Cantor Confusion
in [Math]
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From: Dik T. Winter on 20 Apr 2007 11:00 In article <1176981838.819590.80870(a)l77g2000hsb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 18 Apr., 16:51, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > The mathematical question is: "are all paths available in the infinite > > tree", and the answer is: "yes". A follow-up question can be "is each > > path separated from all other paths", and the answer is, again, "yes". > > And a further question can be, "is there a level such that a particular > > path is separated from all other paths", and the answer to this is "no". > > "exist" is barely a mathematical term. > > Existence of separated paths is not possible without being separated. > But the set of occasions for separations (= nodes) is countable. In what way is that a contradiction? As each path separates at each of its elements (nodes) from uncountable many other paths, I see no contradiction. > > > There cannot be more separated paths in the whole tree than are points > > > of separation in the whole tree, (unless there is more than one > > > separation per point of separation which, however, can be excluded by > > > the construction of the tree). > > > > What is a "point of separation"? > > A node, of course. So you are stating that there are no more paths in the tree than there are nodes? But *that* is obviously false, as any finite tree shows. > > But indeed, in every finite tree with n levels there are 2^n+1 finite > > paths. > > No, there are 2^n finite paths. Now again, you are limiting paths to those that go on to the end. When you do *that*, you get problems when you consider sets of paths. > > And in the infinite tree there are countably many finite paths. But this > > says *nothing* about infinite paths. > > Cantor's diagonal can be defined up to any line n of the list. But > this says *nothing* about an infinite sequence of digits. It is defined for *any* line of the list. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 21 Apr 2007 08:30 On 19 Apr., 14:15, William Hughes <wpihug...(a)hotmail.com> wrote: > It is a set of nodes that separates a path, not a single > node. There are an uncountable number of sets > of nodes. So you do have enough nodes. First we should clear the notation: There are bunches of paths like 0,1 which contains all paths from 0,1000... to 0.111... A searated path is a bunch of paths with only one element like 0,111... In order to get two separateed bunches of paths, there is one bunch required which contains all these paths and one node to split the incomin bunch into wo bunches. The infinite tree contains all separated paths. Therefore there are as many nodes (minus one) required in the tree. Regards, WM
From: mueckenh on 21 Apr 2007 08:34 On 20 Apr., 16:55, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1176981294.085164.20...(a)n76g2000hsh.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 18 Apr., 16:56, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > Without the chance that uncountably many are separated in the whole > > > > tree? > > > > > > They are all separated from each other. I do not know why you need > > > uncountably many separation points for that. > > > > Because one node separates only one (bunch of) path(s) from another > > one. > > It makes sense only if you remove the parenthesis. When you remove the > stuff between parenthesis it makes no sens. That depends on the notation. The "bunch of" makesw it somewhat clumsy. There are bunches of paths like the bunch 0,1 which contains all paths from 0,1000... to 0.111... A searated path is a bunch of paths with only one element like 0,111... In order to get two separateed bunches of paths, there is one bunch required which contains all these paths and one node to split the incomin bunch into wo bunches. The infinite tree contains all separated paths. Therefore there are as many nodes (minus one) required in the tree. Should no separated paths exist, then they would also not exist in Cantor's list as entries or as diagonal number. Regards, WM
From: mueckenh on 21 Apr 2007 08:44 On 20 Apr., 17:00, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1176981838.819590.80...(a)l77g2000hsb.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 18 Apr., 16:51, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > The mathematical question is: "are all paths available in the infinite > > > tree", and the answer is: "yes". A follow-up question can be "is each > > > path separated from all other paths", and the answer is, again, "yes". > > > And a further question can be, "is there a level such that a particular > > > path is separated from all other paths", and the answer to this is "no". > > > "exist" is barely a mathematical term. > > > > Existence of separated paths is not possible without being separated. > > But the set of occasions for separations (= nodes) is countable. > > In what way is that a contradiction? As each path separates at each of > its elements (nodes) from uncountable many other paths, I see no > contradiction. The tree must contain all separated path due to real numbers between 0 and 1. For this sake we need as many nodes as separated paths. > > > > > There cannot be more separated paths in the whole tree than are points > > > > of separation in the whole tree, (unless there is more than one > > > > separation per point of separation which, however, can be excluded by > > > > the construction of the tree). > > > > > > What is a "point of separation"? > > > > A node, of course. > > So you are stating that there are no more paths in the tree than there are > nodes? But *that* is obviously false, as any finite tree shows. Every finite tree contains less separated bunches of path than nodes. > > > > But indeed, in every finite tree with n levels there are 2^n+1 finite > > > paths. > > > > No, there are 2^n finite paths. > > Now again, you are limiting paths to those that go on to the end. When > you do *that*, you get problems when you consider sets of paths. Instead of "path" which seems to be misleading use "bunch of paths". > > > > And in the infinite tree there are countably many finite paths. But this > > > says *nothing* about infinite paths. > > > > Cantor's diagonal can be defined up to any line n of the list. But > > this says *nothing* about an infinite sequence of digits. > > It is defined for *any* line of the list. That does not define it for *all* lines. Here is an example. (You know that all rational numbers obey the laws of trichotomy.) Cantor's (first diagonalization) method allows to enumerate every rational number. So each rational has its place in the sequence. But this sequence does not enumerate *all rationals". Otherwise there must be the largest rational less than 1. The absence of this rational is a natural as he absence of the set {N,k,f} in Hessenberg's proof. Regards, WM
From: William Hughes on 21 Apr 2007 09:00
On Apr 21, 8:30 am, mueck...(a)rz.fh-augsburg.de wrote: > On 19 Apr., 14:15, William Hughes <wpihug...(a)hotmail.com> wrote: > > > It is a set of nodes that separates a path, not a single > > node. There are an uncountable number of sets > > of nodes. So you do have enough nodes. > > First we should clear the notation: > There are bunches of paths like 0,1 which contains all paths from > 0,1000... to 0.111... > A searated path is a bunch of paths with only one element like > 0,111... > > In order to get two separateed bunches of paths, there is one bunch > required which contains all these paths and one node to split the > incomin bunch into wo bunches. So a single node does not separate a single path. However, a set of nodes can. There are uncountable many sets of nodes. Therefore: There are enough nodes to separate uncountable many paths. We have to settle the question of whether there are enough nodes to separate uncountable many paths, before we go on to discuss the question of whether they in fact do separate uncountable many paths. - William Hughes |