Cantor Confusion
in [Math]
|
From: Virgil on 13 May 2007 14:18 In article <1179052221.849088.146980(a)l77g2000hsb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 12 Mai, 21:59, Virgil <vir...(a)comcast.net> wrote: > > In article <1178987932.151519.266...(a)q75g2000hsh.googlegroups.com>, > > > > > > > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 12 Mai, 16:45, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > > Because you need a path that is not p, and any path that is not p > > > > > > will > > > > > > eventually exit. > > > > > > > That is contradicted by the fact that there is always another path p* > > > > > with p. Hence not every path can have exited at a node k(p, n) with a > > > > > finite index n. Why should it be impossible to construct a path from > > > > > these infinitely many nodes, i.e. from the pieces of paths p* =/= p? > > > > > This path p** is not p, because every piece belongs to a foreign > > > > > path. > > > > > > This path p** is not p, because every piece belongs to the same > > > > foreign path. > > > > > (1) Every piece of p belongs to some foreign path. > > > (2) The existence of p proves that these pieces can be put together to > > > one foreign path. > > > > The existence of p only proves that they can be put together to form p. > > A set of paths p* =/= p can be used include them all, i.e., all nodes > of p: > > U{k(p, n) | n in N} c U{k(p*, n) | p* =/= p & n in N} . > > This is only possible, if the right hand side union includes all nodes > of p. This is only possible if > it includes the path p. So that now WM claims to have a path p* containing every node of p and some other nodes as well? > > > > In order to establish existence of your alleged foreign path, you would > > have to prove, contrary to fact, that all the pieces of foreign paths > > which yu claim exist belong to the same foreign path, which they do not. > > > > > > > > > You do not oppose to (1). If you oppose (2) and claim that one foreign > > > path does not suffer, then please prove that at least two foreign > > > paths are required. > > > > In fact, infinitely many foreign are required, but no one of them > > contains all of p. > > If a set of infinitely many foreign paths can do it, then one foreign > path can do it. If any one path contains all the nodes of p, it cannot also contain any nodes not in p, so it must be p. > Otherwise there must be at least two nodes for which > two paths are required. This is not the case. In order to have two paths at all, one must have two nodes at the same level, one in each, which cannot happen with WM's illusionary p*. > > > > In a CIBT, the intersection of the node sets of ANY two paths is always > > finite, containing those nodes in the chain of nodes from the root to > > the last common node of the two paths. This is equally true in finite > > trees. > > > > So that the set of paths containing any infinite set of nodes is either > > a singleton set (one path) or the empty set (no paths at all). > > Yes. So why do you insist that an infinite set of foreign paths could > cover all the nodes of p? If the nth of a sequence of "foreign" paths contains the first n nodes of p then the union of nodes sets for all of them will contain all of the nodes of p. Anyone with any competence should have seen this without help.
From: Virgil on 13 May 2007 14:21 In article <1179057277.295725.155820(a)h2g2000hsg.googlegroups.com>, WM <mueckenh(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?. The number of elements in a set is not greater than the number of elements in that same set, dingbat! > > > > 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? In a sensible system, yes! In WM's system, only WM knows. > > Regards, WM
From: Virgil on 13 May 2007 14:26 In article <1179057660.264712.68500(a)q75g2000hsh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 13 Mai, 00:32, Virgil <vir...(a)comcast.net> wrote: > > In article <1179002100.262022.320...(a)e65g2000hsc.googlegroups.com>, > > > > Mathematician's minds, however, are less limited, in that they can and > > do imagine things that WM says he cannot. > > I have seen and heard about examples of that. Infinite sets and > infinite definitions for instance. > > > > > But why don't > > > you answer my question which is independent of the limits of your mind > > > - or does my question already fall outside of your point of view? > > > > There are things that can 'go on' in mathematicians minds that do not go > > on in WMs more limited one. > > But is that to be considered an advantage of those minds? Very much so, > Wat about > the problems if these things have to come out of those minds and to > meet the hard reality? Mathematics is to "hard reality" a bit like poetry is to prose. There are profound truths in poetry that cannot be expressed in prose.
From: Virgil on 13 May 2007 14:43 In article <1179062054.712048.154380(a)u30g2000hsc.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 13 Mai, 13:48, William Hughes <wpihug...(a)hotmail.com> wrote: > 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. WM keeps claiming the existence of the non-existent as justification for his delusions. Last nodes (leaf nodes) only occur in trees in which some paths are finite and ends in a leaf nodes. In CIBTs no path has a last node. That is what makes them infinite. > 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. There absolutely is a reason! But it is so trivial that we did not think it necessary to state it. If WM's powers are too weak to figure it out for himself, then: Given any path p in a CIBT. For each n in N, Let p_n be a path that has in common with p the first n nodes of p, but no others. There is always such a path, any of the ones branching the other way at level n. Since every node of p is at some finite level n in N, the infinite set {p_n: n in N} of paths "covers" all nodes of p. Is that simple enough for you to grasp, WM?
From: Virgil on 13 May 2007 14:50
In article <1179073448.864312.305650(a)h2g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. Actually every set of paths which has p as a member does that. But there are even sets of paths which do not contain p which do it. For example. for each n in N there is a path which coincides with p up through level n but at level n branches oppositely. Call it p_n. Clearly for each n in n, p_n =/= p. Also clearly, every node of p is in some p_n. So {p_n: n in N} exists as exactly the sort of superset that WM claims does not exist. > > > > > 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. {p_n: n in N}, as defined above, is sufficient reason. |