Cantor Confusion
in [Math]
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From: Dik T. Winter on 15 Mar 2007 11:39 In article <1173954799.919385.61730(a)y80g2000hsf.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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 As long as path-bundles are finite sets of nodes that is right. > This, in the "limit", may yield an infinite number, but certainly not > an uncountable number without having an intermediate countably > infinite number. I do not know. What do you understand under "limit"? > 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. My claim requires nothing of the sort. The number of finite paths is countable. The number of infinite paths is uncountable. There is no jump from countable to uncountable, there is a jump from finite paths to infinite paths. > PS: What about the review of chapter 10? I have not had much time, but I am working on it. -- 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 15 Mar 2007 15:40 In article <1173952316.814267.224320(a)l75g2000hse.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Which is what WM seems to be claiming when he disallows bijection between the set of all paths in the complete infinite binary tree and the set of all subsets of N.
From: Virgil on 15 Mar 2007 15:43 In article <1173952696.371638.224790(a)o5g2000hsb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. It only does so if the set of infinite paths bijects with the power set of the infinite set of levels. > > > > > > > > > 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. It was attempting to match the silliness of the question. Apparently it fell short.
From: Virgil on 15 Mar 2007 19:53 In article <1173953091.121492.222500(a)n59g2000hsh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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). Which carefully avoids my point that the union is not a limit by any standard definition of limit. And if Wm has such a workable definition of limit, other than mere union, he does not make it public. > > > > > lim {n-->oo} n < aleph_0 (that is because every number is finite). Absent any definition for precisely what that limit symbol means, no such inequality statement is justifiable. > > > > 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. The union of /any/ family of sets is the set whose members are just the members of members of that family. This is defined and required by the axioms of both ZF and NBG, and, I have no doubt, holds in most, if not all, other sensible set theories.
From: Virgil on 15 Mar 2007 20:17
In article <1173953509.704127.222770(a)b75g2000hsg.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Assuming all paths are of equal "length", for each path of the binary tree, take the set of levels of those nodes at the end of left bramnches. It is clear that there is one such path for each possible set of levels (beyond the root) and vice-versa. So that the set of paths is equinumerous with the power set of the set of non-root levels, and thus always larger than the set of non-root levels. This is as true for infinite trees as for finite trees, > > > 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? Unless it Oresme proves that there are sets bijectable with their power sets, Oresme is irrelevant. > > 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. Not to anyone who knows what is going on. Unless it Oresme proves that there are sets bijectable with their power sets, Oresme is irrelevant. > > > > > 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). On the contrary, I have shown above that the number of paths always has the cardinal of the /power set/ of the of the set of non-root levels. This equivalence of set of paths with power sets of levels, is trivial to establish, as I have done above, and as it must hold for every finite tree and for the complete infinite binary tree. So that WM is wrong. |