Cantor Confusion
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From: Lester Zick on 19 Dec 2006 17:22 On Tue, 19 Dec 2006 13:45:12 -0700, Virgil <virgil(a)comcast.net> wrote: >In article <1166526723.021088.294160(a)n67g2000cwd.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > >> >> The Earth is round. That can be proved and has been proved - contrary >> to the consistency of set theory. > >Whether the Earth is "round" depends on what one means by round. It is >certainly not perfectly spherical. Well that dependes on what you mean by "spherical" now doesn't it. It also depends on what you mean by "perfectly" now doesn't it. >The earth has been proved not flat. That depends on what you mean by "not flat" doesn't it. It also depends on what you mean by the "earth" doesn't it. >ZFC has not been proved inconsistent. That depends on what you mean by "not proved inconsistent" doesn't it. It also depends on what you mean by "ZFC" doesn't it. ~v~~
From: cbrown on 19 Dec 2006 19:25 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > cbrown(a)cbrownsystems.com schrieb: > > > > > > > > > > > > > > Your edges are not > > > > > > being broken into 2 "shares" each; they are being divided into an > > > > > > infinite number of shares, and that is very different from 2 shares! > > > > > > > > > > For this sake we have mathematics, I mean really correct mathematics, > > > > > not based on dubious infinite sets, which states that 1 + 1/2 + 1/4 + > > > > > ... = 2. > > > > > > > > > > > > > "Really correct mathematics" doesn't simply /state/ that "1 + 1/2 + 1/4 > > > > + ... = 2". Mathematics /defines/ what the series of symbols "1 + 1/2 + > > > > 1/4 + ... = 2" means, and then /proves/ that it is true from these > > > > definitions. > > > > > > > That is the nonsense en vogue today. > > > > So you consider it nonsense that mathematics should consist of clear > > definitions and proofs of theorems. > > No. But I need no proof in order to see that > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > Yes, it appears that you need no proof to make any of your assertions. You simply "see" them; or "feel" them. That's not what I would call "really correct mathematics". Cheers - Chas
From: Dik T. Winter on 19 Dec 2006 21:29 In article <1166526654.547899.146160(a)a3g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > I do not think that link makes much sense for somebody who does not > > > understand the German sentence above. However, I do understand why > > > you post that link. How much did publication cost you? My estimate > > > is about EUR 1000. > > > > For those who wonder. Shaker Verlag is nothing more than a printing > > company. It was started to aid universities to publish their books in > > a nice format and to publish thesises. In addition to the printing it > > does something about ISBN's, some publishing and so on. But they appear > > to be going into the field of vanity press. Whatever you wish to publish, > > they will do it, if you pay what they ask you. If you have a manuscript > > of about 150 pages, they will publish with a print run of 200, for about > > EUR 1000. > > Do you report own painful experience, Dik? No. I can recommend Shaker > warmly. There is *no* fee. Of course, they would be mad if they > published on their own risk a book in German on a topic which is not > taught at any German school or university except my own. But they seem > to work very economically. All they require from the author is to buy > 30 copies of the book on half the selling price. Right. That is the difference between scientific press and vanity press. In the former there is review and editing assistence, in the latter you can publish anything you want to publish. In the former you do not pay anything, in the latter you pay so that the publisher is able to cover his cost. > You may calculate that > your estimation was far too high. My estimate was based on what they stated for a run of 200 copies. But with the selling price of your book and the required buy, I now estimate about EUR 500. Perhaps a good idea for James Harris. > (My costs were somewhat higher > because I required hard cover and additional dust cover which do not > belong to the regular equipment. But I thought that necessary in order > to prevent angry set theorists from tearing it into pieces too easily. > A full metal jacket to prevent burning was too expensive, however.) I think there will not be much sale, except amongst your students for which it will be a required read. (Is that the case?) > > No review, no editing > > and no print allowance by the celebrities and high priests of the > mathelogical church is required. That is correct and very important. So > I am in a better position than, e.g., Galilei was. Oh, well. I have seen such books that described the squaring of a circle and the trisection of an angle. In our library, only for curiosity value. But, indeed, if you pay for it, anything can be published. But that was already true in Galilei's time. > > Oh, and as a nicety, the author will receive parts of the > > sale cost for each book outside the initial print run. It is 10%, but > > I do not know whether that is before taxes or after. > > I would be glad to give these 10 % to any customer who really > understands my ideas. Let me do some calculations. You are required to take 30 books of the initial run for half the price of EUR 28.50. That means 30 times 14.25 or EUR 427.50. My estimate is that the initial run is exactly 30 copies and further printing is only on demand. And so EUR 427.50 is more than their cost to produce 30 copies. It is clear that on futher sales they take in a whole lot of money... But whatever, send me a copy of the book if you want a review in this newsgroup. I promise to donate it to our library when finished. But it still makes me wonder why you are referring to your book in a newsgroup where only a small portion of the readers is able to read German. -- 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 19 Dec 2006 21:31 In article <1166526935.758264.263940(a)48g2000cwx.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1166452045.777425.20310(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > ... > > > > >> > > > All elements that can be shown to exist in the diagonal can be > > > > >> > > > shown to exist in one single line. [(P1)] > > > > > > > > This proposition P1 has _not_ yet been proved (shown). > > ... > > > > Upto now we have no line L_D at all since P1 has not been proved. > > > > > > > There is nothing to prove in a definition. I remember you said a > > > definition cannot be wrong. So it cannot be proved. > > > > P1 is not a definition. > > It is the definition that, in the Equilateral Infinite Triangle, the > diagonal consists of the ends of lines a_nn. A line has elements a_kn > with k =< n and n eps N --- with ALL n eps N. So P1 is not a definition. It is a theorem that is not yet proven. -- 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 19 Dec 2006 21:45
In article <1166527114.881108.181070(a)80g2000cwy.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1166453622.108599.302270(a)j72g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > ... > > > > And what if m is infinite? How do you divide? How do you recombine? > > > > Suppose I have an edge that is divided in countably many shares. Now > > > > suppose I select the even numbered shares and recombine them. Do I > > > > now have an edge? And what when I do the same with the odd numbered > > > > shares, do I now also have an edge? If that is not the case, why not? > > > > And if that is the case, please explain. > > > > > > We need not divide anything by infinite numbers, because there are only > > > finite numbers n in the sum > > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... > > > Every level of the tree has a finite number n. For finite n also 2^n is > > > finite. > > > > I do not understand the connection. Through every edge there go infinitely > > many paths. So to give each path a share of the edge we have to divide the > > edge in infinitely many shares. > > We will find out what is infinite. In any case each level has a finite > number and the sum of edges inherited by one complete path and by no > other path is given by: 1 + 1/2 + 1/4 + ...+ 1/2^n + ... = 2. We were talking about shares of edges. Anyhow, the latter statement is formally false in mathematics. > For the evaluation of this *sum* we need *no limit* but only the > computation > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. We were talking about *shares*. In how many shares is the edge going left from the root divided, and how do you do that division? Pray keep to the subject. > > But each edge is divided in infinitely many shares. > > It is irrelevant what is infinity here. The sum of edges inherited by > one complete path and by no other path is given by 1 + 1/2 + 1/4 + ...+ > 1/2^n + ... >= 2 - 1/2^n. You are not talking about *shares*. This makes no sense at all in establishing a surjection from the edges to the paths. > > > Of course. There is only the finite case for natural numbers. There are > > > no infinite numbers in N. > > > > I do not state so. But through each edge go infinitely many paths. > > > May be we will find out later on how many paths go through each edge. > First let's try to count > 1 + 1/2 + 1/4 + ...+ 1/2^n which is the same as to calculate 2 - 1/2^n. Makes no sense. What are you counting here? > > > > there exists an Hamel basis for the reals as a vector space over > > > > Q. And using that basis it is reasonably simple to well order the > > > > reals. > > > > > > Why does such an ordering definitely *not* exist? No book in any > > > library of the world (incuding all extraterrestrical cultures) contains > > > it, neither a formula, nor a recursion and, obviously, not a list. > > > > Obviously you need AC to make it existing. So only in that version of > > mathematics were AC holds does it exist. > > So let it hold and make it existing. *If* it holds it is existing. > > > > In geometry the > > > > parallel axiom tells us there is only single line. Abandoning that > > > > axiom gives us different kinds of geometry. It is similar with AC. > > > > If we accept it we get a different kind of set theory (where every set > > > > can be well-ordered) from a version where we do not accept AC. .... > I cannot do it (and it has been proved that nobody can do it because > there is no *definable* well-ordering). So I would prefer you > demonstrate it. That is similar to asking me to prove the parallel axiom without the parallel axiom as basis. But I explicitly stated how you could construct a well-ordering when AC is true. You can not do it in general, and nobody can, because AC is not necessarily true. If somebody came up with a construced well-ordering of the reals that would imply that AC is true. In a similar way, if somebody could construct a line parallel to another and show that there are no other parallel lines, that would prove that the parallel axiom is true. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |