An uncountable countable set
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From: mueckenh on 14 Jul 2006 09:03 Randy Poe schrieb: > > In "usual" mathematics we have 10 * 0.999... = 9.999... and from that > > we get easily 0.999... = 1. > > > > I argue that 0.111... does not exist. > > If by "usual mathematics" you mean a mathematics in which the > meaning of infinite decimals have not been defined, then we > don't have 0.999... or 9.999... either. In usual mathematics 0.999... is assumed to exist. > > When you say 0.111... does not exist, do you mean > > (a) the notation is undefined, or > (b) when you define it, you get a number that doesn't exist? > You get a representation that does not exist of a number that des exist as 1/9. > If (a), then that is fixed by defining it, which can be done with > limits. It cannot. You see it by the following argument. Define 0 *+ 0 = 0, 0 *+ 1 = 1 *+ 0 = 1 *+ 1 = 1. Do the following addition *+ 0.1 0.11 0.111 .... What is the result? 0.111... This number is not among the numbers to be added. But it must be, if it is different from any list number and if the result of the addition is 0.111... . Its absence does not matter as its presence does not matter. Add by *+ 0,1 0.11 0.111 .... 0.111... The result is the same. Hence, if 0.111... exists, it is in he list and is not in the list. Regards, WM
From: mueckenh on 14 Jul 2006 09:07 David Hartley schrieb: > In message <1152785777.506058.199660(a)75g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de writes > > > >David Hartley schrieb: > > > >> The transpositions are relevant in so far as WM claims that Cantor wrote > >> that a sequence of transpositions can not change the order-type. Either > >> Cantor was wrong (or using a different idea of the limit), or WM is > >> misinterpreting him, my German is not good enough to tell. > > > >Daraus folgt, daß solche Umformungen einer wohlgeordneten Menge die > >Anzahl derselben ungeändert lassen, welche sich auf eine endliche oder > >unendliche Folge von Transpositionen von je zwei Elementen > >zurückführen lassen, ... > > > >It follows that only such transformations of a well-ordered set leave > >its (ordinal) number unchanged, which can be derived from a finite or > >infinite sequence of transpositions of each two elements, ... > > "of each two elements". That is a strange expression in this context. That is due to my bad translation. Cantor stresses that each transposition involves two elements. (Also the "only" above should have been omitted.) > ",each of two elements" would make more sense, although redundant unless > he hadn't previously defined "transposition". If that is the case, it > would seem Cantor was wrong, you should have no trouble believing that. > Cantor was very often redundant. > >> In any case, > >> it's a proposition WM needs to prove to complete his argument, and he > >> hasn't done so. > > > >What should that be good for? > > You would have finally proven a contradiction within standard set > theory. You've been trying to do that for many years, why give up now, > when you're so close? All you need to do is prove that, in this > quotation, Cantor was right! Listen to Virgil. If the transpositions would lead to a dense order, one of them must be the first. But no transposition can be the first. It is the same as with the Leibniz-series: No fraction finishes it. Always we have a rational number. There is no limit. pi/4 does not exist. > > > > >In the binary tree everyone can see that no path can split without two > >additional edges. So it is by no means possible that there were more > >paths than edges. Nevertheless the set theorists go through eyes wide > >shut and do not care. > > > >Who would accept logic arguments if set theory was concerned? > > Most of us, if you would just provide one. No path in the binary tree can split without two more edges - even in the infinite (which always remains finite, by the way). That should be sufficient. Regards, WM
From: mueckenh on 14 Jul 2006 09:10 Virgil schrieb: > > It is done in zero time. nonsense. >> > Cantor does not need any "limit" process because his rule does a > "parallel processing" for all naturals at one go. Read his paper, most probably available in English: Über unendliche lineare Punktmannigfaltigkeiten: [Math. Annalen Bd. 15, S. 1 - 7 (1879); Bd. 17, S. 355 - 358 (1880); Bd. 20, S. 113 - 121 (1882); Bd. 21, S. 51 - 58 u. S. 545 - 586 (1883); Bd. 23, S. 453 - 488 (1884).] in particular § 19 on ordering of point sets. , "und es erfährt daher der aus unsrer Regel resultierende Zuordnungsprozeß keinen Stillstand." Our mapping process never stands still. There is no parallel processing. Regards, WM
From: mueckenh on 14 Jul 2006 09:16 Virgil schrieb: > > In the binary tree everyone can see that no path can split without two > > additional edges. So it is by no means possible that there were more > > paths than edges. > > As long as paths all end. And if thy do not end, they run without edges? Flying? Regards, WM
From: mueckenh on 14 Jul 2006 09:17
Virgil schrieb: > > The set of naturals shold be the same set as the set of indices of > > 0.111... . But it cannot. > > Maybe not in "mueckenh"'s world but it works fine in mine. It cannot. You see it by the following argument. Define 0 *+ 0 = 0, 0 *+ 1 = 1 *+ 0 = 1 *+ 1 = 1. Do the following addition *+ 0.1 0.11 0.111 .... What is the result? 0.111... This number is not among the numbers to be added. But it must be, if it is different from any list number and if the result of the addition is 0.111... . Its absence does not matter as its presence does not matter. Add by *+ 0,1 0.11 0.111 .... 0.111... The result is the same. Hence, if 0.111... exists, it is in he list and is not in the list. Regards, WM |