Cantor Confusion
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From: Virgil on 20 Dec 2006 15:55 In article <1166616296.692896.64880(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > > No. But I need no proof in order to see that > > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > > > > > > > Yes, it appears that you need no proof to make any of your assertions. > > You simply "see" them; or "feel" them. That's not what I would call > > "really correct mathematics". > > Call it whatever you want. I can calculate 1 + 1/2. But considering the > fact that you would call the following result "correct mathematics", I > prefer to stay with my version. > > Down to level n (for n = 1, 2, 3, ...) we can distinguish P(n) = 2^n > different paths and E(n) = 2*2^n - 2 different edges. So we find > lim{n-->oo} E(n)/P(n) = 2, or, if we are unable to determine limits, > E(n)/P(n) >= 1. Nevertheless "E(n)/P(n) < 1 in the actual infinity". Why not? There is no reason for "continuity" to be required here. By the same limit argument as WM has just claimed, there must be as many leaf nodes in the infinite case as there are paths, since the ratio of number of leaf nodes to number of paths is identically 1 for all finite trees.
From: Virgil on 20 Dec 2006 16:01 In article <1166617425.221440.15570(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > For the evaluation of this *sum* we need *no limit* but only the > > > computation > > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > > > > We were talking about *shares*. In how many shares is the edge going left > > from the root divided, and how do you do that division? Pray keep to the > > subject. > > We are talking about shares, but we do not need more than 3/2 per > path.> How are the edges at the root node to be divided up into shares, and to which paths are those shared to be assigned? Then the same questions need answering about the next level of edges, and then the next, and s on ad infinitum. > The parallel axiom is true on a sheet of paper. Not unless you have an infinite sheet of paper and a perfect protractor.
From: Virgil on 20 Dec 2006 16:05 In article <1166617688.263687.149650(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > I said that there are infinitely many equations of form "1+1+1+...+1 = > > n" with equal finite values on each side, but made no claim of any such > > equation with any infinite number on either side. > > If there are infinitely many different equations of form "1+1+1+...+1 = > n, then there must be infinuitely many different sums of ones. Actual > infinity of he umber of equations implies actual infi8nuity o at least > one sum of ones. That is only true in WM's mythical axiom system, not in any actaul axiom system. Consider the model given within any infinite tree by any of its (infinite) paths. The path is the diagonal and each line is the section of that path from the root node to any other node. Each edge in the path is labeled by (identified with) the natural number of edges in the line of which it is the last edge. In this model, every natural appears in some line. In this model, there is no line in which every natural appears. This model disproves WM's contentions about his EIT. > > > > This statement is in complete compatibility with ZFC and NBG > > An exactly that is the reason why ZFC is wrong. You cannot have an > actually infinite finite number. WM may not be able to, but his inadequacies are personal, not universal.
From: Virgil on 20 Dec 2006 16:09 In article <1166617872.309270.165750(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1166545226.008570.102300(a)n67g2000cwd.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > ... > > > > > So for finte trees there are more edges than paths but for infinite > > > > > trees there are more paths than edges? > > > > > > > > Indeed. > > > > > > LOL. > > > > > > > > So for finte trees there are more edges than paths but for infinite > > > > > trees there are more paths than edges? > > > > > > > > Right. It is taking conclusions about the infinite from the finite > > > > cases > > > > when you think it is otherwise. > > > > > > Perhaps we cannot determine the limit of the series 1 + 1/2 + 1/4 + ... > > > unless we admit some sophisticated calculation of limits. But without > > > any such sophistication we can be sure the sum of arbitrarily many > > > terms of that series will be less than 28.50. And you assert that it is > > > greater than 100! That is what you call mathematics. But it is simply a > > > wrong application of human brain. > > > > The inequality holds for all finite paths. Why it also should hold for > > infinite paths escapes me. > > Because we can prove that 2 - 1/2^n is always, i.e., for all finite > natural n, less than 100. And because even infinity does not supply > other than finite natural numbers. If what holds for finite paths must old in the limit for infinite paths consider the ratio of the number of terminal nodes to the number of paths in finite trees. That ratio is identically 1 for all finite trees, so must be, by WM's arguments, still equal to 1 in the limit. But then all endless paths must end, and WM is rejecting his own models.
From: Virgil on 20 Dec 2006 16:12
In article <1166618383.155418.293700(a)t46g2000cwa.googlegroups.com>, "Albrecht" <albstorz(a)gmx.de> wrote: > Franziska Neugebauer schrieb: > > > Albrecht wrote: > > > > > William Hughes schrieb: > > >> Albrecht wrote: > > >> > William Hughes schrieb: > > >> > > mueckenh(a)rz.fh-augsburg.de wrote: > > >> > > > Franziska Neugebauer schrieb: > > >> > > > > > >> > > > > >> > > > All elements that can be shown to exist in the > > >> > > > > >> > > > diagonal can be > > >> > > > > >> > > > shown to exist in one single line. [(P1)] > > >> > > > > > > >> > > > > This proposition P1 has _not_ yet been proved (shown). > > >> > > > > > >> > > > This proposition is the definition of the diagonal. > > >> > > > > > >> > > > > > You misinterpret L_D. L_D is that line which contains all > > >> > > > > > numbers contained in the diagonal. > > >> > > > > > > >> > > > > The Diagonal is unbounded thus any _assumed_ L_D is not > > >> > > > > bounded, too. > > >> > > > > > >> > > > Correct is: > > >> > > > If the diagonal is unbounded then any _assumed_ L_D is not > > >> > > > bounded, too. > > >> > > > > >> > > L_D is a line and is therefore bounded. > > >> > > > > >> > > Contradiction. > > >> > > Hence, the _assumed_ L_D does not exist. > > >> > > > > >> > > - William Hughes > > - > > >> > > William Hughes > > >> > > > >> > It's always the same game you play: you compare numbers and sets. > > >> > Numbers are always complete, sets are complete only if they are > > >> > finite. Now you claim that the set of the natural numbers contain > > >> > all natural numbers and hence it is complete > > >> > > >> No, the claim is that it is potentially infinite (given > > >> any finite set of natrual numbers one can allways find one more). > > > > > > That's okay. You can say: given *any* set X of natural numbers, one > > ^^^^^^^^^ > > *any* _finite_ set > > > Since there are no other sets, we don't need this constriction. That is an assumption incapable of proof without making other assumptions equally incapable of proof. And it is false in ZFC and NBG. |