Cantor Confusion
in [Math]
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From: Dik T. Winter on 25 Feb 2007 19:52 In article <1172317163.668512.112370(a)k78g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 24 Feb., 03:27, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1172253288.305584.116...(a)j27g2000cwj.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 23 Feb., 14:35, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > Eh? Above you state (properly) that p(oo) is a subset of U(T(N)). > > > > > > I said: "as you say". But how is it possible that a finite tree > > > contains an infinite path? Every tree T(n) contains only finite paths, > > > namely such with n nodes. There is no tree containing an infinite > > > path. > > > > Oh. But the union of all finite trees contains infinite paths. Or are you > > now of the opinion that 1/3 is *not* a path in your infinite tree? > > Why do you say "now". I told you more than once that there is no > infinity. If you believe that infinite paths exist, you must say where > and how. You said, that the infinite path p(oo) is in the union of all > corresponding finite paths p(oo) = U(p(n)). Now we are going to > investigate this case and its consequences. Yes, and you do wrongly. > > Ignoring infinite sequences. > > > > If you want to prove inconsistency of ZF with the axiom of infinity you > > should allow infinite subsets in your proofs. Disallowing them does not > > prove inconsistency of ZF with the axiom of infinity, only inconsistency > > with the way *you* think things should be. Straight out: denial of the > > axiom of infinity. > > I use the infinite set of finite trees T(n). But the paths of these > finite trees are all finite. I can't change it. Indeed, the paths in each individual T(n) are all finite, but the paths in U{T(n)} are all infinite. -- 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 19:55 In article <1172392385.796841.117660(a)s48g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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. By what rule is it forbidden to call them numbers? And so the claim "number field sieve" is wrong? By what rule? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 27 Feb 2007 16:44 On 26 Feb., 01:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1172314973.530995.128...(a)p10g2000cwp.googlegroups.com> mueck...(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. It is the union of the paths of this subset - one of countably many such subsets. > > > 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. Of course there are only finite paths in the union of finite trees. But this union is considered an infinite tree. And the unions of subsets are considered infinite paths. > > > 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. No. The union of the subset {p(3) = 0.000, q(3) = 0.111)} for instance is not at all used because neither the path 0.111... nor the path 0.000... nor any other path contains both of them as subpaths. p(oo) = U{0., 0.0, 0.00, 0.000, ...} Regards, WM
From: mueckenh on 27 Feb 2007 16:46 On 26 Feb., 01:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1172313814.826402.273...(a)v33g2000cwv.googlegroups.com> mueck....(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? You know that every natural number like 2n is finite and therefore less than aleph_0 while |{2,4,5,...,2n}| in the limit is aleph_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. And there is the saying (Extensionalitätsaxiom) that a set which is different from another set must prove this by at least one element. > > > > 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. Let S differ by a finite element from all the finite sets A_1, A_2, A_3, ..., A_n_1. T there is a set A_n, which contains this element, because every finite element is contained in some finite set A_n. So there is A_n and infinitely many sets A_n+1, A_n+2, .... which are not different from S by this element. Further it is clear that a finite set cannot have an element which is not contained in S. Therefore S cannot be distinguished by any finite element n from all the finite sets. If S exists and is different from any finite set, it must cotain an infinite element w as this is not contained in any finite set. Of course, for every set A_n there exists a set which differes from A_n by an element n+1. But the above proof shows that there is no set S (or omega) of finite elements which differs from every finite set A_n. > > > > > 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? To determine the meaning here. Regards, WM
From: mueckenh on 27 Feb 2007 16:51
On 26 Feb., 01:52, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1172317163.668512.112...(a)k78g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 24 Feb., 03:27, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1172253288.305584.116...(a)j27g2000cwj.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > On 23 Feb., 14:35, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > > Eh? Above you state (properly) that p(oo) is a subset of U(T(N)). > > > > > > > > I said: "as you say". But how is it possible that a finite tree > > > > contains an infinite path? Every tree T(n) contains only finite paths, > > > > namely such with n nodes. There is no tree containing an infinite > > > > path. > > > > > > Oh. But the union of all finite trees contains infinite paths. Or are you > > > now of the opinion that 1/3 is *not* a path in your infinite tree? > > > > Why do you say "now". I told you more than once that there is no > > infinity. If you believe that infinite paths exist, you must say where > > and how. You said, that the infinite path p(oo) is in the union of all > > corresponding finite paths p(oo) = U(p(n)). Now we are going to > > investigate this case and its consequences. > > Yes, and you do wrongly. > > > > Ignoring infinite sequences. > > > > > > If you want to prove inconsistency of ZF with the axiom of infinity you > > > should allow infinite subsets in your proofs. Disallowing them does not > > > prove inconsistency of ZF with the axiom of infinity, only inconsistency > > > with the way *you* think things should be. Straight out: denial of the > > > axiom of infinity > > > > I use the infinite set of finite trees T(n). But the paths of these > > finite trees are all finite. I can't change it. > > Indeed, the paths in each individual T(n) are all finite, but the paths > in U{T(n)} are all infinite. The paths are all finite. The unions of paths are infinite. But the cardinality of the set of these unions is countable. Or would you say that 2^omega is uncountable? Regards, WM |