An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: David Hartley on 11 Jul 2006 19:09 In message <virgil-1952BC.12083511072006(a)news.usenetmonster.com>, Virgil <virgil(a)comcast.net> writes >In article <1152628953.131295.202010(a)s13g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > >> Franziska Neugebauer schrieb: > >> > You need a defined limit in order to write about applying *all* of the >> > conditional permutations. >> >> "und es erf?hrt daher der aus unsrer Regel resultierende >> Zuordnungsproze? keinen Stillstand." This wrote Cantor in the same >> context. But the defined limit is clearly stated: It is the order by >> magnitude, because then nothing happens further. > >But it is an unachieved limit, as limits tend to be. >> > >> > Please recall (q i) now. As one easily sees for every j e N the >> > permutation q j q j-1 ... q 1 can be applied to any sequence >> > containing at least 2 members. Nonetheless the application of all q i >> > i e N is not defined: > >> > >> > > All elements which do exist in the well-order will be in the >> > > order by magnitude. >> > >> > Not yet. First you have to define the meaning of applying all >> > conditional permutations and then prove the existence and uniqueness of >> > the result. >> > >> To prove the existence and uniqueness of (|Q +, <)? > >NO! To prove the existence of an end to a process which does not end. > >After each conditional transposition on has a finite initial segment of >rationals in numerical order followed by an infinite sequence of >rationals in which there are infinitely many successive pairs in reverse >order of size. > >This condition does not change for any finite number of transpositions >which are applied. > >Thus if there is a "limit" it must be one with only a finite initial >sequence ordered by magnitude and infinitely many size inversions. > >Otherwise "muecken" must claim that a finite number of transpositions >suffices, which is nonsense. Given a sequence of total-orderings <_n of a set S we can define a partial ordering <_L on S by For any x,y in S, x <_L y iff there exists n in N st. for every m > n, x <_m y This ordering will not necessarily be total, but if it is, it seems reasonable to say that the sequence has a limit, and that it is <_L. Using this definition, Daryl McCullough, Dik Winter and I have all given examples of sequences of well-orderings whose limit is not a well-ordering. WM's original proposal was rather muddled, but his present version works. He *has* described how to construct a sequence of orderings of the positive rationals, starting with that induced by an arbitary enumeration, whose limit is the usual ordering. (In fact, the same technique will give a sequence whose limit is any given total ordering, and a similar, although a bit more complicated, argument shows that there are sequences starting with any given total ordering whose limit is any given ordering isomorphic to N.) What he hasn't proved is his claim that, since each ordering in the sequence is order-isomorphic to the naturals, so is the limit. When challenged on this, he refuses to acknowledge the problem, just claiming that it is "clearly" true. Unless he attempts to prove this there is little point in arguing with him; you can't point out the errors in a non-existent proof. If you do continue, at least attack the part of his argument which fails, not the part that he has - just about - got right. -- David Hartley
From: Virgil on 11 Jul 2006 19:46 In article <oG0E5DCM+CtEFwmW(a)212648.invalid>, David Hartley <me9(a)privacy.net> wrote: > In message <virgil-1952BC.12083511072006(a)news.usenetmonster.com>, Virgil > <virgil(a)comcast.net> writes > >In article <1152628953.131295.202010(a)s13g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > >> Franziska Neugebauer schrieb: > > > >> > You need a defined limit in order to write about applying *all* of the > >> > conditional permutations. > >> > >> "und es erf?hrt daher der aus unsrer Regel resultierende > >> Zuordnungsproze? keinen Stillstand." This wrote Cantor in the same > >> context. But the defined limit is clearly stated: It is the order by > >> magnitude, because then nothing happens further. > > > >But it is an unachieved limit, as limits tend to be. > >> > > >> > Please recall (q i) now. As one easily sees for every j e N the > >> > permutation q j q j-1 ... q 1 can be applied to any sequence > >> > containing at least 2 members. Nonetheless the application of all q i > >> > i e N is not defined: > > > >> > > >> > > All elements which do exist in the well-order will be in the > >> > > order by magnitude. > >> > > >> > Not yet. First you have to define the meaning of applying all > >> > conditional permutations and then prove the existence and uniqueness of > >> > the result. > >> > > >> To prove the existence and uniqueness of (|Q +, <)? > > > >NO! To prove the existence of an end to a process which does not end. > > > >After each conditional transposition on has a finite initial segment of > >rationals in numerical order followed by an infinite sequence of > >rationals in which there are infinitely many successive pairs in reverse > >order of size. > > > >This condition does not change for any finite number of transpositions > >which are applied. > > > >Thus if there is a "limit" it must be one with only a finite initial > >sequence ordered by magnitude and infinitely many size inversions. > > > >Otherwise "muecken" must claim that a finite number of transpositions > >suffices, which is nonsense. > > Given a sequence of total-orderings <_n of a set S we can define a > partial ordering <_L on S by > > For any x,y in S, x <_L y iff there exists n in N st. > for every m > n, x <_m y > > This ordering will not necessarily be total, but if it is, it seems > reasonable to say that the sequence has a limit, and that it is <_L. > > Using this definition, Daryl McCullough, Dik Winter and I have all given > examples of sequences of well-orderings whose limit is not a > well-ordering. WM's original proposal was rather muddled, but his > present version works. He *has* described how to construct a sequence of > orderings of the positive rationals, starting with that induced by an > arbitary enumeration, whose limit is the usual ordering. (In fact, the > same technique will give a sequence whose limit is any given total > ordering, and a similar, although a bit more complicated, argument shows > that there are sequences starting with any given total ordering whose > limit is any given ordering isomorphic to N.) How do you define this alleged sequence of orderings? What is the process by which one goes from one ordering to the next in this sequence? What is your measure of how close to the limit state you have progressed? E.g., what are your analogues to the deltas and epsilons of more prosaic limit processes? > > What he hasn't proved is his claim that, since each ordering in the > sequence is order-isomorphic to the naturals, so is the limit. When > challenged on this, he refuses to acknowledge the problem, just claiming > that it is "clearly" true. Unless he attempts to prove this there is > little point in arguing with him; you can't point out the errors in a > non-existent proof. If you do continue, at least attack the part of his > argument which fails, not the part that he has - just about - got right.
From: mueckenh on 12 Jul 2006 08:34 Dik T. Winter schrieb: > you are arguing that 1/9 does not exist. 1/9 does exist. What does not exist is its representation 0.111... which does not belong to the sequence 0.1, 0.11, 0.111, ... the 1's of which are completely indexed by natural numbers. > Sorry, this ia again not logical reasoning. Again, I ask what logical > steps you take to get from > For all p there is an n such that An[p] = K[p] > to > There is an n such that for all p An[p] = K[p], > I would think you should be able to answer that simple question. I do not do this step! Already your antecedent "for all p there is an n such that An = K[p]" *is wrong*. I say: for those K[p] for which there is an An, one and only one An is sufficient. That is not valid for all p, because for all n we have 0.111... - 0.111...1 = 0.000...0111... such that there are more 1's in K than in any An. > > > > > 0.1 > > 0.11 > > 0.111 > > ... > > it is never necessary to use two elements for the same purpose. > > If you think the sentence "all positions of K = 0.111...are indexed by > > list numbers" is not equivalent to the sentence "K > > is in the list", then you seem to imply that more than one list number > > is required to index the digits of K. > > Yes, each digit position of K can be indexed by a number in the list. And > it is not equivalent to "K is in the list". The implication you mention > I do not understand. All digits which are indexed by smaller list numbers can be indexed by one larger list number. Therefore always only one is required. > Of course to index the digits of K you need each and > every list number. For starters the first digit is indexed by the first > list number (list numbers considered as natural numbers), the second digit > is indexed by the second list number. > > > You do not need the first two list numbers 0.1 and 0.11, because the > > third alone is sufficient: 0.111 does index the first three digits p = > > 1 to 3 of 0.111... > > I am a bit at a loss here what you mean with "indexing". > > > That can be carried on. p = 1 to 4 in 0.111... can > > be indexed by 0.1111. The digits p = 1 to n can be indexed by list > > number n = 0.111...1 with n 1's. If, as you seem to imagine, it may > > happen, that two list numbers are required to index some p, then one of > > the two is smaller than the other. So the other will suffice for the > > task of the smaller. This proves: If a p is finite and, hence, can be > > indexed by finite list number, then all digits < p can be indexed by > > the same list number. Therefore, all digits which can be indexed can be > > indeed by one finite list number alone. > > Pray prove the "therefore", because it does not follow. What is valid > for the finite is not necessarily valid for the infinite. But it is not necessarily just the opposite (like, allegedly, in the binary tree). So let us find a resolution of this dilemma by mathematics. Define "*-" by a_i *- b_i = a_i - b_i if a_i > b_i else a_i *- b_i = 0 (subtraction down to zero but without negative numbers) Consider 0.111... and the list numbers as sequences or as vectors, such that *- can be applied to each term separately. If you are right, you must maintain the result: 0.111... *- (0.1 + 0.11 + 0.111 + ...) = 0.000... I do not agree because 0.111... is different from any list number. Therefore it cannot give the result 0.000... though the difference is smaller than can be expressed. Regards, WM
From: mueckenh on 12 Jul 2006 08:36 Dik T. Winter schrieb: > > When Cantor's proof was published, there was not such an axiom. > > Indeed, at that time it was not yet an axiom, but the thought was present > that infinite sets did exist. There was not such an assumption in general. Infinite sets did not exist (as they do not exist yet), but only the potential infinite was accepted. Cantor was little understood and was blamed to do philosophy or theology but not mathematics. > > > You are not arguing against Cantor's > > > argument, you are arguing against that axiom. > > > > If an axiom states the existence of 100 natural numbers below 20, it > > has to be abolished in order to save mathematics. The axiom of > > infinity, interpreted as you do, is such an axiom. But it was built to > > model Cantor's worldview. > > The first of these two axioms would lead to an inconsistent system. The > axiom of infinity does not. It leads only to an inconsistency with *your* > view. It leads to the inconsistency that you must demand 0.111... *- (0.1 + 0.11 + 0.111 + ...) = 0.000... but 0.111... not being in the sum. It leads to a binary tree having infinitely many path mapped upon the very same edge though no path can split without two new edges. And you must refuse to calculate with fractions in mappings. That is inconsistency enough. Regards, WM
From: georgie on 12 Jul 2006 08:44
mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > Except that after each (1,2), there will only be a finite initial > > subsequence in numerical order and an infinite terminal sequence not yet > > ordered. > > > > Using 1 origin indexing, the nth occurrence of (1,2) occurs at the > > (n^2 + n)/2 th position in the list of "transpostions". > > > > At which (n^2 + n)/2 th operation is the entire list ordered? > > At which number can I find the last element of Cantor's diagonal? I'm pretty sure it's index is the same as the first even number that isn't the sum of two primes. If you work at it, you might prove that the two numbers are equal. Was there anything else you needed to know to be convinced? BTW- I missed the post where someone proved that the axiom of infinity doesn't create an inconsistency with the rest of set theory. Can you show me that post that I'm sure someone posted. I'm sure that nobody here would claim that others are wrong without proof, would they? |