Cantor Confusion
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From: William Hughes on 11 Dec 2006 12:18 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > This is the argument you present. > > > > > > > > 1) The set of lines is also the potentially infinite set of natural > > > numbers. > > > 2) Every element of the diagonal is in at least one line. > > > 3) Every initial segment of the diagonal is (in) a line. > > > > No, there is one initial segment of the diagonal that is > > not in a line. > > Then name the element of the diagonal please, which supports your > claim. There is one initial segment of the diagonal which is not defined by a single element. The potentially infinite sequence {1,2,3 ...} is an initial segment of the diagonal, but it is not in a line, nor does it have a largest element. (It is contained in the union of all lines, but the union of all lines is not a line) > > > > > 4) The limit (n -->oo, for the number n of elements) of the diagonal is > > > oo. > > > > Yes and it attains this limit (i.e the limit is a maximum as > > well as a supremum) > > > > > 5) The limit (m -->oo, for the number m of elements of a line) is less > > > than infinite. > > > > No. The limit is oo. (The limit cannot be any finite number) > > And it cannot be an infinite number. There is no infinite natural > number. Why do you think it is a natural number? The limit of natural numbers does not have to be a natural number. This limit is not a natural number so it can be infinite. - William Hughes
From: Tony Orlow on 11 Dec 2006 13:21 mueckenh(a)rz.fh-augsburg.de wrote: > Tony Orlow schrieb: > > >>>> Repeating a falsehood does not make it any less false. >>> Why then did you repeat always that here are no balls in the vase at >>> time 0? Why do you repeat without reasonable arguments that there are >>> more paths than edges in the binary tree, although no path springs off >>> without its own egde? >>> ========================= >>> >> Again I agree. There are 9 n balls after n iterations, and there are >> half as many paths as edges or nodes in the tree, despite supposed >> bijection hat tricks. Banach-Tarski is ridiculous, and omega is a phantom. > > Correct. The function f(n) = 9n will never yield the value 0. >>>> Why he maintains this this somehow bijects the individual branches with >>>> paths, is not clear. >>> Because every branch consists of two edges, one for the each of the >>> resulting paths. >>> >>> ========================= >>> >> To put it another way, each right branch may be considered a >> continuation of the same original infinite path, like adding a 0 to the >> right of the decimal point - it doesn't change the value. The left >> branch from every node produces a new path, creating a new value with >> the concatenation of a 1 to the string. Therefore, for every two >> branches, there is one additional path. > > Of course that is a possible way to construct a bijection. But set > theorist will argue that by this constructuibn you get only those paths > representing rational numbers. (That is even true, because there are no > other numbers.) But they do not see that there are no paths without > edge. They just believe that "in the infinite" their God will provide > the missing paths. If every value in the tree is generated by an edge, then if 1/3 exists there is an edge where the path achieves that value. But, it cannot achieve that in any finite number of bits. If the depth of the tree is countably infinite, aleph_0 levels, then there are 2^aleph_0 nodes and edges. If not, then 1/3 does not exist in the tree, or anything else that isn't a binary rational (sum of a finite number of powers of 1/2). > >> Is the diagonal longer than any line? Nope. > > By definition, it cannot exist, where no line is. > Agree. >> Perhaps, with a countable number of bits. But, there are more than any >> finite number of reals in the unit interval, and this is an infinite >> number. > > Where *are* they? They canot be addressed. They cannot be imagined. > They cannot be known. And, moreover, the proof proving their existence, > is false. > > Quite a poor existence. > > Regards, WM > Well, the proof is simple. Any finite number of subdivisions of any finite interval will only identify a finite number of real midpoints in that interval, between any two of which will remain more real midpoints. Therefore, there are more than any finite number of real points in the interval. Of course, most of them require infinite strings of bits to specify, but that doesn't mean the point doesn't exist.
From: Michael Stemper on 11 Dec 2006 13:25 In article <1165615427.980137.7670(a)f1g2000cwa.googlegroups.com>, mueckenh writes: >William Hughes schrieb: >> We can now define bijections on potentially infinite sets > >Only if we consider them being actually infinite. Would you please define "actually infinite" or "actually infinite set"? > But that would >exclude them from being potentially infinite. Would you please define "potentially infinite" or "potentially infinite set"? -- Michael F. Stemper #include <Standard_Disclaimer> The FAQ for rec.arts.sf.written is at: http://www.geocities.com/evelynleeper/sf-written Please read it before posting.
From: mueckenh on 11 Dec 2006 16:31 Dik T. Winter schrieb: > > > So why did you state that I erronously believed that nodes represent > > > numbers? > > > > Because you erroneously did. > > I did not. Reread what is stated above. Reread what you wrote on 12 November "What is the node number of 1/3?" and "Do you really think the node 1/3 is finitely far from the root in the tree?" Reread what you wrote on 28 November: "But as I did show in another article, when you assign bits to nodes, you can show that the nodes represent numbers,..." > > > It is easy > > > enough to construct a bijection between the natural numbers and the > > > edges, because the edges are countable. > > > > After all you have grasped that too? Until now you denied. > > Not at all, Look here: Dik T. Winter, Datum: Fr 3 Nov. 2006 16:18 You answered my assertion: > The > mapping N --> {edges} has been established such that every edge knows > its number. You are wrong. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ > > Please give me all the bits of 1/3. Then I will show the bijection. > > I can't so you can not show a bijection. Representations (= paths) which do not exist cannot be part of a bijection. > But I did not even *ask* > for a bijection. You said you had constructed a surjection from the > edges to the paths. To prove that show us a single edge that maps to > 1/3 and also tell us how the construction of the *surjection* works. > I showed you a sum of shares of two edges. Since when does mathematics deny the vailidity of 1 + 1/2+ 1/4 + ... = 2 ? > > Indeed you are unable to understand fractions? > > I understand them well enough. But fractions have nothing to do with > this at all. If you state that part of an edge maps to some number and > another part of that same edge maps to some other number that may be > fine. But that does *not* define a surjection from edges to numbers. No? Even if there are enough parts to gather more than a whole edge to be mapped on every path? > To define such a thing you should be able to tell the single number to > which a specific edge maps. That is what a surjection is (and indeed > is also what a mapping is). So you have not shown a surjection. > -- You cannot add fractions to get a whole edge? Regards, WM
From: mueckenh on 11 Dec 2006 16:33
Virgil schrieb: > As there is an easy and obvious bijection between nodes and node-rooted > finite sub-paths, in which each node is paired with the finite sub-path > connecting it to the root node, the situations are identical. This shows that every node represents a finite path. There is no node left to construct any infinte path. What is the material which the infinite paths consist of? Are they built from the ghosts of void ideas on set theory? Regards, WM |