Cantor Confusion
in [Math]
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From: mueckenh on 15 Mar 2007 05:58 On 12 Mrz., 22:12, Virgil <vir...(a)comcast.net> wrote: > In article <1173715726.469898.72...(a)v33g2000cwv.googlegroups.com>, > > > > Every level has only finite paths, so that the number of nodes and > number of paths are both finite in such finite paths. The number n of levels does not end. If nevertheless this is insufficient to produce infinite paths which cross every level, then there are no infinite paths at all. > > > > > If there are no single paths in the tree, ... then: where are they? > > Married, and raising little paths? That is an apt answer concerning your beliefs. Regards, WM
From: mueckenh on 15 Mar 2007 06:04 On 12 Mrz., 22:28, Virgil <vir...(a)comcast.net> wrote: > In article <1173724460.248046.46...(a)s48g2000cws.googlegroups.com>, > > > 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. No. This union is omega. The cardinal number of this union is aleph_0 (today also denoted by omega). > > > 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. nor the union of all sets of the form {1,2,3,...,n} has any meaning. Regards, WM
From: mueckenh on 15 Mar 2007 06:11 On 13 Mrz., 13:39, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1173715319.243468.87...(a)c51g2000cwc.googlegroups.com> mueck...(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. Not at all. > 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. Remember the proof by Oresme. Why is the number of parentheses of the harmonic series countable and the number of unit fractions too? You try to make a distinction between the results of infinitely many countable parentheses and of infinitely many levels of the tree. This distinction is wrong, because both situations are identical. > > > An uncountable set of paths cannot exist other than outside of the > > tree, i.e., outside of mathematics. > > Why? A proof, please. Inside of the tree the number of paths is restricted to the cardinal number of levels (which easily can be understood as cross sections). Regards, WM
From: mueckenh on 15 Mar 2007 06:33 On 13 Mrz., 14:15, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1173724460.248046.46...(a)s48g2000cws.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > Terminating paths. *Passing* path-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.) A path bundle splits off into two bundles which pass said node. Therefore every set of path-bundles in the tree has a finite cardinal number This, in the "limit", may yield an infinite number, but certainly not an uncountable number without having an intermediate countably infinite number. The tree is continuous because its nodes are connected by paths. There is never more than the factor 2. There are no interruptions possible and no jumps from "finite" to "uncountable". Your claim would require that. Regards, WM PS: What about the review of chapter 10?
From: Dik T. Winter on 15 Mar 2007 11:36
In article <1173953509.704127.222770(a)b75g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 13 Mrz., 13:39, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > 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. > > Not at all. > > > 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. > > Remember the proof by Oresme. Why is the number of parentheses of the > harmonic series countable and the number of unit fractions too? When compared to the tree you can compare each pair of parenthesis with the levels of the tree and each unit fraction with the nodes of the tree. > You try to make a distinction between the results of infinitely many > countable parentheses and of infinitely many levels of the tree. This > distinction is wrong, because both situations are identical. I do not make that distinction. In the harmonic series there are *no* paths defined, so using it in connection with paths is wrong. > > > An uncountable set of paths cannot exist other than outside of the > > > tree, i.e., outside of mathematics. > > > > Why? A proof, please. > > Inside of the tree the number of paths is restricted to the cardinal > number of levels (which easily can be understood as cross sections). 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/ |