Cantor Confusion
in [Math]
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From: mueckenh on 27 Feb 2007 16:55 On 26 Feb., 01:55, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1172392385.796841.117...(a)s48g2000cws.googlegroups.com> mueck...(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? Call them as you like, but these entities are not in trichotomy with really real numbers and not valid as elements of Cantor's list. Regards, WM
From: Virgil on 27 Feb 2007 17:06 In article <1172612682.096653.320720(a)t69g2000cwt.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 26 Feb., 01:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1172314973.530995.128...(a)p10g2000cwp.googlegroups.com> > > 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. But the complete infinite binary tree has demonstrably uncountably many paths, so that WM's pseudo-construction cannot produce that tree, nor all its paths. > > > > 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 the paths of any complete infinite binary tree cannot themselves be finite, as they cannot end. > But this union is considered an infinite tree. Not by any reasonable definitions of what a complete infinite binary tree must be. > And the unions of > subsets are considered infinite paths. Even though WM has declared them finite? > > > > > 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. Yes! > 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. Irrelevant! Your p(3) = 0.000 will be used in uncountably many paths in the complete infinite binary tree, despite not being used in any path using q(3) = 0.111, since there are uncountably many binary extensions of 0.000 to endless strings of binary digits. > > p(oo) = U{0., 0.0, 0.00, 0.000, ...} Which is only one of uncountably many such sequences in which each string, except the first, is a one binary digit extension of its predecessor. There are none so blind as those who will not see. And WM is among them,
From: William Hughes on 27 Feb 2007 17:13 On Feb 27, 4:46 pm, mueck...(a)rz.fh-augsburg.de wrote: > Let S differ by a finite element from all the finite sets A_1, A_2, > A_3, ..., A_n_1. <complicated proof of trival result snipped> No such element exists. This only shows that there is no single element, x, such that x is in S but x is not in in A_1, A_2, A_3 ... Let S be a set that does not have a Mueckenheim cardinality (e.g. the set of all natural numbers, or the set of all even numbers). Then S must differ from any A_i, because every A_i has a Mueckenheim cardinality. The conclusion is that S differs from every A_n, but the element by which it differs may depend on A_n. If A_1 is contained in A_2 is contained in A_3 ..., this can only work if there is no last A_i. - William Hughes
From: Virgil on 27 Feb 2007 17:25 In article <1172612805.121491.277060(a)q2g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Is WM conflating Aleph_0 with 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 (Extensionalit�tsaxiom) that a set which is > different from another set must prove this by at least one element. Does WM have any example of a finite set of naturals which does not differ from the set of all natural;s by at least one element? If so, he should by all means produce it, as without some such example, we will continue to hold that the set of all naturals is not the same set as any finite set of naturals. > > 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 That assumes, falsely, that one must be referring to the same element in every case. Given any finite set of naturals it differs from the set of all naturals by not containing the successor of one of its members, but the set of all naturals does contain the successor or every one of its members. However, the non-member referred to for a given finite set of naturals depends on which finite set of naturals one has chosen, so it not the same from one such set to all other such sets. WM would have it that all finite sets of naturals must have the same natural as non-member, which is obviously both false and foolish. , > 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. Not needed. What is needed, and is the case, is that for each finite set of naturals there is a natural not in that set, which establishes that no finite set of naturals can be the set of /all/ naturals. > 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. Often claimed, never proven, and in some set theories, like ZF and NBG, totally disproven. > > 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. Except that the set of all finite naturals does precisely what WM claims it cannot do, so that WM is WORNG! AGAIN! AS USUAL!
From: Virgil on 27 Feb 2007 17:29
In article <1172613067.239752.152360(a)z35g2000cwz.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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? 2^|omega| is certainly uncountable. From which one can show that for a complete infinite binary tree, however constructed, the 'number' of paths is uncountable. None of WM's legerdemain can hide that fact. |