Cantor Confusion
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From: Virgil on 14 Dec 2006 17:57 In article <1166090594.020341.42340(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > I do not understand why you argue that in > lim{x -> oo} lim{y -> oo} (2x + 3y)/xy = 0 = lim{y -> oo} lim{x -> oo} > (2x + 3y)/xy > interchanging limits is not possible. > > Regards, WM > Nor do I. But for lim{x -> oo} lim{y -> oo} (2x + 3y)/(x + y) = 3 and lim{y -> oo} lim{x -> oo} (2x + 3y)/(x + y) = 2, one cannot exchange the order of the limits without changing the value of the result.
From: Virgil on 14 Dec 2006 18:05 In article <1166090945.526859.60550(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > You misunderstood. What I explained to Dik is another mapping than that > > > discussed by us. It is a mapping of one single edge on the path > > > representing 1/3, the next edge is mapped on the path representing pi, > > > and so on. I claim that for every real number I can name an edge to be > > > mapped on this real number. > > > > Yes, you state that, and you can. But that does *not* provide a surjection > > from the edges to the real numbers. And you stated you had constructed > > such a mapping. > > There is a mapping. This is proven by the fact that there are more > edges than paths. As that "fact" has yet to be proved for infinite trees, you cannot use it to prove itself. > Perhaps we cannot display it in the special way you > wish. But that is the same with the well ordering of the eals. There > you believe in indirect evidence. Absent the assumption of the axiom of choice, or something roughly equivalent, no one claims any well ordering of the reals. Absent some assumption equivalent to assuming the there are more edges than paths in a infinite binary tree one cannot prove that there are more edges than paths. And given such an assumption, one can as easily prove 2 = 1. > > > > RIght. The disprove is in the fact that whenever I state a sequence of > > numbers that sequence inherently does not contain all real numbers. > > But the tree does. And the tree does not contain more paths than edges. That is only true if one assumes it true, it is not deducible from anything less that its own asssumption. > It is pitty that you cannot combine two thoughts. It is a pity you cannot combine one.
From: Virgil on 14 Dec 2006 18:18 In article <1166092128.058778.127950(a)80g2000cwy.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1166035958.305158.282870(a)79g2000cws.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > 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. > > > > Yes. But you stated that you had constructed a surjection. And if you can > > not state to what number a particular edge maps you have *not* constructed > > a sufjection. So you now state that you did *not* construct a surjection? > > I proved the possibility of a surjection. No such proof is possible without making an assumption contrary to any standard set theory, such as ZFC or NBG or NF. > > > But again your statement "it is only necessary that there is no path > > without > > edge mapped to it" is blatantly wrong. Consider the infinite set of > > decimal > > numbers. Map a digit (from 0 to 9) to a number when it has somewhere in > > its expansion that digit. If there are more numbers a digit maps to, use > > shares. By this reasoning (which is your reasoning) there is a surjection > > of the set of digits 0 to 9 to the decimal numbers. > > You have not understood, deplorably. The shares of every edge I use add > to 1 edge. And every share is mapped on one path only. And every path > gets as many shares to restate two full edges on its own. Every path requires an infinite sequence of edges in WM's construction, so it is not a mapping from individual edges but from infinite sequences of edges, which ruins WM's claims. > > No I simply state that the diagonal proof fails in case of the tree. AS the diagonal proof is not allied to trees, so what? > > > We can simply count the edges. They are countable. So we an count the > biginnings of separated parts of paths (because every edge is a > beginnng of the separated part of a path, notwithstanding which it may > be). Every finite beginning of a set of paths, however long, is the beginning of as many paths as the set of all paths. So there are as many paths following any such beginning as ther are paths altogether. > Therefore the beginnings of separated parts of the paths are > countable. It is the endings that are not countable. Cut off as large a finite leading segment of any path as you like , and the number of paths continuing from that point onward is the same as the set of ALL PATHS. > It seems there is a gap > > between 2^1 and 2^2. Why is that not a problem? > > There is no gap in looking at the beginnings of separated parts of > paths. If always counting from the left side, we come to the result 3 > edges WM chooses only to look where the things of interest do not occur. > > Look at the edges. Look at the number of paths sprouting out of any edge. > Do you deny that every edge is the beginning of that > part of a path where it is separated from other paths? I deny that it has any relevance to the "number" of paths. > > Do you believe that paths beginn to run separated wihout any edge being > inolved? I deny that it has any relevance to the "number" of paths.
From: Virgil on 14 Dec 2006 18:24 In article <1166092520.250897.80520(a)f1g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > And the domain is "every edge". Therefore, you claim to have > > constructed a function T such that for every edge e, T(e) is a path. > > And yet you seemed to deny this by saying: "That one which goes left > > from the root is not engaged". > > The domain in my original proof (the rational relation with 1 + 1/2 + > 1/4 + ...edges) is the set of all edges. It is the set of all infinite sequences of edges, which is quite a different matter. > But in fact not all edges are > required. We can waste as (finitely) many as we want to remain enough > to be mapped on every path. Since all of them together will not do it, how will a proper subset suffice? Not by taking individual edges, but only by taking one infinite sequence of edges for each path. > > The mapping paths to edges is an injection. Not in any infinite binary trees. WM is thinking only of finite trees in which every path has a terminal or leaf node.
From: Virgil on 14 Dec 2006 18:31
In article <1166092755.336596.309060(a)l12g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > WM deceives himself over the number of lines versus the number of > > elements in any one line. The number of lines is not finite but the > > number of elements in any one line is finite. > > Each line differs by 1 element from the preceding line. If the number > of lines is actually infinite then the number of differences must be > actually infinite too. So far so good. > Therefore the assumption of an actual infinity > of lines without an actual infinity of elements is a first grade > insanity. Which no one but WM makes. We assume an actual infinity of elements in N in accord with the axiom of infinity. > Here only potental infinity can be accepted. Only if one assumes that only potential infinity can be accepted. Absent such an assumption, one cannot conclude any such thing. > > > > > I think you should consider it as a fact, that an actually infinite set > > > of finite numbers is nonsense. > > > > While any physical infiniteness maybe impossible, until someone can show > > that ZFC or NBG is internally contradictory, actually infinite sets will > > continue to flourish within them. > > > If someone claims to be Napoleon I will not be able to cure him. But I > hope to help all those who are at the point but have not yet fallen > into this sickness. WM cannot prove that in mathematics no actually infinite sets can exist without making it an axiom of his system and assuming it. WM has no more right to impose his own personal set of axioms on everyone that anyone else has. |