Cantor Confusion
in [Math]
|
From: Gc on 10 Mar 2007 18:50 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- All your problems would be solved if you could find a unique finite charaterist for each subset of N. Otherwise. Let`s suppose the nodes of your graph are enumerated by naturals and for each path on level L(n) there is a unique finite set which contains the numerals of the nodes that the path includes. Now if you take union of all the finite paths (unique finite subset), you surely will get a countable set, BUT the set contains the infinite paths as subsets which aren`t discrete.
From: Virgil on 10 Mar 2007 19:52 In article <1173563630.875399.52180(a)p10g2000cwp.googlegroups.com>, mueckenh(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- it only shows that the set of those paths which have a "level" , i.e., end, are countable. It says nothing at all about the number of endless paths. A path starts at the root node and follows by branching left or right from every node except a leaf node if any. Match each infinite path with the set of naturals containing n if and only if the n'th branching is to the left, and one easily sees that this bijects the set of all such infinite paths with P(N). And since P(N) is uncountable, so is every set bijectable with it.
From: Virgil on 10 Mar 2007 20:41 In article <1173563849.198787.252770(a)c51g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. If WM can prove that P(N) is countable then he should have no problem also proving that 2+2 = 5. > I always said that Hessenbergs proof is > insufficient. Saying things does not make them make them true, particularly, as in WM's sayings, when they are presented without proof. A proof has been presented here that P(N) is uncountable, at least in ZF and NBG, and no contrary proof has been presented, so that, as far as mathematics is concerned, P(N) is uncountable, at least in ZF and NBG. > And it is totally clear that the mapping from P(N) into > Q cannot be surjective. Except that it is quite easy to describe such a surjection, at least if Q represents the rationals: The rational value correspondintg to a set of naturals will be determined by the number of elements in that set of naturals as follows: Size Corresponding of rational set number 0 0 1 The member of that set 2 The negative of the first member of the set 3 The quotient of first member divided by second member 4 The negative of the number determined by a 3 element set 5 The reciprocal of that from the 3 element set 6 The negative of that from the 5 element set. Other Any rational, as all rationals are already covered. So once again we see that WM has no comprehension of what he is saying.
From: Virgil on 10 Mar 2007 21:04 In article <1173564286.964652.75250(a)p10g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. WM has yet to produce a valid proof of anything. If a path has a node at every finite distance from the root, as expressed by a natural number, then it has as many nodes as there are naturals. If all such infinite paths are allowed one has a compete infinite binary tree. In a complete infinite binary tree, there are as many paths as subsets of N, as is shown by building a bijection between them. For each infinite path we can pair off the nodes in their natural order starting from the root node with the members of N, with none of either nodes or naturals left out. Then for each subset, S, of of N, there is a unique path in a complete infinite binary tree having left branchings at the nodes corresponding to those naturals in the set, S, and right branchings at nodes whose naturals are not in S. This is clearly a one-to-one correspondence between infinite binary paths and subsets of N. This bijection between the set of all such infinite binary paths and the set of all subsets of N, namely P(N), establishes that they have the same cardinality, and P(N) is known to be uncountable, so the set of paths is also uncountable. > > > > > > > > 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? If your tree is is limited to countably many paths and mine is not, clearly your tree is missing most of its paths. Where they might be hiding I don't know, or care. But your "tree" is incomplete until you find where all those missing paths went, or replace them. > > 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. Some seem people only to want what they cannot have. > Is that distance > infinite for the complete tree? There are infinitely many finite distances, corresponding to the infinitely many finite naturals, none of which measure the length of a path with no definable length. So WM is asking for a length which he knows cannot exist. > > > > > > > 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? No, but WM seems to claim them for his incomplete trees. > > > > 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. But then asks for the index of the "last" one, which falsely presumes what does not exist.
From: Virgil on 10 Mar 2007 21:08
In article <1173569348.823236.258410(a)s48g2000cws.googlegroups.com>, "Gc" <Gcut667(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. Nor is their "level" ennumerated by a finite number. What is true for every finite (natural) number need not be true for things which are not finite (natural) numbers. Otherwise infinite numbers would have to be finite. |