An uncountable countable set
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From: Virgil on 30 Jun 2006 14:38 In article <1151668048.884373.189170(a)i40g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > If actual infinity exists, then my mapping comes to an actual end. > > > > But that "end" does not prove that the well ordering of the rationals > > end up as the natural ordering. > > > > To establish your claim, you must prove that you have an infinite set > > that is well ordered and densely ordered under the same ordering. You > > have not done that. > > Cantor's diagonal proof does not prove that all numbers of the list are > included. All number of the list are certainly included in the list. What else are they supposed to be included in??? > > > > How does sqrt(2) = m/n show that sqrt(2) is rational? It does not show > > > that but it shows that sqrt(2) does not exist as a rational. My proof > > > shows that infinity can never be completed. > > > > Your "proof" hasn't proved it to me, as there are essentials omitted. > > Cantor's diagonal proof does not prove that all numbers of the list are > included. Perhaps not to you, but it does to most. The society of mathematicians is limited of those who agree to accept whatever is proved according to certain generally accepted standards of logical proof and not to insist on things whose proofs cannot meet that standard. You reject things that have been proved within that standard and insist on things whose proof does not meet that standard. So that what you claim is not acceptable as mathematics. > Either equal rights for all or for none. Standards of proof are not democratic. > > > > There are not infinitely many values between two values, because the > > > set is and remains well-ordered, i.e. each element is indexed by a > > > natural number. > > > > Then the set of rationals never becomes naturally ordered, which would > > require exactly what you say never occurs. > > Hence there must be one assumption which is wrong, isn't it? The assumption that any sequence of transpostions can convert a well ordered set into a dense set, or vice versa. > > Hence there must be one assumption which is wrong, isn't it? > The only assumption I made is the existence of infinite sets. You also assumed that you could convert between dense ordering and well ordering by a sequence to transpostions. > > > You are claiming a sequence (or at least a well ordered set) of > > transpositions applied to a well ordered set of all rationals will > > produce the densely ordered set of rationals in their normal ordering. > > > > If that were true, the set S = {x in Q: 0 < x < 1} would have to be > > originally finite but terminally infinite under your process. > > > > Then, by well ordering, there would have to be a first transposition for > > which S is infinite. This cannot happen. > > Do you have an idea why? (What is wrong with my assumptions?) You assumed, in contradiction to what Cantor said, that you could convert between dense ordering and well ordering by a sequence to transpostions. Since you have no proof for that assumption, it is quite reasonable to consider such an assumption to be an error, or at least one of your errors.
From: Virgil on 30 Jun 2006 15:07 In article <1151668384.498485.166640(a)x69g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > If the paths are not more than the nodes > > > > It is only for finite trees, in which every path has a leaf (or > > terminal) node, that the number of paths is less that the number of > > nodes. For infinite binary trees, it is false. > > > We consider only infinite paths without ends in a tree without leafs. > We apply the simple rule which must every path obey: In order to split > into two paths, two nodes are needed. Hence, the number of paths cannot > double unless the number of nodes at doubles at least. The number of child notes IS double the number of their parent nodes, at each generation. But the "number" of infinite paths through any node equals the "number" of infinite paths starting at the root node, and is therefore necessarily infinite. One can easily see that by considering that a given node and its descendant nodes produces a tree isomorphic to the original one. Since half of infinite is still infinite > > The set of paths in an ->infinite<- binary tree, in which no path ends, > > is uncountable, provably of the same cardinality as P(N). > > Therefore, this proof shows that set theory is inconsistent because > another valid (and very lucid) proof leads to the opposite result. The set of infinite paths through any node is easily bijectable with the set of paths through any other node, and both are bijectable with the set P(N). An infinite binary tree is such that each node except the root node is the child end of a branch descending from a single parent node, each node is parent node descending by branches to exactly two child nodes, customarily called a left child and a right child, and the descending branches from a parent node to it cild nodes are called left and right branches respectively. A path in such a tree is an endless sequence of alternating nodes and branches starting at the root node and proceeding from parent to child endlessly. Each path can be identified by the sequence of left verus right branches it takes. So each path can be identified with a subset of S = {1,2,3,...} by including n in the subset if and only if the the n'th branch is a left branch (one could equally well have chosen to include n for right branching, instead of left ones). Thus the set of paths bijects with the set of subsets of {1,2,3,...}. And "Muecken"'s analysis of the cardinality set of paths fails.
From: mueckenh on 1 Jul 2006 09:00 Dik T. Winter schrieb: > I wrote: P(k) = lim{k -> oo} (k,k+1)(k,k+2)...(k,k+n) where I explicitly > wrote that I used some intuitive form of limit here. This is the only > way I can make sense of "infinitely many transpositions". A transposition is an element of the set of transpositions, i.e. a set of pairs of numbers. The set af rationals is also a set of pairs of numbers. Do you raise any questions concerning this infinite set? Do you need some limit in order to well-order the set of rationals? > If what I > wrote is not valid, please provide me with a definition of "infinitely > many transpositions". How do they operate on a set all together? Does > the result depend on the order? So there are many questions. 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 state so (and I think it is true), but a quote from Cantor is not a > > > proof, > > > > Cantor may have been wrong here. But my example is not. > > Eh? Your example just lists infinitely many transpositions without any > definition of meaning. There is the same definition of "infinitely" as in Cantor's diagonal proof. See his original paper "Über eine elementare Frage der Mannigfaltigkeitslehre" , Jahresbericht der Deutsch. Math. Vereing. Bd. I, S. 75-78 (1890-91). There is not a single word said about the meaning of infinitely many comparisons. > So you define a series of mappings. But the series is infinite. How do > you define such an infinite series? Ask Cantor. > 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. I am not an advocate of infinity. Regards, WM
From: mueckenh on 1 Jul 2006 09:04 Dik T. Winter schrieb: > Eh? K(f) is defined regardles the kind of mapping of f. How can you > come to the conclusion that it is not defined for some particular f? > What part of the definition fails for some particular f? 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. > What part of the definition fails? 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 are confused. K(f) is required to be the image of a natural number. > > That is a complete misstatement of the Hessenberg condition. To be clear, > let's have given a set A and a set of sets B (I do not yet state what the > elements of A and B are). Let's also have a mapping f from A to B. > Now construct the set K(f) = {x in A and x !in f(x)}. This set is > well-defined. Remember what I said: The set K e P(|N) which contains all natural numbers which are not mapped on sets containing them. Regards, WM
From: mueckenh on 1 Jul 2006 09:11
Virgil schrieb: > > > 5. Therefore, f is not a surjection. > > > > The existence of one or more non-surjective mappings doesn't tell us > > anyting about cardinalities. > > But the non-existence of any surjective mappiings, combined with the > existence of at least one injective map, does! If this non-existence can be proved. But up to date it has not been proved. > > And in the set of all mappings from N to P(N), there are injections but > no surjections. The latter is an empty assertion. > > > > There is no surjection from any set S to its power set P(S). > Proof: > For every set S and every finction f: S --> P(S), > K(f) = {x in S: not x in F(S)} is a member of P(S). That leads to the same paradox as in my example with |N and P(|N). It does not prove anything. > > You are confused. K(f) is required to be the image of a natural number. > > You are the one confused. In the above, K(f) is only required to be the > image of a real. But K(f) will not, in general, even have to be a member > of P(N), since there is nothing to prevent it from containing > non-natural reals, like pi. Look up *my* definition. I did define the set K(f): 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. > K(f) is not here a subset of N because it contains real numbers not in N. Look up *my* definition of K(f): 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. This K(f) is not defined for any surjective mapping of one or some or all natural numbers on sets containing sets of natural numbers. Regards, WM |