An uncountable countable set
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From: Dik T. Winter on 5 Jul 2006 08:23 In article <1152099458.014628.279080(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > 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. Righ, pretty acute. > > 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. 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. > > 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. Under what ordering? -- 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 5 Jul 2006 10:00 In article <1152100475.267310.35180(a)v61g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > 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. Yup, so there are infinitely many. 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 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. Yes. When you continue, you do not stop, so you do not reach a number, OK? 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. > Hence I do not need more than a finite set of transpositions, though > its cardinality cannot be given. Nope. > 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. > > > 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. THat makes no sense, again. The transpositions have to be execute in the order given. 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. > > > 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). > one of infinitely many steps. Every "if, then" decision is a step. Set There is no dependence of order. It is possible to determine any digit of the result without knowing any other digit. That is sufficient to make a number known. (Not in your sense, yes, I know. But that is a philosophical difference, not a mathematical difference.) > > > 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. > 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. > > 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. If you change the order of those permutations the result will be different after a finite number of such infinite permutations... > > 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. It is true that when you order them consistent with the enumeration you get a well-ordered set (of ty
From: Virgil on 5 Jul 2006 14:05 In article <e8g7t4$ebo$1(a)mailhub227.itcs.purdue.edu>, Dave Seaman <dseaman(a)no.such.host> wrote: > On Tue, 04 Jul 2006 21:31:16 -0600, Virgil wrote: > > > Also, in ZF, ZFC and NBG, any countable union of countable sets can be > > proved countable, so "mueckenh"'s attempts to claim otherwise must fail. > > Not in ZF. Is it the need for C? And if any set is known to be countable, doesn't that imply a known, or at least knowable, bijection from N to that set and thus justify a bijection from N to any such set of such sets?
From: Virgil on 5 Jul 2006 14:26 In article <1152099313.667688.264420(a)v61g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Only if 1000 is in base 4 or less. > > > > > 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. Not of the axiom that says there is a set of all of them. Anyone is quite free to reject any axiom set, but no one is free to impose any prohibition of any axim set on others. Thought control, such as "mueckenh" is advocating, is evil. > > > > 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. Except that "mueckenh" has not shown, and cannot show without a priori assumptions, that any such misinterpretation has occurred. > > > > > 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 elementsf(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.
From: Virgil on 5 Jul 2006 14:30
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 a abbreviation for "as n increases unboundedly beyond any given natural". |