An uncountable countable set
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From: mueckenh on 2 Jul 2006 14:22 Virgil schrieb: > > Of course it is infinite. And the number of nodes in this tree is a > > large as the number in the whole tree. > > What do you mean by "the number in the whole tree"? If you mean the > number of paths, you are wrong. The number of nodes in the subtree is as large as in the whole tree. > > > > > > > Where ever you look at the tree: you will see this pattern: > > > > | > > o > > /\ > > o o > > / \ / \ > > The set of reals between 0 and 1 is uncountable. That is an assumption established by three invalid proofs. You shall look at the pattern above and not repeat preconditioned knowledge like Pawlow's dog. > The set of all their > binary representations is a superset of that, since some reals have dual > representations. > > There is a bijection from the set of binary representations to the set > of all paths in an infinite binary tree, in which a 0 in the nth binary > digit of the real means a left branch from the nth node and a 1 means a > right branch. Bring your nose in front of the center N of the picture, in about D = 25 cm distance from the screen (unless you are short sighted - in that case decrease D to 10 cm). | o /N\ o o / \ / \ And then tell me what you can see. Not what they have told you at school. > > > Obviously there are more edges than paths. > > Infinitely many paths contain each "edge", which I understand to mean a > branch connecting one node to the next. Each path is a set of edges > (branches) connecting succesive parent nodes to child nodes. > > Or can you believe that > > there are more paths than edges? > > Yes, at least for infinite trees. At least? For any finite tree the opposite is easy to prove. "At most" would be better but wrong nevertheless. > > > How can you close your eyes in spite of this fact? > > My question is: How can you close your eyes to the facts of INFINITE > trees. The fact is: Unless a single path contains an edge which no other path goes through, you cannot speak of a single path. Hence there is not even one single path and there are not uncountably many single paths. > > >> > Thus there are as many paths as different subsets of N. The fact is: Unless a single path contains an edge which no other path goes through, you cannot speak of a single path. Hence there is not even one single path and there are not uncountably many single paths. Regards, WM
From: mueckenh on 2 Jul 2006 14:24 David Hartley schrieb: > >Then there must be a ->first<- transposition after which there is no > >longer any first element in the current ordering of the rationals, if > >the final result is to be the usual ordering of the rationals. > > > >But that is clearly impossible. > > I'm afraid this argument is on a par with "each set {1,2,..,n} has a > maximum, therefore N has a maximum". I don't think WM's construction > works, but even if it does, it proves nothing because sequences of > transpositions needn't preserve well-ordering. But my example preserves the bijective mapping between |N and |Q. This is enough to show that its result is in contradiction with set theory. Regards, WM
From: mueckenh on 2 Jul 2006 14:30 Dik T. Winter schrieb: > > Do > > you need some limit in order to well-order the set of rationals? > > No, but that is unrelated. Transpositions operate on a set. A sequence > of transposition hence also operates on a set. There is *no* definition > how an infinite sequence of transformations operates on a set. These transpositions *are* a set. In my example, they do not destroy the bijective mapping between |N and |Q. > >> Well, in the first place, this is false as written. The transpositions > are conditional. But indeed, given the well-ordered set of rationals > Q1, this gives a different ordering of the rationals, say the well-ordered > set Q2. We can call that an infinite permutation. But all transpositions to be applied to a given set are completely defined by the few lines of my definition and they are determined from the scratch. There is not a single one undetermined. > You keep assuming that the definition of the anti-diagonal of a list of > reals is an ongoing steps. That is false. There is only a single step, > the definition. The same is true for my example. There is only one single definition. If you do not like "steps" then drop that word. > After that we can determine each and every digit of > the diagonal number without the need to know other digits. Moreover, > by Cauchy, such a sequence of digits defines a real number. Leave Cauchy out of the play. There is no justification to reach all last digits of the diagonal number by his epsilons. His specification would only show that all columns of the list up to nn have been covered. > > I do not know of any definition of a sequence of overlapping "infinite > permutations" that allows an inifinite sequence. Pray provide one. There are no "overlapping" permutations. > > Because there are not infinitely many comparisons, there is a single > definition. This single definition leads to infinitely comparisons, unless the numbers in the list are all known. My example provides also one single definition. > > Ask Cantor. > > No, I ask you. > I do it just like Cantor does, by one single definition. Regards, WM
From: David Hartley on 2 Jul 2006 15:26 In message <1151864680.790770.34600(a)h44g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de writes > >David Hartley schrieb: > >> >Then there must be a ->first<- transposition after which there is no >> >longer any first element in the current ordering of the rationals, if >> >the final result is to be the usual ordering of the rationals. >> > >> >But that is clearly impossible. >> >> I'm afraid this argument is on a par with "each set {1,2,..,n} has a >> maximum, therefore N has a maximum". I don't think WM's construction >> works, but even if it does, it proves nothing because sequences of >> transpositions needn't preserve well-ordering. > >But my example preserves the bijective mapping between |N and |Q. This >is enough to show that its result is in contradiction with set theory. You have never actually defined what you mean by the limit of your process, so to avoid misunderstanding I state what I mean. If <_1, <_2, ... is a sequence of (total-)orderings of a set X, define <_L by for any x, y in X, x <_L y iff there exists n st. x <_m y for every m > n In general, this is a partial ordering, but if it is a total-ordering we say that the sequence has a limit, and define the limit to be <_L. Similarly, if f_1, f_2, ... is a sequence of mappings A -> B, we can partially define a function f: A -> B by for each x in A, f(x) = b iff there exists n st. for all m > n, f_m(x) = b If this is defined for every element of A, we can say that the sequence has a limit, and it is f. Now, granting that your process works as you hope, you get a sequence of well-orderings of the rationals whose limit is the usual ordering. But, the corresponding sequence of bijections N -> Q does *not* have a limit. (You need only consider the images of 1, these diverge to -oo.) Your example does not "preserve the bijective mapping between N and Q". -- David Hartley
From: Gc on 2 Jul 2006 15:35
mueckenh(a)rz.fh-augsburg.de kirjoitti: > Gc schrieb: > > > 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. > > > No, it does not. P(N) cannot contain K by definition, because K does > not exist. Yes it does. For every function N to P(N) there exists K in P(N). >You can see this by the following argument: 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. > > Here we have not at all any problem with too less elements in the > source. Nevertheless there is no surjection possible. Virgil has already said that your condition is irrelevant. In your mapping there is now reason that a natural number has to mapped with K e P(N). > Regards, WM |