Cantor Confusion
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From: mueckenh on 15 Dec 2006 07:00 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > William Hughes schrieb: > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > William Hughes schrieb: > > > > > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > William Hughes schrieb: > > > > > > > > > > > > > > > > > > > A_1 = {1} > > > > > > > > > > > A_2 = {1,2} > > > > > > > > > > > A_3 = {1,2,3} > > > > > > > > > > > > > > > > > > > > > > B = {1,2,3} > > > > > > > > > > > > > > > > > > > > > > then B is contained in the last A_i. If there is no last A_I, then > > > > > > > > > > > there is > > > > > > > > > > > no A_i that contains B > > > > > > > > > > > > > > > > > > > > That has nothing to do with "last". > > > > > > > > > > > > > > > > > > If A_i contains B, then A_i contains any A_j. > > > > > > > > > Therefore A_i is "last". > > > > > > > > > > > > > > > > > > > > > > > No comment? > > > > > > > > > > > > If B contains any A_i then B is the last. > > > > > > > > > > B cannot be the last A_i, as B is not > > > > > one of the A's (B corresponds to the diagnonal, > > > > > the A_i correspond to the lines) > > > > > > > > In which element does it differ from every A_i? > > > > > > In no element. > > > > > > We know that B is contained in all of the A_i > > > put together. > > > > How can you put all together, if there is no last one? You will never > > be sure that you have all. > > We know that any element that can be shown to be > in B can be shown to be in one of the A_j.. > > The question is: "Is there an A_D, such that > any element that can be shown to be > in B is in A_D?" > > The answer is: "There is an A_D, such that > any element that can be shown to be > in B is in A_D That is correct, because we are dealing with finite linear sets. > if and only if there is a last A_i." No. Correct is: If B is an actually infinite set of finite linear segments, then there must be at least one infinite finite linear segment, which is a contradiction. Therefore B is not actually infinite but only potentially infinite, i.e., B contains only a finite (though not bounded) number of elements. Regards, WM
From: mueckenh on 15 Dec 2006 07:05 cbrown(a)cbrownsystems.com schrieb: > > Every edge is divided into shares, namely in as many shares as there > > are paths to which this edge belongs. > > Therefore /you/ assert that an edge /can/ be subdivided into "as many" > parts as there are paths; which is to say, "as many" parts as there are > real numbers (by the bijection (2)). > > This implies that you assert that the set of all parts of an edge can > be bijected with the real numbers; so "x" is the cardinality of the set > of real numbers. It is the cardinality of the set of paths witch can be obtained by repeated multiplication by 2. > > Note that I have /defined/ what I mean by "uncountable" in this > context. Note that we are investigating how many paths exist in the tree. > > <snip> > > > No. According to ZFC the ordinal 2^omega is a countable set. Even > > omega^omega^omega is a countable set. > > If your assertion is "2^omega is countable and in bijection with the > reals", then we don't /need/ your "rational relation" argument to prove > that there is a bijection between naturals and reals. > > But I am only interested in your "rational relation" argument as a way > to /prove/ that there is a bijection between naturals and reals. I only stated that 2^omega is the ordinal number of a countable set. That is undisputed. Whether it is in bijection with the reals, is just under investigation. > > Remember that you have already /asserted/ that another property of "is > a part of" is that an edge /can/ be subdivided into as many parts as > there are real numbers. > > So, are you now asserting that it is possible to "make 1" by "adding > together" some "value" repeatedly "as many times" as there are real > numbers? What "value" would that be? Is it a real number? I will give you a simple example: The notion "rational relation" is a further development of the notion "relation". While a relation connects elements of a domain to elements of a range, the rational relation connects shares of the elements of the domain to (shares of the) elements of the range. A simple example: A) Consider the relation: 1 ---> a 2 ---> b B) If "1" is divided into two parts, 1#1 and 1#2, we can write (A) also in the form 1#1 ---> a 1#2 ---> a 2 ---> b C) Another rational relation could be: 1#1 ---> a 1#2 ---> b 2#1 ---> a 2#2 ---> b If 1#1 and 1#2 are the only shares 1, i.e., if 1#1 U 1#2 = 1 and if 2#1 and 2#2 are the only shares of 2, i.e., if 2#1 U 2#2 = 2, then one full element of the domain is available for every element of the range. Therefore (C) proves that a surjective mapping like (A) is possible, even if it cannot be found. > > I am only interested in your "rational relation" argument. Fine. Why are you interested only in that argument. And why are you interested in it at all? Regards, WM
From: Bob Kolker on 15 Dec 2006 07:05 Han de Bruijn wrote: > > > I've never seen such useless things as Robinson's infinitesimals. Robinson infinitesimals are exactly as useful as Newtonian and Leibnizian infinitesimals, plus they are well grounded. If Bishop Berkeley were alive today he could not fault the new infinitesimals as he did in his excellent book -The Analyst-. Berkely said: nfinitesimals are the ghosts of departed quantities, scoffingly. Robinson finally grounded an intuitively appealing and -useful- technique. One can now use inifinitesimals with a good conscience and not flog and mortify one's flesh with epsilon delta proofs. I have not yet seen a theorem that can be gotten only with infinitesimals and which is missed by the cauchy time limit proofs. But infinitesimals in many respects are quicker and slicker and we now -know- that they work. In a way it is like the Dirac delta function. As a mathematical artifact employed for the greater glory of quantum physics delta functions (which are not really functions at all) are slicker than Harvard Beets on white bread. It took a mathematician, Laurant Schwarts to show why they work. A delta function is really a distribution and is mathematically well founded. Bob Kolker
From: Bob Kolker on 15 Dec 2006 07:07 Han de Bruijn wrote: > > > Snipped that out of an earlier comment, but it's true. Physicists have > no trouble with Newton's fluxions. Neither do mathematicians. Newton understood what we call the derivative in a mostly physical way. Leibniz with his dy/dx notation was no better grounded than Newton, but dy/dx lends itself to the algebra of quotients. It took the work of cauchy and others to make dy/dx respectable. The notion of limit had to be precisely formulated and grounded. Bob Kolker
From: mueckenh on 15 Dec 2006 07:09
Jonathan Hoyle schrieb: > Bob Kolker wrote: > > Virgil wrote: > > > > > > And it is plain that no sound mathematics can be developed unless based > > > on some axiom system as its solid foundation. > > > > Arithmetic was around long before it was axiomatized and people were > > proving theorems about integers. For example Gauss and Euler. > > > > Bob Kolker > > True, but you are ignoring Virgi's adjective "sound". Calculus existed > way back during the time of Newton and Leibniz, but you could hardly > call their use of the infinite and infinitessimals at all "sound" by > today's standards Do you believe that today's standards are sound by tomorrow's standards? Regards, WM |