Cantor Confusion
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From: Andy Smith on 26 Jan 2007 04:37 davidmarcus(a)alum.mit.edu writes >> I'm not sure what you mean by "permutation". Do you mean enumeration or >> listing? Maybe you mean bijection (a map that is one to one and onto). >> Let's be more precise: >> >> Let's suppose that by "permutation" you mean a bijection . Suppose p is >> a permutation of [0,1] and f: N -> R is a surjection. Let g be defined >> by g(x) = p(f(x)). Then g is a surjection N -> R. > >> On the other hand, if h: N -> R is not a surjection, why should this >> imply that no surjection exists? > Because if there exists a surjection f:N->R for h: N->R not to be a surjection would imply that there is no p such that h=p(f). ? You say surjection - aren't we talking about bijections anyway when we talk about countability? If the reals are countable, you should be able to give any real and I can tell you its natural number index in my catalogue; conversely you can give me an index and I will tell you the corresponding real. Isn't that a bijection - a one-to-one correspondence? -- Andy Smith
From: mueckenh on 26 Jan 2007 04:37 William Hughes schrieb: > > > S can be and is larger than L. S is not countable. > > > That again is an unjustified and provably wrong assertion. > > The sequences belong to the countable union of the finite trees. > > Take any sequence s in S. We know that each initial segment > of a s belongs to a finite tree t(s). However there is no single > finite tree t_D such that every initial segment of s belongs to t_D. > s does not correspond to a single finite tree, > s corresponds to a sequence of finite trees. We are not interested in a finite single tree. Every initial segment s of a path has lengths n, where n is in N. If the union of all initial segments of a path is infinite, then you should try to explain how an infinite number could creep in and can exist unrecognized as a finite natural number among the natural numbers. > > > *Everything* in this union is countable. > > > Yes, but s is a sequence of things from the union. The set of > infinite sequences from a countable set is not countable. That is a different matter. Each of our sequences has a length n. Their union is finite (i.e, not acually infinite) unless there is an infinte number in N, which is impossible. Regards, WM
From: mueckenh on 26 Jan 2007 04:49 On 25 Jan., 17:03, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1169736539.298735.176...(a)v33g2000cwv.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > In article <1169728742.271970.13...(a)v33g2000cwv.googlegroups.com> "William Hughes" <wpihug...(a)hotmail.com> writes: > > > > On Jan 25, 3:19 am, mueck...(a)rz.fh-augsburg.de wrote: > > > ... > > > > > and S is the set of corresponding sequences > > > > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > > > > then L cannot be larger than S, because there cannot be more limits > > > > > than sequences. > > > > > > > > Note that S contains only infinite sequences. > > > > > > S is not a set of sequences. S is an ordered multi-set of the digits > > > 0 and 1. Or, alternatively, S is one of the sets {0, 1}, {0} or {1}. > > > > Sorry. S contains for instance the paths 0.111... and 0,010101... the > > bits (nodes) of which can be indexed by natural numbers. > > Yes? The elements of S as written above are the a_ik for i and k natural > numbers. No. An element of a set S is written as x in S = {x| ...}. > So the elements are elements from {0, 1}. S *does* contain > those paths, but not as elements, but as subsets. S is not the set of > "corresponding sequences". The set of all cars contains every car as an element. But from the elements of this set we can form the set of all brands. The set of all brands contains 10 or 20 elements only. > You specifically wrote: > > However, if L is a set of > > limits L_k > > L = { L_k | k in N } > > and S is the set of corresponding sequences > > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in N } > > then L cannot be larger than S, because there cannot be more limits > > than sequences. > this makes no sense if S is *not* a set of sequences. S is a set of sequences. > Consider the > following: > L = { L_k | k in {0, 1, 2}} > S = { a_1k, a_2k, a_3k, ... | a in {0, 1}, k in {0, 1, 2} } > If S is a set the cardinality of S is at most 2. Cardinality is |S| = 3. Its elements are the sequences number k = 0, 1 , and 2. > If S is an ordered > *multi-set*, the ordinality is omega. The cardinality of L is 3. > S is certainly not an ordered set because it contains multiple identical > elements. S is ordered by the numbers k. Regards, WM
From: Franziska Neugebauer on 26 Jan 2007 04:51 mueckenh(a)rz.fh-augsburg.de wrote: [...] > If you wish we could use different symbols for the tree > and for its set of nodes, but I do not consider this necessary or > useful The set of nodes _is_ not the tree. Not using different names for different things (tree vs set of nodes) is a fallacy of equivocation. Why is it not an appropriate representation of a tree to call its set of nodes the tree? [...] >> So you commit to David's trees (std-trees + tree-union). > > No. I do not read the writings of such writers. I commit to my > definition, again outlined above. Which is not well-enough defined. >> We cannot say so. *In* std-trees there no paths. The set of paths is >> induced by the tree. > > The trees which I use, uniquely define their paths, which are in these > trees. *How* do they? >> > 4) The set of paths in T(oo) is a subset of the countable set of >> > finite sets of all paths in the finite trees. > >> No. The set of paths of T(oo)/G is the full set of paths. > > Counterexample: Paths with length 1 are not contained in any tree T(n) > with n > 1. *Define* path in a tree formaly. Then we can judge. >> > ==> The set of all path is countable. (==> The real numbers are >> > countable.) > >> Non sequitur. Equivocation fallacy. > > Are you really not able to understand that the set of real numbers is > proven countable if the set of real numbers in some interval is proven > countable? You neither proved the former nor the latter. Since I do not have your Third Eye your claim is indeed not immediately obvious. >> > Going on, we can say: >> >> > 6) T(oo) = T contains only finite paths. > >> No! > > Please give a clear statement: Would you say that the union of all > initial segments of paths with length n is a finite path? Define what a path is. Check which definition applies: [ ] A sequence of nodes. [ ] A sequence of edges. [ ] A set of edges. [ ] other, please define: ________________________________ > Yes or no? > If yes, then you should prove your claim by giving an upper bound as > is usual in finity proofs of set theory (where the only alternatives > are finity and actual infinity). > If no, then you should try to explain how an infinite number could > creep in and can exist unrecognized as a finite natural number among > the natural numbers. Define what path means, please. >>The set of paths of T(oo)/G is complete, i.e. contains really all >> paths. > > That is not a contradiction to be introduced by "no". N contains all > natural numbers. Nevertheless all natural numbers are finite. T(oo) > contains all initial segments of paths with length n, where n in N. What does "contain" mean? F. N. -- xyz
From: Franziska Neugebauer on 26 Jan 2007 05:06
mueckenh(a)rz.fh-augsburg.de wrote: > On 25 Jan., 15:34, Franziska Neugebauer > <Franziska-Neugeba...(a)neugeb.dnsalias.net> wrote: context: >> ,----[ <45b5ec2c$0$97243$892e7...(a)authen.yellow.readfreenews.net> ] >> | >> Again: Your notations >> | >> >> | >> T(1) U T(2) U ... >> | >> >> | >> and >> | >> >> | >> U {T(i) | i e N } >> | >> >> | >> are undefined. >> | > >> | > You are in error. The union of the trees T(n) and T(n+1) is >> | > defined. n is a natural number. Therefore the union of all finite >> | > trees is defined. >> | >> | You have misunderstood the induction principle. It is not made for >> | "counting over to the infinite". >> `---- >> >> >> You have misunderstood the induction principle. It is not made >> >> >> for "counting over to the infinite". >> >> >> > It is valid for all existing natural numbers. Counting over to >> >> > the infinite is nonsense.[indentation corrected] Counting occurs >> >> > in the finite. >> > > What you do _is_ "counting over to the infinite" >> >> > By no means! Every n in N is finite. > >> The set of finite trees V* = { T(n) | n e N } is not a finite set. >> Induction over its members does not prove a property of V*. >> This is quite the same as in the case of the infinite set >> N = { n | n e N }. Induction over n for some property p(n) does not >> prove p(N). Maybe you forgot that. > > I am not interested in any property of what you call the set N. I am > interested in a property due to each and every natural number. Believe > it or not, it is sufficient to show by induction the finiteness of > every natural number in order to prove that every natural number is > finite. Whether you believe that the set N has other properties is > your personal opinion. You have obviously been interested in the "union of all finite trees" which still waits for a definition. > It is irrelevant for the consideraton of trees. It is highly relevant to show where your reasoning is malfunctioning. > In particular we can prove by induction that the tree, formally called > T(oo), 1. Define T(oo). 2. Prove that the "formally called" entity T(oo) is really a tree. > contains all paths if it contains all levels enumerated by > natural numbers. 1. Define "path". 2. Define "contains". > We have no use for your set N in the tree (in case it > should differ significantly from the ensmble of all natural numbers.) Define "ens[e]mble of all natural numbers". F. N. -- xyz |