Cantor Confusion
in [Math]
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From: Virgil on 27 Feb 2007 23:50 In article <JE5Gus.BAL(a)cwi.nl>, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > In article <1172612805.121491.277060(a)q2g2000cwa.googlegroups.com> > mueckenh(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. 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. > > > > > > 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. > > 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. > > > > > 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. > > Right. > > > 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. > > Wrong. Show a proof of this. WM is exhibiting his quantifier dyslexia again. Given set S and family A = {A_n: n \eps N}, WM deliberately conflates (ForAll n \eps N) (Exists s \eps S) ( not s \eps A_n), which allows S = Union{A_n:n e N}, with (Exists s \eps S) (ForAll n \eps N) ( not s \eps A_n), which prohibits S = Union{A_n:n e N} For those not compelled to support WM's false arguments, one can have (ForAll n \eps N) (Exists s \eps S) ( not s \eps A_n) without having (Exists s \eps S) (ForAll n \eps N) ( not s \eps A_n). > > > 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 when one untwists one's quantifiers! > > 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. >
From: mueckenh on 28 Feb 2007 03:23 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: > > > On 26 Feb., 01:55, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > > > 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, > > So they can be called numbers after all? I cannot hinder you, but it is misleading. > > > 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. Regards, WM
From: mueckenh on 28 Feb 2007 03:24 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 set of all unions is P(oo) = {p(oo), q(oo), ....}. > > > > > 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. > > 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. > > > > > 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. > > That is *not* what I wrote. I did *not* state that any union of paths forms > a path. > > > 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. Regards, WM
From: mueckenh on 28 Feb 2007 03:29 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. N is the infinite set of all *finite* numbers. Accordinfg to set theory there is no infinite number in it although the limit aleph_0 of numbers is defined by set theory. > 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. > > > > > > 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. > > 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. 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. > > > > > 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. > > 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"? 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. Regards, WM > > > > > > 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. > > I am still unable to determine meaning. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland;http://www.cwi.nl/~dik/
From: Virgil on 28 Feb 2007 03:39
In article <1172650983.733643.316760(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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: > > > > > On 26 Feb., 01:55, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > > > 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, > > > > So they can be called numbers after all? > > I cannot hinder you, but it is misleading. Not to us. And WM manages to mislead himself without our help, so it won't affect him any more then he affects himself. > > > > > 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. So, according to WM, rational numbers are real but real numbers are not. About what can be expected of someone like WM. > > Regards, WM |