Cantor Confusion
in [Math]
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From: Virgil on 28 Feb 2007 03:52 In article <1172651060.321953.253320(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 28 Feb., 02:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1172612682.096653.320...(a)t69g2000cwt.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 26 Feb., 01:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > > > > 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. > > > > Paths of a subset? What does *that* mean? You state: > > (a) The set of finite subsets of a countable set is countable. > > I agree. > > (b) p(oo) is not one of those finite subsets. > > I agree. Now you state what exactly? > > > As all the finite subsets like {p(0), p(1), p(2), ...} which belong to > an infinite path like p(oo) form a countable set, the set of all > unions is counable. This only argues that /one/ path is countable, which no one disputes, not that the set of all paths is countable. This set of all unions is P(oo) = {p(oo), q(oo), > ...}. Wrong! WM indicates it as a list, but it cannot be listed. Does WM claim to be able to list all endless binary sequences? It is trivial that there are as many paths as such sequences, so until WM proves the sequences listable, he cannot claim the paths to be listable. But then, WM is quite incapable of proving anything to mathematical standards, at least on the evidence of his futile attempts to do so. > > I can not follow you anymore. A tree is a set of nodes (with special > > properties), a path is a set of nodes (with special properties). The > > union of trees is an infinite tree, and it contains infinite paths. > > Why do you now state that it does *not* contain infinite paths? > > It is easy to undertand. We are investigating this assertion. > We find that every finite tree contaibns only finite paths. > The union of some finite paths is an ifinite path. > Thee are only countable many unions of finite paths which > yieldinfinite paths. As there is one path for each of the uncountably many endless binary seqeunces in such an infinite tree, WM is wrong. > > > > > p(oo) = U{0., 0.0, 0.00, 0.000, ...} > > > > a(oo) = U{0., 0.1, 0.10, 0.100, ...} > > b(oo) = U{0., 0.1, 0.11, 0.110, ...} > > c(oo) = U{0., 0.1, 0.11, 0.111, ...} > > etc. How many times is the path "0." used? > > As the unions of finite subsets of paths like {0., 0.1, 0.10, > 0.100, ...} form a countable set, the path 0. is used countably often. Once in each of uncountably many paths. The only reason that WM objects to infiniteness is that his mental powers are just not up to dealing with it successfully. WM cannot reason it out, so he has to guess, and he guesses wrong.
From: Virgil on 28 Feb 2007 04:31 In article <1172651370.180739.279270(a)8g2000cwh.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 28 Feb., 02:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1172612805.121491.277...(a)q2g2000cwa.googlegroups.com> > > mueck...(a)rz.fh-augsburg.de writes: > > > > > On 26 Feb., 01:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > 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. > > > > And 2n in the limit is aleph 0, when you define such limits. > > That is what I say, but according to set theory it is wrong. According to WM's view of set theory that is wrong, but WM knows nothing of set theory. > N is the > infinite set of all *finite* numbers. Not at all, there are all those finite rationals and negatives around that are not in N. N is the infinite set of all *natural* numbers, all of which are finite. > Accordinfg to set theory there > is no infinite number in it although the limit aleph 0 of numbers is > defined by set theory. According to set theory there is no infinite NATURAL number, but as aleph_0 is not a natural, it is quite acceptable that it be infinite. WM is getting desperate if he has to conflate naturals, which are all finite, with the more general cardinals or ordinals which are not so limited. > > > So in the > > same way I could state: > > You know that |{2,4,5,...,2n}| is finite and therefore less than > > aleph 0, while 2n in the limit is aleph 0. > > You are actually "proving" here that lim{n -> oo} n/n = 0. Bizarre. > > Yes, it is a mess. But it is exactly what everybody asserts when being > talking about the infinite set of finite numbers. No, it is only the mess WM tries, and fails, to create. > > > > Yes? Where is that in contradiction with what I did state? Give me the > > set of naturals and any finite set of naturals and I will show you an > > element where they differ. > > Give me Cantors diagonal constructed up to any digit d nn and I will > show you the place where d nn can be fuond in the list. You must first give us the list of reals on which the Cantor rule is to be applied. The Cantor proof is in the form of a claim: Cantor says, give me a list of reals and I will give you an real not listed in that list. So that WM, as challenger of that claim, must first produce a list. Only then need Cantor do anything. > > In case of Cantor's argument you apply the principle: "Compare with > all numers simultaneously." In case of N differing from its final > segments you allow only comparison one after the other. That is bad > logic. It is better logic that WM has yet exhibited. Cantor merely says that when you give him a list he can apply a rule which will produce a number not listed. Since his rule produces a number different from any in the list, his claim succeeds. The time it might take to put his rule into effect is immaterial. > > > If S exists and is different from any finite set, it must contain > > > an infinite element w as this is not contained in any finite set. > > > > Wrong. Show a proof of this. > > You say it above. No one but WM and his fellow kooks says anything so stupid. > Why do you use such ridiculous examples to demonstrate your "non- > quantifier dyslexia"? It is WM's actual quantifier dyslexia at issue here. Given N as the set of all naturals and S_n, for n in N, as the set of all naturals less than n, It is true that: For each S_n, there is an x in N such that x is not a member of S_n. It is false that: There is an x in N such that for each S_n, x is not a member of S_n. WM habitually conflates these two statements, which is the logical fault called quantifier dyslexia. > > The initial segments of N are not like your A1, A2, ... as you know > well. Only wrong analogies like that above can lead to the opinion > that " forall A thereis x such that: x notin A and x in S " could be > possible for linear segments. Except that it is necessarily the case in ZF and NBG and the wrongness exists only in WM's head, due to his quantifier dyslexia. For each and every finite initial segment of naturals, there is a natural in the set of all naturals not in that segment. This in no way requires that it be a same one for all segments. The construction: "for all x there is a y" should always read "for all x there is a y whose value in general depends on x". WM wants it to mean "for all x there is a y whose value is independent of x", but it rarely, if ever, does. At least in mathematics. Consider, for example , the delta-epsilon definition of a limit: For every positive epsilon there is a positive delta (depending on epsilon),... If one were to allow WM's meaning, calculus would still be mired in fluxions and fluents. For someone purporting to have produced a text on mathematics, WM's ignorance of his subject is appalling.
From: William Hughes on 28 Feb 2007 07:59 On Feb 28, 3:29 am, mueck...(a)rz.fh-augsburg.de wrote: > On 28 Feb., 02:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1172612805.121491.277...(a)q2g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > > > 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. > > > Right. > > > > If S exists and is different from any finite set, it must contain > > > an infinite element w as this is not contained in any finite set. > > > Wrong. Show a proof of this. > > You say it above. > > > > > > > > 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. > > > Not shown. It only shows that there is no set S that differs from all > > finite sets by a single element. But that is not the requirement. The > > requirement is that for each finite set there is an element where it > > differs. This is you perennial quantifier dislexia. Required: > > forall A thereis x such that: x notin A and x in S > > what you state is: > > thereis x such that forall A: x notin A and x in S > > the two are *different*. Consider: > > A1 = {1, 2, 3, 4} > > A2 = {1, 2, 3, 5} > > A3 = {1, 2, 4, 5} > > A4 = {1, 3, 4, 5} > > A5 = {2, 3, 4, 5} > > and > > S = {1, 2, 3, 4, 5} > > S is different from all of A1 to A5, but there is not a single element where > > it differs from all of A1 to A5. > > Why do you use such ridiculous examples to demonstrate your "non- > quantifier dyslexia"? Because you have a two step approach to showing that S cannot differ from every A_i I: there is no single x such that S differes from every A_i by x II: It is sufficient to show that there is not single x such that S differs from every A_i by x I is trivial, but you spend all your energy on I. You only mention II when someone points out that I is not in general sufficient. And asked for a proof of II you gave a proof of I. Here is a counterexample using sets you have described. Let E be the set of all even natural numbers. Let the A_i be the initial segments of E, (i.e. A_1 ={2}, A_2 = {2,4}, A_3 = {2,4,6}, A_i={2,4,6,...,2i}) Then I: There is no single x such that E differs from every A_i by x. II: E differs from every A_i - William Hughes
From: Dik T. Winter on 28 Feb 2007 19:22 In article <1172650983.733643.316760(a)m58g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 28 Feb., 02:52, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1172613343.331615.163...(a)z35g2000cwz.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: .... > > > > > 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, > > > > So they can be called numbers after all? > > I cannot hinder you, but it is misleading. What is misleading about it if every mathmatician calls them "numbers"? > > > but these entities are not in trichotomy with > > > really real numbers and not valid as elements of Cantor's list. > > > > How do you define "really real numbers"? Before you give a definition I > > refrain from judgement. > > Rational numbers. sqrt(2) is in trichotomy with the rational numbers. So in your mathematical courses you call what we call the "complex numbers", "complex ideas" or something like that? It least those are not in trichotomy with the "really real numbers". -- 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 28 Feb 2007 19:28
In article <1172651060.321953.253320(a)v33g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 28 Feb., 02:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1172612682.096653.320...(a)t69g2000cwt.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: .... > > > > 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. > > > > Paths of a subset? What does *that* mean? You state: > > (a) The set of finite subsets of a countable set is countable. > > I agree. > > (b) p(oo) is not one of those finite subsets. > > I agree. Now you state what exactly? > > As all the finite subsets like {p(0), p(1), p(2), ...} which belong to > an infinite path like p(oo) form a countable set, the set of all > unions is counable. This set of all unions is P(oo) = {p(oo), q(oo), > ...}. Proof, please. Given the set of paths. Each subset of that set defines a union of paths. If the set is countable, each subset is also countable. But the number of subsets can be uncountable. > > > 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. > > > > I can not follow you anymore. A tree is a set of nodes (with special > > properties), a path is a set of nodes (with special properties). The > > union of trees is an infinite tree, and it contains infinite paths. > > Why do you now state that it does *not* contain infinite paths? > > It is easy to undertand. We are investigating this assertion. > We find that every finite tree contaibns only finite paths. > The union of some finite paths is an ifinite path. > Thee are only countable many unions of finite paths which > yieldinfinite paths. A proof, please, that there are only countably many unions of finite paths which yield infinite paths. But what are you now stating, that those infinite paths are *not* in the infinite tree? Where are they? > > > p(oo) = U{0., 0.0, 0.00, 0.000, ...} > > > > a(oo) = U{0., 0.1, 0.10, 0.100, ...} > > b(oo) = U{0., 0.1, 0.11, 0.110, ...} > > c(oo) = U{0., 0.1, 0.11, 0.111, ...} > > etc. How many times is the path "0." used? > > As the unions of finite subsets of paths like {0., 0.1, 0.10, > 0.100, ...} form a countable set, the path 0. is used countably often. Pray give a proof that that set is countable. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |