Cantor Confusion
in [Math]
|
From: mueckenh on 23 Feb 2007 12:54 On 23 Feb., 14:35, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > So the union of finite trees U(T(n)) contains (as subsets) the path > > > > p(oo) and all its co-paths q(oo), ..., i.e., > > > > > > Yes. I never said otherwise. Why do you think I said otherwise? > > > All the p(oo) are subsets of U(T(n)). > > > > > > > it contains P(oo)? > > > > > > And that is wrong. Pray look close at what the elements of the different > > > sets are: > > > U(T(n)) has as elements nodes > > > > and it has paths as subsets. > > > > > P(oo) has as elements paths, i.e. sets of nodes > > > so P(oo) is neither an element of U(T(n)), nor is it s subset of it. > > > > 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. 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. > > > 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. > > > 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. Regards, WM
From: Virgil on 23 Feb 2007 14:06 In article <1172251583.108824.29570(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 22 Feb., 13:26, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > On Feb 22, 2:42 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > > > > > > > > On 20 Feb., 23:26, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > > On Feb 20, 4:40 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > > > 1,2,3,...,n <---> n > > > > > > i: every initial segment of the natural numbers can be mapped > > > > to a natural number. > > > > > > ii: the set of all natural numbers does not contain an element > > > > that is not in an initial segment of the natural numbers. > > > > > > iii: The set of all natural numbers can be mapped to a natural > > > > number > > > > > > No i: and ii: do not imply iii: > > > > > Why not? > > > > Because, as you have noted, there are times when i and ii are true > > and iii is false. > > There are times when it is false and there are times when it is true. There is no way ever that the set of natural numbers can be bijected, or even injected to any natural number. If that is what WM means by (iii) then he is wrong! > The conclusion from the finity of finite segments on the finity of > infinite segments is as false as the conclusion from the evenness on > evenness, because all these are individual properties of segments. GIGO again! > > Another kind of property concerns the fact that every segment of even > positive numbers contains numbers which are greater than the cardinal > number. This is false as stated, as no natural number can ever be of greater cardinality than that of the set of all even naturals. > For sets which have a cadinal number which is n trichotomy with > even positive numbers this holds. Since cardinality is built on injections and bijections, and every even natural, as a set, injects but does not biject to the set of all evens, every even natural is of smaller cardinality than the set of all evens. > 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. WMJ has to explain that since it only is required in his own private world. > It is similar to > create uncountably many paths in the binary tree withoutthe presence > of uncountably many nodes or edges. Which, being equivalent to the existence of uncountably many endless binary strings with only countably many bit positions each andwith only finitely many values possible in each position, is the natural state of affairs.
From: Virgil on 23 Feb 2007 14:16 In article <1172252145.066292.211790(a)p10g2000cwp.googlegroups.com>, mueckenh(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. Perhaps because WM suffers from a form of blindness of his own.
From: Virgil on 23 Feb 2007 14:21 In article <1172253288.305584.116320(a)j27g2000cwj.googlegroups.com>, mueckenh(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, despite WM's many false claims about the nature of such infinite trees. > 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.
From: mueckenh on 23 Feb 2007 16:43
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. Regards, WM |