Cantor Confusion
in [Math]
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From: mueckenh on 11 Feb 2007 11:29 On 10 Feb., 19:39, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > mueck...(a)rz.fh-augsburg.de wrote: > > On 10 Feb., 12:13, Franziska Neugebauer <Franziska- > > Neugeba...(a)neugeb.dnsalias.net> wrote: > >> mueck...(a)rz.fh-augsburg.de wrote: > >> > On 9 Feb., 09:05, Virgil <vir...(a)comcast.net> wrote: > > >> >> > Instead of a path P consider the set S of nodes K which belong > >> >> > to a path P. Do your calculation and arguing. Then substitute P > >> >> > for S to have a brief notation. > > >> >> One cannot determine merely from a set of nodes for a given tree > >> >> whether that set of nodes is or is not a path. > > > >> > One can. > > >> Then please do so: > > >> given tree T := { a, b, c, d, e, f, g } > >> given set of nodes S := { a, b, c } > > >> Tell us whether S is a path in T. And please explain that. > > > Pardon, I overlooked your first question. > > > The tree T(2) in my notation can be given as a chain by > > > a > > bc > > gfed > > > abc is the subtree T(1) with only (the root node and) one level. It is > > not a path. > > Strange. My version of the tree T > > a > / \ > b d > / \ / \ > c e f g > > obviously contains a path having (the node set) S = { a, b, c }. Hence > the attempt to *uniquely* represent trees and/or paths by plain sets of > nodes fails. Your version is incapable of enumerating all the nodes of the infinite tree as one single chain. (As it is MY tree which appears in MY proof, please adhere to MY scheme, when discussion this proof.) Regards, WM
From: Mike Kelly on 11 Feb 2007 12:32 On 11 Feb, 14:58, mueck...(a)rz.fh-augsburg.de wrote: > On 11 Feb., 03:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1171114154.842945.262...(a)s48g2000cws.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > On 9 Feb., 14:53, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1171011850.731985.236...(a)a75g2000cwd.googlegroups.com> mueck.= > ....@rz.fh-augsburg.de writes: > > > ... > > > > > That is a matter of convention. Therefore the unary represantation, > > > > > which is not a matter of convention, is preferable. > > > > > Also unary representation is a matter of convention in my opinion. > > > > But it is requiring the least possible convention. > > > Eh? You first state that it is not a matter of convention and now state > > that it *is* a matter of convention? Are you not contradicting yourself? > > It depends on how you define "matter of convention". It depends on how > well you distinguish between the *numbers 3* given below and the first > word "all" of the verbal explanation. > > > > > > III > > > ooo > > > abc > > > all that is understandable as 3 without other convention than that > > > "all" is the start of this sentence and not another representation of 3. > > > Without convention the first looks like a number of bars, the second like > > a numbers of o's and the third like a sequence of letters from the Latin > > script. The first might suggest the number three, but it is not the > > number three. The second might suggest the number two, but it is not > > the number two. > > I missed one o. I corrected it above. Now it expresses all properties > that the number three can express. Therefore it is the number three > (and in addition it has the special form of o's). > > > The third contains three letters from the Latin script, > > but I do not see a suggestion of the number three at all. > > Each of them expresses everything that the number 3 can express. Each > of them is number 3. I say 111 is number 7. Now what? -- mike.
From: Franziska Neugebauer on 11 Feb 2007 13:24 mueckenh(a)rz.fh-augsburg.de wrote: > On 10 Feb., 19:39, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: >> mueck...(a)rz.fh-augsburg.de wrote: >> > On 10 Feb., 12:13, Franziska Neugebauer <Franziska- >> > Neugeba...(a)neugeb.dnsalias.net> wrote: >> >> mueck...(a)rz.fh-augsburg.de wrote: >> >> > On 9 Feb., 09:05, Virgil <vir...(a)comcast.net> wrote: >> >> >> >> > Instead of a path P consider the set S of nodes K which >> >> >> > belong to a path P. Do your calculation and arguing. Then >> >> >> > substitute P for S to have a brief notation. >> >> >> >> One cannot determine merely from a set of nodes for a given >> >> >> tree whether that set of nodes is or is not a path. > >> >> >> > One can. >> >> >> Then please do so: >> >> >> given tree T := { a, b, c, d, e, f, g } >> >> given set of nodes S := { a, b, c } >> >> >> Tell us whether S is a path in T. And please explain that. >> >> > Pardon, I overlooked your first question. >> >> > The tree T(2) in my notation can be given as a chain by >> >> > a >> > bc >> > gfed >> >> > abc is the subtree T(1) with only (the root node and) one level. It >> > is not a path. >> >> Strange. My version of the tree T >> >> a >> / \ >> b d >> / \ / \ >> c e f g >> >> obviously contains a path having (the node set) S = { a, b, c }. >> Hence the attempt to *uniquely* represent trees and/or paths by plain >> sets of nodes fails. > > Your version is incapable of enumerating all the nodes of the infinite > tree as one single chain. Neither your definition of tree, path or the wording of your claim does contain this constraint. > (As it is MY tree which appears in MY proof, please adhere to MY > scheme, when discussion this proof.) Again: I presented I perfectly acceptabLe tree for which your claim: "One can determine from a set of nodes for a given tree whether that set of nodes is or is not a path." does not hold. So your claim is not universially valid. F. N. -- xyz
From: Virgil on 11 Feb 2007 14:15 In article <1171205918.124082.214920(a)a75g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 11 Feb., 03:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1171114154.842945.262...(a)s48g2000cws.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 9 Feb., 14:53, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1171011850.731985.236...(a)a75g2000cwd.googlegroups.com> > > > > mueck.= > ....@rz.fh-augsburg.de writes: > > > > ... > > > > > That is a matter of convention. Therefore the unary represantation, > > > > > which is not a matter of convention, is preferable. > > > > > > > > Also unary representation is a matter of convention in my opinion. > > > > > > But it is requiring the least possible convention. > > > > Eh? You first state that it is not a matter of convention and now state > > that it *is* a matter of convention? Are you not contradicting yourself? > > It depends on how you define "matter of convention". But since WM is so adverse to having anything defined clearly enough to make things unambiguous, within WMology, definitions are irrelevant. It depends on how > well you distinguish between the *numbers 3* given below and the first > word "all" of the verbal explanation. > > > > > III > > > ooo > > > abc > > > all that is understandable as 3 without other convention than that > > > "all" is the start of this sentence and not another representation of 3. > > > > Without convention the first looks like a number of bars, the second like > > a numbers of o's and the third like a sequence of letters from the Latin > > script. The first might suggest the number three, but it is not the > > number three. The second might suggest the number two, but it is not > > the number two. > > I missed one o. I corrected it above. Now it expresses all properties > that the number three can express. Therefore it is the number three > (and in addition it has the special form of o's). > > > The third contains three letters from the Latin script, > > but I do not see a suggestion of the number three at all. > > Each of them expresses everything that the number 3 can express. Each > of them is number 3. > Then, according to WM, three I's and three o's and three lower case roman letters are all exactly the same thing. Equating three elephants with three nits, only makes WM a nitwit.
From: Virgil on 11 Feb 2007 14:43
In article <1171211394.591754.201520(a)j27g2000cwj.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 10 Feb., 19:39, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 10 Feb., 12:13, Franziska Neugebauer <Franziska- > > > Neugeba...(a)neugeb.dnsalias.net> wrote: > > >> mueck...(a)rz.fh-augsburg.de wrote: > > >> > On 9 Feb., 09:05, Virgil <vir...(a)comcast.net> wrote: > > > > >> >> > Instead of a path P consider the set S of nodes K which belong > > >> >> > to a path P. Do your calculation and arguing. Then substitute P > > >> >> > for S to have a brief notation. > > > > >> >> One cannot determine merely from a set of nodes for a given tree > > >> >> whether that set of nodes is or is not a path. > > > > > >> > One can. > > > > >> Then please do so: > > > > >> given tree T := { a, b, c, d, e, f, g } > > >> given set of nodes S := { a, b, c } > > > > >> Tell us whether S is a path in T. And please explain that. > > > > > Pardon, I overlooked your first question. > > > > > The tree T(2) in my notation can be given as a chain by > > > > > a > > > bc > > > gfed > > > > > abc is the subtree T(1) with only (the root node and) one level. It is > > > not a path. > > > > Strange. My version of the tree T > > > > a > > / \ > > b d > > / \ / \ > > c e f g > > > > obviously contains a path having (the node set) S = { a, b, c }. Hence > > the attempt to *uniquely* represent trees and/or paths by plain sets of > > nodes fails. > > Your version is incapable of enumerating all the nodes of the infinite > tree as one single chain. Until WM invents an infinite alphabet, his is equally incapable. > (As it is MY tree which appears in MY proof, please adhere to MY > scheme, when discussion this proof.) Since other "schemes" can disprove WM's claims, no wonder WM is leery if them. The only "scheme" that is unambiguous is one in which each node explicitly displays all the left or right branchings from the root node to itself. Thus WM's tree, with 0 for left branch and 1 for right branch: a bc gfed with 0 for left branch and 1 for right branch: has a = 0. b = 0.0, c = 0.1 g = 0.00, f = 0.01 e = 0.10 d = 0.11 This also has the advantage of showing that the pseudo-union of all finite trees has uncountably many paths corresponding to the uncountably many endless strings of binary bits. |