An uncountable countable set
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From: mueckenh on 2 Jul 2006 09:01 Virgil schrieb: > In article <1151759780.766369.176910(a)m79g2000cwm.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > Cantor's diagonal proof does not prove that all numbers of the list are > > > > included. > > > > > > All number of the list are certainly included in the list. What else are > > > they supposed to be included in??? > > > > Cantor's diagonal proof does not prove that all numbers of the list are > > included in that sense that all are covered by the diagonal. > If you claim that any member of any list is not differentiated from the > number created for that list by any of uncountably many variations of > Cantor's constructions, then you must be declaring there there exists a > first number in some list which is not differentiated according to some > version of Cantor's rules. > > So Give us an example of this alleged situation. If you can. Nothing easier than that. Exchange 0 by 1 in: 0.0 0.1 0.11 0.111 .... The diagonal up to any line number n is contained in line number n+1. If you think 0.111... is not in the list, remember that any number the digits of which can be indexd by finite numbers is in the list. If 0.111.. is not in the list, then it must have more digits than can be indexed (and hence, can exist). > >Only to > > give you an example: If the list was longer than wide > > As the list is supposed to be endlessly long and the decimal (or other > base) representation of each member is supposed to be endlessly "wide", > why should either endlessness be greater than the other? Why should it not? Remember, there must be an exact equality between width and length (better than +/- 1). Is that precision achievable? Isn't it a fact, that already a list of all rationals is by far longer than wide, even in binary representation? 0.0 0.1 0.01 0.11 .... for instance. I would rather believe in the fairys in your garden and even under your bed than in the Cantor-list to be a precise square. > > On the contrary, Cantor, and others, have specifically defined valid > rules for constructing a real number not in any given list of real > numbers. He used to say in general: Es erfährt daher der aus unsrer Regel resultierende Zuordnungsprozeß keinen Stillstand. [Cantor, collected works, p. 239]. Translated: "The mapping resulting from our rule does never come to a stop." That's it. > > > You reject things that have been proved within that standard and insist > > > on things whose proof does not meet that standard. > > Same did Cantor 130 years ago. > > Only in your opinion, not in fact. And your opinions have proved flawed. "I don't know what predominates in Cantor's theory - philosophy or theology, but I am sure that there is no mathematics there" (L. Kronecker) Regards, WM
From: imaginatorium on 2 Jul 2006 10:12 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)? Brian Chandler http://imaginatorium.org
From: Virgil on 2 Jul 2006 12:00 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. Your objection is no more valid that saying that no one can run a Marathon because quadriplegics cannot run the Marathon.
From: Virgil on 2 Jul 2006 12:11 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. > 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 not willfully blind. Besides which ,that was Cantor's ->second<- proof of the theorem, designed for those incapable of grokking his first proof. You are an example of those incapable of grokking either. > 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. > I don't see a problem You willfully do not see what is actual and you willfully see what is not there.
From: Virgil on 2 Jul 2006 12:28
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. But lack of one form of proof does not falsify the statement. In this case there are other proofs of no surjection, 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. |