Cantor Confusion
in [Math]
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From: mueckenh on 25 Feb 2007 03:33 On 25 Feb., 05:26, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1172315095.706877.313...(a)8g2000cwh.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > When I last looked at it in the Number > > > Field Sieve for factorisation of numbers a big problem was the > > > construction of the square root of an irrational algebraic number. > > > What is the relation with physics? > > > > Irrational numbers do not exist, neither in physics nor elsewhere. > > So the Number Field Sieve does not exist? Strange that is has been > able to factorise numbers. Sorry, I was too apodictic. Irrational numbers do not exist *as numbers* in physics or elsewhere. They exist as ideas. Obviously. Regards, WM
From: William Hughes on 25 Feb 2007 09:42 On Feb 25, 3:22 am, mueck...(a)rz.fh-augsburg.de wrote: > On 25 Feb., 03:20, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Feb 24, 12:28 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > > On 24 Feb., 00:06, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > On Feb 23, 5:49 pm, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > > On Feb 23, 12:26 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > > > > >Otherwise you should explain how the > > > > > > set with n numbers > > > > > > {2,4,6,...,2n} can be filled up by other (namely greater) even > > > > > > numbers , without increasing the sizes of numbers. It is similar to > > > > > > create uncountably many paths in the binary tree without the presence > > > > > > of uncountably many nodes or edges. > > > > > > > Both tasks can only be managed by pure belief in these results and > > > > > > refraining from any attempt to understand it. > > > > > > As nobody has made either claim, this is not a problem. > > > > > You should start reacting to what people claim, not > > > > > what you think follows from the claim. > > > > > Sorry, I misread the second claim. This claim has in fact been made. > > > > (no problem, paths are sets of nodes and edges. Saying > > > > that there are only countably many nodes and edges is > > > > statement i, saying that a path is composed of nodes > > > > and edges is statement ii. Saying that there are a countable > > > > number of paths is statement iii. Statement iii does not follow > > > > from statements i and ii. Statments i and ii are true. Statement iii > > > > is false.) > > > > You misunderstand. Statement iii is not concluded from the other > > > statements. > > > Reprhasing things in terms of unions of finite paths, rather > > than subsets of nodes, changes nothing. > > You are in error. I did not rephrase some claim about the set of > natural numbers, but here I showed that he set of all real numbers is > countable. > > > > > > > > Saying that there are only finitely many paths p(n) in the > > > finite tree T(n) is obviously correct. There are 2^n paths. These > > > paths are finite sequences. > > > > The countable union of all finite trees U(T(n)) is a tree which > > > contains (as subsets) only finite paths > > > Yes. > > > The paths in the infinite tree are the infinite > > paths which are the unions of finite paths. So we have > > > Statement i > > > There are a countable number of finite paths. > > > Statement ii > > > Each infinite path is made up of finite paths. > > You forget: Each infinite path is the union of finite paths. My statement was "The paths in the infinite tree are the infinite paths which are the unions of finite paths" so I did not forget this part. > And all > these unions of finite paths belong to a countable set of unions. And I did not forget this part because it is not true. The set of finite paths is countable. The set of unions of finite paths is not countable. Let P be the set of finite paths. This set is countable. The unions of finite paths belong to the power set of P. There are more elements in the power set of P than in P. Knowning that a set is a subset of the power set of P does not tell you whether or not this set is countable. > > > > > Statement iii > > > There are a countable number of infinite paths. > > > > as the set of all natural > > > numbers contains only finite numbers. Don't you agree? > > > Yes. But I don't agree that you can use this to > > show that the set of all natural numbers has a > > cardinality. > > That is a gross misunderstanding by you. I do not intend to show that > the set of natural numbers has a cardinality. And at this point one might expect a statement of the form "What I intend to show is ..." Well? - William Hughes
From: Virgil on 25 Feb 2007 15:43 In article <1172391737.900395.308180(a)8g2000cwh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 25 Feb., 03:20, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > On Feb 24, 12:28 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > > Reprhasing things in terms of unions of finite paths, rather > > than subsets of nodes, changes nothing. > > You are in error. I did not rephrase some claim about the set of > natural numbers, but here I showed that he set of all real numbers is > countable. In a pig's eye! WM has "shown" nothing. None of what WM has presented as proofs qualify as proofs outside his own mind. > You forget: Each infinite path is the union of finite paths. And all > these unions of finite paths belong to a countable set of unions. Then what you have is not the complete infinite binary tree, as that tree demonstrably has uncountably many paths.
From: Dik T. Winter on 25 Feb 2007 19:44 In article <1172313814.826402.273830(a)v33g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 24 Feb., 03:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > Eh? I would think that 2n - |{2,4,6,...,2n}| equals n, so it increases > > without bounds. Why do you think it drops suddenly drops? > > I do not think that. Those who say that aleph_0 exists being larger > than any natural number claim it. Where? The only reasonably think to do is: lim{n -> oo} (2n - |{2,4,5,...,2n}|) = limt{n -> oo} (2n - n) = = lim{n -> oo} n. So how do you come at the idea that it drops to 0? > > > > The (only) infinite initial segment differs from each finite initial > > > > segment by having an element that is not in the finite initial > > > > segment. > > > > > > How does it differ from all finite segments? Or is here an occasion > > > where the simultaneous consideration of all (segments of) natural > > > numbers is inappropriate (quite contrary to Cantor's diagonal proof)? > > > > There is no single number where it differs from *all* finite segments. > > So there is no set theoretic indication that it exists as a set other > than a finite set? There is, the axiom of infinity. > > But you can do your comparison in parallel, but the comparisons do *not* > > give a single number. Why you think that should be the case escapes me. > > A set S which differs from all sets A, B, C, ... of a set of sets, > either can do so by by differing from set A by at least one element a > and from set B by at least one element b and so on. Here set A may > contain b and set B may contain a. But in a linear order like A c B c > C c ... this is not possible. Here S must differ by an element from > all sets or it does not differ from all sets. Why? Where is your proof? And give a set theoretic proof, please. > > > You need to find the position of such numbers as [pi*10^10^100] in the > > > sequence of natural numbers, but without always having such an easily > > > computable short hand as [pi*10^10^100]. There are at least > > > [pi*10^10^90] natural numbers which cannot be determined other than by > > > counting along the sequence of natural numbers. Therefore inductive > > > counting is required. For reals the same is valid, further you have to > > > count the digit positions behind the point. > > > > This makes not much sense mathematically. At least I can not determine > > any sensible meaning here. > > You should turn your interest more to finite numbers than to infinte > sets. Why? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 25 Feb 2007 19:49
In article <1172314973.530995.128190(a)p10g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 24 Feb., 03:08, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1172252004.929693.56...(a)j27g2000cwj.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 23 Feb., 05:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1172134355.420444.86...(a)k78g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > No? P(oo) = {p(oo), q(oo), r(oo), ...} > > > > > > > > > > p(oo) = Up(n), q(oo) = Uq(m), r(oo) = Ur(i) > > > > > > > > > > How many unions Up(n), Uq(m), Ur(i), ... can be in a union > > > > > of finite trees? > > > > > > > > Uncountably many, > > > > > > How can the set of finite subsets of a countable set be uncountable? > > > > In what way is p(oo) a finite subset? > > Not p(oo) but the elements p(n) of which it is made are finite > subsets. Therefore all combinations form a countable set. And so p(oo) is itself *not* in that countable set of finite subsets of a countable set. Neither as element, nor as subset. > P(oo) is the set of the infinite paths p(oo), q(oo), r(oo). Every such > infinite path is the union of finite paths. The finite paths are > finite sequences. There exist countably many finite paths in the > complete tree U(T(n)). All possible subsets of this countable set of > finite sequences form a countable set. No. All possible *finite* subets of this countable set of finite sequences form a countable subset. You forget a crucial word here. > Some of these subsets are used > to form the paths p(oo) by unions p(oo) = U(p(n)). All of them are used, and all of them are used infinitely many times. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |