An uncountable countable set
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From: Gc on 2 Jul 2006 15:57 Gc kirjoitti: > mueckenh(a)rz.fh-augsburg.de kirjoitti: > > > Gc schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de kirjoitti: > > > > > > > An uncountable countable set > > > > > > > > There is no bijective mapping f : |N --> M, > > > > where M contains the set of all finite subsets of |N > > > > and, in addition, the set K = {k e |N : k /e f(k)} of all natural > > > > numbers k which are mapped on subsets not containing k. > > > > > > > > This shows M to be uncountable. > > > > > > > > Regards, WM > > > > > > The set M exist only if you have already defined exactly funktion f, > > > because there is now certain set K before the funktion is exactly > > > defined. So you have bijection g: N ---> M/K . Then you can add the set > > > K to finite sets and get M. This is no paradox and this is very > > > different than funktion f: N ---> P(N) because P(N) contains K by > > > definition. > > > > > > No, it does not. P(N) cannot contain K by definition, because K does > > not exist. > > Yes it does. For every function N to P(N) there exists K in P(N). I mean for every function which maps integers to the sets that don`t contain them. > > >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. > > > > Here we have not at all any problem with too less elements in the > > source. Nevertheless there is no surjection possible. > > Virgil has already said that your condition is irrelevant. In your > mapping there is now reason that a natural number has to mapped with K > e P(N). > > > > Regards, WM
From: Dik T. Winter on 2 Jul 2006 19:15 In article <1151844299.084352.59780(a)v61g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Daryl McCullough schrieb: .... > > But the assumption here is that f is a surjection from R to P(N), > > not from N to P(N). So we have: > > > > 1. There is no k in N such that f(k) = K(f). > > 2. There exists k in R such that f(k) = K(f). > > Why don't you try the same trick with the mapping (|N and pi) --> R. > Let pi be mapped on the set of non-generators. Again, this makes no sense. Do you mean f: {x | x in N or x = pi} -> P(M)? Well, because K(f) contains pi, it is not an element of P(N), and so that it is not in the image of f does not inhibit f from being a surjection. > Then the mapping can be > surjective. I would not say "can be", but the proof does not show that it is not. > So if one mapping is possible but the other is not, then > there must be something else prohibiting it, isn't it? The reason one might be possible and the other not is because one K(f) is not a member of P(N) and is not necessarily in the image, while the other K(f) is a member of P(N) and so (if f is surjective) *must* be in the image. However this only does shows that the other is not surjective, it does *not* show that one is surjective. But let's have f: {x | x in N or x = pi} -> P(N). Construct the following mapping: g: g(0) = pi, g(n) = n - 1; gf = h. this one is bijective, so the inverse exists. If f(x) is surjective, g(f(x)) is surjective, but that is a mapping from N to P(N) and can not be surjective because we can construct: L(g, f) = {x | x in N and x !in g.f(x)} we can translate it back to: L(g, f) = {g^-1(x) | (x in N or x = pi) and g^-1(x) !in f(x)} and see that that set is a member of P(N) and so should be in the image, but is not in the image. > > Once again, K(f) is *always* defined, for every function f. > > Not for this one: > 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. There is no defined mapping. K(f) is defined for every mapping f. -- 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 2 Jul 2006 19:24 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. > 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 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. > > Do you at least agree for *finite* sets, that there is no > > surjection from a finite set S to P(S)? In particular, if > > S = { 0 }, then P(S) = { {}, {0} }. Do you agree that there > > is no surjection from S to P(S) in this case? > > Of course I do. But infinity obeys laws different from finite numbers > (Galilei). > > > > Is there a finite S such that there is a surjection from S to P(S)? > > If so, what is it? > > > Of course there is none. But infinity obeys laws different from finite > numbers (Galilei). > > > K(f) is defined for any f whatsoever, as long as f is defined. > > No. Pray define an f for which K(f) is not defined, and state *where* the definition fails. > Note that Hessenberg's trick is invalidated by introducing one > unnatural number like (-1). The mapping from N and (-1) onto R is not > disprovable if (-1) is mapped on the set of non-generators. Then the > mapping can be surjective. It does *not* prove that the mapping is not surjective, but it also does *not* prove that such a surjective mapping is possible. -- 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 2 Jul 2006 19:36 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 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. 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. -- 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 2 Jul 2006 19:59
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. > >> 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. > > 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. > > 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 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". > > 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. > > > 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |