Cantor Confusion
in [Math]
|
From: Virgil on 3 May 2007 14:47 In article <1178191761.058003.238240(a)c35g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 1 Mai, 04:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1177941733.864803.29...(a)h2g2000hsg.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 30 Apr., 00:45, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1177674968.281638.41...(a)u32g2000prd.googlegroups.com> > > > > mueck...(a)rz.fh-augsburg.de writes: > > ... > > > > > If it were, than the number of digits were complete. Otherwise > > > > > there > > > > > can be no judgement concerning "all" but at most concerning "each". > > > > > > > > And in mathematics if something is proven for "each", it is proven for > > > > "all. > > > > > > That is wrong in many cases. That's probably why mathematics is wrong. > > > Example: Each natural number is a single number. All natural numbers > > > are not a single number. > > > > All natural numbers *are* a single number. You are confusing "all" with > > "the set of all". > > And are all men a single man too, in your opinion? Dik's English often reflects the idiom of his native tongue, Dutch, I believe, but that does not hide his idea that all natural numbers are individually single numbers, just as all men are individually single men. > > > > > > Meaning that if something is proven for "each" element of a set, it is > > > > proven for "all" elements of a set. And because the set of all > > > > natural > > > > numbers exists by the axiom of infinity, this also holds for natural > > > > numbers. If something is proven for each natural number, it is proven > > > > for all natural numbers (which does *not* mean that it is proven for > > > > the set of all natural numbers, because that is not a natural number). > > > > > > > > > > That means, all the nodes in the tree are passed by path bunches of > > > many paths. There is never a single path in the tree. > > > > Your conclusion is invalid. Indeed, each node is passed by a mutlitude > > of paths. But the path "0.10101010..." is nevertheless separated from > > all other paths. > > No. Only from those few you can ask for. It is different from all of them, except, of course, itself. The set of place values at which it has digit 1 is different from every other subset of N. > > For each two different paths it is possible to state a node where > > they do separate. For each path there is no node where it separates from > > all other paths. There is no contradiction. > > No. Only from those few you can ask for. Each path corresponds to a unique subset of N corresponding to the levels from which that path branches left. > > > But it means that p is never single. If it is never single, then it is > > > at least double, i.e., there is at least one path p' with it. > > > > As long as you remain in the finite part. > > You always remain there. In particular Cantor's diagonal proof always > remains there. Not so. Cantor's diagonal proof that there is no injection from the set of all binary strings to the set of naturals is deducible from the theorem that no power set injects into its base set. And that has nothing to do with finiteness or infiniteness.
From: Virgil on 3 May 2007 14:53 In article <1178196616.357780.9990(a)n76g2000hsh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 3 Mai, 13:18, William Hughes <wpihug...(a)hotmail.com> wrote: > > On May 3, 7:12 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > > > Wrong. If there is always a path p' with p, then for every node of p > > > we have a path p' such that K(p, n) = K(p', n). > > > > Look over there! A pink elephant! > > > > p' never changes. > > > > >Hence p is never > > > different from every other path. > > > > since there is a single fixed path, p', which is not > > the same as p. > > At every node K(p, n) of p there is a single fixede path p', which > has been with p for all nodes K(p, m < n). Yes, without pink > elephants, we can find at every node K(p, n) such a path p'. So p is > never single. If not being completely stubborn, we know or recognize > that. > > Regards, WM Paths which are not "single" do not exist at all, so that WM does not have any paths and so does not have any tree. In the CIBTs of ZF and NBG, there is a single path for every subset of N, the path which branches left at each level in that subset of N and branches right at each level to in that subset of N. That WM carefully ignores this clear model demonstrates that he cannot deal with something that so clearly and unambiguously refutes him.
From: WM on 5 May 2007 07:42 On 3 Mai, 16:05, William Hughes <wpihug...(a)hotmail.com> wrote: > > At every node K(p, n) of p there is a single fixede path p', which > > has been with p for all nodes K(p, m < n). Yes, without pink > > elephants, we can find at every node K(p, n) such a path p' > > Look over there! A pink elephant! > > p' never changes when n changes. > > >So p is never single. > > However, it is not necessary for p to be single for p to be separated > from > a path p'. It is possible for p to be separated from every p' without > p ever being single. That is obviously nonsense for linear sets which, after having been separated, never meet again. I wonder how one can post such a mess. Pink elephants are more realistic. No, p is never separated from those uncountably many paths which you cannot ask for (because you can only ask for countably many paths, in fact you can only ask for a separation at a finite index n, i.e., you can only ask for finitely many separated paths p'). Regards, WM
From: WM on 5 May 2007 07:54 On 3 Mai, 18:13, MoeBlee <jazzm...(a)hotmail.com> wrote: > On May 3, 3:55 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 30 Apr., 20:04, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > On Apr 30, 5:35 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > > On 27 Apr., 22:06, Virgil <vir...(a)comcast.net> wrote: > > > > Why then do you believe that Cantor's diagonal proof is true? > > > > You'd have to define 'true proof'. Meanwhile, with just some basic > > > knowledge of predicate calculus and set theory, one can see that > > > Cantor's diagonal argument is formalizable into a proof in first order > > > logic from the axioms of Z set theory. > > > There is an implicit assumption entering which is wrong. > > The difference of 1 between the digits of two numbers is insufficient > > to distinguish these numbers in the limit n --> oo. > > lim(n-->oo) 10^-n = 0. Only this limit makes power series of digits > > converge and makes real numbers exist. But Cantor's proof forgets > > that. > > There are no implicit assumptions in formal Z set theory. What you say proves that you never have thought about that topic deep enough. One implicit assumption is that, in Hessenberg's proof of Cantor's theorem, the set of all natural numbers which are not mapped on subsets of P(N) which do contain them, does exist together with the mapping pescription (you know that a function consists of formula, domain and range, I hope. Such a set need exists as little as the set which the largest natural number is mapped upon. > > > > If you understand the basic > > > material, there is not much to believing, since it is a simple of > > > matter of observing that a certain sequence of formulas that satisfies > > > the definition of 'first order proof from the Z axioms' does exist. > > > And as easily one can see that the difference between number of paths > > and number of nodes in the binary tree is given by > > 2-1-1+2-1-1+2-1-1+-... which is not much more than 2. If you > > understand the basic material, there is not much to believing. > > So it's still unclear whether you do understand that if you just > studied the basic material you too would observe that there is a proof You are wrong to assume that I did not study set theory. But the formalism veils the problem it is built upon. > in formal Z set theory, with no implicit assumptions, that the set of > real numers is uncountable. Even if this was clear, then the result 2-1-1+2-1-1+2-1-1+-... <= 2 would show that set theory is self contradictive. Regards, WM
From: WM on 5 May 2007 07:58
On 3 Mai, 19:57, Virgil <vir...(a)comcast.net> wrote: > > > What axiom system does WM claim allows him to prove otherwise? > > > 2-1-1+2-1-1+2-1-1+-... is never much more than 2. > > Irrelevant to the existence of bijections between the power set of N and > the set of all paths of a CIBT. Which bijections prove WM wrong. They prove 2-1-1+2-1-1+2-1-1+-... =< 2 wrong, i.e., they prove 2-1-1 =/= 0. Regards, WM |