Cantor Confusion
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From: Virgil on 27 Dec 2006 18:12 In article <1167256736.581223.130600(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > stephen(a)nomail.com schrieb: > > > > In what system can you prove that #edges = 2 * #paths? > > In the infinite binary tree! Not in ZFC or NBG, and not without assuming things false in ZFC and NBG. WM argues his case claiming: Lim n -> oo #edges/#paths = (Lim n -> oo #edges)/Lim n -> oo #paths) But as two of the limits do not exist in any standard sense, and their quotient would not be defined even if infinite limits were defined, WM's claim is false and his proof is invalid. It is also quite legitimately provable in ZFC or NBG that the number of edges in an infinite tree is countably infinite and the number of paths is uncountably infinite. So the best WM could hope for would be a conflict between his as yet unspecified axiom system and the axiom systems of standard set theories.
From: Virgil on 27 Dec 2006 18:17 In article <1167256922.435883.99380(a)a3g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > 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". What is true in bounded cases need not be true in unbounded ones. It is trivial that the set of paths or nodes in an infinite binary tree is at most countable and not much more difficult to show that the set of paths is not.
From: Newberry on 27 Dec 2006 22:05 mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1167082643.909029.231380(a)f1g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > In article <1166921237.502878.48560(a)h40g2000cwb.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > > > > > Dik T. Winter wrote: > > > > > > In article <1166845904.426550.122020(a)48g2000cwx.googlegroups.com> "Newberry" <newberry(a)ureach.com> writes: > > > > > > > > What is level oo? The levels are indexed by natural numbers. > > > > > > > > The completed infinite tree? As long as you are talking about finite > > > > trees, you are welcome. But in that case 1/3 is not a path in the tree. > > > > > > What is an infinite path? > > > > > > Do you think it has some fairy tale character? Some very special > > > properties, different from any thing we can observe in the universe? > > > > > > A path is an infinite path if you can follow it without ever reaching > > > an end - and that's all. > > > > 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. > > > > > Therefore: If you follow some path you will see that whenever it > > > separates itself from another one, this happens by two edges .- one > > > edge for the path, and the other edge for the other path. This process > > > repeats and repeats without end. Nothing else happens. From the > > > unavoidable and "inseparable" connection of separation and appearance > > > of another edge we can conclude that every path which can be > > > distinguished from another path runs through an edge which does not > > > belong to the other path; call it "personalization of an edge. > > > Therefore distinguishability of paths and personalization of edges are > > > unavoidably connected. > > > > 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. But there are algorithms that tell at EACH level whether to go left or right. > > > > > To assert that there are more paths separated by edges than are edges > > > existing at all, is an error which cannot be explained other than by > > > the desastrous and corrupting influence of the "logic" of set teory. > > > > 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. > > Regards, WM
From: Newberry on 27 Dec 2006 22:12 mueckenh(a)rz.fh-augsburg.de wrote: > 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. > > > 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? It depends. You can take the position there are none. Since the sequence never finishes it converges to nothing i.e. there is no number it converges to. But you can also take the position that the infinite sequence itself is a number. If nothing else this terminology is consistent with the mathematical tradition and there is no reason to change it. > > ================================ > > > 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". > ================================= > > > Note that you conclude things about the infinite here based on the finite. > > We cannot conclude anything other than based on the finite. > > Regards, WM
From: Newberry on 27 Dec 2006 23:21
mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1166952847.829226.324240(a)79g2000cws.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > > As not all natural numbers do exist, the set is potentially infinite, > > > > > i.e., it is finite. It has a maximum. L_D. Taking this maximum and > > > > > adding 1 or takin L_D ^ L_D or so yields another maximum. In any cae > > > > > there is a maximum. > > > > > > > > Which is always exceeded by a larger maximum. > > > > > > Which then is the maximum. > > > > As the creation of your "maximum" simultaneously creates its successor, > > the time span for any natural to be maximal is zero. > > There is no creation in zero time, because then, at the same time, > something would exist and would not exist (Aristoteles). > > > ==================================================== > > >> Nobody can name it, because the set is not fixed. > > > > In what set theory are sets not fixed. Wm's theories of volatile > > sets do not work in mathematics. > > It is the only theory about sets which works in reality and does not > require cranks and their beliefs on the infinite, but is based upon > clear facts like: > > The infinite sequence of natural numbers is the union of all finite > initial segments of natural numbers. > An infinite tree is a tree which covers every finite level. > > > ================================ > > >> The absence of more than 10^100 bits limits the existence of objects > >> which require more than 10^100 bits to exist > > > > > In physical existence one may need physixcal bits, but not otherwise. > > Existence = physical existence, unless you have a strong religious > intuition. > (This intuition and its results, however, can be traced back on > physical circumstances like drugs or hormones or electrical currents.) > > ==================================================== > > > The Cantor construction rule produces for any countably infinite list > > a number which necessarily differs from every number in the list. > > The argument produces it only for finite segments of the list. How can > you conclude from the finite domain on the infinite domain where > completely other rules govern the numbers? > > ======================================= > >> You cannot find such a number a_n without having pi already. But this > >> number shall be used to establish the existence of pi. This is > >> impossible. pi does not exist *as a number* (it exists as an idea and > >> in form of close approximations). > > > > > Which idea, in a non-physical world populated only by ideas, suffices. > > But not as a number in Cantor's list, a digit of which has to be > exchanged. > > > > On the contrary, it is WM's claim that the limit of a quotient must > > always equal the quotient of limits that is the problem. > > > WM claims lim n -> oo E(n)/P(n) = (lim n -> oo E(n))/(lim n -> oo P(n)), > > but it does not hold for any form of limit definition in standard > > mathematics. > > The truth is easily stated without quotients: lim n --> oo E(n) is not > less than lim n --> oo P(n)). What on earth are you talking about?? What is wrong with this limit? lim{n-->oo} (2*2^n - 2)/2^n = 2 Why do you need to compute lim{n-->oo} (2*2^n - 2) and lim{n-->oo} (2^n)? > > ==================== > > > The same " limit" argument will conclude the in a tree in which no path > > has a terminal node, every path has a terminal node. > > > > For every finite tree, the number of terminal nodes (leaf nosed) is > > exaclty equal to the number of paths, so that if the limit argument is > > valid for the edge to path ratio, it must equally be valid for the > terminal node to path ratio. > > > No, limit arguments do not hold in the infinite case (but only in the > finite case. Therefore they are called "limit", i.e., something bounded > and bounding.) > > You can see it best by the following example: > > In every finite initial segment of natural numbers, there are as many > even numbers as odd numbers (plus or minus one number). But in the > infinite set N there are 250 times more even numbers than odd numbers. > > Regards, WM |