Cantor Confusion
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From: Dik T. Winter on 28 Dec 2006 21:16 In article <1167256678.283824.142350(a)42g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > Yes. So it is essentially different from a finite path. The same with > > your trees, Each finite tree has a natural number as maximum level and > > has no infinite paths. The completed infinite tree has no maximum level > > and has infinite paths. So any conclusion you can make from the finite > > trees does not necessarily hold for the infinite tree. > > We conclude only that one can follow a path infinitely, and that always > there is an edge which continues it. There is no continuation without > an edge. ok? That is unavoidable and it is enough to prove that there > are not less edges than paths. Well, in that case, pray, show a proof. > > For each two paths you can identify an edge where they separate. But there > > is *no* edge that separates (for instance) the path leading to 1/3 from > > all other paths. And through each and every edge run infinitely many > > paths. > > Yes. That shows that there are no infinite sequences of edges (or > digits) which individually represent nunmbers like 1/3 or ideas like > irratinal numbers. Why? > > You are extending conclusions valid for finite trees to the infinite tree > > without proof, it is just your intuition. > > Not at all. Forget about all finite trees. In the infinite tree there > is always an edge which continues a path. There is no continuation > without an edge. That is unavoidable and it is enough to prove that > there are not less edges than paths. In that case: prove it. As I have already shown, all edges can be made to represent the rational numbers where the denominator is a power of 2. That is because all edges are at a finite distance from the root. -- 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 28 Dec 2006 21:21 In article <1167256922.435883.99380(a)a3g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > What is an infinite path? > > > > Why do you not answer my question? (Note: rood -> root.) > > The second left branch is assigned to half of all paths. If the number > of all paths is a meaningful notion, then half of it is a meaningful > notion too. Oh. How do you *define* aleph_0 / 2? > > The point is that you think that also in the infinite tree you can assign > > shares of edges as 1 + 1/2 + 1/4 + ... to a path. But that is false. > > There is *no* edge that is shared by finitely many paths. > > So there is no irational number? Can you give your reasoning behind this statement? > > In the finite case all paths end at a terminal node. So when you compare > > number of edges to number of paths at finite levels you are comparing > > edges to paths with terminal nodes. > > But even without terminal nodes, the paths consist of edges. Where no > edges are, there is no path. Even "in the infinite" that is true. The > number of paths cannot "overtake" the number of edges "in the > infinite". Why not? > > Note that you conclude things about the infinite here based on the finite. > > We cannot conclude anything other than based on the finite. So you think. -- 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 28 Dec 2006 23:14 In article <1167338300.513980.203010(a)a3g2000cwd.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > Virgil wrote: > > In article <1167316205.832225.238540(a)48g2000cwx.googlegroups.com>, > > "Newberry" <newberry(a)ureach.com> wrote: > > > > For NO "levels of an infinite tree" is it true, but it is true for > > > > finite trees. > > > > > > It is true for ALL nodes, blue, green, red, finite, infinite, all nodes > > > of the infinite path, simply all. > > > > Then prove it. > > Proof: > 1) lim{n-->oo} (2*2^n - 2)/2^n = 2 > 2) for n = 1 we get (2*2^1 -2)/2^1 = 1 > 3) (2*2^n - 2)/2^n is a monotonic function of n > Hence An(1 <= (2*2^n - 2)/2^n <= 2), that is for all n 1 <= > #edges/#paths <= 2 > All n means ALL levels and ALL edges in any infinite paths. Actually, "For all n in N, #edges(n)/#paths(n) >= 2", not "<= n". "For all n" means for all n in some set. and in this case the set is the set of finite naturals, N. But For n not in N, #edges(n)/#paths(n) is not defined, And #edges(N)/#edges(N) = 0, if it has any definable value at all. > > > > I can prove that in an infinite tree there are a countably infinity of > > nodes or edges but an uncountable infinite of paths. > > > > Since every edge terminates in a unique node, there is at least one node > > for each edge. > > > > One can biject the nodes with the naturals by: > > root node <--> 1 > > 2nd level nodes <--> 2,3 > > 3rd level nodes <--> 4...7 > > 4th level nodes <--> 8...15 > > ... > > nth level nodes <--> 2^n ...(2^(n+1)-1) > > ... > > > > One can biject the the set of infinite binary sequences with the set of > > infinite paths by matching each 0 in a sequence with a left branching > > edge of a path and 1 with a right branching edge. > > > > And while there are injections from the set of naturals to the set of > > binary sequences, a simple diagonal argument shows that there are no > > surjections from the set of naturals to the set of binary sequences. > > > > Thus there are "more" paths than nodes (or edges). Proof not refuted!
From: Virgil on 28 Dec 2006 23:27 In article <JB0Jw4.9L0(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > In article <1167256922.435883.99380(a)a3g2000cwd.googlegroups.com> > mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > What is an infinite path? > > > > > > Why do you not answer my question? (Note: rood -> root.) > > > > The second left branch is assigned to half of all paths. If the number > > of all paths is a meaningful notion, then half of it is a meaningful > > notion too. > > Oh. How do you *define* aleph_0 / 2? Actually, half of all paths in an infinite tree would be c/2 or aleph_1/2.
From: Newberry on 29 Dec 2006 01:02
Virgil wrote: > In article <1167338300.513980.203010(a)a3g2000cwd.googlegroups.com>, > "Newberry" <newberry(a)ureach.com> wrote: > > > Virgil wrote: > > > In article <1167316205.832225.238540(a)48g2000cwx.googlegroups.com>, > > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > > For NO "levels of an infinite tree" is it true, but it is true for > > > > > finite trees. > > > > > > > > It is true for ALL nodes, blue, green, red, finite, infinite, all nodes > > > > of the infinite path, simply all. > > > > > > Then prove it. > > > > Proof: > > 1) lim{n-->oo} (2*2^n - 2)/2^n = 2 > > 2) for n = 1 we get (2*2^1 -2)/2^1 = 1 > > 3) (2*2^n - 2)/2^n is a monotonic function of n > > Hence An(1 <= (2*2^n - 2)/2^n <= 2), that is for all n 1 <= > > #edges/#paths <= 2 > > All n means ALL levels and ALL edges in any infinite paths. > > > Actually, "For all n in N, #edges(n)/#paths(n) >= 2", not "<= n". > > "For all n" means for all n in some set. and in this case the set is > the set of finite naturals, N. Yes, but there is one-to-one mapping between the naturals and the edges in each infinite path. > > But For n not in N, #edges(n)/#paths(n) is not defined, > And #edges(N)/#edges(N) = 0, if it has any definable value at all. > > > > > > > > > I can prove that in an infinite tree there are a countably infinity of > > > nodes or edges but an uncountable infinite of paths. > > > > > > Since every edge terminates in a unique node, there is at least one node > > > for each edge. > > > > > > One can biject the nodes with the naturals by: > > > root node <--> 1 > > > 2nd level nodes <--> 2,3 > > > 3rd level nodes <--> 4...7 > > > 4th level nodes <--> 8...15 > > > ... > > > nth level nodes <--> 2^n ...(2^(n+1)-1) > > > ... > > > > > > One can biject the the set of infinite binary sequences with the set of > > > infinite paths by matching each 0 in a sequence with a left branching > > > edge of a path and 1 with a right branching edge. > > > > > > And while there are injections from the set of naturals to the set of > > > binary sequences, a simple diagonal argument shows that there are no > > > surjections from the set of naturals to the set of binary sequences. > > > > > > Thus there are "more" paths than nodes (or edges). > > Proof not refuted! |