Cantor Confusion
in [Math]
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From: William Hughes on 13 May 2007 08:18 On May 13, 7:54 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 12 Mai, 22:54, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On May 12, 4:28 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > <snip> > > > > You should know: > > > The union (as I defined it for trees) of all finite trees is an > > > infinite tree with countably many nodes and countably many finite > > > paths. (If you like, you can union the nodes and the paths > > > separately.) > > > > If we switch to the CIBT, then there is not a single node added. Only > > > the number of paths and their lengths changes from countable to > > > uncountable and from finite to infinite, respectively. > > > This is nothing more than the statement. > > > Let R be a countably infinite set. > > > the set of finite subsets of R is countable > > the set of infinite subsets of R is uncountable. > > > Note the number of elements in R does not change. > > The number of elements in any infinite subset is greater than the > number of elements in any finite subset, or not?. > > > > > When you change the size of the sets (the lengths of the > > paths) you change the number of sets (the number of paths) > > But in order to increase the length of the paths you have to add > nodes, haven't you? > If you have finite paths in the union of all infinite trees (a Wolkenmuekenheim special) then you have to have all that all paths do not end at leaf nodes. You can increase the length of any path without adding a node. - William Hughes
From: WM on 13 May 2007 09:14 On 13 Mai, 13:48, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 13, 6:30 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > If a set of infinitely many foreign paths can do it, then one foreign > > path can do it. Otherwise there must be at least two nodes for which > > two paths are required. This is not the case. > > There do not exist two nodes for which two > foreign paths are required. > > There does not exist a finite set of nodes > for which two foreign paths are required. > Look! Over there! A pink elephant! There does exist a set of nodes for which two foreign paths are required. No, the elephant does not help. You state: There is no last node of p, therefore there is no single p** which covers all the nodes of p. But only by this last node p is different from any other path in the tree. A very fine trick! But you forgot to say why an infinite set of paths should be able to cover all nodes of p. There is absolutely no reason. Regards, WM
From: William Hughes on 13 May 2007 09:47 On May 13, 9:14 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > There does exist a set of nodes for which two foreign paths are > required. > Correct. One such set is the set of all nodes of p. > No, the elephant does not help. > > You state: There is no last node of p, therefore there is no single > p** which covers all the nodes of p. But only by this last node p is > different from any other path in the tree. Piffle. There is no last node but despite this p is different from any other path in the tree. The reason why p is different from any other path is the tree is that for every other path in the tree, there is a node where this path and p are different (this node may be different for different paths) > A very fine trick! But you > forgot to say why an infinite set of paths should be able to cover all > nodes of p. There is absolutely no reason. The simplest way to do this is one path per node. For every node kn, of p, there is a path not equal to p which contains kn, call this p(kn). Let R be the set of all paths p(kn). R is an infinite set of paths that covers all the nodes of p. (No, R does not cover the nonexistent last node of p. This is not a problem). - William Hughes
From: WM on 13 May 2007 12:24 On 13 Mai, 15:47, William Hughes <wpihug...(a)hotmail.com> wrote: > Piffle. There is no last node but despite this p > is different from any other path in the tree. > The reason why p is different > from any other path is the tree is that for every other path in the > tree, there is a node where this path and p are different > (this node may be different for different paths) Therefore there is no set of paths p* the nodes of which form a superset of the nodes of p. > > > A very fine trick! But you > > forgot to say why an infinite set of paths should be able to cover all > > nodes of p. There is absolutely no reason. > > The simplest way to do this is one path per node. > > For every node kn, of p, there is a path not equal to p > which contains kn, call this p(kn). > Let R be the set of all paths p(kn). > R is an infinite set of paths that covers all the nodes of p. No. The infinite set R cannot be used because its supremum is p and, therefore, not a path different form p. On the other hand, every path p* of R misses to include at least one node of p. Further, every finite set of paths of R can be substituted by a single path, namely its maximum. And the infinite set R can be substituted by p. If you do it, then the condition of "paths different from p" is not satisfied. If you do it not, then not every node of p is an element of paths of R because that set has no supremum. Regards, WM
From: William Hughes on 13 May 2007 13:53
On May 13, 12:24 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 13 Mai, 15:47, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > A very fine trick! But you > > > forgot to say why an infinite set of paths should be able to cover all > > > nodes of p. There is absolutely no reason. > > > The simplest way to do this is one path per node. > > > For every node kn, of p, there is a path not equal to p > > which contains kn, call this p(kn). > > Let R be the set of all paths p(kn). > > R is an infinite set of paths that covers all the nodes of p. > > No. The infinite set R cannot be used because its supremum is p and, > therefore, not a path different form p. We need a set for which every element is different from p. Every element of R is different from p. The supremum of R is not different from p. No contradiction. R does not contain the supremum of R. - William Hughes |