Cantor Confusion
in [Math]
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From: David Marcus on 10 Feb 2007 18:51 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. -- David Marcus
From: Dik T. Winter on 10 Feb 2007 20:45 In article <1171113679.295629.41750(a)a34g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 9 Feb., 14:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > So a statement like "every element is a set > > > > in ZFC" is false. > > > > > > It is used by most authors of text books on set theory. Talk to them > > > about their errors. I am no interested in what they use to denote the > > > elements (or sets) of their erroneous theory. > > > > But they give a proper foundation for that statement. > > So you see now that you were wrong? No, only that you were incomplete by not giving the foundation for it. > > You do not. You > > pull that statement out of thin air. > > I did not support it because to support all familiar knowledge hre > would lengthen these contributions overdue. That is irrelevant. You pull a statement out of thin air. > > Read what I wrote. P(i) (uppercase P) is a set of paths, p(i) (lowercase > > p) is a path. (This is from your notation from an earlier article.) I > > state: > > (1) p(0) U p(1) U p(2) U ... = p(oo) > > (2) P(0) U P(1) U P(2) U ... != P(oo) > > see the difference? The union of paths establishes the infinite path. > > The union of *sets* of paths does not establish the set of infinite paths. > > The union of sets of paths P(0) U P(1) U P(2) U ... contains all > finite the segments of the path p, Right. > p(0) U p(1) U p(2) U ... and, therefore, establishes p(oo). But *not* their unions. > This is the same for every infinite path p, q, r, .... of the tree. > P(0) U P(1) U P(2) U ... contains all finite the segments of the path > q, Right. > q(0) U q(1) U q(2) U ... and, therefore, establishes q(oo). But *not* their unions. 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? > And so on: every infinite path r(oo) , s(oo), t(oo), ... of the tree > is established. But now you state a new term. What do you *mean* with the term "established" here? > 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. -- 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 10 Feb 2007 21:06 In article <1171114154.842945.262850(a)s48g2000cws.googlegroups.com> mueckenh(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.= > .=2E(a)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? > III > oo > 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. The third contains three letters from the Latin script, but I do not see a suggestion of the number three at all. > > > I believe that without any other cultural contact two persons could > > > start communicating by unary numbers. > > > > Yes, your belief. > > Even my firm belief. Even when in a previous article I already stated that there is a culture where it would make no sense at all. > > But, in (1) you state "is a natural number", in (2) you use "is in N", > > without stating anywhere in that article that the two things mean the > > same. So the combination of those two is *not* a proper definition. > > In the articles presented here one cannot state every self-evident > convention. But when I ask you to define something, I espect a proper definition, not something that relies on conventions that are used when the definition is stated. > > > > We are going to define natural numbers: > > > > (1) 1 is a natural number > > > > (2) if a is a natural number, the successor of a is also a natural > > > > number > > > > These two together define the natural numbers. > > > > > > Not yet. According to your definition also -7 and pi could be natural > > > numbers. > > > > How do you come to that conclusion? At which step is either -7 or pi > > generated by (2)? > > It is not excluded. Why should something be excluded in a definition? > > Right. But not as you did: > > (1) 1 is a natural number > > (2) if a is in N, the successor of a is also in N. > > These two together define the natural numbers. > > By this definition only 1 is a natural number, because you can not > > even start with statement (2). > > I copied these statements from my book without the explanation which > is familiar to everyone discussing here. I know, but that was the reason I answered to it. You were asked for a definition, and you copied something from your book without context. BTW, it is my opinion that that page would improve if you did *not* use that convention. > > > > 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? > > What is a parabola? > > An invention by Archimedes. I am sure you know the definition in terms > of distances. Yes, I know. But not in terms of "measurable" distances. Neither in terms of lines that are "measurable distances" between points. -- 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 11 Feb 2007 09:58 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. > > > > > > 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. Regards, WM
From: mueckenh on 11 Feb 2007 10:11
On 11 Feb., 02:45, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > Read what I wrote. P(i) (uppercase P) is a set of paths, p(i) (lowercase > > > p) is a path. (This is from your notation from an earlier article.) I > > > state: > > > (1) p(0) U p(1) U p(2) U ... = p(oo) > > > (2) P(0) U P(1) U P(2) U ... != P(oo) > > > see the difference? The union of paths establishes the infinite path. > > > The union of *sets* of paths does not establish the set of infinite paths. > > > > The union of sets of paths P(0) U P(1) U P(2) U ... contains all > > finite the segments of the path p, > > Right. > > > p(0) U p(1) U p(2) U ... and, therefore, establishes p(oo). > > But *not* their unions. > > > This is the same for every infinite path p, q, r, .... of the tree. > > P(0) U P(1) U P(2) U ... contains all finite the segments of the path > > q, > > Right. > > > q(0) U q(1) U q(2) U ... and, therefore, establishes q(oo). > > But *not* their unions. > > 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. I do not require more than set theory allows, but also not less. > > > And so on: every infinite path r(oo) , s(oo), t(oo), ... of the tree > > is established. > > But now you state a new term. What do you *mean* with the term > "established" here? > > > 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. Regards, WM |