Cantor Confusion
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From: Dik T. Winter on 15 Oct 2006 22:02 In article <1160858083.663838.195390(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1160675848.377420.163220(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > In article <1160646886.830639.308620(a)c28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > ... > > > > > If every digit position is well defined, then 0.111... is > > > > > covered "up to every position" by the list numbers, which > > > > > are simply the natural indizes. I claim that covering "up > > > > > to every" implies covering "every". > > > > > > > > Yes, you claim. Without proof. > > > > > > 1) "Covering up to n" > > > means > > > 2)"covering n" > > > and "covering the predecessors of n". > > > Therefore we need not prove (2) if (1) is true. > > > > Yes. But you claim: (3) "covering 0.111...", not covering n. > > Yes. But you claim 0.111... consists merey of finite n. That is the > error. Yes, I know, you detest the axiom of infinity. > "...classical logic was abstracted from the mathematics of finite sets > and their subsets...Forgetful of this limited origin, one afterwards > mistook that logic for something above and prior to all mathematics, > and finally applied it, without justification, to the mathematics of > infinite sets. This is the Fall and original sin of [Cantor's] set > theory ..." (Weyl) Oh, perhaps. What is the relevance to mathematics? > > > > You state it is true for each finite > > > > sequence, so it is also true for the infinite sequence. > > > > > > It is true *for every finite position*. I do not at all care how many > > > such positions there are. The obvious covering of (2) by (1) does not > > > depend on frequency. > > > > But were is the "covering up to 0.111..."? > > If there are actually infinitely many positions, then 0.111... is not > completely covered, hence not defined , Still unproven. > hence not existing, hence the > "if there are actually infinitely many positions" contradicts itself. Pray, first show a *valid* mathematical proof of your statement above. -- 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 15 Oct 2006 22:09 In article <1160857746.680029.319340(a)m7g2000cwm.googlegroups.com> Han.deBruijn(a)DTO.TUDelft.NL writes: > Virgil schreef: .... > > I do not object to the constraints of the mathematics of physics when > > doing physics, but why should I be so constrained when not doing physics? > > Because (empirical) physics is an absolute guarantee for consistency? Can you prove that? -- 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 15 Oct 2006 22:34 In article <1160932984.428777.57670(a)f16g2000cwb.googlegroups.com> Han.deBruijn(a)DTO.TUDelft.NL writes: > Dik T. Winter schreef: .... > > But pray state when you are arguing in anything other than standard > > mathematics. > > But, ah, Dik and others, you should _know_ that by now! After having > endured my sci.math postings since 1989. I've posted essentially the > same message over and over again for more than 15 years! "it would be good if physicists started to comprehend the basics of mathematics." is the best I can come up with today. A physicist only wants to have a method to calculate something. A mathematician wants to know *why* the particular method does work. And if the mathematician does not find a good reason, he will warn the physicist that it is possible that he method will fail. There are typical examples in iterations. Newton-Raphson will in general have quadratic convergence to the intended result. This is based on the ideal mathematical model. For a physist that is sufficient, but a mathematician will also consider under what conditions it may fail. And one of the reasons can be (and frequently is) reduced precision in the actual operations. Can you precisely state under what conditions, using finite precision mathematics (with truncated results), N-R will fail to converge to a particular floating-point number when calculating an approximation of the square root of a number? Or is that not interesting enough to you, so you use it without any knowledge about the workings of the algorithm? -- 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 15 Oct 2006 22:36 In article <1160933229.072292.316580(a)e3g2000cwe.googlegroups.com> Han.deBruijn(a)DTO.TUDelft.NL writes: > Dik T. Winter schreef: > > In article <1160856895.115824.134080(a)b28g2000cwb.googlegroups.com> > > Han.deBruijn(a)DTO.TUDelft.NL writes: .... > > > Come on, guys! You all know that, in the world of approximations, > > > 2 _is_ the square of a rational and the circle _is_ squared. > > > > I thought you were talking mathematics? > > I thought approximations were a part of mathematics? They are. Numerical mathematics in particular. But also in them, 2 is *not* the square of a rational number. The best you can state is that there is a rational number whose square approximates 2 with a certain precision. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Alan Morgan on 16 Oct 2006 02:01
In article <1160944143.122919.243860(a)h48g2000cwc.googlegroups.com>, <mueckenh(a)rz.fh-augsburg.de> wrote: > >William Hughes schrieb: > >> However, you wish to do more. You want to show >> that claiming "N does not have an upper bound and >> N exists as a complete set" leads to a contradiction.] >> >That is true too. And it is easy to see: If we define Lim [n-->oo] >{1,2,3,...,n} = N, then we can see it easily: > >For all n e N we have {2,4,6,...,2n} contains larger natural numbers >than |{2,4,6,...,2n}| = n. Agreed. >There is no larger natural number than aleph_0 = |{2,4,6,...}|. >Contradiction, because there are only natural numbers in {2,4,6,...}. That would be a contradiction only if Aleph0 e N, but it isn't. Your statement above is true for finite n. Showing that it isn't true for infinite n (or in the limit or whatever terminology you choose to use) does not produce a contradiction. Alan -- Defendit numerus |