An uncountable countable set
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From: mueckenh on 3 Jul 2006 09:45 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. Regards, WM
From: mueckenh on 3 Jul 2006 09:47 Virgil schrieb: > In article <1151843785.711371.113080(a)75g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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. > > > If K does not exist, there is no bar to f being bijective. > > 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. > > Does every potential bijection from R to P(N) have to map some natural > to K? why can't it map , say , pi to K? Then K can exist, and IS in P(N) > and f(pi) = K with no problems, and f CAN be a bijection. > > There are all kinds of surjections from R to P(N), though none of them > with the specific artificial restrictions you would impose. There are all kinds of surjections from N U {pi} to P(N). You cannot prove that there is none. Regards, WM
From: mueckenh on 3 Jul 2006 09:51 Virgil schrieb: > In article <1151844027.410356.228200(a)m79g2000cwm.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > David Hartley schrieb: > > > > > > > >The set of rationals is taken to be well-ordered. The following > > > >transpositions operate on that set simultaneously (given are the > > > >indices): > > > >(1,2), (3,4), (5,6), ... > > > >where Elements q_2n-1 and q_2n are interchanged, if they deviate from > > > >order by size. > > > >In the next step the pairs > > > >(2,3), (4,5), (6,7), ... > > > >are ordered by size, in the next step the pairs (1,2), (3,4), (5,6), > > > >... are ordered by size, and so on. There are exactly as many steps as > > > >are required to define the diagonal of a Cantor list. And there is the > > > >same definition of "infinitely" as in Cantor's diagonal proof. > > > > > > You have not proved - or even given any justification at all - that this > > > process has any meaningful limit, nor that if it does, it is ordered as > > > you wish. > > > > This is exactly the same with Cantor's diagonal proof which not valid > > for an infinite list. > > > That you do not choose to admit the difference is not evidence that > there is no difference. There is none. > > > He has not proved - or even given any > > justification at all - that this process has any meaningful limit. > > Cantor has, at least in the eyes of those who are willful believers of the absurd. > > > In fact, it has not a meaningful limit, because all digits of an > > unending number like 0.111... (to take a very simple case) which can > > be identified are contained in the list of numbers: > > > 0.1 > > 0.11 > > 0.111 > > ... > > 0.111...1 > > ... > > > > But in this list the number 0.111... is not contained. Hence not all of > > its digits can be identified. > > You have just proved that your example supports Cantor's theorem by > producing a number, 0.111..., not in your own list. All digits of a number must be indexed by natural numbers. Otherwise they cannot be identified. All digits which can be identified are pesent in the list which contains all unary representations of naturlal numbers. 0.111... does not belong to this set. Regards, WM
From: mueckenh on 3 Jul 2006 10:01 Virgil schrieb: > In article <1151844299.084352.59780(a)v61g2000cwv.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Daryl McCullough schrieb: > > > > > > > >Remember what I said: 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. You should be able to find out that in case > > > >of a surjective mapping, K is undefined. > > > > > > Yes, you said that, but it is convoluted nonsense. For every > > > mapping that exists, K(f) is defined. K(f) is only undefined if f > > > is undefined. > > > > It is nonsense for any mapping where a number has to be mapped on a > > set non containing it, if it contans it. Yes. It is nonsense. And I > > hope that by now it has become obvious. Therefore this condition > > cannot appear in a valid proof. > > Let us change "mueckenh"'s condition to one in which a non-natural, say > pi, has to be mapped to K in P(N) and there is no objection to now > having a surjection, or even a bijection, from R to P(N). > > Thus "mueckenh"'s objections to surjections or bijections from R to P(N) > are frivolous. > > > > > > > f is not a surjection from N to P(N) > > > > > > 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. Then the mapping can > > be surjective. Would you deny that (|N and pi) and |N have same > > cardinality? So if one mapping is possible but the other is not, then > > there must be something else prohibiting it, isn't it? > > Since R is not a subset of N union {pi}, here is no K possible in this > case. Take (N and pi) --> P(N) is not provable non-surjective. N --> P(N) is apparently provable non-surjective. You do not wonder why that is so? > But lack of one form of proof does not falsify the statement. In this > case there are other proofs of no surjection, wrong > the obvious one is that > N bijects with N union {pi} by the function g:N <--> (N union {pi}) > with g(0) = pi, g(n) = n+1, so that any f:N union {pi} --> R induces > fog:N --> R which would also have to be surjective if f were, but cannot > be by Cantor's multiply proved theorem. That is the same with Cantor's diagonal. The set of list numbers bijects with N. The set of list numbers & diagonal bijects with the set of list numbers. Hence, no proof of uncountability. Please stay with one single arguing, and do not judge with different measures as you like to do. Regards, WM
From: imaginatorium on 3 Jul 2006 10:16
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...) Brian Chandler http://imaginatorium.org |