Cantor Confusion
in [Math]
|
From: Virgil on 11 May 2007 16:32 In article <1178876103.813398.237830(a)w5g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 11 Mai, 00:25, William Hughes <wpihug...(a)hotmail.com> wrote: > > On May 10, 6:32 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 9 Mai, 22:43, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > >...there must be such a path. > > > > >Why not call it p''. > > > > > > because the name p'' has already been used > > > > for a path that does branch off from p. > > > > > No, it has not. p'' is, by definition, that path which is with p also > > > at the next node, wherever you rest. > > > > Funny, I thought it was different. > > No. That was my original definition which you changed. But if you > claim to want to use p'' only according to your definition, then I'll > take p* as: > > Definition: p* is a path in the binary tree which is with path p at > the next node K(p, n+1) in the binary tree, whatever node K(p, n) you > investigate. Then if I choose to investigate I choose to investigate the successor node in p* to the one at which p* has separated from p? Since for every path p and every path p' in a CIBT, with p != p', there must be a node at which they separate. If p* is actually a path, then there must also be one for p*. > > Lemma: In the binary tree for every path p there exists at least one > path p* as defined. Not when I choose to investigate the successor node in p* to the one at which p* has separated from p.
From: Virgil on 11 May 2007 16:40 In article <1178878942.895369.182760(a)e65g2000hsc.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Mai, 22:30, Virgil <vir...(a)comcast.net> wrote: > > > > > > > In fact, to establish the difference between the diagonal and any > > > > > > one > > > > > > other only requires finitely many digits, which WM knows but > > > > > > chooses to > > > > > > ignore. > > > > > > > This is also true for the numbers 1.000... and 0.999... > > > > > > > Why must this case be excluded from the proof? > > > > > > Because they are both excluded from appearing after the point in the > > > > diagonal. The diagonal rule excludes both 0's and 9's after the decimal > > > > point, so automatically cannot be equal to any number whose decimal > > > > expansion can have them there. > > > > > I ask: Why must this case be excluded from the proof? > > > You answer: Because they are excluded. > > > > It was Cantor, or someone of his time, who created that rule, not me. > > No, it was not Cantor. It was someone who recognized that lim[n-->oo] > 10^-1 = 0 in case of dual representation, i.e., 1.000... = 0.999... As Cantor DID recognize dual representations, that is insufficient reason to be sure it was not Cantor. > > Unfortunately he did not recognize that he same holds for the > difference between the diagonal number and a list entry in case of > this limit. The reason is clear. Nobody ever considered the case that > a real number is changed in the limit lim[n-->oo] except the natural > "dual representation". Whatever does it mean to say that " a real number is changed in the limit"? Wm seems t be losing what little contact with actuallity he still has to propound such a footless argument. > And Cantor's proof deceives the reader by > stating that every digit has a finite index. In fact, that is true So WM accuses Cantor of confusing the reader by stating the truth. That is far, far better than WM's practice of trying to confuse the reader by a compilation of falsehoods and nonsense. nt to yield one and the same number. > > > > The various Cantor-type rules for proper decimals all say, among other > > things, that the diagonal shall be made up digits other than 0 or 1. > > > > This is specifically to avoid the problem of dual representations. > > > > Why is there a problem if only digits at finite places are changed? > > Why is there a problem with lim[n-->oo] 10^-1 = 0 in one case but not > in another one? If WM cannot see why on his own, then he has no grasp of the meaning of Cantor's proofs.
From: WM on 11 May 2007 16:41 On 11 Mai, 18:22, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 11, 11:46 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 11 Mai, 14:36, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > On May 11, 5:35 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > Definition: p* is a path in the binary tree which is with path p at > > > > the next node K(p, n+1) in the binary tree, whatever node K(p, n) you > > > > investigate. > > > > > Lemma: In the binary tree for every path p there exists at least one > > > > path p* as defined. > > > > > Proof: Choose a node K(p, n) of path p in the binary tree. There is a > > > > path p* which is with p at node K(p, n+1) too such that K(p, n+1) = > > > > K(p*, n+1). This path p* has been with p at all nodes m < n too. The > > > > choice of n is arbitrary. Therefore the Lemma holds for all nodes K(p, > > > > n) of p with n in N. > > > > > Remark: This proof is independent of the number of nodes. > > > > Look! Over there! A pink elephant! > > > You should consult an eye doctor. > > > > Remark: When you change n, p* never has to change. > > > It cannot change? > > Nope. By definition p* is a path. A path is > something that cannot change. > It need not change. Take just the path constructed in my last posting. My question concerned your idea always to use that path p* which takes the next exit (irony warning). Why did you make this poor choice? Regards, WM
From: Virgil on 11 May 2007 16:44 In article <1178898811.693238.9190(a)q75g2000hsh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > > At each node you get a path that is with p up to that node. > > Why should I use that path for any applications? > > > However, at no node do you get a path that remains with p forever. > > Then there must be a node wat which p is single. Such nodes are all leaf nodes. If WM finds leaf nodes in CIBT's, he should learn to be more careful with his medications.
From: MoeBlee on 11 May 2007 17:02
On May 11, 2:07 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 11 Mai, 00:22, Virgil <vir...(a)comcast.net> wrote: > If a set of ordered pairs is defined, there must be a formula or > whatever you may call it, defining it. Otherwise it is undefined. CORRECT. To define a PARTICULAR set of ordered pairs, we need a formula to do it. But to define the PREDICATES 'ordered pair', 'relation' and 'function' we only need the formula that is the definition of the PREDICATE and THOSE definitions do not themselves mention formulas, domains, or ranges. You are conflating defining a particular set with defining a predicate (actually a predicate symbol) ABOUT sets. I think you can finally see that difference now, right? Moreover, even as we use a formula to define a PARTICULAR set, (generally) the formula ITSELF is not a part of the set, not a component of the set, and the set itself does not consist of any formula used to define it. And you can just see that, if by nothing else, seeing that for any given set that is defined, there are MANY formulas that define that set. For example: Take the set p = {x | x is a natural number & x>0}. That set is defined by the formula: xep <-> x is a natural number & x>0} but it is also defined by: xep <-> (x is a natural number & x >= 1) and there are an INFINITE number of formulas that define p, and ANY set that is defined can be defined by INFINTELY many different formulas. So there is no formula that is THE formula is THE defining formula of p. So a set is NOT associated with a particular formula that we can say that set is that formula or whatever. Look at as about as simple a function we can define: {<0 0>} is a function. Now, what is THE definining formula of {<0 0>}? There is not one formula that is THE defining formula of {<0 0>}. For example be{<0 0>} <-> b = <0 0> is one defining formula fro b but be{<0 0>} <-> (b is an ordered pair & Ax(x is a coordinate of b -> x=0} is another defining formula for b. MOREOVER, notice that the formulas that define b are NOT MEMBERS of b nor are the domain or range of b members of b, and MOREOVER is is NOT the case that b = <f d r> where f is a defining formula for b, d is the domain of b, and r is the range of b. b is {<0 0>}. And it has many formulas that define it and its domain and range are not part of it. It's domain and range EXIST and the memberships of the domain and range of b are ENTAILED by what b is or how b was defined, but the domain and the range of b are not part of b ITSELF since the only thing that is a member of b ITSELF is <0 0>. You see it now, right? > Further the first elements and the second elements form sets. Without > such sets there are no triples aRb etc. Right. And that does not contradict that nieither a relation nor a function is itself a formula with a domain and range. > > Once defined, one can then define other related sets, such as a domain, > > codomain or range, but these are only consequences of the original > > definition, as are such properties as being injective or surjective. > > > I said that a function consists of formula, domain and range. Then how is 'consists of' defined in the LANGUAGE OF SET THEORY? > > There need not be a 'formula', in any formal sense, in evidence in order > > to have a function, but one does need a set of ordered pairs. > > And just this set is defined by the formula. See above in this post. > > > This is true and provable > > > It is only provable when it is assumed. > > It is provable that domain and range exist, if you have a binary > relation. Nothing else has to be assumed. Well, it's provable from certain axioms of set theory, which are what we assume. But it is correct that from those axioms we prove that every function has its domain and range. > See Exercise 2.1 of H&J: Let > R be a binary relation; show that dom R c U(UR), ran R c U(UR). > Conclude from this that dom R and ran R exist. We don't need to see that exercise. We did it a LONG TIME AGO and we remember it perfectly any time it is mentioned and we can explain that exercise in greater detail than you have ever thought about and we do dozens and dozens and dozens of exercises that are just like it in the sense of proving that certain sets exist having a certain property by finding a set from which to take a subset. MoeBlee |