Cantor Confusion
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From: William Hughes on 17 Jan 2007 08:03 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > > If there is no difference in terms of nodes, why does T1 not contain > > > all possible paths? > > > > Because the paths define the nodes, not the other way round. > > Wrong. My trees are defined by nodes. In this case T1 is not a tree. There is no set of nodes which contains all finite paths but does not contain an infinite path. T1, the union of all finite trees, contains only finite paths. You cannot make T1 contain an infinite path by playing with definitions. > > > > > > > > What must be changed so that T1 gets T2? Concerning the hard ware (not > > > the breath of God). > > > > > > > > > If you cannot distinguish T1 and T2 by > > > > > finding a node which is in T2 but not in T1, then T2 = T1. > > > > > > > > Incorrect. Two different trees can have the same nodes. > > > > > > What must be changed in order to make T1 being T2? > > > > A tree is a set of paths. (there are other possible > > definitions, however, we need T1 to be a tree, > > so we can compare the trees T1 and T2) > > > > The tree contains paths, the paths > > contain nodes. Therefore a tree contains nodes. However, > > the fact that a set of nodes M is in the tree does not > > mean that path, P1, corresponding to the nodes M is in the tree. > > M may also be composed of nodes from more that one path. > > > > Call a tree 'node-complete' if > > > > Whenever the nodes M corresponding to a path > > P are in the tree, then path P is in the tree. > > > > Given any tree (set of paths) we can add paths to > > it to make it node-complete. > > > How can we add? > > > T1 is not node-complete. T2 is node-complete. > > To make T1 into T2 we have to add paths to > > T1 to make it node complete. > > That is a nice story. Can you say what has to be done in order to make > a tree node complete? Using your definition of tree, nothing. Using your definition, there is no such thing as a tree that is not node-complete. However, using your definition, T1 is not a tree. If you define a tree to be a set of paths, then two different trees can have the same nodes. Start with one set of paths A, these paths contain nodes M. The nodes M contain the set of paths A, and as well they contain the set of paths B. Add the set of paths A to the set of paths B to get the set of paths C. C contains the same nodes as A, but C is not the same set as A. > > What must be done to make Cantor's matrix non-node-complete (i.e., to > remove the diagonal without changing the hard ware)? Or what has to be > done to make any tree non-node-complete, for instance a complete tree > or a tree which contains 0.010101... but not 0.010110110...? How can we > recognize whether a tree is (partially) node-complete? We cannot. If you define a tree to be a set of paths then just looking at the nodes will not tell you which paths are in the tree. If you define a tree to be a set of nodes, and any path through the nodes is in the tree, then looking at the nodes will tell you what paths are in the tree but T1 is not a tree. > > Try to find the difference by following simple construction: > > 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.) > > The union of all finite binary trees contains all levels which can be > enumerated by natural numbers: > Correct > 0 0. > / \ > 1 0 1 > / \ / \ > 2 0 1 0 1 > ............................... > However, this is not T1. Using your definition of tree, the union of all finite trees is not a tree. > But it does not contain the complete path 0.111... (Otherwise Cantor's > proof of the uncountability of the real numbers was contradicted). Correct. A union of finite paths does not contain an infinite path. Note, however, that the union of finite paths, does contain every node of the infinite path. Let L_D be the single finite path that contains every element of the (potentially) infinite path 0.111... L_D is a finite path 0.111... is a potentially infinite path L_D does not exist. > > > Now, the path 0.111... in fact would look like a diagonal > > 0 0. > 1 1 > 2 1 > 3 1 > ................. > As noted the EIT contains every node needed to make up the diagonal. Whether the EIT contains the diagonal is a matter of definition. > > Are the different notations "line" and "level" the reason for the > difference? > No. The difference is whether you say that a tree must contain any path defined by its nodes. - William Hughes
From: Carsten Schultz on 17 Jan 2007 08:38 mueckenh(a)rz.fh-augsburg.de schrieb: > Virgil schrieb: > >> In article <MPG.2016e6f7746f9f07989b49(a)news.rcn.com>, >> David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: >> >>> mueckenh(a)rz.fh-augsburg.de wrote: >>>> No. What I am using is this: >>>> 1) The union of all finite trees contains the union of all finite paths >>>> and this set is countable. >>>> 2) The union of all finite trees contains all nodes. >>>> 3) You cannot add another tree to the set of all finite trees without >>>> adding at least one node. >>>> (The paths are subsets of the set of nodes. Two different sets must be >>>> distinguished by at least one element.) >>>> 4) But there is no node to add. >>>> 5) Therefore the union of trees contains all possible paths. >>>> 6) We know that the set of all paths contained in this union is >>>> countable. >>> How does 6 follow from 1-5? >> It doesn't, > > It follows from the theorem that a countable union of finite sets is > countable. > Every finite tree contains only a finite set of paths. > The union of finite trees is countable. Still making the same mistake after all these years. Let N_i be the node sets of the subtrees, which we will denote by T_i), and let us assume that they are all induced subtrees. Let us also denote the set of paths in T_i by P_i. Now let p be a path in the big tree. Then the following statements hold: 1) The path p is a path in the union of the trees T_i (ie the induced subtree on the union of the N_i) if and only if every node that p visits is in some N_i. 2) The path p is an element of the union of the P_i if and only if there is an N_i that contains every node that p visits. Now you might think that these conditions are the same and derive from the countability of the union of the P_i that there are only countably many paths in the union of the T_i. But that would not prove this statement but just the fact that you are incapable of arguing mathematically. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: Dik T. Winter on 17 Jan 2007 08:37 In article <MPG.20174de97ffb81ac989b5e(a)news.rcn.com> David Marcus <DavidMarcus(a)alumdotmit.edu> writes: > Dik T. Winter wrote: > > In article <1168962527.282591.33370(a)l53g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Did you receive and enjoy my book which I sent off on 10. Jan.? > > > > Not yet. > > You didn't receive it or you didn't enjoy it? I didn't receive it and I didn't enjoy it (how could the last happen if the first did not happen?). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Han de Bruijn on 17 Jan 2007 08:46 mueckenh(a)rz.fh-augsburg.de wrote: > Han de Bruijn schrieb: > >>stephen(a)nomail.com wrote: >> >>>All you have proven is that you do not know what the Continuum Hypothesis is. >> >>LOL! Should I add a dozen smileys ? >>:-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) :-) > > Add several dozens! > > The problem with your example is that set theorists distinguish very > sharply between 2^omega and 2^aleph_0. On the other hand, they use > omega instaed of aleph_0. Cantor already did so. > There are some ways to show that there is no difference between 2^omega > and 2^aleph_0. Compare my recent arguing which forces set theorists to > assert that two binary trees with identical nodes contain different > paths. This is obviously such a wishful thinking that I hope, most > newcomers will soon recognize that set theory is as useless as an > intellectual game as it is with respect to applications. http://groups.google.nl/group/sci.logic/msg/1cb9c1859bf71201 Set theory is quite useful as a limited, relatively unimportant, part of mathematics. I've seen very nice applications of it, like for example in Constructive Solid Geometry, music, and (the visualization of) graphs: http://en.wikipedia.org/wiki/Constructive_solid_geometry http://groups.google.nl/group/sci.math/msg/254b1b0fa9564ed5 http://hdebruijn.soo.dto.tudelft.nl/www/programs/delphi.htm#gd Pascal source code comes with the latter reference. In this code, there are several Units with the phrase "set of Byte". Meaning that _finitary_ set theory may be employed, rather sparsely though, in applications. Han de Bruijn
From: Dik T. Winter on 17 Jan 2007 08:59
In article <1169033684.287474.208720(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1168962527.282591.33370(a)l53g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > > > > 0. > > > / \ > > > 0 1 > > > / \ / \ > > > 0 1 0 1 > > > | | | | > > > 0 1 0 1 > > > | | | | > > > 0 1 0 1 > > > ........................... > This finite tree I call: Trauerweide > > > > So it contains infinitely many nodes? You are again shifting your paradigm. > > And I would not think about that as a finite tree, only as a tree that > > splits at finitely many places. > > This was my first definition. The property "finite" belongs to the tree > because there are splittings only between level 0 and level n. This > shows that the union of paths can really be fomed. So your recent > objections are removed (therefore I do not answer your recent > contribtions.) Ok. But this does not help either. The complete tree contains the path 0.0101010101...; none of the 'finite' trees contains that path. So that path is *not* in the union of the sets of paths of the finite trees, while it is in the union of trees. > > > > You do when you want to prove that the set of paths is countable. > > > > Remember what you wrote: > > > > (1) In each finite tree the set of paths is finite. > > > > (2) A countable union of finite sets is countable. > > > > (3) So the set of paths in the complete tree is countable. > > > > (3) is not justified because (2) is not about the set of paths in the > > > > complete tree. Now try to do this proof *without* using sets of paths. > > > > No answer to this? > > The answer is: We can use sets of levels, sets of edges, sets of nodes > or sets of paths. Look at the "Trauerweide". Yes. But still the union of the sets of paths in the finite trees is not the set of paths in the complete tree. There are paths in the complete tree that are in *none* of the finite trees and so not in the union of those sets of paths. > > > > > In the union of all finite squares > > > > > > > > > > (11) > > > > > > > > > > (11), (12) > > > > > (21), (22) > > > > > > > > > > etc. > > > > > > > > > > there is no infinite diagonal. > > > > > > > > Why not? I only state that that diagonal is not in the union of the sets > > > > of diagonals of the finite squares. > > > > > > Which digit is missing? > > > > 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. > 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. > 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). 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. > > > There are more limits than sequences? > > > > Eh? Why do you think so? Each sequence has only a single limit (when the > > limit is properly defined and when it exists). > > Correct. So why your question? > > > What we need is: All paths > > > in the union tree are countable. All possible nodes are in the union > > > tree. > > > > Yes, you desperately need it. I do not need it, but you do. > > 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... . > > > A path can only then differ from all paths of the union tree if > > > there is another node (or if it is running in other directions --- try > > > that argument?). > > > > What is the argument here? I think you are back to the old argument, that > > is not valid either. > > (1) For each pair of paths there is an edge were they differ. > > in the union of all finite trees. Accepted. > > > (2) For each path there is no edge where it differs from all other paths. > > in the union of all finite trees. Accepted. But you state above "A path can only then differ from all paths of the union tree if there is another node". What do you mean with that? (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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |