Cantor Confusion
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From: Virgil on 19 Dec 2006 16:00 In article <1166526935.758264.263940(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1166452045.777425.20310(a)79g2000cws.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > ... > > > > >> > > > 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). > > ... > > > > Upto now we have no line L_D at all since P1 has not been proved. > > > > > > > There is nothing to prove in a definition. I remember you said a > > > definition cannot be wrong. So it cannot be proved. > > > > P1 is not a definition. > > It is the definition that, in the Equilateral Infinite Triangle, the > diagonal consists of the ends of lines a_nn. A line has elements a_kn > with k =< n and n eps N --- with ALL n eps N. Within ZFC of NBG, any construction of WM's EIF satisfies: (1) For each "line" of EIF there is an n eps N such that a_nn is NOT a member of that line. (2) For the diagonal of EOF every a_nn, with n eps N, IS a member of that diagonal. If WM wants to construct an EIF outside ZFC or NBG or any similar system allowing an infinite N, he must first produce an axiom system in which such a construction can take place.
From: Virgil on 19 Dec 2006 16:06 In article <1166527114.881108.181070(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1166453622.108599.302270(a)j72g2000cwa.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > ... > > > > And what if m is infinite? How do you divide? How do you recombine? > > > > Suppose I have an edge that is divided in countably many shares. Now > > > > suppose I select the even numbered shares and recombine them. Do I > > > > now have an edge? And what when I do the same with the odd numbered > > > > shares, do I now also have an edge? If that is not the case, why not? > > > > And if that is the case, please explain. > > > > > > We need not divide anything by infinite numbers, because there are only > > > finite numbers n in the sum > > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... > > > Every level of the tree has a finite number n. For finite n also 2^n is > > > finite. > > > > I do not understand the connection. Through every edge there go infinitely > > many paths. So to give each path a share of the edge we have to divide the > > edge in infinitely many shares. > > We will find out what is infinite. In any case each level has a finite > number and the sum of edges inherited by one complete path and by no > other path is given by: 1 + 1/2 + 1/4 + ...+ 1/2^n + ... = 2. > > 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. Absent any definition of convergence to a limit, there is no convergence to a limit. And then an infinite sequence of partial sums is just an infinite set of reals. > > Given the Hamel basis, do some lexicographic ordering of the reals as a > > vector space over the rationals. > > > I cannot do it Ignorance is no excuse.
From: Virgil on 19 Dec 2006 16:12 In article <1166527560.520380.49820(a)n67g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > > > > Element n+1 can be shown to exist in L_D (which is obviously a > > > > > > > line > > > > > > > containing n+1). > > > > > > > > > > > > No. L_D is bounded. The largest element of L_D is n. > > > > > > L_D does not contain n+1. > > > > > > > > > > You misinterpret L_D. L_D is that line which contains all numbers > > > > > contained in the diagonal. If your L_D does not contain them, then > > > > > you > > > > > have the wrong L_D. > > > > > > > > Assume that there exists an L_D which contains all the numbers > > > > contained in the diagonal. L_D is bounded, > > > > > > If an unbounded diagonal exists, then obviously an unbounded line must > > > exist. > > > The conclusion is false, so the antecedent cannot be true. > > > > The diagonal is the potentially infinite set of natural numbers. > > This exists and is unbounded. > > Yes, but not in the way you understand "to exist". Potentially infinite > means: always finite. The diagonal is always finiter though not > bounded. If it existed in the way you understand by "to exist", then > also a line had to be infinite (by the definition: in the Equilateral > Infinite Triangle, the diagonal consists of the ends of lines a_nn. A > line has elements a_kn with k =< n and n eps N --- ALL n eps N), which > is obviously wrong. One can construct in ZFC or NBG, based upon an 'actually infinite' ordinal N = omega, each "line" as a finite ordinal and the diagonal as the unique actually infinite limit ordinal of the lines. If WM rejects such systems as ZFC and NBG, WM has no system in which his model of the "triangle" can be constructed at all.
From: Virgil on 19 Dec 2006 16:13 In article <1166527648.094014.311530(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > Albrecht wrote: > > > > but in the same time |N isn't complete > > > since |N is bijectable to the diagonal and the diagonal is never > > > complete since there is no complete line (number) which covers the > > > diagonal. > > > > No, the claim is that there is no line that covers the diagonal. > > Which results in the conclusion that there is no diagonal. Not in any logically coherent system. In ZFC and NBG, that conclusion is provably false. As WM has presented no system in which it is true, he loses by default.
From: Virgil on 19 Dec 2006 16:27
In article <1166527989.937543.147010(a)t46g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1166449661.044809.225940(a)73g2000cwn.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > > > All elements that can be shown to exist in the diagonal can be shown > > > > > to > > > > > exist in one single line. > > > > > > > > Which line is that? > > > > > > That one which guarantees that a diagonal can contain all the elements > > > which it contains. > > > > Why does such a supposed line need to exist? In ZFC no such line can > > exist. > > > Because each element of the diagonal beongs to a line which contains > all peceding elements. WM is conflating things again. That each element of the diagonal must belong to SOME line does not mean that they all must belong to the same line. Consider a new triangle each "line" containing only one element, what was in the former triangle the last element, and the diagonal containing all these elements. It is still true that every element in the diagonal is in some line, but now it is obvious that there is no line which contains all elements of the diagonal. Since what Wm claims for his triangle is clearly not true for all triangles, he neds to PROVE his claim. Which he has not yet done, and cannot do without the added assumption that there is a last natural, or some equivalent assumption. > > > > That presumes a last line, which presumption is > > > > unwarranted in ZFC and NBG and most other set theories. > > > > > > It presumes that all elements of the diagonal exist in the EIT. > > > > But WM's EIT presumes a last line, which is contrary to fact in ZFC. > > There are no "facts" in ZFC. Sure there are. It is a /fact/ in ZFC that any the union of all finite ordinals is an ordinal which is not finite. > > > > > > > > > ZFC and > > > NBG and most other set theories which wish to make us believe that all > > > elements of the diagonal do exist although not in one line but > > > distributed over many lines, are obviously wrong. > > > > They are only wrong if one imposes an assumption which makes them wrong. > > > > Absent such an assumption, there is noting to prevent them from being > > right. > > > > > > > > > For linear sets it is > > > impossible that an element exists in line m but not in line n > m. > > > > But quite possible in some line k with k > n > m. > > whch thebn can be called line m. So now WM claims that m > m? Not in ZFC or any sane system. > > > > > Threfore all elements which exists in smaller lines exist in a larger > > > line. > > > > That every element is in some line does not require that some line > > contain every element. > > In a linearly ordered set with only finite elements this is required. > Otherwise give me a counterexampe. Every line of your triangle, when modeled in ZFC or NBG, is a finite ordinal but the diagonal is the first limit ordinal which is not finite. QED. |