Cantor Confusion
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From: mueckenh on 8 Feb 2007 03:07 On 7 Feb., 16:02, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Feb 7, 9:29 am, mueck...(a)rz.fh-augsburg.de wrote: > > > For all for finite natural numbers > > > > n = |{2,4,6,...,2n}| < 2n > > > > Now take the limit as n-->oo (the < becomes <= in a limit, > > > No. n < 2n is always true for natural numbers (which property excludes > > 0). There is no reason to assume that 1/2 becomes 1 in "the limit". > > > > lim[n-->oo] n = oo, lim[n-->oo] 2n = oo) > > > > oo <= oo > > > > There is no contradiction. > > > Wrong for all finite numbers. > > And since the limit is not a finite number Every number which can appear in this inequality is a finite number. If something can appear in place of n which is not a finite number, then you agree that N contains infinite numbers. But that is wrong for any set N, be it actually or potentially infinite. > the fact that there > is a contradiction for all finite numbers does not mean that > there is a contradiction for the limit. Potentially infinite means here only that the value represented by n can become as large as we like. > > > > > Can you answer the following question yes or no? > > > > Is the potentially infinite set of finite even numbers > > > a set of finite even numbers? > > > Yes, I can. The answer is yes. Probably set theory gives another > > answer. > > Let E be the potentially infinite set of finite even numbers > > You have now made three claims > > -every set of finite even numbers contains numbers which > are larger than the cardinal number of the set True. > - E is a set of finite even numbers Yes, but not a complete or finished set as set theory requires. > - the statment "E contains numbers which are larger > than the cardinal number of the E" is false No. The answer is not so simple. The statement can be proved correct for an actually existing set E with a cardinal number |E| which is a number that can be compared by size with natural numbers. The statement is false for a potentially infinite set E because such sets do not have cardinal numbers. Regards, WM
From: Virgil on 8 Feb 2007 03:21 In article <1170920049.191525.283730(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 6 Feb., 22:30, cbr...(a)cbrownsystems.com wrote: > > > Do you now return to your claim that there exists a surjection from > > 1/13 onto 1/64, but not from 1/64 onto 1/13? > > > > Cheers - Chas- Zitierten Text ausblenden - > > Didn't you receive my last answer? > > > What are the /members/ of the set 2 that you mean to act as the domain > > of the surjection? Or is the set 2 the empty set? > > Let m, n be natural numbers. > If n =/= m, then there is no one-to-one mapping of n onto m. > If If n > m, then there is no injective mapping of n onto m. > If If n < m, then there is no surjective mapping of n onto m. According to WM's definition of natural numbers, a natural number is the set (class?) of all sets bijectable with some given finite set. Under WM's definition, it would be quite difficult to compare any two non-zero naturals using his above rule. > > > I don't mean some finite collection of /subsets/ of the set 2, but the > > /members/ of the set 2. You agree that there is in general a difference > > between "x is a subset of Y" and "x is a member of Y", yes? > > n = {0,1,2,...,n-1} > > The members of the set 2 = {0, 1} are 0 and 1. > > Perhaps there is a gap in your understanding of numbers and their > representation in set theory? If you mean he is missing all those things that WM believes which are not so, it is a gap every mathematician cultivates.
From: Virgil on 8 Feb 2007 03:27 In article <1170920549.402271.300960(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 7 Feb., 14:35, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > The potentially infinite set of even numbers is *constructed* by its > > > segments > > > > > {2,4,6,...,2n} > > > > > Every time we increase n by 1 we increase 2n by 2. This cannot be > > > avoided. Therefore it is impossible to have for finite natural numbers > > > > > lim[n-->oo] |{2,4,6,...,2n}| > 2n > > > > The limit of finite natural numbers is not a finite natural number. > > There is no limit. The union of every family of sets is a set. If each natural is a set then the union of a all naturals exists. and is a set. and is a set containing every finite natural as a member, and containing nothing else. > The notation lim[n-->oo] says nothing but that > there is *no* limit. The size of n is not bounded. But the set of all n exists. > > > What *is* true is that it is impossible to have for finite natural > > numbers > > > > n= |{2,4,6,...,2n}| > 2n > > The set of natural numbers has only finite natural numbers. But it has more than any finite natural number of natural numbers in it. > > Of course not. How could there be a contradiction? In the infinite "<" > becomes even ">" if required to keep ZFC free of contradictions. How > could it be otherwise? It is only in WM's eyes that such miracles occur. In mathematics, one is not allowed such freedom from logic and reason as WM habitually indulges in.
From: Virgil on 8 Feb 2007 03:29 In article <1170920789.176126.27540(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Cantor wrote in German. If someone really wants to understand the > origins of set theory, she should learn German. > (If you have forgotten it, look at my post to Virgil. There I > translated the relevant paragraph.) > Current set theories are quite adequately expressed in English.
From: mueckenh on 8 Feb 2007 03:34
On 7 Feb., 16:48, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > mueck...(a)rz.fh-augsburg.de wrote: > > On 6 Feb., 13:47, Franziska Neugebauer <Franziska- > > Neugeba...(a)neugeb.dnsalias.net> wrote: > >> mueck...(a)rz.fh-augsburg.de wrote: > >> > It can be proven. {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } > >> > contains {p(0), p(1), p(2), ... } > > >> What does contain mean? > > > The tree is a set of nodes. > > This is not true. A tree is "more" than a set of nodes. If the structure is defined separately, the only variable is n, the number of levels. A tree is then a set of levels as well as a set of paths as well as a set of nodes. This set of nodes can be enumerated by a "chain" in the manner of Cantor's first diagonalization 0. 12 6543 789... The tree is completely defined by one single natural number. This could be a number of the chain, but as we will not use trees with different pathlengths, we will use only the number n of the last level of the tree. This can appropriately be expressed by T(n). > > > It contains paths as subsets. > > 0. In the usual definition of tree T = { v, e } this is obviously not > true since neither of the non-trivial subsets > {v}, {e}, {v, e} is a path at all. Forget this usual definition. We need not define the structure because we talk about one structure only, a complete binary tree. If you are too inflexible to swallow this then call my tree as you like it, but do me a favour and retranslate your choice to tree when responding to my letter. > > 1. Only in Virgil's view of tree T it is true that a tree is a set of > paths { p1, ... } . It is true in any case. Among others a tree can be viewed as a set of paths. > > 2. A set s is a subset of a superset S iff > > A x (x e s -> x e S) > > 3. I interpret your conjecture "It contains paths as subsets." as > "there is (at least) a path which is a subset of the tree", formally > > E p (p c T) (*) > > which means > > E p (A x (x e p -> x e T)) > > Assume now that there is some p = p' for which (*) holds. We have then > > A x(x e p' -> x e T) (**) > > Now the definition of path comes into play. As explained below the > elements of p' are ordered pairs, i.e. have structure (number, node) or > (number, edge). As one easily sees T only has paths as elements, i.e. > has sequence structure { ..., (number, node), ... }. Therefore there is > no (non-trivial) p' which is a subset of a tree. Froget this complicated picture. I does not lead ahead. A path is an ordered set of nodes. But we need not use this order. If we apply the chain sketched above, then there is a path [0, 1, 6, 7]. This path is well defined by its nodes. The order is unimportant. Therefore we can denote by {0, 1, 6, 7} and consider it a set of nodes, a subset of the tree. If you do not agree with this simple picture, please say where it becomes undefined. > > Hence your conjecture "there are paths which are subsets of a tree" is > wrong. See above. > > > The special meaning of "contain" is irrelevant as long as we refer to > > a unique meaning, i.e., > > either: something which is contained in the tree is really in the > > tree, > > or: something which is contained in the tree does belong to the tree > > as a supremum. > > It is absolutely necessary to carefully distinguish between membership > (element-relation "e") and the subset-relation ("c"). If the paths p(1), p(2) , p(3) ... (all outmost left hand side) are in the union of all finite trees as members then the union of all finite trees contains the path p(oo) (if this path exists). If the paths p(1), p(2) , p(3) ... (all outmost left hand side) are in the union of all finite trees as subsets then the union of all finite trees contains the path p(oo) too (if this path exists). > >> X is not finite -> there must be an x in X which is infinite > > >> which is alas not part of ZFC or any other modern set theory. > > > I did not start off with this assumption. > > You do persistently refer to that axiom but you don't grasp that you do. I start off with the observation that *for even positive numbers* |{2, 4, 6, ..., 2n}| < n. Further I assume that the set of all even positive numbers contains nothing but even positive numbers. There is no axiom involved up to now. The conclusion is that aleph0 is not a number larger than every even positive number. Regards, WM |