 04:13, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1171205918.124082.214...(a)a75g2000cwd.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 11 Feb., 03:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > Well, I would concede that the above three things are representaions of the > > number three, using some convention. Anyhow, they are *not* the number > > three. > > What has "the number three" that is not expressed above? Universality. Just as no single man represents mankind, no single instance of a set with three objects represents all sets of three objects, at least without common consent.
From: Virgil on 12 Feb 2007 15:35 In article <1171284657.431713.275560(a)l53g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 12 Feb., 04:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1171206703.911869.175...(a)l53g2000cwa.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > You repeatadly misunderstand: > > (1) The union of sets of finite elements does contain an infinite element > > with > > (2) The union of finite sets is infinite. > > The first is false, the second is true. > > > > But you want to conclude (1) from (2). > > No. I do not conclude that p(oo) is in the union of P(0) U P(1) U P(2) > U ... and I do not conclude that p(oo) is in the union of {p(0)} U > {p(1)} U {p(2)} U .... > > What I conclude and what reasonably must be concluded is: > {p(0)} U {p(1)} U {p(2)} U .... c P(0) U P(1) U P(2) U ... If p(0) ={A} and p(1) = {B} then p(0) U p(1) = p(1) = {A,B} but {p(0))} U {p(1)} = {{A}, {B}}, which can only be the same if. e.g., A = {A}, which is impossible is any well formed set theory. > If p(oo) is established by {p(0)} U {p(1)} U {p(2)} U .... > Then p(oo) is also established by P(0) U P(1) U P(2) U ... So that what WM insists is "established" in his version of set theory, can only occur in such naive set theories as are self-contradictory. > > In my opinion the path p(oo) does not exist. As WM's opinions have so often been shown to be worthless, we need not worry about what he thinks.
From: Michael Press on 12 Feb 2007 19:10 In article <45d097fc$0$97230$892e7fe2(a)authen.yellow.readfreenews.n et>, Franziska Neugebauer <Franziska-Neugebauer(a)neugeb.dnsalias.net> wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > > On 12 Feb., 13:59, Franziska Neugebauer <Franziska- > > Neugeba...(a)neugeb.dnsalias.net> wrote: > >> mueck...(a)rz.fh-augsburg.de wrote: > >> > Set theory, however, says: If the union of all finite paths, > >> > 0., 0.0, 0.00, 0.000, ... , > >> > is in the tree, then the infinite path p(oo) = 0.000.... is in the > >> > tree too. > >> > >> No contemporary set theory or graph theory does posit such a > >> nonsensical claim. It is entirely *your* claim, better known as > > > > It is easy to recognize from the union of all finite segments of N. > > Set theory says: If the union of all finite segments, > > {1,2,3,...,n} is present, > > "Present"? Where? > > > then the infinite set N = {1,2,3,...} is present. > > Where? I don't know of any contemporary set theory which states that > entities are "present". Perhaps you borrowed that work from > M�ckenmatics. Hey, I saw that advertised last night on television. It slices, it dices; it schmooshes, it whooshes. It's the Muckematic. Turn all those carefully prepared ingredients into mush! Is it a union or is it composition? We don't care! Is it an offense unto thine eye? Is it shunned by all right-thinkers? Is it anathema? Muckematic is the answer. Only $79.95 plus shipping and handling. -- Michael Press
From: Dik T. Winter on 12 Feb 2007 20:02 In article <1171283208.696664.25590(a)a75g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 12 Feb., 04:13, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > > III > > > > > ooo > > > > > abc > > > > > all that is understandable as 3 without other convention than that > > > > > "all" is the start of this sentence and not another > > > > > representation of 3. > > > > Well, I would concede that the above three things are representaions of > > the number three, using some convention. Anyhow, they are *not* the > > number three. > > What has "the number three" that is not expressed above? The abstract number three is not a representation. > > > I missed one o. I corrected it above. Now it expresses all properties > > > that the number three can express. Therefore it is the number three > > > (and in addition it has the special form of o's). > > > > Numbers can express properties? You have lost me here. > > To have three elements is a property of a set. Oh. Your terminology is unfathomable. Indeed the number 3, when seen as a set, can have three elements. In the von Neumann model. However, I remember to also having seen another model, where the number three was {{{}}} (but I think now you can only model the natural numbers). Anyhow, within this model 3 when seen as a set has only one element. > > > > The third contains three letters from the Latin script, > > > > but I do not see a suggestion of the number three at all. > > > > > > Each of them expresses everything that the number 3 can express. Each > > > of them is number 3. > > > > Well, except that in a number of languages using the Latin script, "abc" > > are *not* the first three letters. And I have *no* idea what you are > > meaning with a statement that numbers can express something. > > If you think that numbers cannot expres anything, then you can use the > empty set for all of them. I have *no* idea what you are meaning with a statement that numbers can express something. I do not deny it, because I do not understand the meaning. > A number can express how many wheels your car has, for instance. Perhaps, but I would not use this strange terminology. > > > > > The set of points with a fixed distance from a given point. > > > > > > > > I think you mean "measurable" distance here? And how can measurable > > > > distances be fixed? > > > > > > By some physical means (body, wavelength, ...) and by nothing else. > > > > So it can not be fixed, but only approximated, and all mathematics should > > depend on the level of approximation you can obtain. > > Quantized properties like "having 50 students" or "containing 50 units > of length" can be fixed. So in your opinion the points that have a distance to a given point of, say, 205 +- 0.5 units is a circle? > > Can you, please, > > start your own form of mathematics? > > Nothing to start. It has been existing for 5000 years. Yes, and because nobody gave proper definitions for some parts, because they were unable to do so, lead to problems. Leibniz, for instance, in his talk about infinitesimals gave no proof of his formula's because he saw no way to define the infinitesimals. A proper definition could only be given with limits, at that time (later Anders Kock, c.s., has shown that there is at least one other possible way). However, limits have the "for all" requirement that was not well-founded, and founding that lead to set theory. Moreover, in a finitistic view we would have that Gau� conjecture that pi(x) < Li(x) is true. It is (in this finitistic view) impossible to calculate an x for which it is false. -- 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 12 Feb 2007 20:17
In article <1171284657.431713.275560(a)l53g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 12 Feb., 04:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > You repeatadly misunderstand: > > (1) The union of sets of finite elements does contain an infinite element > > with > > (2) The union of finite sets is infinite. > > The first is false, the second is true. > > > > But you want to conclude (1) from (2). > > No. I do not conclude that p(oo) is in the union of P(0) U P(1) U P(2) > U ... and I do not conclude that p(oo) is in the union of {p(0)} U > {p(1)} U {p(2)} U .... > > What I conclude and what reasonably must be concluded is: > {p(0)} U {p(1)} U {p(2)} U .... c P(0) U P(1) U P(2) U ... But, indeed, that is true. > If p(oo) is established by {p(0)} U {p(1)} U {p(2)} U .... > Then p(oo) is also established by P(0) U P(1) U P(2) U ... But again, I ask you what you mean with "established" here. It is neither an element of those unions, nor is it one of those unions. So what is your meaning? As I have stated again and again: p(oo) = p(0) U p(1) U p(2) U ... but that one is quite different from {p(0)} U {p(1)} U {p(2)} U ... > > Great. I state that I do not know what you mean with the term "establish", > > and you follow up with a paragraph having that word all over the place, > > and ask me to explain it. With all those occurrences of the word > > "establish" it makes no sense at all. And all that is followed by QED. > > What do you *mean* with "establish"? > > In my opinion the path p(oo) does not exist. Set theory, however, > says: If the union of all finite paths, > 0., 0.0, 0.00, 0.000, ... , > is in the tree, then the infinite path p(oo) = 0.000.... is in the > tree too. Yes, because the union *is* the infinite path p(oo). > I do not know how this is accoplished. The axiom of infinity, and from that some simple theorems. > Therefore I leave it to you and > I say only: "p(oo) is established" by the set of finite paths, i.e., > p(oo) belongs to the tree which contains all the finite paths. > And if this is correct, then it is also correct if we have the paths > 0., 0.0, 0.00, 0.000, ... , and some others like 0.10 or 0.111. Then > the binary tree contains also the path 0.000.... But I never denied that. But I now think I understand what you mean with that word "establish": A path is established if it is the union of a collection of a sequence of paths with the property that each path is a subset of its successor. Correct me if I am wrong, and if so, please state what you actually *do* mean with that word (not examples, a clear definition, please). Bit with this definition above, p(oo) is *not* established by: {p(0)} U {p(1)} U {p(2)} U ... because the paths are not united, it is the sets containing paths as a single element that are united. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |