An uncountable countable set
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From: mueckenh on 1 Jul 2006 09:16 Virgil schrieb: > > 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??? Cantor's diagonal proof does not prove that all numbers of the list are included in that sense that all are covered by the diagonal. Only to give you an example: If the list was longer than wide, the diagonal could not cover all list numbers. Further there is no limit consideration at all for the infinite. > > Cantor's diagonal proof does not prove that all numbers of the list are > > included. > > Perhaps not to you, but it does to most. That is not an argument for truth, rather the opposite. > > 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. The diagonal proof is not valid for an infinite list. There is no definition at all by Cantor. He only stated continuation. > You reject things that have been proved within that standard and insist > on things whose proof does not meet that standard. Same did Cantor 130 years ago. Therefore this arguing is invalid. > > So that what you claim is not acceptable as mathematics. > > Either equal rights for all or for none. > > Standards of proof are not democratic. Correct. And your arguing few lines above about the society of mathematicans is nonsense. But you seem not even to be able to recognize that you contradict yourself within few words. > > > > > > 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. That is the result. Which is the wrong assumption? > > > 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. That is the result. Which is the wrong assumption? > > > > > > Do you have an idea why? (What is wrong with my assumptions?) > > You assumed, in contradiction to what Cantor said, In *agreement* with what Cantor said. > that you could > convert between dense ordering and well ordering by a sequence to > transpostions. That is the result. Which is the wrong assumption? Regards, WM
From: mueckenh on 1 Jul 2006 09:22 Virgil schrieb: > 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. Of course it is infinite. And the number of nodes in this tree is a large as the number in the whole tree. > One can easily see that by considering that a > given node and its descendant nodes produces a tree isomorphic to the > original one. Where ever you look at the tree: you will see this pattern: | o /\ o o / \ / \ Obviously there are more edges than paths. Or can you believe that there are more paths than edges? How can you close your eyes in spite of this fact? > > An infinite binary tree is such that each node except the root node is > the child end of a branch Customarily they are called edges. > 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). This analysis is wrong. You must show that each path is different from each other path. In order to do so they must differ by nodes and edges. > Thus the set of paths bijects with the set of subsets of {1,2,3,...}. And therefore you need again Hessenberg's proof which, as we have seen, fails because of the imprdicable definition of K. So nothing is shown. > > And "Muecken"'s analysis of the cardinality set of paths shows that consistent calculations with the infinite are impossible. Can't you think with your own brain, but only repeat preconditioned stuff? Where ever you look at the tree: you will see this pattern: | o /\ o o / \ / \ Obviously there are always more edges than paths. Regards, WM
From: Gc on 1 Jul 2006 09:35 mueckenh(a)rz.fh-augsburg.de kirjoitti: > An uncountable countable set > > There is no bijective mapping f : |N --> M, > where M contains the set of all finite subsets of |N > and, in addition, the set K = {k e |N : k /e f(k)} of all natural > numbers k which are mapped on subsets not containing k. > > This shows M to be uncountable. > > Regards, WM The set M exist only if you have already defined exactly funktion f, because there is now certain set K before the funktion is exactly defined. So you have bijection g: N ---> M/K . Then you can add the set K to finite sets and get M. This is no paradox and this is very different than funktion f: N ---> P(N) because P(N) contains K by definition.
From: David Hartley on 1 Jul 2006 10:10 In message <1151758813.109076.7170(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de writes >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 have not proved - or even given any justification at all - that this process has any meaningful limit, nor that if it does, it is ordered as you wish. To see that this is not straightforward, consider the negative naturals, indexed by the corresponding positives, and apply your process. The first few steps give 2 1 4 3 6 5 8 7 ... 3 1 5 2 7 4 9 6 ... 4 2 6 1 8 3 10 5 ... 5 3 7 1 9 2 11 4 ... (Each step disrupts the "size" orderings set up by the previous step, so all possible transpositions are needed at each step.) The least element after n steps is n+1, so what's the least element of the limit ordering? 2 and 1 alternate indefinitely with each successive step, which comes first in the limit ordering? Can you give any reason why your process should be any more successful with an enumeration of the rationals? -- David Hartley
From: Daryl McCullough on 1 Jul 2006 10:11
mueckenh(a)rz.fh-augsburg.de says... >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. Yes, you said that, but it is convoluted nonsense. For every mapping that exists, K(f) is defined. K(f) is only undefined if f is undefined. You are asking (in a convoluted way) Find a map f from R to P(N) such that K(f) is in the image of f. If K(f) = { x in N | x is not in f(x) }, then there certainly *is* such a map. There is no contradiction. What we can prove about K(f) is this: There is no k in N such that f(k) = K(f). So f is not a surjection from N to P(N) But the assumption here is that f is a surjection from R to P(N), not from N to P(N). So we have: 1. There is no k in N such that f(k) = K(f). 2. There exists k in R such that f(k) = K(f). What follows from that is that k must be a real number that is *not* a natural. Once again, K(f) is *always* defined, for every function f. There is no k in N such that f(k) = K(f). So f is not a surjection from N to P(N). But f may very well be a surjection from some set *other* than N. -- Daryl McCullough Ithaca, NY |