Cantor Confusion
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From: Dik T. Winter on 24 Jan 2007 09:58 In article <SH63Oi8og1tFFwVA(a)phoenixsystems.demon.co.uk> Andy Smith <Andy(a)phoenixsystems.co.uk> writes: .... > So why are you happy with 0.0111... and not ...1110 ? (Because natural > numbers are all finite, even though the set N is infinite.) We are happy with ...1110. Only, it is not a natural number. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 24 Jan 2007 10:33 On 23 Jan., 20:52, Virgil <vir...(a)comcast.net> wrote: > In article <1169548200.625843.205...(a)v45g2000cwv.googlegroups.com>, > > > an awful lot of new paths > > > that are brought into light, namely non-terminating paths. > > > Finite trees have absolutely zero (non-repeating) non-terminating > > > paths, > > > The union of all finite trees has no terminating path because the union > > of all finite segments {1,2,3..., n} (with in N) does nowhere > > terminate. >By any mathematical definition of "union" a union of two or more > distinct trees is not a tree at all. > What is the union of {a, b, c} and {a, b, c, d, e, f, g} in your opinion? Regards, WM
From: G. Frege on 24 Jan 2007 10:33 On 24 Jan 2007 07:33:13 -0800, mueckenh(a)rz.fh-augsburg.de wrote: > > What is the union of {a, b, c} and {a, b, c, d, e, f, g} in your > opinion? > {c, r, a, c, k, p o, t} ? F. -- E-mail: info<at>simple-line<dot>de
From: Dik T. Winter on 24 Jan 2007 10:40 In article <1169641853.359139.100270(a)j27g2000cwj.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > > > 2} U {1, 2, 3} U... U {1, 2, 3, ..., n} U ... is the domain of an > > > > > infinite seqeunce {1, 2, 3, ...}. > > > > > > > > Pray explain what you understand under "domain of", and what you > > > > understand under "union of domains". That is complete non-standard > > > > terminology. > > > > > > The domain of a sequence is a natural number. The domain of an infinite > > > sequence is omega. > > > > That is not an explication. That is obfuscation. That is, it is still > > clear as mud. > > It is from a book on set theory. The domain of a sequence is also > called its pre-image, in German Definitionsbereich. I don't know f > other Enlish words for it. I am beginning to understand. You equate a sequece with a function f: N -> objects and talk about the domain and range of that function. > > > Therefore there is no uncuntable set of path in the union tree. But > > > there are all paths in the union trees. > > > > Conclusion is false. > > Summary > > 1) Every complete infinite binary tree T (containing all nodes and > edges) contains all paths. > 2) The union tree T(oo) of all finite trees is well defined (as I have > shown elsewhere) and yields the complete infinite binary tree > containing all nodes and edges: T = T(oo). No, you have not shown it. But I have given some definitions that *do* show it. This can only be done if a tree is *defined* as a set or as a collection of sets. For other obejcts than set the concept of union is not defined. > 3) The union of all finite trees includes the union of all nodes and, > with it, the union of all such subsets which are paths (because every > path is a well defined subset of the set of nodes if the structure of > the tree is well defined). If a tree consists (amongst others) of the set of nodes in it, the union of two trees indeed consists (amonst others) of the union of the sets of nodes. > 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. As worded this is trivially wrong: it states that "the set of paths in T(oo) is one of the set of finite sets of all paths in finite trees", or "the set of paths in T(oo) is the set of paths in one of the finite trees". I think you mean: 4) The set of paths in T(oo) is a subset of the countable union of finite sets of all paths in the finite trees. But this is also wrong. An easier example. P(x) denotes the powerset of set x, i.e. the set of subsets. We have: N = U[n in N] {1, 2, ..., n} P(N) !subset U[n in N] P({1, 2, ..., n}) so the union of the sets of subsets is not the set of subsets of the union. (The reason is that the set on the left hand side contains infinite subsets, while the set on the right hand side does *not* contain infinite subsets.) > 5) A countable union of countable sets is a countable set (according to > ZF with AC). > ==> The set of all path is countable. (==> The real numbers are > countable.) The statement is correct, but what you state does follow, does not follow. The set of all paths is *not* the countable union of countable sets of paths. > Going on, we can say: > > 6) T(oo) = T contains only finite paths. > 7) T(oo) = T contains all paths including all infinite paths. By your definition (paths are specific subsets of nodes), T contains infinite paths. Conclusion (6) is false. On the other hand, the union of the sets of paths contains only finite paths. And they are countable indeed. > > > > No. T1 is a set of paths, it is not a tree. T1 does not contains > > > > nodes or edges as elements. Only paths. > > > > > > The nodes can be enumerated in various ways. > > > > That does not matter. T1 as you defined above is a set of paths. You defined as follows: > T1 as the countable union of all finite sets of finite paths contains > only a countable set of finite paths. With that definition, T1 is a set of paths. > No. I *define* T(oo) is an ordered set of nodes, the order being > expressed by the edges. So now you have a different definition *again*. > > *Not* > > a tree. And a set of paths does not have nodes or edges as elements, it > > has paths as elements. > > The union of paths is a set of nodes as a path is set of nodes. But where am I talking about union of paths? And where are you talking about union of paths? You are talking about union of *sets* of paths. Something quite different. > > > And if you consider a path as a set of node, then > > T1 *still* does not have nodes as elements but sets of nodes. I refer here to your definition above of T1 (the countable union of all finite sets of finite paths). > T(oo) is the union of the T(n) which are sets of nodes. Therefo T(oo) > is a set of nodes. And that is something *very* different. > > > The union of trees is a union of their nodes. The paths are merely > > > subsets which are defined by the nodes and the special kind of tree. > > > > But you have to be careful when uniting sets of paths, that is uniting > > sets of sets of nodes. That union contains all paths from both tree, > > you can only omit duplicates from that union, so only paths that are > > *identical*, i.e. consist of the same set of nodes. > > The union of trees is defined by nodes! Paths are maximal subsets > following some prescription. Therefore in long trees there are no short > paths, but there are long paths only. In that case you are *not* uniting sets of paths. And your reasoning about a set of paths being the union of sets of paths does not hold. > > Suppose {a, b, c} > > is a path in tree T1, and {a, b, c, d, e} is a path in tree T2, than > > the union of the sets of paths of T1 and T2 contains *both* paths. > > But not the union of the elements (nodes a, b, c, d). Indeed. But that is not relevant when you are talking about the union of sets of paths, as you do. Quote from above, see (4) above as corrected by me. > > Finally now you define a the domain of a sequence. So it is now also > > clear that a path is an ordered set of nodes. And the domain is the > > initial subset of the natural numbers that terminates at the cardinality > > of the path (as set of nodes). Still if T1 contains the path {a, b, c} and > > T2 contains the path {a, b, c, d, e}, the union of the sets of paths > > contains both. > > But not the unin of the trees T1 and T2. Tghe union contains only the > path {a, b, c, d, e}. But you are talking about the union of sets of paths. > > > If all nodes and edges are in the union, then all subsets are in the > > > union too. > > > > You misread again. The set of subsets of the complete tree is *not* > > equal to the union of the set of subsets of the finite trees. > > No. The set of subsets (which are maximal, i.e., which are paths) of > the complete tree is only a subset of the union of the set of subsets > of the finite trees Also wrong. > > The > > reason is clear: that union does not contain an infinite subset, while > > the complete tree *does* contain infinite subsets. > > The union contains all we have. There is nothing remaining. No, the union contains less than what we have. See my example with sets of integers. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 24 Jan 2007 10:44
On 23 Jan., 21:54, Virgil <vir...(a)comcast.net> wrote: > In article <1169551459.707075.42...(a)v45g2000cwv.googlegroups.com>, > > > Correct. This shows that there are no other paths. It is the core of my > > contradiction of set theory.Then the infinitely many paths that from some node onwards alternate > left branch then right branch do not exist? And none of the infinitely > many other infinitely alternating patterns exist? They do not exist as completed entites, as I have shown by my union (or however you'd like to call it without destroing its definition) of all finite trees. > > > Everything else claimed by you is belief in ghost paths (vibrations of > > vacuum? small black holes in a green tree?), but has nothing at all to > > do with mathematics.There are all sorts of proper fractions with odd denominators which have > binary expansions corresponding to paths that WM says do not exist. > It would seem that these "vibrations in a vacuum" are taking place > inside WM's head of course, they are always everywhere. > > > No, he said that for a_nn we can put b_nn and claimed that this is > > valid for every n. I claim that a path touching a node will also touch > > the next one and that all possible combinations are realized in an > > infinite tree with all nodes, > > > Why do you believe that in Cantor's matrix there is only one fixed kind > > of diagonals?Why does WM keep making these wild and totally unfounded claims that we > believe things we do not believe? Because you believe things which are obviously wrong. To assert that a binary tree allows for other paths than the same tree is simply killing of mathematics. Regards, WM |