Cantor Confusion
in [Math]
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From: Virgil on 12 May 2007 14:46 In article <1178959303.279713.119490(a)h2g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. And always a node at which p* leaves p, provided p* != p. > Hence not every path can have exited at a node k(p, n) with a > finite index n. Each path other than p exits p at some node, but in an infinite tree in which each p has infinitely many nodes, there is no node of p by which all other paths have exited. > Why should it be impossible to construct a path from > these infinitely many nodes, i.e. from the pieces of paths p* =/= p? Because all paths share the root node.
From: Virgil on 12 May 2007 15:11 In article <1178979741.537850.290630(a)w5g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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: > > > 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? Do you thing that everything that can go on in any mind is limited to what goes on in your mind? > > > > > 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. Why bother to reconstruct what is already there? Every set of nodes in a CIBT which can represent a path already does so. > If this is > impossible, then there must be a node which does not belong to another > path. For each 'other' path there already is a node of p that does not belong to that other. But there is no single node of any path from which there is no branching, as such a node would have to be a leaf node, and CIBT's do not have any leaf nodes. > 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. Any set of paths which does not contain p does not contain p. So what? > > > > > > > > > > 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 We can sum up for each, but to do it for 'all' would require parallel processing on an infinite scale. > > 1+2+3+.... But the result is not an n(n+1)/2, but an aleph0, Where does 1+2+3+... appear in S = { sum{ x in N: x <= n}: n in N} ? I do not find it anywhere there. > > > > > > > > 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. Okay, math fantasy is MathReaism. > The idea to be able to determine *in principle* > arbitrarily many digits of an irrational number is fantasy. The idea that what may be deduced in an axiom system may not be deduced in that axiom system unless it fits WM's math fantasies is WM's math 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. No, it is required to prevent the diagonal from being equal to an alternate representation of any real. > Hence your arguing that only finite positions are relevant, is proven > invalid. Non Sequitur, as usual. If the diagonal decimal differs from any other decimal at any one position (place value) then it is not equal to that other decimal. So that one need only find one differing position in any comparison. WM again claims what he has not proved.
From: Virgil on 12 May 2007 15:18 In article <1178980444.675977.258960(a)e65g2000hsc.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 11 Mai, 21:52, Virgil <vir...(a)comcast.net> wrote: > > In article <1178870415.724641.38...(a)n59g2000hsh.googlegroups.com>, > > > > > H&J could have written: "it does not matter that the set of these > > > ordered pairs is described by a property" in your opinion? > > > > They could have, with no loss of correctness, said that. > > What set can exist without being described by a property? > > > > > In dealing with sets > > > > of ordered pairs in general, one need not describe it as anything but a > > > > set of ordered pairs. > > > > > Did I specify the rule, process, mapping > > > > You claimed the necessity of one without which no function can be > > defined. > > That is correct. > > > > > And without having a domain, the function is not defined at all. > > > > If one can deduce a domain, it is not necessary to specify one. > > I did not specify one. I said not more than: "you know that a function > consists of formula, domain and range, I hope." > > Then this little know-all replied: That you even think a set theoretic > function must have a FORMULA (!) as a component shows, again, your > lack of understanding even the basics of set theory. WM has been unable to find any set theory in which the formal definition of a function /requires/ a rule or formula, at least none published in the last 50 years or so, because mathematicians know better. > And now it is definitely enough about this silly topic. It is WM who has provided all the silliness. Mathematicians go by formal definitions. What is not specifically required by those formal definitions is not required at all. And "rules" or "formulae" are not required by any formal definition.
From: Virgil on 12 May 2007 15:22 In article <1178980528.276730.305790(a)e51g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 11 Mai, 21:59, Virgil <vir...(a)comcast.net> wrote: > > > Ad infinitum there is a path p' with p, because ad infinitum there is > > > no node where p is single. Therefore, ad infinitum, there is a path > > > which shares with p the node in question and all earlier nodes too. > > > > There is an ambiguity in your argument. > > > > While it is true that at every node there is more than one path through > > that node, it is also true that for any two paths there is a ladt node > > that they have in common. > > We have seen that these two "truths" are incompatible. Only in systems in which WM's unwritten laws are imposed over and above any explicit axioms. If one limits oneself to ZF or NBG, for example, excluding all of WM's assumtions beyond them, no such incompatibilities occur.
From: William Hughes on 12 May 2007 15:35
On May 12, 1:59 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: <snip> >My proof stands. You assert that p** is not equal to p. Note that for path p every node of p belongs to another path as well as to p. So saying that every node of p~ belongs to another path as well as to p, is not enough to show that p~ is not p. The only support you offer for you assertion is every node of p** belongs to another path as well as to p. A proof? You can't handle a proof! - William Hughes |