An uncountable countable set
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From: Dik T. Winter on 25 Jun 2006 23:21 In article <1151135467.925502.14460(a)b68g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > This mapping is impossible. (See my article > > > "On Cantor's Theorem", arXiv, math.GM/0505648, 30 May 2005.) > > > > Consider two cases: > > 1. > > a -> {} > > 1 -> {1} > > K =3D {} > > so K is in the image > > 2. > > a -> {1} > > 1 -> {} > > K =3D {1} > > so K is in the image. > > K is not in the image of *a natural number*. a is not a natural number! Why should it be? The question was about a bijection between {a, 1} and {{}, {1}}. The pre-image does not consist of natural numbers, so why should K be the image of a natural number? > > > In both cases there is this impredicable request {f, k, K} which is > > > impossible to satisfy. But in the proof by Hessenberg, you insist, it > > > would prove non-surjectivity? > > > > Wrong. In the first case the condition can be satisfied, in the second > > case it can not be satisfied. > > The request {f, k, K} cannot be satisfied in any case,. That is > completely independent of surjectivity. It is not. It is surjectivity in the case N -> P(N) that requires such a triple to exist. > > Give a quote please. Where did he state that? I will translate, please correct me if I am wrong. > p=2E 214: "Die Frage, durch welche Umformungen einer wohlgeordneten Menge > ihre Anzahl ge=E4ndert wird, durch welche nicht, l=E4=DFt sich einfach so > beantworten, da=DF diejenigen und nur diejenigen Umformungen die Anzahl > unge=E4ndert lassen, welche sich zur=FCckf=FChren lassen auf eine endliche > oder unendliche Menge von Transpositionen, d. h. von Vertauschungen je > zweier Elemente." The question through which transformations of a well-ordered set can change the number of elements, and through which not, is easily answered as follows. Any transformation that leaves the number of elements unchanged can be brought back to those that allow a finite or infinite number of transpositions (e,g, interchange of two elements). Can you give a source where Cantor states that a well-ordered set remains well-ordered after infinitely many interchanges? You stated he did. Where? -- 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 25 Jun 2006 23:24 In article <1151153602.449685.154410(a)u72g2000cwu.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > Such a function exists from |N --> |R for all real numbers which can be > individualized, i.e. which are really real numbers. It was my aim to > show this with your arguments. Back to the start. Start your own mathematics with your own axioms, and see what you get. The term "really real numbers" has no definition in mathematics. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 26 Jun 2006 06:36 Virgil schrieb: > > Any countable alphabet has only countably many finite words (cp. the > > set of finite subsets of |N). Any alphabet that we can use is countable > > (in fact it is finite). Hence we have only countably many symbols, > > names, notations. We can distinguish only countably many numbers. > > > > Further the set of elementary cells of the whole universe is finite. > > Do you have some personal line from God's mouth to your ear which allows > you to "know" things that no one else knows? We may strongly suspect > that the universe is finite, but we do not know it. We know that that part of the universe, which can be used by us to store bits, *is finite* and will remain so forever. > > > But mathematicians in general do not like to be frustrated by physical > > needs. > > We get hungry and thirsty like anyone else. > > > > > >> It shows that every list of reals, there is at least one > > > >> real that is not on the list. From that, it follows that > > > >> there is no list containing all real numbers. > > > > > > > >No. > > > > > > Yes. > > > > The list of all names contains all we can denote as a number. > > The list of all reals which can be represented by unending strings of > digits. We cannot write or read an unending string. The assumption, such strings existed, is wrong. > > > > >which can be remedied by switching the mapping from f to g and this > > > >is possible for every real number constructed. Therefore there is no > > > >uncountable set of reals. > > You are saying that because there is a function on any list of infinite > strings of digits which will produce a string not in that list, that > there is another rule that will always produce a string in that list. > > I am not at all sure that your claim is true, but I am quite certain > that it is irrelevant, even if true. It is true (include that very string into the list) and it is very relevant. > Of what relevance is the fact that one can show that certain numbers ARE > listed in considering whether certain numbers are NOT listed? All numbers you will ever produce can be listed. That is enough. > Now for every subset, S, of N (member of P(N)), the rule is to use rule > 1 above if n is a member of S and rule 2 if N is not a member of S. > > It should be clear that each subset of S produces a non-member of a > given list and that futhermore different subset produce different > non-members. > > Thus for any list of reals, we have shown that there is 1 non-member of > that list for each member of P(N). > > Thus removing any list of reals from the set of all reals leaves as many > reals as before. And they all can be bijected with |N. > If you want to do only constructionist mathematics, you are in the wrong > place. Above, you constructed something. > > > > forall f from N to any set, K(f) is not in the image of f . > > That is OUR point! Which you conceded only with great reluctance. That is exactly the same point as in Cantors diagonal argument. K(f) is not in the image of f. <==> The diagonal number is not in the list (image of f: |N --> |R). Regards, WM
From: mueckenh on 26 Jun 2006 06:49 Virgil schrieb: > > Cantor's general proof breaks down by one counter example like this > > one: Construct the diagonal number of the following list by digit > > replacement 0 --> 1 (only the digits behind the point are to be > > considered): > > > > 0.0 > > 0.1 > > 0.11 > > 0.111 > > ... > > > > It is D = 0.111... > > > > Do you really believe that actually infinitely many ones (1) can be > > contained in 0,111... without exactly as many appearing in the list > > numbers? I think in the limit of infinity, there is no difference between aleph_0 and aleph_0 -1, but I know that the limit of a monotonic sequence is not a member of that sequence. > Only finite sequences of 1's are included in your list. > You are making the common error of assuming that the limit of a sequence > is necessarily a member of the sequence. I do not believe that 0.111... is in the list. But I do not judge at all. If it were in the list, then Cantor's proof would break down. If it is not in he list, which contain every sequence of ones, which can be indexed by natural numbers, then there are digits not defined by natural numbers, hence undefined. > > You are making TO's vulgar false assumption that in order for a set of > naturals to be infinite, some of its members must be infinite. That is > no more true of sets of naturals than it is for sets of rationals or > sets of reals. Of course it is true, but we need not dicuss it here. Either 0.111 has more ones than any list number, then it is undefined, or it has not, then Cantor's proof breaks down. There are only these two alternatives. > > > > So, whatever is your choice: There is a contradiction: Either 0.111... > > differs from any list number and, therefore, does not exist at all, or > > it does not differ from any list number; then the diagonal proof fails. > > You left out the only valid case in which 0.111... = 1/9. > > Non-terminating decimals are defined as the limits of sequences of > terminating decimals as the number of digits increases without bound. Cantor's proof requires hat the diagonal number is a number which differs at least at one place which can be identified from each of he list numbers. There is no room for cloudy epsilons. > > Your reasoning does not hold when one regards non-terminating decimals > as infinite series. Cantor's proof requires hat the diagonal number is a number which differs at least at one place which can be identified from each of he list numbers. There is no room for cloudy epsilons. Regards, WM
From: mueckenh on 26 Jun 2006 06:53
Virgil schrieb: > In article <1151225575.137720.193510(a)u72g2000cwu.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > So either "mueckenh" must be declaring the set of rationals uncountable > > > or he must admit that they can be well-ordered. > > > > They are neither well-orderable nor "countable". Because they just do > > not exist. > > They do in my world. You may believe it, but it is wrong. The number [pi*10^10^100] does not exist in any wordl. > > > I don't want to start a new discussion here. But in case you are > > interested: Elements of a set must be distinguished from one another. > > In the universe there are less than 10^100 bits which can be use to > > distinguish two elements. > > > That assumes that we live in a finite universe, which may be suspected > but has not been proved. The universe which we can utilize for storing bits it finite. > > In any case, it does not disrupt the purely mathematical theory of > infinite series. It does disrupt the arbitrarily small epsilon and the arbitrarily close approximation of pi etc. Regards, WM |