Cantor Confusion
in [Math]
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From: Dik T. Winter on 18 Apr 2007 11:04 In article <1176726344.265077.211810(a)n59g2000hsh.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 16 Apr., 02:52, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > Of course. Definitions are not wrong, in principle. But by the common > > definitions natural numbers are limited to be finite. Just a case of > > defining things. There is nothing in nature, or reality, that (at least > > for mathematics) *mandates* that they should be finite. It is their > > mathematical definition that makes them finite. > > No. It is that they are obtained from natural sets of distinguishable > elements. But there are no infinite sets in nature or reality. That has *no* mathematical meaning. The mathematical definition is that N is the smallest inductive set. From this it is easily proven that all elements of N are finite. Nature or reality does not come in at all. > > > Physicists will never look for a red quark because of wavelength > > > problems. > > > > So what is the distinction between a red quark and a blue or green one? > > Moreover, quarks change colour all the time. Are the terms 'red', 'blue' > > and 'green' in this case not so much chosen because of intrinsic properties > > but more based on similarities? But you critique mathematics for this. > > Quarks are far too small to emit or absorb light of 700 nm wavelength. > So there is no red quark. The colour labels of QCD are only attached > because the system of three propertiesf its the observations very > well. Hrm, not really. The observations tell us only that there are three distinctive properties, that may change by "emission" or "absorption" of gluons. And I wonder what the reality is of, say, anti-red. What is a "green - anti-red" gluon? The labels are just chosen because someone wanted three catchy names, there is no relation to colour at all. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 18 Apr 2007 15:16 In article <1176892818.034849.119620(a)d57g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 17 Apr., 23:40, William Hughes <wpihug...(a)hotmail.com> wrote: > > On Apr 17, 4:20 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > > > On 17 Apr., 18:11, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > On Apr 17, 11:11 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > > there are only countably many separations and > > > > > countably many points of separation. > > > > > > No, each "point of separation" is a set of nodes. There are > > > > uncountable many sets of nodes. > > > > > Each "point of separation" is a single node. > > > > Piffle. Even in Wolkenmuekenheim a "point of separation" must be > > able to separate a path from all other paths. > > > Piffle. A point of separation increases the number of paths which can > be distinguished. By construction of the tree X nodes can create X+ 1 > separated paths. A set of X nodes can create, at best, X+1 uncountable sets of separated paths, but cannot separate any path from all others unless X is not finite. While any two paths can be separated by a single node, to separate any node from all others requires infinitely many nodes. So each "separated path" requires a different infinite set of nodes. > > > Since a single node > > cannot separate > > a path from all other paths > > a single node is not a point of separation. > > > > A set of nodes can separate a path. Therefore a set of nodes > > can be a "point of separation". > > Yes, a set of separation points can separate a path from all other > paths. Not all nodes of the tree are required for this purpose. > Anyhow, only countably many nodes are avaivable. Only countably many, any infinite subset of the nodes of a given path are required to separate that path from all others. But the number of such infinite sets of nodes is uncountable. So if one such set is one "point of separation" there are uncountably many points of separation as subsets of a countable set of nodes.
From: mueckenh on 19 Apr 2007 07:11 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. 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. The number of paths leaving an element without having being bijected with a node is 2-1-1=0. The number of elements is countable. Therefore, we can set up the series giving the paths without nodes: 2-1-1+2-1-1+-... This series is divergent. But we know that no partial sum is greater than 2 (or less than 0). Therefore we have sum of paths being not mapped to nodes is 2-1-1+2-1-1+-... < 3 and therefore sum of paths being not mapped to nodes is 2-1-1+2-1-1+-... < 2^aleph0. Regards, WM
From: mueckenh on 19 Apr 2007 07:14 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. Regards, WM
From: mueckenh on 19 Apr 2007 07:23
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. > > > > > > This sentence is quite difficult to pars. Yes, each real number is > > > > > represented by a path which is completely within the infinitely many > > > > > levels of the tree. > > > > > > > > But not all real numbers coexist in the tree? Only a countable number > > > > of them is admitted simultaneously? > > > > > > Why? > > > > 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. > But indeed, in every finite tree with n levels there are 2^n+1 finite paths. No, there are 2^n finite 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. Regards, WM |