Cantor Confusion
in [Math]
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From: MoeBlee on 11 May 2007 17:05 On May 11, 3:22 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > dual representation. I explained to you already why dual representation is not at issue. And if you understood such presentations as in ordinary set theory textbooks such as, e.g., Suppes's, then you'd understand exactly why your objection regarding dual representation is a red herring. MoeBlee
From: William Hughes on 11 May 2007 18:20 On May 11, 4:41 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > 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. This path does not exists. As usual, your construction made use of an unjustified assumption that a set of paths could be chosen in such a way that *all* paths were the same. > My question concerned your idea always to use that path p* which takes > the next exit In a counterexample I gave I used a path p* that takes the next exit. However in general I do not claim that this must be true. I do claim that you always have to choose a path with will eventually exit. >(irony warning). Why did you make this poor choice? Because you need a path that is not p, and any path that is not p will eventually exit. (given that the path will eventually exit, it makes little difference where, so in my counterexample I took a path which takes the next exit). - William Hughes
From: Virgil on 11 May 2007 19:30 In article <1178916088.249412.260910(a)u30g2000hsc.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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? > I choose to "investigate" the second node of p* that is not a node of p. So the poor choice is in WM's definition of p*.
From: WM on 12 May 2007 04:41 On 12 Mai, 00:20, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 11, 4:41 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > It need not change. Take just the path constructed in my last posting. > > This path does not exists. As usual, your construction made use > of an unjustified assumption that a set of paths could be chosen > in such a way that *all* paths were the same. It can. See below. Or say why it cannot, according to your opinion. > > > My question concerned your idea always to use that path p* which takes > > the next exit > > In a counterexample I gave I used a path p* that takes the next exit. > However > in general I do not claim that this must be true. I do claim that you > always > have to choose a path with will eventually exit. > > >(irony warning). Why did you make this poor choice? > > Because you need a path that is not p, and any path that is not p will > eventually exit. That is contradicted by the fact that there is always another path p* with p. Hence not every path can have exited at a node k(p, n) with a finite index n. Why should it be impossible to construct a path from these infinitely many nodes, i.e. from the pieces of paths p* =/= p? This path p** is not p, because every piece belongs to a foreign path. This path p** is a path of the tree, because every infinite connected chain of nodes is a path. The connection, however, is guaranteed by the fact that the nodes belong to p too. Regards, WM
From: WM on 12 May 2007 10:22
On 11 Mai, 21:45, Virgil <vir...(a)comcast.net> wrote: > In article <1178866486.658963.253...(a)y5g2000hsa.googlegroups.com>, > > > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 10 Mai, 21:21, Virgil <vir...(a)comcast.net> wrote: > > > In article <1178795907.957410.94...(a)o5g2000hsb.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 10 Mai, 03:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > So in your opinion there are only finitely many paths in the > > > > > > > infinite > > > > > > > tree. > > > > > > > I did not say that. > > > > > > Quote: "Only from those few you can ask for". I can ask only for > > > > > finitely > > > > > many paths. > > > > > That is correct. But that does not mean that I said that there are > > > > only finitely many paths in the tree, as you implied. > > > > Since your other comments imply that what one cannot ask for does not > > > exist, it is reasonable to conclude you meant only finitely many. > > > My personal opinion is not the starting point here. Since mathematical > > entities exist only in the mind, such entities, which cannot exist in > > the mind, cannot exist at all. > > Since your own mind is the only one for which you can speak, what goes > on it other minds is beyond your competence to decide. Do you think mathematical entities exist other than in minds? > > > > > > > > > > > > Not at all. I ask the single question: "are all other paths different > > > > > from > > > > > a particular path?" > > > > > That is circular, because by "other" you imply that path p can be > > > > distinguished by a node from all the others. > > > > Not so. It only implies that any two distinct paths can be distinguished > > > from each other at some node. > > > > And they can, for if they are not the same it is only because at some > > > node they branch differently. > > > Then all different paths should be distinguishable. This is > > contradicted by the fact that every node of p which you can ask for is > > shared by the same path p'. > > Only when p = p'. No. Every node of path p in the tree is shared by some other path. Why not use all these nodes and construct some other paths. If this is impossible, then there must be a node which does not belong to another path. Otherwise it would be possible. After having constructed a set of paths which does not contain p, but only paths of a set P*, we can ask, how many are necessarily needed. My opinion is that only one is needed. If you do not share this opinion, let me know why at least two are needed. > > > > > > > It is valid for those cases I did prove. I did not prove it (and one > > > > cannot prove it) for "all natural numbers which obey this formula", > > > > because all natural numbers which obey this formula do not obey this > > > > formula. > > I fail to see any possible justification for "because all natural > numbers which obey this formula do not obey this formula" If it holds for all natural numbers which we can sum up, then we can sum up them all 1+2+3+.... But the result is not an n(n+1)/2, but an aleph0, implying that at the left-hand side there exists an infinite number. > > > > > > So that. Apparently, one of WM's axioms is that finite induction does > > > not hold. > > > In mathematics it holds for the potentially infinite set of natural > > numbers > > MatheRealism shows that it holds only for few natural numbers. > > MathRealism is math fantasy. Just the opposite. The idea to be able to determine *in principle* arbitrarily many digits of an irrational number is fantasy. > > > > > > > My argument is: Why do you have to exclude the case 0.999... = > > > > 1.000... if you remain always in the finite? For every finite index > > > > both numbers are different. > > > > We, and Cantor, exclude them by having a diagonal rule that never uses > > > either 0 or 9. > > > Why is that rule required? > > To exclude the diagonal from having any alternate representation. No, it is required to exclude the diagonal from being identical to an entry. Hence your arguing that only finite positions are relevant, is proven invalid. Regards, WM |