An uncountable countable set
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From: Dik T. Winter on 11 Jul 2006 08:09 In article <1152604358.062964.188890(a)s13g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. So you have first to define addition and multiplication from infinite strings. How do you do that? > > 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. In mathematics it is necessary to define things. > > > 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. I see no proof below, only vague remarks: > If all places of representations like 1.000... and 0.999... would exist > and could be indexed by natural numbers, Well, we have the set of natural numbers, the first number is represented as d0 = 1, d_i = 0 for i > 0, the second as d0 = 0, d_i = 9 for i > 0. > then there were only those n > by which these numbers differed: 10(-n) > 0. As the representations differ in *every* digit position. > As they do not differ, we > would have 10(-n) = 0. This is incomprehensible nonsense. The numbers they represent are the same, it is the representations that are different. So how I defined the number represented by such a representation above. > 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... . This is again incomprehensible. > Hence K = 0.111... does not exist. As I defined that notation above as meaning: sum{i = 1..oo} 1/10^i = lim{n -> oo} sum{i = 1..n} 1/10^i = = lim{n -> oo} 10^(-1) * (10^(-n) - 1) / (10^(-1) - 1) = = lim{n -> oo} (1 - 10^(-n))/9 = 1/9 - lim{n -> oo} 10^(-n) / 9 = 1/9 you are arguing that 1/9 does not exist. (When you want to attack something in methematics it is better to use the definitions mathematics supplies.) > > > > 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 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. > > 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. 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 11 Jul 2006 08:11 In article <1152604493.362026.129000(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. Indeed, at that time it was not yet an axiom, but the thought was present that infinite sets did exist. > > 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: georgie on 11 Jul 2006 08:40 mueckenh(a)rz.fh-augsburg.de wrote: .... > You completely misunderstand mathematics as you misunderstand my proof. I think you mean that he misunderestimated you.
From: georgie on 11 Jul 2006 08:57 Dik T. Winter wrote: .... > 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. Really? I would like to see the proof that the axiom of infinity does not lead to an inconsistency. Would you please explain?
From: Dik T. Winter on 11 Jul 2006 09:12
In article <1152605227.458575.197240(a)35g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > 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. That makes no sense as a definition, as you will never reach that stable state. > > 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. But it is for the rationals. And I think it can also be found for the algebraic numbers, but it will take a bit more care. > > > 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. Ok, perhaps. > > In the second place I have no idea what you mean > > with "diagonal element". > > A digit of the diagonal number. What do you mean with a sentence like "in which line of Cantor's list a certain digit of the diagonal number will be placed"? > > 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. Well, an interesting question, perhaps, but irrelevant to the discussion. > > > 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. The mapping from the naturals to the list gives the n-th element. It is in the *definition* of the list. > > 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. The mapping gives that. > 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. Makes no sense. The natural numbers can not be exhausted by a sequential process. > > 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. No, you do not show anything of the sort. You only show that in the limit well-order is destroyed (by some definition of limit). But there is no problem with that. I have already shown (with definitions of limit) that you can destroy well-order of the naturals by an infinite sequence of transpositions. Nobody has a problem with that, as it is well known that what is the case in the limit is not necessarily what is the case outside the limit. > > > > 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. I wonder whether it was a proof or just some handwaving. > > > Interesting is that such sequences are > > > uncountable. > > > > A sequence is *never* uncountable. By definition of the words sequence > > and uncountable in mathematics. > > I said "sequences". Ah, I misread. Interesting that is just what the first version of Cantor's proof did show, without any reference to numbers. > > > But in order to prove that the diagonal differs from every entry, there > > > a limit is required but not available. > > > > No, it is just that place where a limit is not required. By definition, > > for every n in N, the n-th digit in the diagonal will be different from > > the n-th digit of the n-th element of the list. I do not see why limits > > are needed here. > > The reas |