Cantor Confusion
in [Math]
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From: Dik T. Winter on 28 Feb 2007 19:48 In article <1172651370.180739.279270(a)8g2000cwh.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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: > > > > 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. Where is that stated? Set theory does not define limits, so it is not stated in set theory. (And do not come with the term "limit ordinal", that is not intended as "limit" but a a specification for some kind of ordinals, namely those that have no predecessor.) > 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. No, that is *not* defined as "limit". It is stated that from the axiom of infinity the set N does exist (and does contain finite elements only, because of the way it is defined). It further *defines* the cardinality of that set as aleph-0. It does not say anything about a limit at all. > > 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. You are making a mess of it because of lack of understanding. Assume we define limits of sequences of natural numbers such that if the elements grows without bound the limit is aleph-0. But if we do that, the old theorem: lim [ a_n / b_n ] = lim a_n / lim b_n if both limits exist and lim b_n != 0 no longer holds, which you assume. And that is easy to show, because in the proof of the theorem use is made of the fact that all limits are finite (in the context of that theorem a limit exists only if it is finite). So in the context of the extended definition we have to adjust the condition on the limit to: if both limits exist and are finite and lim b_n != 0. You are making a mess of it because you use a theorem unchanged while a basic definition has been changed. > > > 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. > > 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. The comparisons can be done all together, but they will not all yield the same result, which is not needed, they yield all yield a result, and that is needed. > > > 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. No. I do not say that. I say that S can not be distinguished by any finite elemen n from all the finite sets. This does inhibit S to exist and being different from all finite sets, that is why I ask for a proof. > > 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"? What is ridiculous about it? It clearly shows that for a set S to exist and being different from a collection of other sets it is *not* necessary that there is a single element where it differs from all those other sets. > 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. Prove that it is *not* possible. Until now you have only asserted it is wrong because it would imply: "thereis x such that for all A: x notin A and x in S". It is clear that both statements are *not* equivalent, so it is *your* task to prove the equivalence in this case. -- 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 1 Mar 2007 04:07 On 1 Mrz., 01:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1172650983.733643.316...(a)m58g2000cwm.googlegroups.com> mueck....(a)rz.fh-augsburg.de writes: > > > What is misleading about it if every mathmatician calls them "numbers"? It veils the physical restrictions. > > sqrt(2) is in trichotomy with the rational numbers. In fact not even all rational numbers are numbers. You know, there are less than 10^100 bits ... > So in your mathematical courses you call what we call the "complex numbers", > "complex ideas" or something like that? I use the current words. I do not talk about these things there. > At least those are not in > trichotomy with the "really real numbers". Yes. I would prefer another word, but that will be impossible to introduce. You may have noticed that I talk of "irrationale Zahl" even in my book. I had considered to use only "Irrationalität". But it sounds too heavy-handed. Regards, WM
From: mueckenh on 1 Mar 2007 04:31 On 1 Mrz., 01:28, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1172651060.321953.253...(a)v33g2000cwv.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > > 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 countable. 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. Not every subset of the set of paths defines a union which is an infinite path. There are only very few such subsets of he set of paths. Here we have the same situation as with the number 2^omega which has cardinality aleph_0. There is a one-to-one map to the cross sections of the trees and hence to the cross section of the unit tree. > > > > 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 understand. We are investigating this assertion. > > We find that every finite tree contains only finite paths. > > The union of some finite paths is an infinite path. > > Thee are only countable many unions of finite paths which > > yield infinite paths. > > A proof, please, that there are only countably many unions of finite paths > which yield infinite paths. The set of unions cannot be larger than the cross section of the tree U(T(n)) which is Card(2^omega). > But what are you now stating, that those > infinite paths are *not* in the infinite tree? Where are they? I said: The union of some finite paths is an infinite path. There are only countable many unions of finite paths which yield infinite paths Regards, WM
From: Franziska Neugebauer on 1 Mar 2007 05:45 mueckenh(a)rz.fh-augsburg.de wrote: > On 1 Mrz., 01:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: >> In article <1172650983.733643.316...(a)m58g2000cwm.googlegroups.com> >> mueck...(a)rz.fh-augsburg.de writes: >> >> What is misleading about it if every mathmatician calls them >> "numbers"? > > It veils the physical restrictions. There is no objective in mathematics to "unvail" physical (material) restrictions of the world. F. N. -- xyz
From: G. Frege on 1 Mar 2007 06:43
On 1 Mar 2007 01:07:21 -0800, mueckenh(a)rz.fh-augsburg.de wrote: "In fact not even all rational numbers are numbers." (W. M�ckenheim) An all-time classic! :-) F. -- E-mail: info<at>simple-line<dot>de |