Cantor Confusion
in [Math]
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From: WM on 13 May 2007 06:30 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. > > 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. Otherwise there must be at least two nodes for which two paths are required. This is not the case. > > 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? Regards, WM
From: William Hughes on 13 May 2007 07:48 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 not exist a set of nodes for which two foreign paths are required. - William Hughes
From: WM on 13 May 2007 07:54 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? Regards, WM
From: WM on 13 May 2007 07:57 On 13 Mai, 00:29, Virgil <vir...(a)comcast.net> wrote: > > 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.) > > Then it is incomplete, as in a complete infinite binary tree there must > be a separate path for every separate subset of N. The set of nodes is complete. > > > > > 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. > > So that WM's tree must have omitted uncountably many paths. The union has countably many nodes and paths. Nevertheless, there is no node of the CIBT missing. Regards, WM
From: WM on 13 May 2007 08:01
On 13 Mai, 00:32, Virgil <vir...(a)comcast.net> wrote: > In article <1179002100.262022.320...(a)e65g2000hsc.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 12 Mai, 21:11, Virgil <vir...(a)comcast.net> wrote: > > > In article <1178979741.537850.290...(a)w5g2000hsg.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 11 Mai, 21:45, Virgil <vir...(a)comcast.net> wrote: > > > > > In article <1178866486.658963.253...(a)y5g2000hsa.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > My personal opinion is not the starting point here. Since mathematical > > > > > > entities exist only in the mind, such entities, which cannot exist in > > > > > > the mind, cannot exist at all. > > > > > > Since your own mind is the only one for which you can speak, what goes > > > > > on it other minds is beyond your competence to decide. > > > > > Do you think mathematical entities exist other than in minds? > > > > Do you thing that everything that can go on in any mind is limited to > > > what goes on in your mind? > > > No, most minds are trapped in by far narrower limits. > > 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 But is that to be considered an advantage of those minds? Wat about the problems if these things have to come out of those minds and to meet the hard reality? Regards, WM |