Cantor Confusion
in [Math]
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From: Dik T. Winter on 23 Feb 2007 21:16 In article <1172252145.066292.211790(a)p10g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 23 Feb., 05:42, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > > So "nine apples" divided by "three apples" is "three I"? A strange > > > > formulation. What is "nine oranges" divided by "three apples"? > > > > > > In mathematics we usually do not use apples and oranges. > > > In physics, the result is 3 orange/apple. > > > > You are starting to see the light? Mathematics is not physic. > > Therefore in mathematics the unit "of something" is sufficient. Well, but that is completely new terminology. And if you use it in mathematics, you should define it. > > It is not even a subset of it. > > Really not??? No. > > What is the physical relevance of the factorisation > > of large numbers? There is none. Nevertheless that is a quite important > > field in mathematical research. > > And could not be done without physical means. No idea (of numbers, > factors, relations or else) would have appeared without the > possibility of physically visualizing its basics. I do not question the physical implementation, I ask about the physical relevance. But anyhow, in factorisation, how do I physically visualise irrational algebraic numbers? 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? -- 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 23 Feb 2007 21:27 In article <1172253288.305584.116320(a)j27g2000cwj.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 23 Feb., 14:35, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > Every element of P(oo) is a path and as such a subset of U(T(n)), as > > > you say. > > > U(T(N)) has only finite paths as subsets (as N has only finite initial > > > segments --- both U(T(n)) and N have infinitely many such subsets, > > > but these subsets are finite). > > > > 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? > > Are > > you contradicting that now? And if we consider: N subset N, N contains > > one infinite initial subset. Or do you mean something different with > > U(T(N)) from U(T(n))? If so, what do you mean with it? > > I mean that U(T(n)) is the union of all finite trees with n levels and > hence with paths with n nodes. I can't believe that there is an > infinite path in this union. But when I argued before that 1/3 was not in your tree, you contradicted me. (I thought so because there was a lack of a proper definition.) But now you apparently have switched to a completely different view. > > > There are countably many unions of finite subsets of U(T(n)). > > > > Prove it. There are countably many *finite* unions of finite subsets, > > not countably many arbitrary unions of finite subsets. Consider N. > > Each subset of N is a union of finite subsets of N, namely the union > > of all singleton sets that contain an element of that set. > > > > > There are countably many unions like U(p(n)). > > > > Prove it. > > Every set of finite subsets of N is countable. The set of all finite > subsets of N is countable. Ignoring infinite subsets. > > > > > There are countably many unions like U(p(n)). > > > > Prove it. > > If A is countable, then the set of all finite sequences of elements of > A is countable. 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. -- 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 24 Feb 2007 05:43 On 24 Feb., 03:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1172251940.593004.240...(a)h3g2000cwc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 23 Feb., 05:31, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > On 21 Feb., 16:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > ... > > > > > > like infinity. But it is impossible that both follows, infinity > > > > > > and naturality. The definition a = z for an infinite chain of > > > > > > inequalities a < b < c < ... < z is a *wrong definition*. > > > > > > > > > > What are you *talking* about? > > > > > > > > Simply about 2n - n < 2(n+1) - (n+1). > > > > > > I do still not understand. What are you *talking* about? > > > > The difference 2n - |{2,4,6,...,2n}| steadily inceases with incresing > > n. Only after infinitely many increasings it drops to -aleph_0. > > 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. > > > > 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? > 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. > > > > > Wrong, there is no induction involved. > > > > > > > > Either induction or belief. > > > > > > I think you are on "induction and belief". What induction is there in > > > f(n) = 2 * n? Moreso, what induction is there in (when considering the > > > reals) f(x) = x^3? > > > > 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. Regards, WM
From: mueckenh on 24 Feb 2007 06:02 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. 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. Some of these subsets are used to form the paths p(oo) by unions p(oo) = U(p(n)). Regards, WM
From: mueckenh on 24 Feb 2007 06:04
> 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. Regards, WM |