An uncountable countable set
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From: Dik T. Winter on 6 Jul 2006 06:37 In article <1152181648.546076.208700(a)s26g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > No, of course not. The cardinal number of q will not change. What changes > > is ordering of q and also ordinal number and possibly well-ordering. > > The ordering is omega and remains so, as long as the first index is 1 > and there is no limes-ordinalnumber omega between two indices. And that is the case as long as the number of transpositions is finite. > > > And that is completely sufficient for the set of all positive > > > rationals. > > > > Under what ordering? > > The ordering is omega and remains so, as long as the first index is 1 > and there is no limes-ordinalnumber omega > between two indices. Indeed, so as long as the number of transpositions is finite. -- 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 6 Jul 2006 06:45 Dik T. Winter schrieb: > > All sequences, even all single transpositions which I apply can be > > enumerated by natural numbers, just like the lines of Cantor's list. > > Yup, so there are infinitely many. Are yo sure? How could an enumeration by natural numbers supply infinity? > But you have moreover an infinite > sequence of such infinite sequences. > > > OK? And as long as I can enumerate, the number reached is finite. OK? > > I do not know what you intend to say here. But you can enumerate, yes, and > when you stop you have done a finite number of transpositions you have > done. And when I do not stop I nevertheless have done a finite numer of transpositions anyhow. > Yes. When you continue, you do not stop, so you do not reach a > number, OK? I can never reach the number oo or omega or aleph_0 if you mean that. > Moreover, this way you will only do the transpositions in > the first sequence of transpositions, so you will never even start with > the second sequence of transpositions. If I do not get finished with this first infinity, then also Cantor does get get finished with his list. That is the consequnce I wnated to point out. > > > Hence I do not need more than a finite set of transpositions, though > > its cardinality cannot be given. > > Nope. There are only finite natural numbers. > > > 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". > > In addition to cardinality we have to do here with ordinality. The sequence > of transpositions you give has order type w * w. No. There is a first element and there remains a first element. > > 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. > > That makes no sense, again. The transpositions have to be execute in the > order given. Of course. But that does not exclude that these transposition can be executed and finished (if Cantor's list can be finished). > If you do them in any other order the result can be different. > So that you *can* enumerate them does not mean that the result is independent > from the order in which you perform them. That is true but has no relevance. > > > > > > 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. > > In fact: given sum{i = 1, oo} d_i/10^i, where 0 <= d_i < 10, by Cauchy > it follows that that denotes a real number. What the number is is > irrelevant, but it *is* a real number. That is the theory of Meray and Cantor. > > > 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). > > Perhaps. Please use my definition of the diagonal number above, which > does not definie a "limit process", but uses the definition of limit. Cauchy requires the epsilon. The theory of irrationals was made by Meray (C. MERAY: Remarques sur la nature des quantités définies par la condition de servir de limites à des variables données, Revue des Sociétés Savants 4 (1869) 280-289.) and Cantor (G. CANTOR: Über die Ausdehnung eines Satzes aus der Theorie der trigonometrischen Reihen, Math. Ann. 5 (1872) 123-132.) > > > > 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 problem. Everything is determined. > > If you change the order of those permutations the result will be different > after a finite number of such infinite permutations... But why should I change the order??? > > > > 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. > > Yes, you can enumerate them. But enumerating does *not* mean ordering. Enumerating by 1,2,3,... does mean well-ordering with order type omega. > It is true that when you order them consistent with the enumeration you > get a well-ordered set (of type w). This does *not* mean that any > ordering gives a well-ordered set (of type w). A straight counter-example > is the integers ordered in reverse order. That is not a well-ordered set. >From that example we see that it is impossible to maintain the well-order and we see why: Because the actual infinite does not exist. Would it exist, there was always a first element, a second, an so on. Regards, WM
From: mueckenh on 6 Jul 2006 06:52 Virgil schrieb: > > Anyone is quite free to reject any axiom set, but no one is free to > impose any prohibition of any axim set on others. Unless there appears a conradiction. > > > 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) } > > This can only fail if the elements f(x) are sets. Suppose a,b, and c are > not sets (or classes) and thus cannot have members, then any function > from X to X is a function with codomain X = K = {x in X : x not in > f(x)}, since x in f(x) is now impossible. There is only one element K, which, accordng to its definition, is a set - it is as impossible a set as the set of all sets. Regards, WM
From: mueckenh on 6 Jul 2006 06:55 Virgil schrieb: > In article <1152099458.014628.279080(a)m79g2000cwm.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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. > > To say n --> oo does not require that there exist any "oo". > > It is merely an abbreviation for "as n increases unboundedly beyond any > given natural". But always remaining a natural itself, yeah. And the number of naturals < n is always a finite number, so that it never does reach infinity too. Regards, WM
From: mueckenh on 6 Jul 2006 06:58
Virgil schrieb: > In article <1152100475.267310.35180(a)v61g2000cwv.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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? > > But the numbers which are reachable include values larger than any given > natural. To assume a limit on what is achievable is counterfactual. > > > Hence I do not need more than a finite set of transpositions, though > > its cardinality cannot be given. > > You, in fact, need more than any finite number can supply. > > > > 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". > > WE do not require that it be proved for "oo", which is not a natural, > but that it be proved for "for all n in N". > > You method produces exceptions to "for all n in N", by implying that > there is some remote member of N beyond which it is not necessary to > proceed in order to establish "for all n in N". There is not a remote member, but each natural is finite. Therefore, how far we may go, there is never an infinite number of transpositions accumulated. And the Cantor diagonal has not an infinite number of digits. Regards, WM |