Cantor Confusion
in [Math]
|
From: mueckenh on 24 Feb 2007 06:39 On 24 Feb., 03:27, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1172253288.305584.116...(a)j27g2000cwj.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 23 Feb., 14:35, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > Every element of P(oo) is a path and as such a subset of U(T(n)), as > > > > you say. > > > > U(T(N)) has only finite paths as subsets (as N has only finite initial > > > > segments --- both U(T(n)) and N have infinitely many such subsets, > > > > but these subsets are finite). > > > > > > Eh? Above you state (properly) that p(oo) is a subset of U(T(N)). > > > > I said: "as you say". But how is it possible that a finite tree > > contains an infinite path? Every tree T(n) contains only finite paths, > > namely such with n nodes. There is no tree containing an infinite > > path. > > Oh. But the union of all finite trees contains infinite paths. Or are you > now of the opinion that 1/3 is *not* a path in your infinite tree? Why do you say "now". I told you more than once that there is no infinity. If you believe that infinite paths exist, you must say where and how. You said, that the infinite path p(oo) is in the union of all corresponding finite paths p(oo) = U(p(n)). Now we are going to investigate this case and its consequences. > > > > Are > > > you contradicting that now? And if we consider: N subset N, N contains > > > one infinite initial subset. Or do you mean something different with > > > U(T(N)) from U(T(n))? If so, what do you mean with it? > > > > I mean that U(T(n)) is the union of all finite trees with n levels and > > hence with paths with n nodes. I can't believe that there is an > > infinite path in this union. > > But when I argued before that 1/3 was not in your tree, you contradicted me. > (I thought so because there was a lack of a proper definition.) But now > you apparently have switched to a completely different view. No. The tree is but a picture of the reals. If 1/3 exist as a real number, then it is in the tree. > > > > > There are countably many unions of finite subsets of U(T(n)). > > > > > > Prove it. There are countably many *finite* unions of finite subsets, > > > not countably many arbitrary unions of finite subsets. Consider N. > > > Each subset of N is a union of finite subsets of N, namely the union > > > of all singleton sets that contain an element of that set. > > > > > > > There are countably many unions like U(p(n)). > > > > > > Prove it. > > > > Every set of finite subsets of N is countable. The set of all finite > > subsets of N is countable. > > Ignoring infinite subsets. > > > > > > > > There are countably many unions like U(p(n)). > > > > > > Prove it. > > > > If A is countable, then the set of all finite sequences of elements of > > A is countable. > > Ignoring infinite sequences. > > If you want to prove inconsistency of ZF with the axiom of infinity you > should allow infinite subsets in your proofs. Disallowing them does not > prove inconsistency of ZF with the axiom of infinity, only inconsistency > with the way *you* think things should be. Straight out: denial of the > axiom of infinity. I use the infinite set of finite trees T(n). But the paths of these finite trees are all finite. I can't change it. Regards, WM
From: mueckenh on 24 Feb 2007 06:44 On 23 Feb., 20:16, Virgil <vir...(a)comcast.net> wrote: > In article <1172252145.066292.211...(a)p10g2000cwp.googlegroups.com>, > > mueck...(a)rz.fh-augsburg.de wrote: > > On 23 Feb., 05:42, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1172134605.880848.197...(a)q2g2000cwa.googlegroups.com> > > > mueck...(a)rz.fh-augsburg.de writes: > > > > What is the physical relevance of the factorisation > > > of large numbers? There is none. Nevertheless that is a quite important > > > field in mathematical research. > > > And could not be done without physical means. No idea (of numbers, > > factors, relations or else) would have appeared without the > > possibility of physically visualizing its basics. > > WM is clearly unaware of mental arithmetic, which can be done by those > blind from birth and incapable of "physically visualizing" anything. Even done by those lacking 10 fingers, two hands and one head? Regards, WM
From: mueckenh on 24 Feb 2007 06:50 On 23 Feb., 22:54, Virgil <vir...(a)comcast.net> wrote: > In article <1172267009.642042.165...(a)k78g2000cwa.googlegroups.com>, > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > On 23 Feb., 20:21, Virgil <vir...(a)comcast.net> wrote: > > > In article <1172253288.305584.116...(a)j27g2000cwj.googlegroups.com>, > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > > On 23 Feb., 14:35, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > Every tree T(n) contains only finite paths, > > > > namely such with n nodes. There is no tree containing an infinite > > > > path. > > > > Then there can be no infinite trees. > > > Correct! > > > > > Every set of finite subsets of N is countable. The set of all finite > > > > subsets of N is countable. > > > > > > > There are countably many unions like U(p(n)). > > > > > > Prove it. > > > > > If A is countable, then the set of all finite sequences of elements of > > > > A is countable. > > > > But neither extends to the infinite subsets of A or infinite sequences > > > of elements of A. > > > The theorem extends to the finite subsets or finite sequences of > > elements of of U(T(n)) . As in case of the tree all elements of P(oo) > > like p(oo) are unions of finite pathes, p(oo) = Up(n), q(oo) = Uq(m), > > r(oo) = Ur(i), and the sets of finite paths like p(n), are countable, > > the unions of sets of finite paths like U(p(n)) = p(oo) belong to a > > countable set. > > > In short: Everything in the countable union of finite nodes and finite > > paths of U(T(n)) is countable. Therefore the set of all possible > > unions of finite paths is countable too. P(oo) is a subset of this set > > of all possible unions of finite paths. > > Then the complete infinite binary tree has paths not in the union of > those finite binary trees, It seems to be th consequence. There must be indexes of infinite, super-natural size to explain such super.natural paths. As I said already casually: There is no set of infinitely many numbers unless there are infinite numbers. >as the set of paths of the complete infinite > binary tree has a distinct path for every possible endless binary > string, and there are uncountably many such strings. "Such" paths, which are not in the union of all finite trees, may be there as many as you like to imagine. Alas, I am not at all interested in thier presence. Regards, WM
From: Carsten Schultz on 24 Feb 2007 07:25 mueckenh(a)rz.fh-augsburg.de schrieb: > A set S which differs from all sets A, B, C, ... of a set of sets, > either can do so by by differing from set A by at least one element a > and from set B by at least one element b and so on. Here set A may > contain b and set B may contain a. But in a linear order like A c B c > C c ... this is not possible. Here S must differ by an element from > all sets or it does not differ from all sets. Interesting proposition. I see how it is possible to derive lots of interesting consequences from it. But can you prove it? Let me help here by stating the proposition exactly so that we know what we are talking about. Claim 1 (Mückenheim?). Let F be a set of sets and S a set. If for all A,B in F we have A subset B or B subset A then one of the following holds. (i) S in F. (ii) There is an x in S such that for all A in F we have x not in A. (iii) There is an x not in S such that for all A in F we have x in A. [Here `in' means `element of', `subset' does not rule out equality.] Maybe you want to claim something slightly weaker and replace (i) by (i') There is an A in F such that S subset A. Then you also might feel that (iii) can be dropped, at least if S or F is non-empty. So your claim might be Claim 2 (Mückenheim?). Let F be a set of sets and S a set. If for all A,B in F we have A subset B or B subset A then one of the following holds. (i') There is an A in F such that S subset A. (ii) There is an x in S such that for all A in F we have x not in A. (iii') S is empty. So what exactly is your claim and how do you prove it? Since I can derive Claim 2 from Claim 1 let me ask more succinctly: Is Claim 2 something you would claim? If so, how do you prove it? Best regards, Carsten -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: mueckenh on 24 Feb 2007 12:28
On 24 Feb., 00:06, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Feb 23, 5:49 pm, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > On Feb 23, 12:26 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > >Otherwise you should explain how the > > > set with n numbers > > > {2,4,6,...,2n} can be filled up by other (namely greater) even > > > numbers , without increasing the sizes of numbers. It is similar to > > > create uncountably many paths in the binary tree without the presence > > > of uncountably many nodes or edges. > > > > Both tasks can only be managed by pure belief in these results and > > > refraining from any attempt to understand it. > > > As nobody has made either claim, this is not a problem. > > You should start reacting to what people claim, not > > what you think follows from the claim. > > Sorry, I misread the second claim. This claim has in fact been made. > (no problem, paths are sets of nodes and edges. Saying > that there are only countably many nodes and edges is > statement i, saying that a path is composed of nodes > and edges is statement ii. Saying that there are a countable > number of paths is statement iii. Statement iii does not follow > from statements i and ii. Statments i and ii are true. Statement iii > is false.) You misunderstand. Statement iii is not concluded from the other statements. Saying that there are only finitely many paths p(n) in the finite tree T(n) is obviously correct. There are 2^n paths. These paths are finite sequences. The countable union of all finite trees U(T(n)) is a tree which contains (as subsets) only finite paths, as the set of all natural numbers contains only finite numbers. Don't you agree? Regards, WM |