An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: mueckenh on 2 Jul 2006 08:40 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. He has not proved - or even given any justification at all - that this process has any meaningful limit. 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. > To see that this is not straightforward, consider the negative naturals, > indexed by the corresponding positives, -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). > 3 1 5 2 7 4 9 6 ... > 4 2 6 1 8 3 10 5 ... > 5 3 7 1 9 2 11 4 ... > > (Each step disrupts the "size" orderings set up by the previous step, so > all possible transpositions are needed at each step.) The transpositions of each step can be executed simultaneously, as I tried to show above. But, in fact, as for a given initial order all transpositions of all steps are predetermined, I don't see a problem of lacking simultaneity at all. > The least element after n steps is n+1, so what's the least element of > the limit ordering? What is the last natural number enumerating an element of Cantor's diagonal? > 2 and 1 alternate indefinitely with each successive > step, which comes first in the limit ordering? That is a misunderstanding. There is no indefinite alternation possible. See my correction above. > > Can you give any reason why your process should be any more successful > with an enumeration of the rationals? As you can see from the corrected sequences above, the process will be successful if infinity does exist. Regards WM
From: mueckenh on 2 Jul 2006 08:44 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. > > > 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? > > What follows from that is that k must be a real number that is *not* > a natural. > > 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. You may repeat this as often as you like without making it a truth. > There is no k in N such that f(k) = K(f). So f is not a > surjection from N to P(N). But f may very well be a surjection > from some set *other* than N. That does not make Hessenberg's proof valid. Regards, WM
From: mueckenh on 2 Jul 2006 08:51 Daryl McCullough schrieb: > >> And in the set of all mappings from N to P(N), there are injections but > >> no surjections. > > > >The latter is an empty assertion. > > It is a provable fact. Yes, you don't accept the proof, because > you don't accept the basic principles of mathematics, but the > proof certainly follows from those principles. "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). >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 principle is not mathematics, but matheology at most. 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. > 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. > > There is a surjection from R to P(N), but there is no > surjection from N to P(N). So if f is a surjection from > R to P(N), what we can conclude is > > If f(k) = K(f), then k is not a natural number. We do not need the surjection from N to P(N) but only the mapping from an n e N on the set K which does contain n if it does not contain it and does contain n if it does not contain n. 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. Would you deny that (|N and (-1)) and |N have same cardinality? Regards, WM
From: mueckenh on 2 Jul 2006 08:53 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. > > > > > Again you are starting with a conclusion about the finite and extending it > > > to the infinite. You have to prove that you can do that. > > > > 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? Regards, WM
From: mueckenh on 2 Jul 2006 08:57
Virgil schrieb: > If one sets conditions on the mappings allowed, perhaps, but what if > there are no such restrictions, other than having domain R and codomain > P(N). In that case nothing prohibits surjection, and, in fact, > surjections have been constructed. I have, myself, even constructed a > bijection. Note that Hessenberg's trick is inavlidated 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. Would you deny that (|N and (-1)) and |N have same cardinality? So, what only can destroy it in one case while leaving it in the other? > > > > Why not map it without that restriction? In order to show that it this insane restriction is the only argument against surjectivity in case of N --> P(N). To show you the answer to my question above. > Regards, WM |