Cantor Confusion
in [Math]
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From: mueckenh on 7 Mar 2007 05:40 On 6 Mrz., 23:22, Virgil <vir...(a)comcast.net> wrote: > In article <1173215257.272668.121...(a)64g2000cwx.googlegroups.com>, > > > > C(oo) is a part of T(oo). > > Then it cannot contain all paths of a complete infinite binary tree, as > has been repeatedly proved here. It does not contain the paths, but it contains a part of every path. > > Show us just how any edge is a "continuous" connection. By what > definition of continuity? By the definition that every pair of nodes of the tree is connected by edges. And by the definition that every pair of nodes belonging to one path is connected by edges which belong to this path. But above all continuity guarantees that every path is connected t the root node. For every number of paths (-bundles) which may exist somewhere, there is a level of the tree which hosts exactly half of this number. Therefore no unexpected "explosion" of path number can happen. > > WM likes to throw big words around without regard for their meaning. But > in the context of the edges connecting nodes, and edge is no more than > an ordered pair of nodes, and there is nothing "continuous" about an > ordered pair. An edge is a connection. The pair defines the same connection. > > > Have you really been blinded for the fact that most of the subsets of > > nodes of the complete cannot serve as paths? > > As there are uncountably many which CAN serve, Wrong. As the cross sections show, these uncountably many are not in the tree, neither in U(T(n)) nor in T(oo). Regards, WM
From: mueckenh on 7 Mar 2007 05:48 On 6 Mrz., 23:40, Virgil <vir...(a)comcast.net> wrote: > In article <1173215723.210725.37...(a)v33g2000cwv.googlegroups.com>, > WM keeps trying to ignore that the construction of an infinite binary > path requires an infinite sequence of binary choices, which Cantor > proved leads to more that countably many number of paths. This proof is wrong. Hessenberg's triple {N_k, k, f} simply does not exist. Therefore it cannot not be in the mapping f: k |--> N_k of f: N --> P(N). By the way, the mapping from Q to P(N) does not and cannot contain the smallest positive rational number. Is this a proof that Card(Q) > Card(P(N))? If you should deny, you are very biased. But we know that already. > > That simple and unrefuted proof has just been refuted for another time. Regards, WM
From: Dik T. Winter on 7 Mar 2007 10:01 In article <1173215257.272668.121480(a)64g2000cwx.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 6 Mrz., 15:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1173176849.548735.27...(a)n33g2000cwc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: .... > > > Of course. And we need not use the limit definition (of Hrbacek and > > > Jech or else) in order to get this result, because the whole tree > > > T(oo) has only a countable set of nodes. Therefore C(oo) cannot be > > > larger than countable. > > > > The latter is an invalid conclusion. > > Did you ask the referee? Or are you the arbiter yourself? C(oo) is a > part of T(oo). What do you *mean* with "is a part of T(oo)". How do you *define* C(oo). You continuously *fail* to give a proper definition of it. And do not refer to "cross section" of T(oo), because that is undefined either. Neither do refere to limits, unless you *define* those limits. > No, Dik. I know that the functions in set theory are not continuous > and perform unpredictable jumps when leaving the finite and entering > the infinite. One of these tricks is the set of all finite sets being > countable and the set of all infinite sets being uncountable. Another > trick is the famous observation: > Forall n in N: |{2,4,6,...,2n}| < 2n > while > lim {n-->oo} |{2,4,6,...,2n}| > lim {n-->oo} 2n, and similar jokes. That is tricky of you. You have defined *neither* limit. If we use the definition by Hrbacek and Jech on page 193, and assume that cardinals are also ordinals (as is commonly done), the two limits are equal (aleph-0). So you apply your own very special definition of limit that you do not supply. Show your definition of limit, and we will show that your definition is inconsistent. > Therefore I devised the tree. Its nodes are connected by continuous > paths. They prevent any discontinuity. If we have a cross section less > than the number of nodes for any finite tree, then the cross section > cannot surpass the number of nodes of the infinite tree. The structure > of the tree forces the function C(n)/|T(n)| to remain continuous under > any circumstances. But you do *not* define the cross-section of the infinite tree, neither do you define the limits you are using. > > > > Wrong. Each path is a set of nodes, hence a subset of the complete > > > > set of nodes, hence an element of the powerset of this set of nodes. > > Have you really been blinded for the fact that most of the subsets of > nodes of the complete cannot serve as paths? That does not matter. Some subsets are paths. Of these some are infinite and some are finite. The finite ones are a subset of a countable set, the infinite ones are a subset of an uncountable set. But there *do* exist uncountable subsets of uncountable sets. > > > > We know that this powerset is not countable, so the set of paths can > > > > be (and actually is) uncountable. > > > > > > This argument should be recognizable as invalid. > > > > What is invalid? I may note that the parenthetical remark is *not* > > part of the argument. > > Part of which argument? I cannot see any argument at all. You state > (1) that Cantor's argument is valid (at this place I do no judge about > it) and you state (2) that set theory is free of contradictions (I > don't know who told you). From (1) and (2) you conclude that contrary > to all mathematical evidence (I remind you of the ration of 2 nodes > per path and the cross section of the tree which counts its paths) my > result cannot be true. As you do not *define* the cross section of the infinite tree, your argument is empty. > > > It could only be > > > applied if ALL combinations of nodes were paths. But this is not the > > > case! Not every set of nodes is a path, as you well know. In order to > > > count the paths (-bundles) which pass through a certain level of the > > > tree, it is sufficient to know the cross section of the tree. More > > > than C(n) paths (-bundles) cannot fit into the tree. This proves > > > without any doubt that the complete tree T(oo) cannot contain more > > > than countably many paths (-bundles). > > > > This proves that in no way. > > Have you some argument which at least calms down your own doubts? At each finite level, the cross-section C(n) is the number of path-bundles. This does not tell us anything about the infinite tree. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 7 Mar 2007 13:48 In article <1173264035.284564.126790(a)j27g2000cwj.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 6 Mrz., 23:22, Virgil <vir...(a)comcast.net> wrote: > > In article <1173215257.272668.121...(a)64g2000cwx.googlegroups.com>, > > > > > > > > C(oo) is a part of T(oo). > > > > Then it cannot contain all paths of a complete infinite binary tree, as > > has been repeatedly proved here. > > It does not contain the paths, but it contains a part of every path. The root node is part of every path. Is C(oo) the root node alone? A set of paths which contains all of every path in a complete infinite binary tree contins uncountably many paths. > > > > Show us just how any edge is a "continuous" connection. By what > > definition of continuity? > > By the definition that every pair of nodes of the tree is connected by > edges. That is not a definition of continuity. And is certainly not a mathematical definition in any sense. An edge, being no more that an ordered pair of nodes, does not constitute anything continuous. > And by the definition that every pair of nodes belonging to one > path is connected by edges which belong to this path. What is "continuous" about pairs of nodes? > But above all > continuity guarantees that every path is connected t the root node. Connectedness and continuity are topological qualities, so what is the topology of a path? > For every number of paths (-bundles) which may exist somewhere, there > is a level of the tree which hosts exactly half of this number. What is "half" of an infinite number? > Therefore no unexpected "explosion" of path number can happen. Those who understand what trees are, expect an "explosion" from finiteness to infiniteness when trees become infinite. > > > > WM likes to throw big words around without regard for their meaning. But > > in the context of the edges connecting nodes, and edge is no more than > > an ordered pair of nodes, and there is nothing "continuous" about an > > ordered pair. > > An edge is a connection. The pair defines the same connection. My parents and I are "connected" in the same way as nodes, or even more closely, but we are not continuous. > > > > > Have you really been blinded for the fact that most of the subsets of > > > nodes of the complete cannot serve as paths? > > > > As there are uncountably many which CAN serve, > > Wrong. As each infinite binary tree bifurcates infinitely many times and each path chooses one of those bifurcations infinitely many times, there are at least as many possible such paths as there are infinite binary strings, and there are, therefore at least an uncountable number of infinite binary strings in each complete infinite binary tree. One wonders which branchings WM denies in his incommplete trees. > As the cross sections show, these uncountably many are not in > the tree, neither in U(T(n)) nor in T(oo). Then neither of them can contain all the paths in the complete infinite binary tree, since THAT tree must have uncountably many paths, as has been repeatedly proven and never refuted.
From: Virgil on 7 Mar 2007 13:58
In article <1173264512.141292.313040(a)p10g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 6 Mrz., 23:40, Virgil <vir...(a)comcast.net> wrote: > > In article <1173215723.210725.37...(a)v33g2000cwv.googlegroups.com>, > > > WM keeps trying to ignore that the construction of an infinite binary > > path requires an infinite sequence of binary choices, which Cantor > > proved leads to more that countably many number of paths. > > This proof is wrong. Hessenberg's triple {N_k, k, f} simply does not > exist. Therefore it cannot not be in the mapping f: k |--> N_k of f: N > --> P(N). Given f:S -> P(S), then for every s in S, either s in f(s) or s not in f(s), so both {x in S: x in f(x)} and {x in S: x not in f(x)} are well-defined whenever f is well dfinied/ Given f:S -> P(S), which element s of S is such that f(s) = {x in S: x not in f(x)}? > By the way, the mapping from Q to P(N) does not and cannot > contain the smallest positive rational number. It cannot contain a pink elephant either, which has a far better chance of existing that does a "smallest positive rational number". > > If you should deny, you are very biased. I am strongly biased in favor of valid arguments, which bias WM unfortunately does not share. |