Cantor Confusion
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From: mueckenh on 18 Jan 2007 05:43 Dik T. Winter schrieb: > > If as you (finally) defined, a tree consists only of the sets of nodes > and edges, and a path in a tree is defined as a sequence of connected > edges starting at the root of the tree, in that case the infinite > tree contains the path 0.010101... . Yes. > Further with the finite trees as > you currently define them (with all having infinitely many nodes, so I > the only finiteness is the number of splitting), the union of sets of > paths in two finite trees is the set of paths in the union of those > two trees. *But* that does not hold when you unite the whole collection > of finite trees. No? 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 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 their union as well as in WWT. > > > But *all* paths are in the union of the > > finite trees, because it is simply silly to suspect different sets of > > paths in a tree with all nodes. > > Eh? There is a whole collection of different sets of paths in a tree with > all nodes (assuming the definition of tree as I stated above, that is the > pair [set of nodes, set of edges]). Of course it is silly to suspect > different complete sets of paths. So it is. > > > Therefore the only conclusion is that > > the path 0.010101... does not exist. > > Wrong. If it exists, then it exists in every tree with all nodes and edges. > > > So you say. If you were right, there should be some indication for the > > difference. > > What makes the path 0.010101... be in a tree with all nodes but not in > > a tree with all nodes? > > Again, I do not state that. What I state is that that path is (with the > definitions above) in the complete tree. It is in every tree which contains all nodes and edges (as defined by me), because there is no end when we follow the path always swithing betwen 0 and 1. > And I also state that that path > is not in the union of the sets of paths of the finite trees (also again > according to the above definition). And I say so (and am allowed to say > so) because that path is in none of the finite trees. That is the same as saying N is not in the union of all finite segments {1,2,3,...,n}. N is this union. The complete tree is the union of all finite trees. The complete tree contains all nodes and edges. Therefore it contains all paths. > > > > > > None. But all sets of diagonals of finite squares contain only finite > > > > > diagonals, so there can not be an infinite diagonal in their union > > > > > (which is a set of infinite size). > > > > > > > > So there cannot be an infinite diagonal at all? > > > > > > I do not state that. I state that in the union of the sets of diagonals > > > of the finite squares only finite diagonals do exist (infinitely many > > > of them). This does *not* prevent an infinite diagonal to exist. > > > > What can we do to switch its existence on and off? > > I do not do that. Where do I state that it does not exist? I only state > that it exists but that it is not in the union of the sets of diagonals of > finite squares. Is it or part of it outside of this union? > > > > > The Equilateral Infinite Triangle (EIT) contains all lines which can be > > > > enumerated by natural numbers: > > > > > > > > 1 0.1 > > > > 2 0.11 > > > > 3 0.111 > > > > ........... > > > > > > > > In addition it contains the complete diagonal 0.111... (Otherwise, > > > > Cantor's diagonal proof would fail.) > > > > > > Right. But the complete diagonal is not in the union of the sets of > > > diagonals of the finite triangles. > > > > Where is it? Or is it nowhere? > > Why should it be in that union? What reasoning are you going to assume that > it *must* be in that union? It is simply not in that union, and that is > easily enough to prove. But it exists, but not in that union. If it is not in the union, then it must be outside of that union. At least part of it. Which part is outside? You say it exists. It exists not in the union. Where does it exist? > > > > > 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 > > > > ............................... > > > > > > > > But it does not contain the complete path 0.111... (Otherwise Cantor's > > > > proof of the uncountability of the real numbers was contradicted). > > > > > > It does contain that path (according to your definition above). > > > > Yes, that is obvious. But, for a moment, switch to the cut trees, > > please. Does the union of all cut trees contain the paths 0.111...? You > > would say no. > > And again, I would say yes. That is remarkable. Are you really sure? >I even provided a definition of union of trees > for that case and with that definition that was easily enough to prove. But > then we have that the union of the sets of paths of two different finite trees > is not the set of paths of the union of those two trees. What is lacking? > > > > But > > > the union of the sets of paths in the finite trees is not the set of > > > paths in the infinite trees. The first does not contain the path > > > 0.01010101..., but it is in the second. > > > > If this path exists, then it must be constructible from all the nodes > > present. There is a bit, 0 or 1, for every index n in N. That is not > > sufficient? > > Not sufficient to be in the union of the sets of paths of the finite trees. All nodes and all edges are in the union. There can be no path being utside. > The reason is that a path is in that union *if and only if* it is in one > of the sets that are used to form the union. The reason is: If every subset of a path in the union, then the whole path in that union. Every subset is in the union. Or not? > > > So why your question? > > > > There are only countable many sequences. > > I disagree. There are uncountably many limits. You are trying to prove > that there are countably many sequences and limits. If there are countably > many sequences there are countably many limits, and the other way around. > But until now you have failed to prove either the first or the other. Every initial segment of a sequence in the union of trees. All initial segments are countable. > > Your proof of countably many sequences hinges on the assumption that the > union of the sets of paths in the finite trees is the set of paths in the > completed trees. But that one is false, as I did show above. No. If every subset of a path in the union, then the whole path in that union. This is valid for every path. > > > > > The countablity of all paths in the union of finite trees is a theorem > > > > of set theory. > > > > > > No, it is not. Prove it. You use the *false* assumption that the > > > union of the sets of paths in the finite trees is the set of paths > > > in the complete tree. It is not because the first does not contain > > > the path 0.01010101... . > > > > I use the fact that the union of all nodes contains the union of all > > paths. > > Now you are talking about unions of paths. The theorems of set theory are > about unions of *sets* of paths. There is quite some difference. What > is the union of all paths? How do you *define* the union of two paths? The union of all paths is the tree, because the paths are subsets of the tree. > > > I mean that in an infinite tree (= union of all finite trees) with all > > nodes, there is a set of paths. Paths which are not therein, must > > indicate why they are not therein - not be a spell but by a bit (or > > node). > > Again, I never stated that there are paths missing from that set of paths. > I only stated that there are paths that are not in the union of the sets > of paths of the finite trees. And it is simple to state why, those are > the paths that are not present in the finite trees. Every subset of a path in the union of trees. ==> The path in the union of trees. > > > > (1) and (2) together state that while all paths are different, there > > > is *no* single edge that makes a single path stand out from all other > > > paths. > > > > But paths which are not determined by the set of all nodes must stand > > out by some indication, > > There are no such paths. > -- Correct. All paths are in the union of all finite trees. Regards, WM
From: mueckenh on 18 Jan 2007 06:00 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. Therefore, both trees contain the same set of paths. Therefore the "complete" set of all path is countable. Therefore the set of all real numbers is countable. Therefore ZFC is inconsistent. Regards, WM
From: mueckenh on 18 Jan 2007 06:09 Franziska Neugebauer schrieb: > > The union of {1} and {1,2} is {1,2}. This holds for initial segments > > of N as well as for finite trees, cut or as Trauerweide. > > { 1 } U { 1, 2 } = { 1, 2 } is simply true, but does not "hold". No? > > But there is no infinite path by definition? > > In a tree of finite hight: no. In the union of all finite trees: yes. (If every subset of an infinite path (= set of nodes) is in the uninon of finite trees, then the complete path in the union.) > > >> > If there is an infinite diagonal, however, then there is no reason > >> > for the path 0.111... of the union of finite trees to be finite. > >> > >> The question is whether 0.[01] is in the "union of all finite paths". > > > > If it exists, > > It exists and it is 0.[01]. Then it exists in the union of all finite paths. (Every segment of 0.010101... exists there, so the whole number does.) > > Define "union of all finite trees". We are currently talking about > finite paths. I did. > > > and all finite paths which could extend any of its paths. > > > > The union of all finite paths is an infinite path, > > Don't see that even a union of two paths is a path. A path is an ordered set of nodes. The union of two paths is he union of these nodes. > > > because for every end node n of a path, there is a path crossing n. > > Please elaborate. For every node on level n of a path, there is a node at level n+1 which also belongs to the path. if you do not agrre, then please say where a path ends in the union tree. Regards, WM
From: mueckenh on 18 Jan 2007 06:17 Carsten Schultz schrieb: > > You assume that the union P_i of paths contains more paths than can be > > constructed from finite initial segments? > > I do not assume anything. I just note that being a path in the union of > the T_i and being an element of the union of the P_i are a priori > different things and that you would have to prove their equivalence in > your setting should you claim this equivalence. The union of all finite trees is an infinite tree. 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. Therefore, both trees contain the same set of paths. > >> the countability of the union of the P_i that there are only countably > >> many paths in the union of the T_i. > > > > Is Cantor's diagonal longer than the union of all of its initial > > segments? > > Can you stay on the subject? That is the subject. Try to get it. > > > Is N more than the union of all of its initial segments {1, 2, 3, ..., > > n}? > > Are there subsets of N that are not subsets of a proper initial segment? Of course. The set of primes, for instance. > > > Yes? > > Then N is certainly also more than the union of all ends {n} of its > > initial segments? > > No? The union of ends is more than the union of segments? > > > > Next time, please switch on your brain before posting. > > I should not be surprised that replying to cranks leads to abuse. So I will no longer reply to you. By the way, for your further education, mst probably in vain, even the subset of all even numbers is not in a proper initial segment of N. Regards, WM
From: Carsten Schultz on 18 Jan 2007 06:30
mueckenh(a)rz.fh-augsburg.de schrieb: > Carsten Schultz schrieb: > >>> You assume that the union P_i of paths contains more paths than can be >>> constructed from finite initial segments? >> I do not assume anything. I just note that being a path in the union of >> the T_i and being an element of the union of the P_i are a priori >> different things and that you would have to prove their equivalence in >> your setting should you claim this equivalence. > > > The union of all finite trees is an infinite tree. > 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. > Therefore, both trees contain the same set of paths. > > >>>> the countability of the union of the P_i that there are only countably >>>> many paths in the union of the T_i. >>> Is Cantor's diagonal longer than the union of all of its initial >>> segments? >> Can you stay on the subject? > > That is the subject. Try to get it. >>> Is N more than the union of all of its initial segments {1, 2, 3, ..., >>> n}? >> Are there subsets of N that are not subsets of a proper initial segment? > > Of course. The set of primes, for instance. Wow. Even though the set N is identical to the union of its proper initial segments and identical sets cannot have different sets of subsets? -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page. |