Cantor Confusion
in [Math]
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From: mueckenh on 10 Mar 2007 06:17 On 9 Mrz., 20:52, Virgil <vir...(a)comcast.net> wrote: > In article <1173438120.383110.280...(a)h3g2000cwc.googlegroups.com>, > > > > > One wonders which branchings WM denies in his incomplete trees. > > > > > None. Alas the number of branchings is countable. > > And simultaneoulsy uncountable? In ZF and NBG they are uncountable. WM > has yet to produce an axiom system in which they are countable. The number of branchings of the binary tree is uncountable??? The number of nodes is countable, isn't it? A node is a branching. > > > But the number of possible sequences of branchings is what we need to > > > count. And that is not countable. > > > The number of paths which are generated by branchings is not counted > > by sequences of branchings but by branchings. > > If it could be counted at all then the set of all infinite binary > sequences could be counted to, as they are easily seen to be bijectable. I is easy, in fact, to show tat there are twice as many branchings than there are paths. > > And after seeing Cantor's proof of the latter's uncountability, If the proof was correct, then we had a contradiction in set theory. But the proof is based upon the assumption that there are infinitely many finite indexes, which is wrong. You will see this yourself, if you consider the Waft Maximum (WM) of the natural numbers, then the WM of the paths of all finite trees, and so on. But for this sake you must think a bit, not only bicker about the meaning of "function". Regards, WM
From: mueckenh on 10 Mar 2007 06:30 On 9 Mrz., 21:10, Virgil <vir...(a)comcast.net> wrote: > In article <1173464401.633116.124...(a)q40g2000cwq.googlegroups.com>, > > The cross section C(n) = |L(n)| is the number of nodes of the level > > L(n). > > Which is only the case when n is a natural number. What else is required? > > This number of > > nodes is the countable cross section C(n) = |L(n)|. It is sufficient > > to have proved this in infinity, that is for EVERY level L(n), > > But it is also necessary to show that it holds for non-natural cardinals > or ordinals, and this has not been done. Is it? Are there levels which are not indexed by finite numbers n? Are there infinite natural numbers involved. > > > If you disagree: What part of a path should not be covered by a node > > of a level L(n) with a finite n? > > The square root of Newton's apple. > > Not that my answer makes as much sense as your question. .... Don't worry. Nobody expected that. Regards, WM
From: Virgil on 10 Mar 2007 12:03 In article <1173524975.470534.275640(a)q40g2000cwq.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Does an infinite tree possess any level which is *not* enumerated by a > finite natural number n, i.e., which has an infinite distance to te > tree? Does the infinite set of naturals need to contain an infinite natural? No more does an infinite set of levels need to contain an infinite level.
From: Virgil on 10 Mar 2007 12:18 In article <1173525477.490852.229630(a)s48g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > I is easy, in fact, to show tat there are twice as many branchings > than there are paths. Only in finite trees. In in a complete infinite binary tree, there are as many paths as subsets of N: We can umber the nodes in any infinite path with the members of N. Then for each subset of N, there is a unique path in a complete infinite binary tree having left branchings at those nodes and right branchings at all other nodes. So that the 'number' to paths is the 'number' of members of P(N). > > > > And after seeing Cantor's proof of the latter's uncountability, > > If the proof was correct, then we had a contradiction in set theory. Wrong. If the proof is correct, which it is, we have a contradiction of WM's beliefs. Which is quite acceptable to everyone, except possibly WM himself. > But the proof is based upon the assumption that there are infinitely > many finite indexes, which is wrong. Not in ZF or NBG. But what is wrong in WM's world is not necessarily wrong in those worlds which do not require his axioms. > You will see this yourself, if > you consider the Waft Maximum (WM) of the natural numbers, then the WM The WM is, in every sense, irrelevant to mathematics. > of the paths of all finite trees, and so on. But for this sake you > must think a bit, not only bicker about the meaning of "function". I have though a bit, and find WM still to be wrong and his WM still to be irrelevant to mathematics, as are his other complaints about how mathematicians do things. In other words, I see no benefit and possible harm, in mathematicians paying any heed to WM.
From: Virgil on 10 Mar 2007 12:28
In article <1173526215.959130.304560(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 9 Mrz., 21:10, Virgil <vir...(a)comcast.net> wrote: > > In article <1173464401.633116.124...(a)q40g2000cwq.googlegroups.com>, > > > > The cross section C(n) = |L(n)| is the number of nodes of the level > > > L(n). > > > > Which is only the case when n is a natural number. > > What else is required? For complete infinite binary trees, in which the number of nodes in a path cannot be a natural number, your definition cannot apply. > > > > This number of > > > nodes is the countable cross section C(n) = |L(n)|. It is sufficient > > > to have proved this in infinity, that is for EVERY level L(n), Except that it carefully omits any representation of any complete infinite binary tree. > > > > But it is also necessary to show that it holds for non-natural cardinals > > or ordinals, and this has not been done. > > Is it? Are there levels which are not indexed by finite numbers n? Are > there infinite natural numbers involved. There are paths which cannot be "indexed" by natural numbers in the sense of having a natural representing the number of nodes in the path. > > > > > > If you disagree: What part of a path should not be covered by a node > > > of a level L(n) with a finite n? What natural is the index for the "last" node of a path? How does WM claim to index the last node when there isn't one? |