Cantor Confusion
in [Math]
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From: mueckenh on 10 Mar 2007 16:53 On 10 Mrz., 18:03, Virgil <vir...(a)comcast.net> wrote: > In article <1173524975.470534.275...(a)q40g2000cwq.googlegroups.com>, > > mueck...(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. Therefore it suffices to show that the number of paths of the tree is countable for every level L(n) which is enumerated by a finite number. This proof shows that all path in the tree are countable- Regards, WM
From: mueckenh on 10 Mar 2007 16:57 On 10 Mrz., 18:18, Virgil <vir...(a)comcast.net> wrote: > In article <1173525477.490852.229...(a)s48g2000cws.googlegroups.com>, > > mueck...(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). Then P(N) is countable. I always said that Hessenbergs proof is insufficient. And it is totally clear that the mapping from P(N) into Q cannot be surjective. Regards, WM
From: mueckenh on 10 Mar 2007 17:04 On 10 Mrz., 18:28, Virgil <vir...(a)comcast.net> wrote: > In article <1173526215.959130.304...(a)v33g2000cwv.googlegroups.com>, > > mueck...(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. M yproof is valid for every path which is composed of nodes which have a distance from the root node to be expressed by a natural number. Other paths do not express real numbers. > > > > > > 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. What is the difference between the complete infinite binary tree and a tree which covers only all levels L(n) with finite 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. > > 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. i d no want to represent the number of nodes in the path. I only want to represent the distance from the root node. Is that distance infinite for the complete tree? > > > > > 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? Is there a last node in the complete infinite tree? > > How does WM claim to index the last node when there isn't one? I claim to index all nodes - each one by a natural number. Regards, WM
From: Gc on 10 Mar 2007 18:29 On 10 maalis, 23:53, mueck...(a)rz.fh-augsburg.de wrote: > On 10 Mrz., 18:03, Virgil <vir...(a)comcast.net> wrote: > > > In article <1173524975.470534.275...(a)q40g2000cwq.googlegroups.com>, > > > mueck...(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. > > Therefore it suffices to show that the number of paths of the tree is > countable for every level L(n) which is enumerated by a finite number. > This proof shows that all path in the tree are countable- What proof? Those paths aren`t discrete.
From: Gc on 10 Mar 2007 18:34
On 11 maalis, 01:29, "Gc" <Gcut...(a)hotmail.com> wrote: > On 10 maalis, 23:53, mueck...(a)rz.fh-augsburg.de wrote: > > > On 10 Mrz., 18:03, Virgil <vir...(a)comcast.net> wrote: > > > > In article <1173524975.470534.275...(a)q40g2000cwq.googlegroups.com>, > > > > mueck...(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. > > > Therefore it suffices to show that the number of paths of the tree is > > countable for every level L(n) which is enumerated by a finite number. > > This proof shows that all path in the tree are countable- > > What proof? Those paths aren`t discrete. I mean, for every finite level L(n) every path is a root for (uncountable) many infinite long paths. So you can`t take an union of all finite paths L(n) and obtain set which contains every path as a discrete subsets. |