Cantor Confusion
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From: Virgil on 21 Jan 2007 03:03 In article <u236r29nhp24hqheijcv9id8tfo7ca8s0t(a)4ax.com>, G. Frege <nomail(a)invalid> wrote: > On Sun, 21 Jan 2007 01:32:15 -0500, David Marcus > <DavidMarcus(a)alumdotmit.edu> wrote: > > >> > >> A quote from M�ckenheim's oeuvre: > >> > >> <begin quote> > >> > >> ...it is inconsistent to speak of /an infinite set of finite numbers/. > >> Finite numbers can only form a potentially infinite set. An actually > >> infinite set cannot exist other than including its cardinal number > >> aleph_0 or the number "ordinal infinity", denoted by 'omega' This had > >> been unconsciously acknowledged by Cantor himself already: "/Every > >> number/ smaller than omega is a finite number, and its magnitude /is > >> surpassed by other finite numbers/." Here the phrase "by other finite > >> numbers" is obviously to be interpreted as "by such finite numbers > >> which did not yet belong to the set containing /every finite number/". > >> > >> <end quote> > > > > The word "obviously" is a nice touch. Always good to toss in a few > > "obviously"'s when saying things that are wrong or nonsensical. > > > Indeed! (Since it allows for very smooth "argumentation" concerning > crank stuff. I mean, it is really helpful when dealing with things > which might interrupt your flow of ...well... "thoughts".) > > Another one in that vain: > > "As there is only a countable set of intervals of length 1 in R, the > denumerability of the whole set R is obvious." (WM, The Meaning of > Infinity) > > Obvious, indeed. Obviously circular, as are most of his thought processes.
From: mueckenh on 21 Jan 2007 05:29 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > > > UT is not a tree. > > > > > > > Call it as you like. Simply call it T1. (If I speak of "tree" blow, > > simply read "eert".) It may be what you like. In any case it is the > > same by nodes, edges > > Yes > > > and paths as T2. > > No. The directions of the paths are the same in T1 and T2. If you insist on a difference, then it can only result from the length of the paths. > The paths in T1 are not determined by the nodes in T1. By its nodes and edges every path is determined. > The paths in T2 are determined by the nodes in T2. In fact? Why that? > The paths > in T1 are not the same as the paths in T2. > > T1 contains only finite paths. This is in fact the crucial question. If you are right, then I have been working in vain for four years. If you are wrong, then thousands of mathematician have been working in vain for 130 years. So let us decide this question > T1 does not contain a single infinite > path. The fact that there is an infinite path that can be contructed > using the nodes from T1 is not relevant. The paths in T1 are > not determined by the nodes in T1. What are the nodes and edges for, in your opinion? Why do you think the EIT has an infinite diagonal? I say: The union of indexes of paths {1} U {1,2} U {1,2,3} U ... is an infinite index union {1,2,3,...} corresponding to n infinite path. If you disagree, please prove your assertion. Where is the end of one path of T1? Please don't come up with answers like: "I don't know the end but I am sure it exists." A sequence is a function whose domain is either an initial segment of natural numbers or all natural numbers N. A sequence whose domain is some initial segment is called a finite sequence. In mathematics a sequence is proven finite if and only if there is a natural number n_0 which not belonging to its domain. Otherwise a sequence is infinite. > > > > An index denotes the level to which a node belongs. The union of all > > indexes of nodes of finite paths is the union of all initial segments > > of natural numbers > > {1} U {1, 2} U {1, 2, 3} U... U {1, 2, 3, ..., n} U ... = {1, 2, 3, ...}. > > This is also the set of all last elements of the finite segments, i.e., > > it is the set of all natural numbers N. This is the set of all indexes > > - there is no one left out. With respect to this observation we > > examine, for instance, all finite paths of the tree T1 which always > > turn right: 0.1, 0.11, 0.111, ... If considered as sets of nodes, their > > union is the infinite path representing the real number 0.111... = 1. > > No. Their union is a set of finite paths. The union of domains {1} U {1, 2} U {1, 2, 3} U... U {1, 2, 3, ..., n} U ... is a domain {1, 2, 3, ...}. 0.111... is an > (potentially) infinite path. 0.111... can be seen as a limit > of a sequence of finite paths. It is not the union of a sequence > of finite paths. Union and limit are not the same thing. > (The union of {1/2},{3/4},{4/5} .... is not {1}). But the union of the intervals (0, n/(n+1)) has length 1. And the union of singletons {1} U {2} U {3} U... U {n} U ... is not an infinite number omega? Two trees which are identical by all nodes and edges are identical by all paths. T1 as the countable union of all finite sets of finite paths contains only a countable set of finite paths. T1 = T2. Regards, WM
From: mueckenh on 21 Jan 2007 05:38 Dik T. Winter schrieb: > > According to set theory (including the axiom of choice) a countable > > union of countable sets is a countable set. The set of paths in the > > union tree T1is merely a countable union of finite sets, > > And that is wrong. The set of paths is *not* the union of the sets of > paths in the finite trees. And that is easy to prove. The path > 0.010101... is not in that union of sets of paths because it is in > *none* of the sets of paths, so it also can not be in the union of > sets of paths. Hi Dik, I hope not to disturb you. If the union of singletons {1} U {2} U {3} U... U {n} U ... is an infinite number omega, then the union of domains of sequences {1} U {1, 2} U {1, 2, 3} U... U {1, 2, 3, ..., n} U ... is the domain of an infinite seqeunce {1, 2, 3, ...}. > > >From this we can conclude that also every other infinite path belongs > > to the union T1 of all finite trees. > > Yes, it is in the union of finite trees. But *not* in the union of sets > of paths in finite trees. How could a path be in the union of finite trees if it were not a path of a finite tree? Two trees which are identical by all nodes and edges are identical by all paths. T1 as the countable union of all finite sets of finite paths contains only a countable set of finite paths. And T1 = T2. > > We see, the trees T1 and T2 are identical with respect to all nodes, > > all edges, and all paths (which would already have been implied by the > > identity of nodes and edges). But the set of all paths is countable in > > the tree T1 and uncountable in the same tree T2. > > It is not countable, you have not given a proof. Here it is (for trees of the kind weeping willow): Taking the union of two trees corresponds to taking the union of their sets of paths Then we have:The set of paths in the union is the set of paths in the larger tree. This holds if it holds for domains of finite seqeuences. The domain of two finite sequences is the domain of the longer sequence. The domain of all finite sequences is omega. > > > > Then it exists in the union tree - together with all paths of same > > length. > > But I am not contradicting that. Again, I state that it is not in the union > of sets of paths in finite trees. There is a huge difference. See the proof above. The sets of paths can be unioned as the trees can be unioned. There is a bijecton between the set of finite trees T(n) and the sets of paths P(n) in finite trees. > > > > > You say it exists. It exists not in the union. Where does it exist? > > > > > > Outside the union. Let's illustrate. Call the diagonal of the triangle > > > of order n: d_n. The union of the sets of diagonals of the triangles is: > > > {d_1, d_2, d_3, d_4, ...} > > > the diagonal d of the infinite tree is not in it. > > > > It is the union. > > Yes, so it is not *in* the union. The cardinal number of the union (in general 1) cannot be larger than the cardinal number of the set of elements of the union (often an aleph). The cardinal number of the union of path segments cannot be larger than the cardinal number of all path segments. > > > So if we have the union of all digits, then we have > > the diagonal. > > The union of the digits is *not* the union of the sets of diagonals. If we have the ordered union of all digits, then we have the diagonal. > > > > Consider a tree of level 1 and a tree of level 2. The tree of level 1 > > > contains only paths going through two nodes, the tree of level 2 only > > > paths going through three nodes. So none of the paths in the level 1 > > > tree is a path in the level 2 tree. So the union of the sets of paths > > > contains both all paths of the level 1 tree and all paths of the level > > > 2 tree. The set of paths of the level 2 tree contains only the paths > > > of the level 2 tree. > > > > This is a wrong approach. (It is insufficient to observe the problem.) > > It is the only possible approach when you consider unions of sets of paths. No. The directions of paths are given. If you assert that in T2 and T1 there are different paths, then they can differ by length only. But you do not assert that, contrary to William, for instance. > > The elements of our sets are nodes. The paths are subsets. If you do > > not observe this fact, you cannot find a contradiction. > > You are using a countable union of *sets* of paths. The tree is an (ordered) set of nodes. The nodes can be used to form paths, i.e., subsets of the tree. Regards, WM
From: mueckenh on 21 Jan 2007 05:44 Virgil schrieb: > > Trees which are identical with respect to nodes and edges cannot have > > different *paths*. > > While that may be true for finite trees, it is demonstrably false for > infinite trees. > This limited infinite tree contains every node and every edge that the > complete binary tree contains, If there are all nodes and all edges but only some of the paths: what prevents the other paths from being there? > but it is clearly not the same tree. That is your wrong assertion. Knowing that you also think the vase it empty after infinitely many transaction, I will not bother further. > \So WM is trivially and totally wrong in his claim above. > > It is WM's thoughtless assumption that every property of finite trees > automatically also holds for infinite trees that leads him to make a > fool of himself in this way. Personal insults will not really support your claims. ======================= > The contradiction is in WM's false assumption that nodes and edges are > eniugh. But unless (4) is applied and can be shown to generate ALL > infinite paths, one cannot guarantee a complete infinite binary tree by > any such unioning. Correct is: Two trees which are identical by all nodes and edges are identical by all paths. T1 as the countable union of all finite sets of finite paths contains only a countable set of finite paths. T1 = T2. Regards, WM
From: mueckenh on 21 Jan 2007 05:55
Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > > > >> Whether a _meaning_ of > >> > >> T(1) U T(2) U T(3) U ... (inf-u) > >> > >> exists does _not_ depend on such question. It depends on a definition > >> you have to provide. So please answer the question, what (inf-u) > >> means and prove that it exists. > >> > >> > This is connected > >> > >> Whether that is "connected" is irrelevant to a possible definition of > >> (inf-u). > > > > That is not a true statement. Nevertheless: > > > > Definition of the infinite union of trees by induction: If the finite > > tree with n levels exists, the finite tree with n+1 levels exist. The > > tree with 1 level exists. > > By induction you only define (if at all) every _finite_ union > > T(1) U ... U T(n) That is completely sufficient as long as there is no upper bound, but n --> oo (potential infinity). But if you, personally, think it is necessary, then simply use the axiom of infinity. There is a set of all natural numbers, a set of all levels, a set of all finite trees. Regards, WM |