Cantor Confusion
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From: Virgil on 18 Jan 2007 16:02 In article <1169113948.875219.279820(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1169030073.463923.305630(a)v45g2000cwv.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > > > > > > > > This is obviously not the union in the sense of set theory. Set > > > > theoretically a tree is a directed graph which is defined as an ordered > > > > pair (V, E). The union of two trees is in general not a tree at all. > > > > > > What I defined a sthe union of finite trees is the union in the sense > > > of set theory. > > > > The set theory union of trees is not a tree at all, but with suitable > > modifications one can construct something that works somewhat in the way > > WM intends. Unfortunately for WM, it disproves his claims: > > The union of two finite trees T(m) and T(n) with m and n levels, > respectively, where m < n, is the tree with n levels. This definition > unites sets of nodes (and sets of edges, respectively) and it is valid > for Cut Trees (CT) as well as for trees of type Weeping Willow (WWT). The union of any finite number of such trees is a finite tree. In a finite binary tree of order n, if it contains the maximal number of edges for an order n binary tree, it must contain a terminal edge for each n-th level node and so it will be a complete tree. This is not true for an infinite tree, and is not true for infinite unions of finite trees. > > The union of all finite trees is the union of all trees with n levels > where n is a natural number: > UT = T(1) U T(2) U T(3) U ... > > The paths in a tree are completely defined by the sequences of nodes > (or edges) which can be followed to an end in CT or without an end in > the union of all CTs as well as in the WWT and the union of all WWTs. But only by the full sequence, which is, in essence the path itself. So that paths define paths.
From: Virgil on 18 Jan 2007 16:07 In article <1169114242.694192.53400(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1169031719.407811.187320(a)a75g2000cwd.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > So far, I hope WM and I agree. > > > > > > > > The issue between us is whether that union will be a proper subtree of > > > > the complete infinite binary tree or will be the entire tree. > > > > > > > > Since every path in every one of the incomplete infinite binary trees in > > > > the union is 'eventually constant', every path in the union of all those > > > > trees will also be eventually constant. > > > > > > Yes, this is valid for a finite union of finite trees. And perhaps, > > > without giving a thought to it, may be claimed for the infinite union. > > > > And when one does give it more thought it is even more obviously valid. > > So the union of all finite segments {1,2,3,...,n} with n in N is > finite? > > Think again! Every finite segment in N terminates and every finite segment in the union of all of those finite segments also terminates, so thinking again confirms the truth of what I said. > > > > > > > > But only countably many of the uncountable paths in the complete binary > > > > tree are eventually constant. > > > > > > > > So that, for example, the union will not contain any path which > > > > alternates between branching left and branching right. > > > > > > Which alternation will be missing? > > > > For one, the path which repeats one left then one right. > > For a second, the path which repeats one left then two rights. > > For a third, the path which repeats one left then three rights. > > And so on. > > And for the union tree, i.e. a tree with an ordinal number being > transfinite? Trees do not, in any ordinary way, /have/ an ordinal number. The set of nodes and the set of edges of a complete infinite binary tree are both countable, but the set of paths is not.
From: David Marcus on 18 Jan 2007 16:12 Virgil wrote: > Since according to WM's "definition" one is unioning sets of nodes and > sets of edges and sets of paths to form the "union" of two trees, one > must note that the union of all those sets of paths is like the union of > all the finite initial sets of naturals in that it only contains finite > objects as members (that it is an infinite set of paths does not make it > a set of inifinite paths). It is impressive how many different ways WM has of making the same mistake. > So WM's "union" still does not contain any infinite paths, and certainly > not all infinite paths as he wrongly claims. -- David Marcus
From: Virgil on 18 Jan 2007 16:14 In article <1169114652.572824.126050(a)11g2000cwr.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1169032407.857382.32680(a)51g2000cwl.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > > > Since every path in every one of the restricted infinite binary > > > > > > trees in > > > > > > the union is eventually constant, every path in the union of all > > > > > > those > > > > > > trees will also be eventually constant. > > > > > > > > > > That would be correct for a finite union of finite trees. What is > > > > > correct for the finite case need not be correct for the infinite > > > > > case. > > > > > > > > If every path in every tree in a union of trees has a certain property > > > > then every path in the resulting tree has that property. > > > > > > If every finite initial segment of N has the property of being finite, > > > then the union of all initial finite segments cannot be an infinite > > > segment. N does not exist. > > > > False analogy. > > > > The correctly analogous statement is": > > > > if every member of every finite initial segemtn of the naturals > > is finite then every member of the union of thsoe finite initial > > segments is finite, i.e., every member of N is finite. > > Of course every member is finite. It is the union of infinitely many > members which is an infinite set and an infinite tree and an infinite > path in the infinite tree. > > > > > > > > > We have infinitely many finite sets of eventually constant paths. We > > > > form the set-union of all those sets of paths. And WM requires that the > > > > union of those sets contain something not in any of the separate sets. > > > > > > > > What does WM think a "union" of sets is? > > > > > > I think the union of all initial finite segments {1,2,3,...,n} is the > > > union of all natural numbers, namely N. > > > > Which, as above, has no infinite members. > > Correct. But the union is infinite. Of course no path in the infinite > tree has an infinite initial segment. The path *is* the infinite > iinitial segment. (Just like Canor's diagonal.) Each path in the complete infinite binary tree is essentially a function from N to {0,1}, i.e., an infinite binary sequence of left(0) or right(1) branchings. But Cantor's diagonal proof, as originally stated, proved that the set of all such sequences not to be countable. So if WM's tree has only countably many paths, it must be leaving most paths out. An several people have presented analyses showing where WM's omissions occur.
From: Virgil on 18 Jan 2007 16:22
In article <1169118050.668265.56330(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Andy Smith schrieb: > > > I > > > > > > The union of all finite binary trees contains all levels which can > > > > > > be > > > > > > enumerated by natural numbers: > > > > > > > > > > > > 0 0. > > > > > > / \ > > > > > > 1 0 1 > > > > > > / \ / \ > > > > > > 2 0 1 0 1 > > > > > > ............................... > > > > > > > > > > Out of interest, aren't the set of all numbers defined by the union of > > all paths through a finite binary tree with N levels just all the > > numbers addressed by the first N bits? If so, why do you bother with > > the tree construction - does it have some special significance? > > The real numbers are represented as infinite paths in the "complete" > infinite tree. Some even twice. > > The union of all finite trees is an infinite tree. > Every finite tree contains only a finite set of paths. > The countable union of all paths of the finite trees is therefore the > countable union of all finite paths. > The countable union of all finite paths is in the union of all finite > trees. > The "complete" tree containing all paths is identical to the union of > al finite trees, with respect to nodes and edges. > Identical trees cannot contain different sets of paths. Finite trees with the same node sets and edge set are identical, but this is not true for infinite trees. For example, the incomplete infinite binary tree containing only those paths which are eventually constant has the same node set, sat Ti, and the same edge set as the complete infinite binary tree, say Tc, but has a different set of paths. > Therefore, both trees contain the same set of paths. False for infinite trees as proven by the example above, Ti != Tc. > Therefore the "complete" set of all path is countable. True for Ti but false for Tc, so false. > Therefore the set of all real numbers is countable. False for the reasons above. > Therefore ZFC is inconsistent. Actually it is WM who is inconsistent by making false arguments. > > Regards, WM |