An uncountable countable set
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From: Virgil on 3 Jul 2006 18:46 In article <1151961303.406453.116760(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Consider a surjective mapping from N on the set of even positive > numbers & K. Only if we consider a function g and K = K(f). > Or consider a surjective mapping from R on the set of rational numbers > & K. Only if we consider a function g and K = K(f). > Or Consider a surjective mapping from R on the set {1, 2, 3, K}. Only if we consider a function g and K = K(f). > > But without readjusting f by g. That is forbidden in case of Cantor's > diagonal as well as in our case. Then "mueckenh" does not understand either 'Cantor's diagonal' or your case. In the Cantor "diagonal" on has a choice of uncountably many diagoals rules to choose from, any one of which is sufficient to prove the incompleteness of the list. In the "mueckenh" example, when K is tied to f, one must show that f's K also prevents any other function from surjecting, which it cannot do unless the codomain of the function is the power set of the domain. That "mueckenh" does not see this is evidence of his incompetence. > 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: Virgil on 3 Jul 2006 18:49 In article <1151961443.469526.135010(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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), ... Since any line and the following line both act on the same elements, they cannot be applied simultaneously, and they do not "commute", but must be applied sequentially. "Mueckenh" is innumerate.
From: Virgil on 3 Jul 2006 18:52 In article <1151961619.723466.21750(a)v61g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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). Learn to read deadhead. Having infinitely many does not require that any one of them be infinitely large. And each of the infinitely many naturals is only finitely large.
From: Virgil on 3 Jul 2006 18:57 In article <1151962063.059935.176450(a)75g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Gc schrieb: > > > > > 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). > > Not for any surjective function. As there are no surjective functions, that is a non-issue. For any function, f, that can exist from N to P(N), there is a K(f) which shows that function not to be a surjection. it "mueckenh" thinks he can suspend the laws of logic long enough to produce such a surjection, let him present it here. I guarantee he will make headlines if he succeeds.
From: David Hartley on 3 Jul 2006 18:51
In message <1151961927.534777.45250(a)v61g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de writes > >David Hartley schrieb: > > >> If this is defined for every element of A, we can say that the sequence >> has a limit, and it is f. >> >> >> Now, granting that your process works as you hope, you get a sequence of >> well-orderings of the rationals whose limit is the usual ordering. But, >> the corresponding sequence of bijections N -> Q does *not* have a limit. >> (You need only consider the images of 1, these diverge to -oo.) > >As I only consider positive rationals, these go to the smallest >positive rational, but that is the same problem. So you acknowledge that your process doesn't give, in the limit, a bijection N -> Q+ ? > >But to answer you argument of the missing limit: Cantor's diagonal has >also this very same problem. In case of the Cauchy-limit of an >irrational number, we see (a_k)*10^-k disappear, because it becomes as >small as desired. In the Cantor-diagonal however, there is always a >digit with the same weight and importance as the first one, even for k >diverging to oo. This is a complete non sequitur, as well as nonsense. Even if there were a problem with the diagonal argument, it would not alter the simple fact that: >> Your >> example does not "preserve the bijective mapping between N and Q". >Remember, here we have the same answer as in Cantor's diagonal: All >natural numbers are finite. Though there are infinitely many, each one >is finite. Hence even the smallest positive rational is in one-to-one >correspondence with a finite natural number. There is no natural number >oo (like in Cantor's list). I thought your reference above to a "smallest positive rational" was a joke, you don't really believe there is such a thing, do you? The rest of the statements in this last paragraph are simple truths (unless you're claiming "Cantor's list" contains oo), but don't answer anything, as far as I can see. -- David Hartley |