Cantor Confusion
in [Math]
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From: mueckenh on 12 Feb 2007 07:04 On 11 Feb., 20:15, Virgil <vir...(a)comcast.net> 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. > > > > Without convention the first looks like a number of bars, the second like > > > a numbers of o's and the third like a sequence of letters from the Latin > > > script. The first might suggest the number three, but it is not the > > > number three. The second might suggest the number two, but it is not > > > the number two. > > > 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). > > > > 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. > > Then, according to WM, three I's and three o's and three lower case > roman letters are all exactly the same thing. Aren't |{a,b,c}| and |{1,2,3}| and |{Jumbo,Mumbo,Cumbo}| the same thing? Regards, WM
From: mueckenh on 12 Feb 2007 07:26 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: > ... > > > Eh? You first state that it is not a matter of convention and now state > > > that it *is* a matter of convention? Are you not contradicting yourself? > > > > It depends on how you define "matter of convention". It depends on how > > well you distinguish between the *numbers 3* given below and the first > > word "all" of the verbal explanation. > > > > > > > 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? > > > > Without convention the first looks like a number of bars, the second like > > > a numbers of o's and the third like a sequence of letters from the Latin > > > script. The first might suggest the number three, but it is not the > > > number three. The second might suggest the number two, but it is not > > > the number two. > > > > 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. > > > > 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. A number can express how many wheels your car has, for instance. > > > > > > > > Such lines also do not have physical existence. Each and every > > > > > > > physical line has a width smaller than anything measured yet. > > > > > > > And I do not think that physical lines are really straight either. > > > > > > > > > > > > The physical existence of a line is "a measurable distance" between > > > > > > two points. The points exists as sets of coordinates. > > > > > > > > > > Oh. What is a circle? > > > > > > > > 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. > Can you, please, > start your own form of mathematics? Nothing to start. It has been existing for 5000 years. Regards, WM
From: mueckenh on 12 Feb 2007 07:50 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(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 ... > > 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. I do not know how this is accoplished. 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.... Regards, WM
From: Franziska Neugebauer on 12 Feb 2007 07:59 mueckenh(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 X is not finite -> there must be an x in X which is infinite. > I do not know how this is accoplished. By the fallacy of composition? F. N. -- xyz
From: mueckenh on 12 Feb 2007 08:13
On 12 Feb., 13:53, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > > Your arguing resembles the following: A straight line in the Cartesian > > coordinate system cannot be described by the set of points {x,0,0} > > with x in R because in a rotated system it is {x,x,x} with x in R. > > I don't use any "coordinate system" at all. Enumerating (of the nodes in form of a chain) is using a coordinate system with only one coordinate. Didn't you learn that in one of your first lectures on mathematics? > > > Please use the coordinate system which I have defined. > > Reducing a tree or of a path to a simple set of nodes is accompanied > with a loss of information, namely the loss of the information of the > connectedness of the nodes. The connectedness is given by the stucture of the tree. The connectedness of elements of a matrix is given by the structure of the matrix. It is independent of the method of indexing like 11, 12, 13 21, 22, 23 or a,b,c d,e,f or a,c,e b,d,f. In principle we could also use your system, alas it would fail for the infinite path 0.000.... Therefore my proof in your system is as impossibe as the proof of 2 < 3 when proceeding via 2*0 and 3*0. Intentionally? > Hence your "coordinate system" does not > allow binary trees to be "located" at a unique position. In other > words: A set of nodes is not sufficient to uniquely *determine* trees > or paths. No, but it is sufficient to determine uniquely all paths of my tree. See the simple example above. > > As I have shown in my example solely from the set of nodes of a tree and > a set of nodes of a path you cannot generally decide whether these sets > have been originally derived from a tree which contains that path or > does not contain it. Given my fixed cordinate system, there remains no ambiguity at all. Really no longer valid arguments available? Regards, WM |