Cantor Confusion
in [Math]
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From: Lester Zick on 11 Feb 2007 18:07 On Sat, 10 Feb 2007 18:51:22 -0500, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: >Lester Zick wrote: >> On Sat, 10 Feb 2007 15:36:45 -0500, David Marcus >> <DavidMarcus(a)alumdotmit.edu> wrote: >> >> >Lester Zick wrote: >> >> On Sat, 10 Feb 2007 12:06:02 -0500, David Marcus >> >> <DavidMarcus(a)alumdotmit.edu> wrote: >> >> >> >> >mueckenh(a)rz.fh-augsburg.de wrote: >> >> >> On 9 Feb., 21:26, "MoeBlee" <jazzm...(a)hotmail.com> wrote: >> >> >> >> >> >> > > Reality. >> >> >> > >> >> >> > What reality? Empirically testable physical reality? A mathematical >> >> >> > statement is not of that kind. >> >> >> >> >> >> Every mathematical statement is of that kind. You would recognize this >> >> >> (if you could, but you couldn't) if all reality ceased to exist. >> >> > >> >> >Are all statements (not just mathematical ones) about reality? >> > >> >Something from the "Lord of the Rings" trilogy. >> >> Not that I'd necessarily disagree but I just wonder if you could be a >> little more specific? > >Take anything fictional. E.g., "Gandalf has a long beard". There is no >Gandalf in reality. Of course, Gandalf is a character in movies and >books, and the latter are real. But, the statement "In the movie, >Gandalf is portrayed as having a long beard" is different from "Gandalf >has a long beard". The latter might appear as dialog in the book. If it >did, I can't quite see how it is a statement about reality. How about if we just take "unicorn" instead or a "one sided triangle"? What makes a thing real is not so much whether it exists or could exist as its predicate combinations with which to assess the nature of its existence. And as long as it has predicate combinations it's real. At least in my book. If predicate combinations are self contradictory though I consider that makes it unreal as well and a limiting case. For example a "one sided triangle" would be such a combination. A "unicorn" though would problematic since we really don't know whether its predicate combinations are self contradictory or not. All we really know is they don't exist in combination as they stand. Fiction is a complex involved predicate combination which is why I asked for a more specific example. You can easily have a mixture of real and unreal predicate combinations which need to be assessed in terms of each other to decide the way in which they are real or not. So I think the answer to your original question is that all statements are about reality to the extent they represent predicate combinations. However the kind of reality can be further categorized according to the properties of predicate combination involved from "unreal" real characteristics to "mathematical" real characteristics to "fictional real and unreal" real characteristics and so on. But the statement itself is perforce about reality whatever kind of reality is entailed. ~v~~
From: Dik T. Winter on 11 Feb 2007 22:13 In article <1171205918.124082.214920(a)a75g2000cwd.googlegroups.com> mueckenh(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. > > 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. > > 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. > > > > > > 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. Can you, please, start your own form of mathematics? -- 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 11 Feb 2007 22:34 In article <1171206703.911869.175250(a)l53g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 11 Feb., 02:45, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > I ask you again, and I hope you finally will answer this question: > > How is it possible that P(0) U P(1) P(2) U ... contains an infinite > > element if none of the P(i) contains an infinite element? > > Where does that infinite element come from? Out of thin air? > > It comes from the same source as p(oo) is "established" by its finite > initial segments. It comes from the same meachnism which allows all > initial finite segments {1,2,3,...n} of N to create the infinite set > N. You are babbling. P(0) U P(1) U P(2) U ... contains all those elements that are in at least one of the P(i). That is the *definition* of such a union. So if there is in that union a particular element, it should be in at least one of the constituting sets that create the union. I do not talk about that union being infinite, I talk about one of the elements of that union being infinite. Pray explain that. 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. > I do not require more than set theory allows, but also not less. And apparently you do not understand it at all. In the above (1) is false and (2) is true. But you want to conclude (1) from (2). > > > Therefore every infinite path of the set P(oo) is established by the > > > union of the sets P(n): > > > With every member, the set P(oo) is established by P(0) U P(1) U P(2) > > > U ... > > > More is not required. > > > > I do not know what you mean with the term "established", but I *do* know > > that it is not a subset of the union, and that is what you need in your > > proof. > > No. What I need to "establish" (I do not know how this should work - > therefore I leave it to you to explain it) - what I need to > "establish" the path p(oo) and all its co-paths q(oo), r(oo), ... , > i.e., the whole set P(oo), is all finite paths belonging to p(oo) as > intial segments and to q(oo) and to r(oo) and so on. These finite > paths are in the union of the finite trees. And if you say the union > of the finite paths p(n) "establishes" p(oo), then the union of the > finite trees establishes p(oo) too. So there is only the choice to > have no infinite paths at all or to have countably many of them. QED. 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"? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 12 Feb 2007 06:51 On 11 Feb., 18:32, "Mike Kelly" <mk4...(a)bris.ac.uk> wrote: > On 11 Feb, 14:58, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > > On 11 Feb., 03:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1171114154.842945.262...(a)s48g2000cws.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > On 9 Feb., 14:53, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > In article <1171011850.731985.236...(a)a75g2000cwd.googlegroups.com> mueck.= > ....@rz.fh-augsburg.de writes: > > > > ... > > > > > > That is a matter of convention. Therefore the unary represantation, > > > > > > which is not a matter of convention, is preferable. > > > > > > Also unary representation is a matter of convention in my opinion. > > > > > But it is requiring the least possible convention. > > > > 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. > > > > 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. > > I say 111 is number 7. Now what? The binary meaning is a convention and does not refute or change the unary meaning of 111. It does not change the geographic position of New York if you describe it in polar coordinates or spacial coordinates and if you use this or that origin of the coordinate system. Regards, WM
From: mueckenh on 12 Feb 2007 07:01
On 11 Feb., 19:24, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > mueck...(a)rz.fh-augsburg.de wrote: > > On 10 Feb., 19:39, Franziska Neugebauer <Franziska- > > Neugeba...(a)neugeb.dnsalias.net> wrote: > >> mueck...(a)rz.fh-augsburg.de wrote: > >> > On 10 Feb., 12:13, Franziska Neugebauer <Franziska- > >> > Neugeba...(a)neugeb.dnsalias.net> wrote: > >> >> mueck...(a)rz.fh-augsburg.de wrote: > >> >> > On 9 Feb., 09:05, Virgil <vir...(a)comcast.net> wrote: > > >> >> >> > Instead of a path P consider the set S of nodes K which > >> >> >> > belong to a path P. Do your calculation and arguing. Then > >> >> >> > substitute P for S to have a brief notation. > > >> >> >> One cannot determine merely from a set of nodes for a given > >> >> >> tree whether that set of nodes is or is not a path. > > > >> >> > One can. > > >> >> Then please do so: > > >> >> given tree T := { a, b, c, d, e, f, g } > >> >> given set of nodes S := { a, b, c } > > >> >> Tell us whether S is a path in T. And please explain that. > > >> > Pardon, I overlooked your first question. > > >> > The tree T(2) in my notation can be given as a chain by > > >> > a > >> > bc > >> > gfed > > >> > abc is the subtree T(1) with only (the root node and) one level. It > >> > is not a path. > > >> Strange. My version of the tree T > > >> a > >> / \ > >> b d > >> / \ / \ > >> c e f g > > >> obviously contains a path having (the node set) S = { a, b, c }. > >> Hence the attempt to *uniquely* represent trees and/or paths by plain > >> sets of nodes fails. > > > Your version is incapable of enumerating all the nodes of the infinite > > tree as one single chain. > > Neither your definition of tree, path or the wording of your claim does > contain this constraint. > > > (As it is MY tree which appears in MY proof, please adhere to MY > > scheme, when discussion this proof.) > > Again: I presented I perfectly acceptabLe tree for which your claim: > > "One can determine from a set of nodes for a given > tree whether that set of nodes is or is not a path." > > does not hold. So your claim is not universially valid. 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. Please use the coordinate system which I have defined. I am not about to prove some essentials of the theory of binary trees and their notational details but I use a very special binary tree to show that set theory is self contradictive. Regards, WM |