Cantor Confusion
in [Math]
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From: Virgil on 30 Apr 2007 13:44 In article <1177936811.325960.278110(a)y80g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 27 Apr., 22:20, Virgil <vir...(a)comcast.net> wrote: > > In article <1177671404.952318.309...(a)r35g2000prh.googlegroups.com>, > > > > > > > > > > > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 24 Apr., 20:55, Virgil <vir...(a)comcast.net> wrote: > > > > These uncountability proofs are wrong. > > > > > > Saying that they are wrong does not make them wrong. You must show that > > > > some step in their proof is wrong in order to establish that they even > > > > can be wrong, much less that they are wrong > > > > > > Here it is again: > > > > (1) The set of levels in a CIBT is indexed by N. > > > > (2) For each subset of N there is a unique path which branches left > > > > at all the levels indexed by that set and right elsewhere. > > > > (3) this establishes a bijection between the set of all paths in the > > > > CIBT and the set of all subsets of N > > > > (4) Card(set of all subsets of N) = Card(P(N)) > Card(N) > > > > (5) Card(set of all paths in a CIBT) > Card(N). > > > > (6) The set of all paths in a CIBT is uncountable, > > > > meaning of cardinality greater than that of N. > > > > > > Which step is wrong, WM? > > > > > Wrong is the assumption (2) that all paths can be distinguished. I have not assumed that. What I have assumed is: (1)that every set has a power set, as is guaranteed by axiom in ZF and NBG, and (2) the cardinality of a power set is always greater than the cardinality of the original set, which is provable in ZF and NBG. Thus, at least in ZF and NBG, there exist complete infinite binary trees and such trees have uncountably many paths. If WM chooses to assume that is not the case, there are systems in which he is wrong. There > > > are no unique paths but only unseparated bunches. The path always branching left does not exist as a unique path? The path always branching right does not exist as a unique path? > > > > If there are no unique paths, there is no tree at all. > > > > Does WM deny that there is a countably infinite set of levels? > > Does WM deny uncountably many subsets to that set of levels? > > > > The number of critics is unimportant, in particular if they are biased > > > and unable to think in logic terms like: > > > If there are X separated paths in the tree, then there must e X-1 > > > nodes separating them in the tree. > > > > That may hold for finite trees, > > There cannot be more splitting results than splittings. That is true > in any case which is object to sensible thinking. If it is not true in > infinity, then infinity is not subject to sensible thinking. Mathematics is not about "sensible thinking" but about what are the consequences of a given set of axioms. Mathematics is in no way constrained by any limitations of what can occur in the merely physical world.
From: Virgil on 30 Apr 2007 14:01 In article <1177939040.660107.315400(a)y5g2000hsa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 30 Apr., 00:25, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1177672072.574980.291...(a)n15g2000prd.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > If you look from the one side, then the result is different from the > result you get from the other side. Why do you think that only your > side is the correct one? We have several "sides", called axiom systems, in each of which CIBTs have uncountably many paths, WM has no discernible axiom system, but merely randomly states his opinion on what should or should not be allowed in his "system". > > At *every* node in the tree there is a bunch of paths existing. Hence > your point of view is completely irrelevant. There is no specific path > p in the tree. Otherwise the existence of p would imply the existence > of {p}. In our trees, there is one path existing for every subset of the set of levels. What sort of structure WM's pseudo-trees have is not at all clear. > > > > > > A bunch like 0,0 is split into two bunches 0.00 and 0.01 by the due > > > node. > > > > No. The bunch 0.0 terminates and the bunches 0.00 and 0.01 continue. > > But only till the next node. Then they are split off. Which subset of the set of all levels does not represent the unique path branching left from just those levels and branching right from all other levels? Failure to find such a subset shows that WM's position is untenable. > > > > > > > That is a matter of convention of notation. In any case a node > > > increases the number of separated path bunches by 1. > > > > No. You are confusing bunches with edges. The number of paths does > > *not* increase at a node. The number of paths remain the same, but > > some go right and others go left. > > Definition: A bunch of paths is a nonempty set of paths which run > together through a common set of nodes, as long as they do so. Red herring! Which subset of the set of all levels does not represent the unique path branching left from just those levels and branching right from all other levels? Failure to find such a subset shows that WM's position is untenable. > > A bunch of paths is a set of nodes which have a node in common. Still a red herring. Nobody but WM is concerned with his "bunches". Which subset of the set of all levels does not represent the unique path branching left from just those levels and branching right from all other levels? Failure to find such a subset shows that WM's position is untenable. > > Definition: A bunch of paths is a nonempty set of paths which run > together through a common set of nodes, as long as they do so. Still a red herring. Nobody but WM is concerned with his "bunches". Which subset of the set of all levels does not represent the unique path branching left from just those levels and branching right from all other levels? Failure to find such a subset shows that WM's position is untenable. > > In every node 1 path bunch comes in and one more paths bunch goes out. Still a red herring. Nobody but WM is concerned with his "bunches". Which subset of the set of all levels does not represent the unique path branching left from just those levels and branching right from all other levels? Failure to find such a subset shows that WM's position is untenable.
From: MoeBlee on 30 Apr 2007 14:04 On Apr 30, 5:35 am, mueck...(a)rz.fh-augsburg.de wrote: > On 27 Apr., 22:06, Virgil <vir...(a)comcast.net> wrote: > Why then do you believe that Cantor's diagonal proof is true? You'd have to define 'true proof'. Meanwhile, with just some basic knowledge of predicate calculus and set theory, one can see that Cantor's diagonal argument is formalizable into a proof in first order logic from the axioms of Z set theory. If you understand the basic material, there is not much to believing, since it is a simple of matter of observing that a certain sequence of formulas that satisfies the definition of 'first order proof from the Z axioms' does exist. MoeBlee
From: Virgil on 30 Apr 2007 14:20 In article <1177941733.864803.29600(a)h2g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 30 Apr., 00:45, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1177674968.281638.41...(a)u32g2000prd.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 25 Apr., 04:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > > > Cantor's diagonal can be defined up to any line n of the list. > > > > > > > But > > > > > > > this says *nothing* about an infinite sequence of digits. > > > > > > > > > > > > It is defined for *any* line of the list. > > > > > > > > > > That does not define it for *all* lines. > > > > > > > > It does. > > > > > > If it were, than the number of digits were complete. Otherwise there > > > can be no judgement concerning "all" but at most concerning "each". > > > > And in mathematics if something is proven for "each", it is proven for > > "all. > > > That is wrong in many cases. That's probably why mathematics is wrong. > Example: Each natural number is a single number. All natural numbers > are not a single number. "Each...is..." implies "all...are...". And ALL natural numbers are natural numbers. > > That means, all the nodes in the tree are passed by path bunches of > many paths. There is never a single path in the tree. Is the single path always branching left not in the tree? is the single path always branching right not in the tree? They certainly exist, so if they are not in the tree that tree is incomplete as an infinite binary tree, and is totally irrelevant to the question of how many paths exist IN a COMPLETE infinite binary tree, CIBT. > That is irrelevant, because there is no node passed by less than > infinitely many paths. Uncountably many, in a CIBT. > > Wrong. If you could name all, then no one would remain unnamed. > (Otherwise you had not named all.) In a completed, i.e., finite set, > there must be a largest one. If one can define a rule which names them all then they are all named by that rule, and one can define such rules. For example the rule that creates a unique finite binary string of "0"s and "1"s for each natural. The rule is complete even if the set of names is, in WM's world, not. > > > > > > All means "completed". Completed means, there is nothing to be added. > > > > All means indeed completed. The set of all natural numbers is completed > > by the axiom of infinity. > > But it is not completed by the existence of all natural numbers. It is in ZF and NBG. In what axiom system is it not completed? > > > > > In case of a set the elements of which obey trichotomy, there is a > > > largest one. > > > > Prove it. The law of trichotomy only states that given two numbers it > > is possible to determine that the first is larger, the second is large, > > or they are equal. > > Given all numbers, it is possible to determine the largest. In what axiom system? > > > It is your simplified finitistic view where that > > means that there is a larger number. > > It is the meanig of "completed". Then "completed" has no mathematical relevance. > Deplorably this mathematics leads to the result that there are more > splitting results than splittings in the binary tree. It certainly lead to more paths than nodes in such straightforward systems as ZF and NBG. We have yet to see any adequate description of a system in which it doesn't. > > As far as mathematics is concerned, 0.000... is unique (as long as we use > > the standard definition for that notation). > > 0.000... is not unique in the tree. The number of unique numbers is > countable. Actually it is the non-unique ones (having more than one representation) which are countable.
From: Virgil on 30 Apr 2007 14:23
In article <1177942248.683128.181240(a)p77g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > If {p} is impossible even in the infinite tree, then the nonexistence > of {p} implies the nonexistence of p. Every subset of the infinite set of levels n a CIBT is /possible/, and for each one there is a unique path which branches left from just those levels and right from all others. Thus as many paths as subsets of an infinite set. |