Cantor Confusion
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From: mueckenh on 14 Dec 2006 06:16 cbrown(a)cbrownsystems.com schrieb: > > > (1) Edges are countable. > > > (2) Paths can be bijected with the reals. > > > (3) B(e) is the (uncountable) set of all paths to which edge e belongs. > > > (4) Shares (of edges in the original diagram) are denoted (p,e) for > > > some path p and some edge e where e belongs to p. > > > (5) S(e) is the (uncountable) set of all shares of edge e. > > > (6) C(p) is the (countable) set of all shares which belong to path p. > > > (7) 1 + 1/2 + 1/4 + ... +1/2^n + ... = 2 > But that is precisely the problem here: you are mixing up two different > definitions of "part of an edge", because you have not been precise in > stating what you mean by "part of an edge". An edge can be subdivided in x parts. If we collect x parts together, then we have the same value as if we take one full undivided edge. > > > > > > seems to be > > > that there are two "full edges" whose "shares" are exactly the shares > > > of edges composing a particular path p. But a path p is composed of a > > > countable number of shares; so a "full edge" can have at most a > > > countable number of shares. > > > > You are aware of the number 2^omega being a countable one? If so, why > > then do you speak of an uncountable set of shares? > > > > Here's what I am "aware" of for (1)-(7): there is a bijection between > paths and reals. There is a bijection between shares of edges and > paths. Therefore, there is a bijection between reals and shares of > edges of any particular edge. > > By "set X is uncountable" I mean there is a bijection between the > elements of X and the reals. By "set Y is countable" I mean there is a > bijection between Y and the naturals. > > You claim that you can /prove/, using the 7 statements above upon which > we have agreement, that there is a bijection between the reals and and > the naturals; and that therefore countable = uncountable. > > When you say "You are aware of the number 2^omega being a countable > one?", I would respond : that's what you're trying to /prove/. You > can't just assert it that it's true for other reasons, and claim that > therefore you have proven it using your "rational relation". No. According to ZFC the ordinal 2^omega is a countable set. Even omega^omega^omega is a countable set. > . > > Do you think that lim {n-->oo}1/2^n * 2^n consists of uncountably many > > parts? > > > > lim {n->oo}(2^-n*2^n) = 1. If your question is "does the real number 1 > consist of uncountably many parts", I must ask: what do you mean by a > "part" of the real number 1? There are many interpretations of the > term, each yielding a /different answer/. You talked about different parts of an edge. Start with 1/3 + 1/3 + 1/3. Do you think that these 3 parts of 1 make 1? > > > > Meanwhile, an edge in the original diagram > > > consists of an uncountable number of shares. So full edges can only be > > > said to be "the same as" edges in the original diagram if you /assume/ > > > that countable is the same as uncountable. > > > > > > But that is what you are trying to prove - you can't simply assume it > > > in a proof of it. > > > > How can 1 + 1/2 + 1/4 + ... +1/2^n + ... yield the value 2 if we add > > only countably many shares which are uncountably small? > > > > Why would I think that the real number 1 in the above series is a > "share of an edge" in the /same sense/ that you have defined "shares of > an edge" as being equinumerous with the real numbers? > > A "share of an edge" is associated with a particular edge and a > particular path. No. In my proof we need only the value, not the individual edge. Because it is sure from other considerations that any part is used for one path only., > A path is associated with a real number. But the sum > of the real numbers that correspond to the paths to which edge e > belongs is not 1; it is instead undefined (the sum does not converge). > > > Would you mind to think about that topic again? > > > > Sure. I think that because you fail to make clear definitions, you > don't even realize that you are using the term "share of an edge" to > mean two completely different things; and this is the source of your > mathematical error. Please inform you: 2^omega is not an uncountable set. I do not divide edges in uncountably many parts. Any book on set theory including ordinals will be sufficient. Regards, WM
From: mueckenh on 14 Dec 2006 06:19 Virgil schrieb: > > One can define such functions called sequences, but that does not > > guarantee that all the pairs do actually exist. The pair (x, f(x)) with > > x = floor(pi*10^10^100) does not exist, for instance. > > If f(x) = 1 for all x then the pair (x,1) exists for all x. No. The pair (x,1) exists only for all those x which exist. > > But what do you mean and understand by "to exist"? In *every* set > > theory, starting from Cantor, it means to exist actually. > > I am not aware of any set theory in which any set is declared to > "actually" exist, though there are those in which sets are declared to > exist. You are not aware of many things in set theory. If Cantor's list does not actually exist, then there is no diagonal number, because the diagonal process is never finished. You cannot use the quantifier "forall". If not all digits of pi exist, then it is not an irrational number. Regards, WM
From: mueckenh on 14 Dec 2006 06:29 Virgil schrieb: > > There is no line with every natural number. > > > > Similarly you cannot find every elment of the diagonal. > > So that one can find any but not every? Correct, because there is not every. > > > > If you name an > > element, then the diagonal is constructed up to that element at least. > > And does it stop there? Until any one bothers to continue it. > > > > As there is no line with every number there can be no diagonal with > > every number. > > That is an assumption which WM imposes, not a conclusion which can be > drawn from anything incontrovertible. It is not an assumption but it is clear from the fact that the diagonal is a subset of the IET. > > If WM will admit that he is taking it as an axiom, that is one thing, > but he keeps insisting the he can deduce it from something which > requires that he assume it. The diagonal is a subset of a matrix. It cannot exist outside. Yes, that is an axiom. And after having cleared our positions it is useless to talk about that any further. Regards, WM
From: mueckenh on 14 Dec 2006 06:37 Virgil schrieb: > In article <1166035958.305158.282870(a)79g2000cws.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Dik T. Winter schrieb: > > > > > > > > > > That one which goes left from the root is not engaged, because we > > > > > > need > > > > > > only half of the set of edges. > > > > > > > > > > So your construction is not a surjection. > > > > > > > > Of course it is! It is a surjection from the set of edges onto the set > > > > of paths. > > > > > > Than again. To what number maps the edge that goes left from the root? > > > If it is a surjection, there should be one. > > > > > A surjection onto the paths covers all elements of the *range* = every > > edge. It is not necessary that every edge is mapped on a path, to show > > that there are not less edges than paths. It is only necessary that > > there is no path without edge mapped on it. > > Which necessity WM has not accomplished. All Wm has done is to map > infinite sequences of edges onto paths, but that is trivial since an > infinite sequence of edges is essentially what a path is. > > > > You had said that you had constructed a surjection. A surjection is a > > > mapping with specific properties. It is necessary to know what a mapping > > > is in order to construct a surjection, which you did not do. And I have > > > *no* idea what you mean with the second half of your sentence. > > > > If someone makes an asserted list of all real numbers and hands it out > > to Cantor, then Cantor can construct a diagonal number which is not in > > the list, contradicting the assertion. > > > > If someone makes a binary tree of all real numbers and hands it out to > > Cantor, then Cantor cannot construct a diagonal number which is not in > > the tree, so no contradiction of completeness is possible. > > > Actually anyone can. One does it by dealing with successive pairs of > binary digits to make the representation equivalent to a base 4 > representation, in which the Cantor diagonal method works quite nicely. Not in the binary (or any other) tree. There is any combination realized. So no real number (path) is mssing. > > > > The problem with Cantor's theorem aleph0 < 2^aleph0 is that the > > function f(x) = 2^x must be discontinuous: > > Since the domain and codomain of that function are neither of them > topological spaces, or even metric spaces, the notion of continuity is > impossible to consider. Not in the tree with countably many beginnings of separated parts of paths. > > > By means of the binary tree I can exclude any discontinuity. > > What is your topology or metric on the binary tree? > > Absent metrics, or at least a topologies, one cannot even speak of > continuity or discontinuity of a function. Look at the tree. Continuity is guaranteed by the edges between two nodes. Regards, WM
From: William Hughes on 14 Dec 2006 07:15
Han de Bruijn wrote: > William Hughes wrote: > > > Therefore, as there are at least as many digit positions in 0.111... > > as there are primes, there are an infinite number > > of digit positions in 0.111... > > Binary or decimal? > > Binary: 0.111... = lim(n->oo) 1/2 + 1/4 + .. + 1/2^(n-1) = > = lim(n->oo) (1 - 1/2^n)/(1 - 1/2) - 1 = 2/1 - 1 = 1 > > Decimal: 0.111... = lim(n->oo) 1/10 + 1/100 + .. + 1/10^(n-1) = > = lim(n->oo) (1 - 1/10^n)/(1 - 1/10) - 1 = 10/9 - 1 = 1/9 > > Please make your choice. But .. what is your problem, then? Piffle. This was a transparent deliberate misinterpretation. - William Hughes |