Cantor Confusion
in [Math]
|
From: Virgil on 12 Mar 2007 17:12 In article <1173715726.469898.72890(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 11 Mrz., 17:21, "Gc" <Gcut...(a)hotmail.com> wrote: > > On 11 maalis, 10:33, mueck...(a)rz.fh-augsburg.de wrote: > > > > > On 11 Mrz., 00:29, "Gc" <Gcut...(a)hotmail.com> wrote: > > > > > Please stick to one pseudonym. Otherwise it is tedious to killfile > > > you. I don't want to read you. > > > > > Regards, WM > > > > I don`t have other pseudonyms, but fine: I will leave you alone with > > your hallucinations. > > Sorry if I have I have insulted you, but I have mixed you up with > someone else. My error. To answer your question: > > If there are single paths in the tree, then their set is subject to > the restriction imposed by the cross sections (= numbers of nodes) of > the levels. These single paths cannot exist outside of any level. > Every level has a countable number of nodes. Every level has only finite paths, so that the number of nodes and number of paths are both finite in such finite paths. > > If there are no single paths in the tree, ... then: where are they? Married, and raising little paths? > > Regards, WM
From: Virgil on 12 Mar 2007 17:28 In article <1173724460.248046.46960(a)s48g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 12 Mrz., 16:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1173464401.633116.124...(a)q40g2000cwq.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 9 Mrz., 14:05, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1173430532.626624.124...(a)c51g2000cwc.googlegroups.com> > > > > mueck...(a)rz.fh-augsburg.de writes: > > > > > > > > > > So, according to *your* definitions it is >. Not according to > > > > > > common > > > > > > definitions. Would it not be possible that your definitions are > > > > > > not > > > > > > consistent? It is your last paragraph which is inconsistent with > > > > > > the > > > > > > other definitions. See how you *did* define: > > > > > > > lim {n-->oo} |{1,2,3,...,n}| = aleph_0 > > > > > > which means (as |{1,2,3,...,n}| = n: > > > > > > lim {n-->oo} n = aleph_0? > > > > > > > > Do you not have a comment on this? > > > > > > It is wrong. A limit is either approached to any positive eps or it > > > isn't a limit. > > > > We are not doing analysis here. But you now state that your definition: > > lim {n->oo} |{1,2,3,...,n}| = aleph_0 > > is wrong? Strange as you use it. > > It is not my definitin, but it is the definition of set theory. False claim. If it were a set theoretical definition, surely, WM, with all his resources and other citations wold be able to provide at least one citation from a set theory source to support his claim. That he has never done so is some evidence that there are no such definitions to be found. > > The actually infinite set of finite numbers (without their supremum) > is by set theory defined as: > > lim {n-->oo} |{1,2,3,...,n}| = aleph_0 (that expresses the actually > infinite set) One can make a case for the union of all sets of form {1,2,3,...,n} being aleph_0, but that is quite different from any limit being anything. > lim {n-->oo} n < aleph_0 (that is because every number is finite). Which only shows that neither "lim {n-->oo} |{1,2,3,...,n}|" nor "lim {n-->oo} n" exist. > aleph_0 is an upper bound for any set of separated paths at a finite > level. It is equally an upper bound on the set of terminating proper decimal fractions, but the set of non-terminating decimals is uncountable. > No. The infinite paths do not terminate at any level. They cross the > level L(n) and then there are 2^n paths (-bundles). In any finite evel > there are countably many paths (-bundles). At any finite level, the set of proper decimals cut off at that point is finite. > > No. Up to the level L(1) there are only two separated path (-bundles). But if those path-bundles include all of the infinite-length paths through the appropriate nodes, then each bundle is equinumerous with their union, and, in fact, all bundles of infinite paths are equinumerous, so looking at bundles doesn't help. The number of bundles does not affect the number of paths per bundle, which, for any complete infinite binary tree, is uncountably many paths per bundle.
From: Dik T. Winter on 13 Mar 2007 08:39 In article <1173715319.243468.87220(a)c51g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 12 Mrz., 16:41, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > And this tells us precisely nothing about C(oo), which you *do* use. > > > > > > Forget it. Use only level L(n) and the countable number of nodes C(n) > > > for every n in N to determine the number of paths crossin this level. > > > If you find this insufficient, then tell me what after every n may be > > > imagined. > > > > The paths in finite trees terminate at soe level n. The paths in the > > infinite tree do not terminate. Do you not see the difference? > > The levels which have a countable number of nodes do not terminate > either. Therefore there is no difference: As long as paths can exist > the cardinality of them is restricted to a countable number. Sorry you are confusing two things. Lengths of paths and number of paths. The lengths do not terminate and are countable. The number is *not* countable. Show a *proof* for once, that it *is* countable. > An uncountable set of paths cannot exist other than outside of the > tree, i.e., outside of mathematics. Why? A proof, please. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 13 Mar 2007 09:15 In article <1173724460.248046.46960(a)s48g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 12 Mrz., 16:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1173464401.633116.124...(a)q40g2000cwq.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: .... > > > > > > It is your last paragraph which is inconsistent > > > > > > with the > other definitions. See how you *did* define: > > > > > > > lim {n-->oo} |{1,2,3,...,n}| = aleph_0 > > > > > > which means (as |{1,2,3,...,n}| = n: > > > > > > lim {n-->oo} n = aleph_0? > > > > > > > > Do you not have a comment on this? > > > > > > It is wrong. A limit is either approached to any positive eps or it > > > isn't a limit. > > > > We are not doing analysis here. But you now state that your definition: > > lim {n->oo} |{1,2,3,...,n}| = aleph_0 > > is wrong? Strange as you use it. > > It is not my definitin, but it is the definition of set theory. Of > course it is wrong. Pray provide a reference from set theory where that definition of limit is defined and used. If you can not provide such you are simply telling an untruth here. The closest I can come up with is Hrbacek and Jech page 193 from which you can find: lim{n->oo} |{1,2,3,...,n}| = lim{n->oo} n = aleph_0. So either show a source where lim{n->oo} |{1,2,3,...,n}| is defined and where lim{n->oo} n is not defined or admit defeat. > > > lim {n-->oo} n = aleph_0 is neither stated by set theory -nor by > > > anyone else. > > > > See page 193 of Hrbacek and Jech where you will find a definition of > > limit from which you can derive precisely that. > > I read that book. Therefore I know their definitions and I knew > already that set theory works with limits when you disputed that. I really have my doubts. Your first reference to limits in that book were from a section where they explicitly had stated that the theorems and definitions from that section were from topology. > Nevertheless their definition does not apply to the infinite set of > finite numbers. Why does it not apply to the infinite set of finite numbers? Did you really read the book? First a definition from page 147: Functions whose domain is an ordinal alpha are called transfinite sequences of length alpha. Note that the function whose domain is the finite integers by this definition is a transfinite sequence of length omega. Further, from page 193: Let <alpha_nu | nu < theta> be a transfinite sequence of ordinal numbers of length theta. ( this applies to <n | n < omega> ) We say that the sequence is increasing if alpha_nu < alpha_mu whenever nu < mu < theta. ( this also is applicable here ) If theta is a limit ordinal number and if <alpha_nu | nu < theta> is an increasing sequence of ordinals, ( both applicable ) we define alpha = lim{nu -> theta} alpha_nu = sup{alpha_nu | nu < theta} and call alpha the limit of the increasing sequence. So that definition is applicable in this case, and by this definition: lim{n -> omega} n = omega or (with slight rewording): lim{n -> oo} n = aleph_0. > (Further it is wrong because what they "call the > limit" is not a limit. But this is irrelevant here.) Irrelevant, but interesting. Why is it not a limit? I think you focus to much on the definition of limit from analysis, disallowing the definition of limits from other branches. > The actually infinite set of finite numbers (without their supremum) > is by set theory defined as: > > lim {n-->oo} |{1,2,3,...,n}| = aleph_0 (that expresses the actually > infinite set) Pray show a source giving that definition. > lim {n-->oo} n < aleph_0 (that is because every number is finite). See above how the definition on page 193 of the book by Hrbacek and Jech does apply. And as that is the only place in set theory texts where I actually *have* found a set theoretic definition of limit, I wonder where you did find *your* versions of limit. > > > > No, they are *not*. The cross sections are sets of nodes (by your own > > > > definition), except for C(oo), which is a cardinal number (again by > > > > your own definition). And I see *no* relation between aleph-0 and > > > > the paths. > > > > > > The cross section C(n) = |L(n)| is the number of nodes of the level > > > L(n). > > > > I still see no relation between aleph-0 and the paths. > > aleph_0 is an upper bound for any set of separated paths at a finite > level. *finite paths*. > Shopuld there be more paths, then they had to cross at least > one level L(alpha) with an infinite number alpha. Why? > > But you have not proven that. You have only proven that there are as many > > paths that *terminate* at nodes at that level as there are nodes at that > > level. > > No. The infinite paths do not terminate at any level. They cross the > level L(n) and then there are 2^n paths (-bundles). Those are *terminating* paths. (A path-bundle can be seen as a terminating path: it is a set of nodes containing a finite number of nodes.) > In any finite evel > there are countably many paths (-bundles). Terminating path. > > > If you disagree: What part of a path should not be covered by a node > > > of a level L(n) with a finite n? > > > > If you disagree, are there only two paths in the tree because the > > cross-section C(1) contains only two nodes? > > No. Up to the level L(1) there are only two separated path (-bundles). Terminating paths. > Up to level L(n) ther are 2^n. Terminating paths. > And only if there are levels with non- > natural number there can be more than countably many paths (-bundles). Terminating paths. This says nothing about non-terminating paths. -- 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 15 Mar 2007 05:51
On 12 Mrz., 22:10, Virgil <vir...(a)comcast.net> wrote: > In article <1173715319.243468.87...(a)c51g2000cwc.googlegroups.com>, > > > > The paths in finite trees terminate at some level n. The paths in the > > > infinite tree do not terminate. Do you not see the difference? > > > -- > > > The levels which have a countable number of nodes do not terminate > > either. Therefore there is no difference: As long as paths can exist > > the cardinality of them is restricted to a countable number. > > Then the alleged "union" of those infinitely many finite trees does not > produce the complete infinite binary tree, This is only possible if the alleged "union" of all lines of a Cantor- list does not produce the complete list. Regards, WM |