An uncountable countable set
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From: mueckenh on 5 Jul 2006 07:31 imaginatorium(a)despammed.com schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > imaginatorium(a)despammed.com schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > <snip> > > > > > > > No, it does not. P(N) cannot contain K by definition, because K does > > > > not exist. You can see this by the following argument: Map |R > > > > (including |N) on P(|N) with the only condition that a natural number > > > > has to be mapped on that set K e P(|N) which contains all natural > > > > numbers which are not mapped on sets containing them. > > > > > > Perhaps I missed it, but can someone explain what |N is and how (if at > > > all) it differs from N (which I assume means the set of naturals)? > > > > It is the same. In Germany it is usual to denote N by |N, R by |R. > > Aha! Thanks.. (I suppose you mean "In German", though...) Sorry, I don't believe that German language has an official definition of these symbols. It is only a habit of mathematicians in Germany (and other European countries). Regards, WM
From: mueckenh on 5 Jul 2006 07:35 Dik T. Winter schrieb: > In article <1151844660.650801.186480(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Daryl McCullough schrieb: > ... > > >They aren't *your* > > > principles, but why should anyone besides you care about that? > > > > Your principle is that the complete set N including the largest element > > does exist. > > This is a complete misrepresentation. The complete set N does exist but > has no largest element. How you conclude that the thinking is that there > is a largest element escapes me. The existance of the complete set N is > an axiom. An equivalent axiom (to actual infinity) would be the existence of 1000 natural numbers with two decimal digits. > > > If, however, the largest element does not exist or is not included in > > the mapping, then there is no complete set N and, hence, no complete > > set P(N) to be mapped upon, and there is definitely no completed set K > > of non-generators which could used to prove something. > > This is not mathematics, at most matheology. The complete set N exists, > due to an axiom, but it does not have a largest element (and that is > easily proven). It is easily proven that the largest is missing. The axiom does not state that all are there, but only that n+1 is there if n is. The assumption that all elements are existing and available for bijections is a gross misinterpretation. > > It may be that you do not like that axiom, if so, why do you not set up > your own mathematics without that axiom? This is similar to the > parallel postulate. You can do geometry with it or with its negation. > In the same way you can do set theory with the axiom of infinity or > without it. Start some new set theory, and see where that leads you. The axiom is not the problem. It is its actual misinterpreation. > > Pray define an f for which K(f) is not defined, and state *where* the > definition fails. For any f it fails in f : X --> K, where X is any non-empty set and K = { x in X | x is not an element of f(x) } For any surjective f it fails in f : X --> (Y U {K}), where X is any non-empty set, Y is any set, and K = { x in X | x is not an element of f(x) } Regards, WM
From: mueckenh on 5 Jul 2006 07:37 Dik T. Winter schrieb: > In article <$UHhNrCdjAqEFwc5(a)212648.invalid> David Hartley <me9(a)privacy.net> writes: > ... > > Indeed, I agree that it is possible to apply a sequence of > > transpositions and change a well-ordering of the rationals to the usual > > ordering. (I posted my own example last night.) However, I take this to > > imply simply that such transformations do not preserve well-ordering, > > not that there is a contradiction in standard set theory. > > Indeed. The simplest example (using indices rather than numbers) is the > sequence of transpositions on the natural numbers in there standard order > is the sequence > for n -> oo (0, 1)(1, 2)(2, 3)(3, 4)...(n, n+1) > Applying this will lead (when we properly define what "-> oo" means) to It means that all natural numbers, which according to set theory do exist, are included as indices. Not more and not less. In particular there is no number oo. > the ordered set: > (1, 2, ..., 0) > applying it again will lead to > (2, 3, ..., 1, 0) > and applying it again and again will lead to > (..., 3, 2, 1, 0) > So the first sequence will lead to a well-ordered set with a different > ordinal number (contradicting what Cantor stated according to the quotes > supplied). Applying an infinite number of such sequences will destroy > well-orderedness. It will (and can) not destroy the one-to-one correspondence between q's and n's. > > Note also that I do not see how to apply transpositions using indices > when you start with a set that is not well-ordered. You can only do it > with a well-ordered set and they operate only on an initial segment of > order type w or smaller. And that is completely sufficient for the set of all positive rationals. Regards, WM
From: mueckenh on 5 Jul 2006 07:54 Dik T. Winter schrieb: > In article <1151865025.167976.75740(a)v61g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > you need some limit in order to well-order the set of rationals? > > > > > > No, but that is unrelated. Transpositions operate on a set. A sequence > > > of transposition hence also operates on a set. There is *no* definition > > > how an infinite sequence of transformations operates on a set. > > > > These transpositions *are* a set. In my example, they do not destroy > > the bijective mapping between |N and |Q. > > This makes no sense. Yes, there is a set of transpositions, but the > transposition operate in sequence on an ordered set Q1 (which is a > well-ordered set to start with). After each finite sequence of > transpositions there is still a bijective mapping between N and the > resulting ordered set. You have not proven that that is still the case > after an infinite sequence of transpositions. Much less that it is > still the case after an infinite sequence of infinite sequences of > transpositions. You have first to define what that means. All sequences, even all single transpositions which I apply can be enumerated by natural numbers, just like the lines of Cantor's list. OK? And as long as I can enumerate, the number reached is finite. OK? Hence I do not need more than a finite set of transpositions, though its cardinality cannot be given. This case is in general denoted by "countably infinite" - but the number oo does *not* appear. Please consider these facts before you demand something to be proved for "oo". > > > >> Well, in the first place, this is false as written. The transpositions > > > are conditional. But indeed, given the well-ordered set of rationals > > > Q1, this gives a different ordering of the rationals, say the well-ordered > > > set Q2. We can call that an infinite permutation. > > > > But all transpositions to be applied to a given set are completely > > defined by the few lines of my definition and they are determined from > > the scratch. There is not a single one undetermined. > > No, but the "infinite permutations" have to be executed in order. Or can > you prove that they need not? Moreover, you have to show that an infinite > sequence of such "infinite permutations" leave well-orderedness. You have > not shown that. I have shown that the transpositions can be enumerated by natural numbers. There is no number oo. There is neither such a number in Cantor's list. All we do is enumerated. Othewise it would not be defined at all. Neither in Cantor's list. Therefore it is irrelevant whether or not something has to be "executed in order". We are in the countable domain and do not leave it. > > > > You keep assuming that the definition of the anti-diagonal of a list of > > > reals is an ongoing steps. That is false. There is only a single step, > > > the definition. > > > > The same is true for my example. There is only one single definition. > > If you do not like "steps" then drop that word. > > You *explicitly* say: apply aleph-0 times. Consider the Cantor diagonal > on a list l_n of reals in the range [0, 1): > I define two subfunctions (only to shorten the definition line): > remainder(x, a): x - entier(x / a) * a > and > digit(k): if k = 4 than 5 else 4 > and now: > D = sum{k = 1...oo} digit(remainder(entier(l_n * 10^k), 10)) / 10^k. > there are no steps involved. If k = 4 means to consider whether k = 4. If you get a solvable system of 100 linear equations, then the result is determined, but there are rather a lot of steps involved to obtain the solution. The somewhat simpler task to distinguish 4 and not 4 is in principle the same. It is one of infinitely many steps. Every "if, then" decision is a step. Set theorists prefer to deny this, because otherwise it would become too obvious that set theory is based upon Humbug (nonsense). > > > > After that we can determine each and every digit of > > > the diagonal number without the need to know other digits. Moreover, > > > by Cauchy, such a sequence of digits defines a real number. > > > > Leave Cauchy out of the play. There is no justification to reach all > > last digits of the diagonal number by his epsilons. His specification > > would only show that all columns of the list up to nn have been > > covered. > > Apparently you do not understand what Cauchy involves. I understand that Cauchy requires precision on a positive epsilon, arbitrarily small. I understand, that an infinite number of n follow the n_0. It is completely differenet from Cantor's requirement. Cantor does not define any limit process (because he considers finite natural numbers for enumeration purposes only). > > > > I do not know of any definition of a sequence of overlapping "infinite > > > permutations" that allows an inifinite sequence. Pray provide one. > > > > There are no "overlapping" permutations. > > There are. The first sequence of transpositions defines the first > "infinite permutation". Indeed, the transpositions do not overlap. > The second sequence of transpositions defines the second "infinite > permutation", but that one overlaps the first "infinite permutation". No prblem. Everything is determined. > > > > Because there are not infinitely many comparisons, there is a single > > > definition. > > > > This single definition leads to infinitely comparisons, unless the > > numbers in the list are all known. > > By representing a list it is assumed that all members are known. If they > are not all known, it is not a list. And that all are enumerated by finite numbers. The same is valid for the set of all positive rationals. > > > > > Ask Cantor. > > > > > > No, I ask you. > > > > > I do it just like Cantor does, by one single definition. > > No. Give a single definition as I have presented above for the anti-diagonal. See above. Regards, WM
From: Dik T. Winter on 5 Jul 2006 08:21
In article <1152099313.667688.264420(a)v61g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1151844660.650801.186480(a)75g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > Your principle is that the complete set N including the largest element > > > does exist. > > > > This is a complete misrepresentation. The complete set N does exist but > > has no largest element. How you conclude that the thinking is that there > > is a largest element escapes me. The existance of the complete set N is > > an axiom. > > An equivalent axiom (to actual infinity) would be the existence of 1000 > natural numbers with two decimal digits. Eh? Do you know the axiom of infinity? > > > If, however, the largest element does not exist or is not included in > > > the mapping, then there is no complete set N and, hence, no complete > > > set P(N) to be mapped upon, and there is definitely no completed set K > > > of non-generators which could used to prove something. > > > > This is not mathematics, at most matheology. The complete set N exists, > > due to an axiom, but it does not have a largest element (and that is > > easily proven). > > It is easily proven that the largest is missing. The axiom does not > state that all are there, but only that n+1 is there if n is. No. The axiom states that N does exist. But not in so many words, it can merely formulated to stated that an inductive set exists, or, equivalently, that an infinite set exists. > > It may be that you do not like that axiom, if so, why do you not set up > > your own mathematics without that axiom? This is similar to the > > parallel postulate. You can do geometry with it or with its negation. > > In the same way you can do set theory with the axiom of infinity or > > without it. Start some new set theory, and see where that leads you. > > The axiom is not the problem. It is its actual misinterpreation. You are not talking about the axiom I am talking about. > > Pray define an f for which K(f) is not defined, and state *where* the > > definition fails. > > For any f it fails in > f : X --> K, where X is any non-empty set and > K = { x in X | x is not an element of f(x) } Again, this makes actually no sense, as you take the definition out of the context where it was given: 1. Let's have a set A and a set of sets B. 2. Let's have a mapping f from A to B. 3. Define K(f) = {x | x in A and !in f(x)} What part of that definition fails? In your case you assume the definition (3) before (1) and (2). But (3) is defined only *after* (1) and (2) are given. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |