Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Virgil on 24 Jan 2007 16:40 In article <1169644109.860794.220700(a)k78g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1169545975.865624.242740(a)d71g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > In article <1169396052.939963.194070(a)q2g2000cwa.googlegroups.com>, > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > Virgil schrieb: > > > > > > > > > > > > > > > Therefore, by induction, all finite n-level trees for all n in N > > > > > > exist. > > > > > > But that is all that standard induction allows one to conclude. > > > > > > > > > > It is enough. Or should there one level be missing? Please specify > > > > > which remains to be included. > > > > > > > > Anything that concludes anything about infinite trees. > > > > > > Conclusions are not members or subsets of trees. > > > > Standard induction does not justify your /conclusion/ that your > > conjunction of finite trees is an infinite tree. > > > Induction covers all natural numbers and, hence, all finite trees. I > don't need more for the union of all finite trees. Yes you do! Induction can say no more than that the pseudo-union, as defined by WM, of a finite set of finite trees is finite tree. The appropriate induction is as follows: (1) the pseudo-union of binary trees of maximal path length 1 is a finite binary tree. (2) If the pseudo-union of finite binary trees up to max path length n is a finite binary tree, THEN the pseudo-union up to max path length n+1 is also a finite binary tree. Conclusion: By finite induction, for all n in N, the pseudo-union of binary trees of max path length up to n is a finite binary tree. One never gets to any infinite pseudo-unions in such standard inductive arguments.
From: Virgil on 24 Jan 2007 16:45 In article <1169652793.081817.69430(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 23 Jan., 20:52, Virgil <vir...(a)comcast.net> wrote: > > In article <1169548200.625843.205...(a)v45g2000cwv.googlegroups.com>, > > > > > an awful lot of new paths > > > > that are brought into light, namely non-terminating paths. > > > > Finite trees have absolutely zero (non-repeating) non-terminating > > > > paths, > > > > > The union of all finite trees has no terminating path because the union > > > of all finite segments {1,2,3..., n} (with in N) does nowhere > > > terminate. > > >By any mathematical definition of "union" a union of two or more > > distinct trees is not a tree at all. > > > What is the union of {a, b, c} and {a, b, c, d, e, f, g} in your > opinion? Which, if either, of {a, b, c} and {a, b, c, d, e, f, g} is a binary tree in WM's opinion? Now if the set of nodes is M = {a,b,c} and the set of edges is E = {(a,b), (a,c)} then T = (M,E) /is/ a binary tree.
From: Virgil on 24 Jan 2007 16:55 In article <1169653459.278098.283110(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 23 Jan., 21:54, Virgil <vir...(a)comcast.net> wrote: > > In article <1169551459.707075.42...(a)v45g2000cwv.googlegroups.com>, > > > > > > Correct. This shows that there are no other paths. It is the core of my > > > contradiction of set theory.Then the infinitely many paths that from some > > > node onwards alternate > > left branch then right branch do not exist? And none of the infinitely > > many other infinitely alternating patterns exist? > > They do not exist as completed entites, as I have shown by my union (or > however you'd like to call it without destroing its definition) of all > finite trees. Then your alleged tree is not a tree at all. Your pseudo-union will be identical in nodes and edges to the union of finite trees having all paths of equal length n, for all n in N, in which latter case, the paths in your union will also all be of equal length. If this path length is required to be finite, then your pseudo-union is also finite. But if it is not finite then every path will be non-finite and every possible non-finite path will be present. > > > > > Everything else claimed by you is belief in ghost paths (vibrations of > > > vacuum? small black holes in a green tree?), but has nothing at all to > > > do with mathematics.There are all sorts of proper fractions with odd > > > denominators which have > > binary expansions corresponding to paths that WM says do not exist. > > It would seem that these "vibrations in a vacuum" are taking place > > inside WM's head > > of course, they are always everywhere. Which concedes that the inside of WM's head is a vacuum. > To assert that a binary tree allows for other paths than the same > tree is simply killing of mathematics. What might very well kill mathematics is WM's Procrustean straight jacketing of it, were he to succeed.
From: Virgil on 24 Jan 2007 17:03 In article <1169654203.478565.170700(a)v45g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > {a,b,c} U {a,b,c,d,e,f,g,h,} = {a,b,c,d,e,f,g,h,} > > Ordering of the nodes is defined by the type of tree. "Ordering" of the nodes is defined by the set of edges, and is normally only the partial ordering of parent preceding progeny. > 1) Every complete infinite binary tree T (containing all nodes and > edges) contains all paths. Then, unless it contains leaf nodes, it must contain endless paths. But every node that is a leaf node in one finite tree is not a leaf in a larger tree, so that can be no leaf nodes. > 2) The union tree T(oo) of all finite trees is well defined (as I have > shown elsewhere) and yields the complete infinite binary tree > containing all nodes and edges: T = T(oo). Except that "union" is not a legitimate name for the operation as WM defines it, agreed. One might note that it also contains all possible paths. > 3) The union of all finite trees includes the union of all nodes and, > with it, the union of all such subsets which are paths (because every > path is a well defined subset of the set of nodes if the structure of > the tree is well defined). Actually paths would be "subsets" of edges, as it is only sequences of nodes connected by edges that can be paths. > 4) The set of paths in T(oo) is a subset of the countable set of finite > sets of all paths in the finite trees. Trivially false. Since the "union" tree cannot contain any leaf nodes, no path can be finite. > 5) A countable union of countable sets is a countable set (according to > ZF with AC). True but irrelevant. > ==> The set of all path is countable. False, and proven false. > (==> The real numbers are > countable.) False, and proven false. > > Going on, we can say: > > 6) T(oo) = T contains only finite paths. False, and proven false. > 7) T(oo) = T contains all paths including all infinite paths. True but irrelevant. > ==> There are no infinite paths. (There are no irrational numbers.) False, and proven false. > > > Going on, we can say: > > 6) T(oo) = T contains only finite paths. Then it is a finite tree and not WM's pseudo-union of all finite trees. > 7) T(oo) = T contains all paths including all infinite paths. OK. > ==> There are no infinite paths. (There are no irrational numbers.) Garbage. > > Nothing further remains to say. WM has lied his best, but since he assumes a priori what he wishes to prove, his attempts at proof all fail.
From: Virgil on 24 Jan 2007 17:10
In article <1169654536.816807.257410(a)k78g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 24 Jan., 01:07, Virgil <vir...(a)comcast.net> wrote: > > > Of course, "winning" does not require that one convinces the crank > > himself of anything, it only requires convincing the vast majority of > > lurkers of the crank's crankhood. > > If the vast majority consists of lurkers believe in the finished > infinity, this result is excluded. Since, according to WM, all such things as infinite sequences and functions having infinite domains require "finished infinities", that includes almost everyone dealing with standard mathematics at above grade school level, WM would bar almost everyone interested from judging him. > > > > In which task the crank often unconsciously cooperates > > in particular if he sees different sets of paths realized by same sets > of nodes and edges. What cranks see that exists only in their own minds is not the responsibility of their critics. > But I cannot believe, in fact, that you argue unconsciously. I don't, but I do not have that same confidence about WM's arguments. Considering the quality of WM's logic, one cannot be sure that he does argue consciously. |