An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: mueckenh on 11 Jul 2006 03:39 Virgil schrieb: > > The *only* criterion about countable infinite sets is whether one can > > determine *precisely* at which natural number something happens: After > > how many steps in a > > well-order of |Q the fraction 4711/235537 will appear, for instance, or > > Given an arbitrary bijection, f: N --> Q, "mueckenh" claims to be able > to tell us the value of n such that f(n) = 4711/235537, even though that > will differ from one such bijection to another. The initial well-order is arbitrary, but fixed. Are you really unable to understand that much? Regrads, WM
From: mueckenh on 11 Jul 2006 03:45 Virgil schrieb: > > > > Excuse me. There is not a single exchange carried out. All are > > simultaneously defined. More is not required. > > You have to show (1) that they can be carried out Of course any two different rationals can be compared by size. More is not required to determine whether they have been transposed or not. >and (2) that after > they are carried out the result will be what you claim it should be. > You can do neither. > You said Cantor "proves" the existence of a well-ordering of the rationals by proving the existence of a bijection from N to them. That should be sufficient for me and my result too. > > Have you ever tried to > > well-order the set of rationals for more than, say, 20 elements. > > One can do it for all of Q by establishing a rule for what rational > follows any given rational. This works simultaneoulsy for all of Q, and > has often been done. And I establish a rule which determines what well-order follows any given well-order. > > > You are right. I responded only to Virgil's very naive belief that > > there something really should be carried out. Of course it is > > sufficient to give the principle by a short expression like this > > > > (1,2) > > (2,3) > > (1,2) > > ... > > > > to intelligent people, and all is instantaneously ordered by magnitude. > > The effect of transpositions (1 2)(2 3)(1 2) applied successively to > the first 3 elements of any list of 3 or more is simply to reverse the > order of those first 3. Transpositions are carried out only if the elements are not yet in order by magnitude. > This means that unless one has the exact reverse of the order by > magnitude to start with, the result will not be ordered by magnitude. > You completely misunderstand mathematics as you misunderstand my proof. They need thinkers like you to uphold set theory. Regards, WM
From: mueckenh on 11 Jul 2006 03:52 Dik T. Winter schrieb: > > > > > > If that were true, there must be some n such that 10^(-n) equals 0. > > > > Indeed, if you consider 1.000... - 0.999... = 0 > > That is true because of the *definition* of that notation. If we assume 0.999... to be a well-defined (i.e. really existing) expression, then we need no "special definition" because (9+1) * 0.999... = 9 + 0.999... The only definition required (besides + and *) is how to express real numbers in the decimal system. > Without a > special definition that notation has no meaning. (There do not exists > things like infinite sums in arithmetic.) So the definitions: > 1.000... =(def) 1 + sum{i = 1..oo} 0/10^i > and > 0.999... =(def) 0 + sum{i = 1..oo} 9/10^i > where sum{i = 1..oo} is defined as lim{n -> oo} sum{i = 1..n}. > Note, again, the limit implied in the notation. And, at no place > do we leave the finite real. Archimedes already knew the infinite sum. It exists whether you think it is necessary to define a limit or not. And at no place it leaves the finite real. > > > Indeed, if you consider 1.000... - 0.999... = 0 and if you claim that > > all positions are enumerated by natural numbers, then there must be > > natural number n with 10^(-n) = 0. > > Why? Can you *prove* that? Because that would mean that there is a > natural number that is not finite, but there is no such number. If all places of representations like 1.000... and 0.999... would exist and could be indexed by natural numbers, then there were only those n by which these numbers differed: 10(-n) > 0. As they do not differ, we would have 10(-n) = 0. But you are correct, as there is no infinite natural number, there is no such place. And that is precisely the reason, why there are less than aleph_0 places available in 0.999... as well as in K = 0.111... . Hence K = 0.111... does not exist. > > > > There is an n such that for all p An[p] = K[p] No, there is not such a list number An which can index every p. But every p which can be indexed, can be indexed by one single list number Ap which also can and does index all p' < p. > What logical reasoning do you use to come from the first to the second? I > asked already twice before, and have not yet seen an answer. The reason lies in the fact, that in a linear sequence like this sequence of list numbers 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. But that is wrong, as you can see from the following proof: 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... 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. To put it in other words: If two list numbers are capable of indexing the first p digits of 0.111..., then always one of them is sufficient. If a list number An is capable of indexing some digit of 0.111... , then no other list number is required to index smaller indices. And if another list number Am with m > n is capable of indexing even more digits of 0.111..., then An is no longer required. If you believe that it could be of any use to employ more than one list number An for indexing any number, then you should be able to give an example where more than one An are required. Of course it must be a finite example, because there are, by definition, only finite list numbers An = natural numbers. Regards, WM
From: mueckenh on 11 Jul 2006 03:54 Dik T. Winter schrieb: > In article <1152450704.938408.69690(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Virgil schrieb: > > > What we have said, and what is quite true, is that for every natural > > > there is a larger natural. > > > > Of course. And therefore it is impossible to exhaust all of them or to > > find a set which is larger than all naturals together. > > But nowhere an attempt is made to exhaust them. There is only an axiom > from which it can be derived that there is a set that contains them all. > With that axiom, Cantor's argument is a proof. Without that axiom, > Cantor's argument is meaningless. When Cantor's proof was published, there was not such an axiom. > 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. > > > > Right! There are more than any finite number of finite naturals, indeed, > > > an endless supply of them which collectively form an infinite set. > > > > There is no "supply". The elements of a set do exist, instantaneously > > and immediately. Sets are static. > > Yup. So an axiom was needed to assert the existence of such a set. That was only a joke. Regards, WM
From: mueckenh on 11 Jul 2006 04:07
Dik T. Winter schrieb: > > And that is all you have to object against my proof? We have a fixed > > scheme of transpositions. > > It is my objection to what you see as the similarity. But it is not justified. If a fixed rule is given which determines for each rational when it will be included in the set ordered by magnitude (and if this is determined for every rational), then the order by magnitude is established. Remember: You believe in the existence of a well order, although the smallest rational larger than 1/2 and the largest rational smaller than 1/2 and so on do not appear. > > > But we could apply the transpositions even without any law, applying > > only the rule that a pair which is already ordered shall not be treated > > for a second time. > > I thought that was part of the rule (they are really conditional > transpositions), so what? "If, then"-decisions occur, just like in Cantor's diagonal proof. Yes, it was part of the rule. But the rule also includes a schedule which transpositions will be considered first. This schedule could be abolished. The transpositions could be chosen by chance as long as the order by magnitude was not yet achieved. If the infinite existed, then Tristram Shandy would complete his diary and he would complete the well-order by size too. From that you can obtain that the infinite does not and cannot exist. > > > Then we can apply countably many arbitrary > > transpositions until all elements are ordered by magnitude. The limit > > to which every path will lead is always the same: The set of rationals > > ordered by magnitude and by natural indices. > > Again, you use the word "limit" here, without definition. How do you > *define* the limit of a sequence of transpositions? How do you *define* > the limit of the set on which the transpositions are applied? The limit is the stable state where no further transpositions are applied, because there are no two elements remaining, which were not ordered by magnitude. > > > But there is no limit process. The *only* criterion about manipulations > > on countable infinite sets is whether one can determine *precisely* at > > which natural number something happens: After how many steps in a > > well-order of |Q the fraction 4711/235537 will appear, > > There are no "steps" in a well-order of Q. There is no notion of at > which natural number something happens. A well-ordering of Q means a > precise set of rules that determine the natural number to which a > rational corresponds. It is your re-ordering that requires some notion > of limit. "Step" was meant here as counting from n to n+1. A precise set of rules is not yet established for the algebraic numbers, as far as I know. > > > for instance, or > > in which line of Cantor's list a certain diagonal element will be > > placed and so on. > > This makes no sense to me. In the first place, it is not Cantor's list, > it is a given list. The first list was given by him. > In the second place I have no idea what you mean > with "diagonal element". A digit of the diagonal number. > But what is known is that the n-th digital > place of the diagonal is derived from the n-th element in the given list. The question is in which line the 20th 5 will appear as a digit of the diagonal number, for instance. > > > And I can determine *precisely* after how many steps > > the number 4711/235537 will be inserted in the order by magnitude with > > all of its predecessors of the initial well-order. > > Indeed, you can. But now you are talking about steps. In the previous > no steps were involved. So, you are talking about a sequential process, > which was not he case in the previous things. The well-ordering of the algebraic numbers without repetitions is also a sequential process. What is the problem of such a process if the sequences are determined? Even Cantor's diagonal proof is a sequential process because to find line number n you have to count from 1 to n - and counting is a sequential process. You cannot know line number n without knowing line number n-1. > A mapping that well-orders > the rationals is *not* a sequential process. The determination of the > n-th digit of the diagonal from a given list is *not* a sequential process. Wrong. You cannot know line number n without knowing line number n-1. But even if you were right, your argument would be void. If infinity would exist, then an infinite set could be exhausted by a definition like that of Cantor or that of mine. > > > This is fixed and > > can be calculated for any rational number. Therefore all rational > > numbers are covered and will successively appear in the well-order. The > > argument that there remain always infinitely many other rationals is > > wrong, because by definiton the fate of each and every rational is > > determined and can be calculated. > > Yes, for each rational number in the well-ordered list you can calculate > the step when it comes in place in a numerically ordered segment of the > rationals. But this does *not* mean that the final result is a well-order. > Because at no time can you calculate the place where that rational number > will be at the end. You need to show (at least) that there is a first > element in the final ordering. I show that infinity does not exist, because there will never be the first element of the order. > > But that is not interesting. It is easy to see that the diagonal is a > > sequence of the same sort as are the list entries. Whether they are > > real numbers is uninteresting. > > But it is just that part that is interesting. Try the same with a > sequence of algebraic numbers. You need to prove that what you get is > also an algebraic number (and you cannot). So for algebraic numbers > the proof fails. For real numbers the proof goes through, because you > can prove that the resulting number is also a real number. I proved in my special list even that the diagonal number is a rational. No, it was definitely not Cantor's intention to show that the diagonal is a real. He even emphasized the indepence of his arguing from real numbers: "Es läßt sich aber von jenem Satze ein viel einfacherer Beweis liefern, der unabhängig von |