An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: mueckenh on 3 Jul 2006 17:12 Dik T. Winter schrieb: > In article <1151759090.073603.122700(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > Eh? K(f) is defined regardles the kind of mapping of f. How can you > > > come to the conclusion that it is not defined for some particular f? > > > What part of the definition fails for some particular f? > > > > 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. > > Yes, such a map is impossible, but that proves nothing about the impossibility > to have a surjective mapping from R to P(N). If a mapping is surjective > the requirement is that *each* element of the target should be the image > of an element in the source. There is nothing that requires that element > in the source to be a natural number, *unless* the source is the set of > natural numbers (or a subset of it). So what does it prove that such a > required mapping is impossible? Try to map R on the set {1, 2, 3, K} and you will see it. > > Next you come with something completely different: > > 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. > Yup, a mapping with such a condition can not be found. In the case of > N -> P(N) this proves that there is no surjective mapping. In the case > of R -> P(N) this proves nothing like that. And what is proved in case of a surjective mapping from N on the set of even positive numbers & K? And what is proved in case of a surjective mapping from R on the set of rational numbers & K? > > > > You are confused. K(f) is required to be the image of a natural number. > > > > > > That is a complete misstatement of the Hessenberg condition. To be clear, > > > let's have given a set A and a set of sets B (I do not yet state what the > > > elements of A and B are). Let's also have a mapping f from A to B. > > > Now construct the set K(f) = {x in A and x !in f(x)}. This set is > > > well-defined. > > > > Remember what I said: The set K e P(|N) which contains all > > natural numbers which are not mapped on sets containing them. > > That makes no sense. It depends on the map. So you should talk about > K(f) (with f the mapping assumed). But what is wrong with what I wrote? Wrong is the assumption that K(f) = { x in N | x is not an element of f(x) } in case of a surjective f is an element of P(N). It is not. Regards, WM
From: mueckenh on 3 Jul 2006 17:15 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. Consider a surjective mapping from N on the set of even positive numbers & K. Or consider a surjective mapping from R on the set of rational numbers & K. Or Consider a surjective mapping from R on the set {1, 2, 3, K}. But without readjusting f by g. That is forbidden in case of Cantor's diagonal as well as in our case. Or why do you believe that there was a difference? > > > > > > > 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. Readjusting f by g is forbidden in case of Cantor's diagonal. Why do you believe that there was allowed in our case? Regards, WM
From: mueckenh on 3 Jul 2006 17:17 Virgil schrieb: > In article <1151844816.810459.198700(a)75g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > The difference being that Cantor's sequence of non-diagoal digits need > > > not be created seqeuentially, but can be done independently and > > > simultaneously, whereas yours cannot. > > > > Poor arguing! Is that all you have? (Euler thought a function must be > > defined by a single expression.) > > > > We have infinitely much time. Further I can supply you with a > > single-step creation of all transpositions required. But that is > > unimportant. > > You need an infinite sequence of such infinite sequences of > transpositions, and two successive such sequences of transposition must > be applied in a particular order to achieve a particular result. They do > not commute and they cannot be applied simultaneously. Wrong. The scheme is a fixed matrix and each line can be applied simultaneously. (1,2), (3,4), (5,6), ... 1, (2,3), (4,5), (6,7), ... (1,2), (3,4), (5,6), ... 1, (2,3), (4,5), (6,7), ... (1,2), (3,4), (5,6), ... 1, (2,3), (4,5), (6,7), ... .... For a given well-order of Q, the result is determined. > > > > > Ask Cantor, how he can be sure that his diagonal is different from > > > > every number of the list in case of infinitely many numbers. > > > > > > One does not have to ask Cantor to explain what is transparently clear. > > > All Cantor needs do is specify a rule which can be applied > > > simultaneously to ever number in a given list to produce something that > > > cannot be a member of the list. > > > > > And the last number is included with certainty? > > There is no such "last" in ZF or NBG. > > What system are you messing up in? So how can you be sure that all elements are included, if you never get the "end of file" report? Regards, WM
From: mueckenh on 3 Jul 2006 17:20 Virgil schrieb: > In article <1151845298.388602.107330(a)a14g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > 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. > > Since every number in your list terminates, butyour "diagonal" 0.111... > does not, it differes from each list member in infinitely many digit > positions, and it is clearly not in your list. So YOU have provided a > nonmember to prove your own error. > > > > If > > 0.111.. is not in the list, then it must have more digits than can be > > indexed (and hence, can exist). > > They are satisfactorily indexed by the infinite set of finite natural > numbers, N. You just proved that there are infinitely digit positions which are not indexed by natural numbers (*all* of which are given in the list). > > > > 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). > > If "mueckenh" insists on exact equality and "mueckenh" simultaneously > insists on a difference, "mueckenh" is in trouble. Therefore he does not insist on a difference, but doubts that this precision can be achieved. > Regards, WM
From: mueckenh on 3 Jul 2006 17:22
David Hartley schrieb: > >-1, -2, -3, ... indexed by > > 1, 2, 3, ... > > > >> and apply your process. The > >> first few steps give > > > >the indices after the first step are > >> > >> 2 1 4 3 6 5 8 7 ... > > > >This is correct, but the next four steps would supply > >2 4 1 6 3 8 5 10 7... > >4 2 6 1 8 3 10 5 ... > >4 6 2 8 1 10 3 ... > >6 4 8 2 10 1... > >because we must always alternate between pairs (2n-1, 2n) in one step > >and (2n, 2n+1) in the next (for all n e |N). > > > I see we have different ideas of what these transpositions mean. I took > (1 2) to mean swap the pair originally indexed by 1 and 2, wherever they > now are. You seem to mean swap the current occupants of positions 1 and > 2 in the list. This is certainly more likely to have the effect you > desire. It is a very simple rule and it determines the result for any well-order we are starting with. > 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. The bijection (one-to-one corespondence) between rationals and naturals is preserved. This is if not a well-order so at least more then set theory can supply for the natural order of Q. Regards, WM |