Cantor Confusion
in [Math]
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From: William Hughes on 19 Apr 2007 08:15 On Apr 19, 7:11 am, mueck...(a)rz.fh-augsburg.de wrote: > On 18 Apr., 13:32, William Hughes <wpihug...(a)hotmail.com> wrote: > > [See if you can reply to this > > > without mentioning "separation point" or changing > > the subject] there are an uncountable number > > of subsets of a countable set. So a countable set of nodes can > > separate > > an uncountable number of paths. > And you attempt a change of subject Your claim A countable number of nodes cannot separate an uncountable number of paths because you do not have enough nodes. My reply 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. Rather than adressing this, you ignore it and bring up another argument. Try again - William Hughes
From: Virgil on 19 Apr 2007 14:41 In article <1176981117.885305.5450(a)y80g2000hsf.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 18 Apr., 13:32, William Hughes <wpihug...(a)hotmail.com> wrote: > > [See if you can reply to this > > without mentioning "separation point" or changing > > the subject] there are an uncountable number > > of subsets of a countable set. So a countable set of nodes can > > separate > > an uncountable number of paths. > > This is an element of the tree: > | > N > / \ > > The number of paths entering an element (see above) is 1. The number > of paths leaving an element is 2. The number of nodes of an element > is > 1. The number of paths entering an 'element' had better be equal to the number of paths leaving it if one is to have a tree at all (otherwise one has paths starting at non-root nodes). And unless each edge leaving an 'element' ends n a leaf node node, there are more than such two paths. > One path is the continuation of the incoming path having been > mapped to a node above the element. The other path is mapped to the > node of the element. Every path starts at the root node and ends, if at all, in a leaf node. and through every 'element' in between, as many paths enter it as leave it. And for infinite trees, uncountably many paths enter every 'element'. So that WM's new attempt to prove his old falsehood fails as completely as all his previous ones.
From: Virgil on 19 Apr 2007 14:45 In article <1176981294.085164.20320(a)n76g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 18 Apr., 16:56, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1176724396.229576.101...(a)p77g2000hsh.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 16 Apr., 02:43, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1176300393.408149.165...(a)w1g2000hsg.googlegroups.com> > > > > mueck...(a)rz.fh-augsburg.de writes: > > > > > > > > If all paths exist simultaneously, then there must exist > > > > > uncountably > > > > > many in the infinite tree. > > > > > > > > But there are. > > > > > > 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. But the maximal 'bunches' of paths being separated from one another by a single node are always uncountable, as has several times been proved and which WM has never successfully refuted, despite his many futile attempts to do so.
From: Virgil on 19 Apr 2007 14:56 In article <1176981838.819590.80870(a)l77g2000hsb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 18 Apr., 16:51, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1176723918.670619.273...(a)q75g2000hsh.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 16 Apr., 02:19, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1176298903.609533.227...(a)d57g2000hsg.googlegroups.com> > > > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 11 Apr., 03:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > Do you claim that all paths exist as separated entities? If > > > > > > > so: > > > > > > > how or where do they exist? > > > > > > > > > > > > Philosophy? I would expect mathematics in this newsgroup. > > > > > > > > > > I think that numbers belong to mathematics. > > > > > > > > But the question "where do they exist" is not a mathematical question. > > > > > > If you consider the tree, it is as mathematical as if you consider a > > > "list". > > > > 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. No single node separates any path from all others, it can at most partition the uncountable set of all paths into two disjoint uncountable subsets, those branching left from that node from those branching right. > > > 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. No single node separates any path from all others, so one node cannot be a "point of separation' of anything except one uncountable set of paths from another. > > > But indeed, in every finite tree with n levels there are 2^n+1 finite > > paths. > > No, there are 2^n finite paths. If one counts the root node as level 1, then there are 2^(n-1) paths for n levels. > > > 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. Cantor's 'diagonal' is actually a rule for creating a 'diagonal', and that rule applies equally well to any of infinitely many digits. To prove that an element is not in a set, even an infinite set, it is enough to show that the element cannot be equal to any member of that set. That is what Cantor's proof does. And that is what WM cannot comprehend.
From: Dik T. Winter on 20 Apr 2007 10:55
In article <1176981294.085164.20320(a)n76g2000hsh.googlegroups.com> mueckenh(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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |