Cantor Confusion
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From: mueckenh on 17 Jan 2007 05:42 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > The trees are sets of nodes. > > Trees usually consist of a set of nodes _and_ a set of *edges*. The set > of paths in a particular tree is defined only if _both_ of the sets are > given. > Every edge is related to a node. The set of nodes is the set of ends of edges united with the set of root nodes (which is only one). Edges can but need not be used. One can use only nodes as well as one can use only edges. 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: 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). Now, the path 0.111... in fact would look like a diagonal 0 0. 1 1 2 1 3 1 .................. Are the different notations "line" and "level" the reason for the difference? Regards, WM
From: mueckenh on 17 Jan 2007 05:51 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. > > > > > 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? Just say "I wish the path 0.010101... be in here"? Harry Potter number omega? 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? 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: 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). Now, the path 0.111... in fact would look like a diagonal 0 0. 1 1 2 1 3 1 .................. Are the different notations "line" and "level" the reason for the difference? Regards, WM
From: mueckenh on 17 Jan 2007 06:13 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 propery of being finite, then the union of all initial finite segments cannot be an infinite segment. N does not exist. Or do you caim that the union of all initial finite segments {1,2,3,...,n} is different from the union of all the ends of such segments, i.e., the natural numbers n??? > 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. > > The proper definition of a union of sets is a set that contains x as a > member if and only if at least one of the sets being unioned contains x > as a member. Whether the union is of finitely many sets or infinitely > many makes no difference to that definition. > > So whatever WM's idea of "union" is, it is not the mathematical union of > sets. No? Then I will quickly withdraw my statements above and say: There cannot be an infinite union N unless there is at least one infinite number in N. But, alas, that would not be a finite number and, hence, would not belong to N. Therefore there is no N. I had been advocating this position already, once upon a time. Regards, WM
From: Franziska Neugebauer on 17 Jan 2007 06:16 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> > Franziska Neugebauer schrieb: >> >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > Franziska Neugebauer schrieb: >> >> [...] >> >> >> > Obviuosly yoou intermingle "unary" and "binary". >> >> >> >> >> >> Indeed I mean "binary represenation". To resume: If your set of >> >> >> binary representations does not contain the representation of >> >> >> 1/3 your proof is meaningless already before starting to write >> >> >> it down. >> >> > >> >> > The representation of 1/3 is in the union of all finite trees. >> >> >> >> The path of 0.[01] is not a member of Sigma* with binary alphabet >> >> Sigma = { 0, 1 }. >> > >> > But the path of 1/3 = 0.[01] is a member of a binary tree which >> > contains all possible nodes --- infinitely many. >> >> 0.[01] (the alternating path) is a member of the infinite binary >> tree. >> >> > The union of all finite binary trees contians all nodes --- >> > infinitely many. >> >> 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. I remember that you refused to use usual graph theoretical approach where a tree is a directed graph. A graph is an ordered pair of a set of nodes _and_ a set of edges. Please show that in general a union of trees (ordered pairs of sets of nodes and sets of edges) is a tree. >> So obviously you mean a different tree representation in terms of >> sets and/or a different definiens in the definition of "union". >> >> > Now apply your logic. >> >> To which definitons? > > I gave them. Look them up or leave it. I will not repreat them hre. Maybe you could put your definitions into wikipedia, SCNR. >> >> > It is the limit of a set of finite paths like N is the limit of >> >> > all of its finite initial segments. >> >> >> >> This should read: The binary representation of 1/3 is _not_ a >> >> member of the set of all finite paths _as_ N is _not_ a member of >> >> the set of all natural numbers. >> > >> > Correct. >> >> Aha. >> >> > One could add: Cantor's diagonal number is not a member of >> > the set of all finite initial segments of Cantor's diagonal number. >> > (And only such initial segments occur in Cantor's diagonal proof). >> >> Since no infinite string is a member of the set of all finite >> sequences of strings this is trivially true. So what? > > The Equilateral Infinite Triangle (EIT) contains all lines which can > be enumerated by natural numbers: > > 1 0.1 > 2 0.11 > 3 0.111 > ........... This is simply a(n infinite) list (sequence) of unary representations. Actually this sequence lists Sigma^+ = Sigma* \ { epsilon }. It's cardinality is card(omega). > In addition it contains the complete diagonal 0.111... Since Sigma^+ does not contain an element "1..." your "EIT" sequence does not contain it as element either. So your claim is _wrong. > (Otherwise, Cantor's diagonal proof would fail.) 1. Non sequitur. 2. Cantor's proof does not "fail" in that case cause 0.1... is not _in_ the "EIT"-list (i.e. there is no pair (n, 0.111...) e EIT, n e N). F. N. -- xyz
From: mueckenh on 17 Jan 2007 06:19
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. Regards, WM |