Cantor Confusion
in [Math]
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From: Dik T. Winter on 19 Feb 2007 09:23 In article <1171890647.125846.84650(a)h3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 19 Feb., 01:35, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > The limit {1,2,3,...} = N does exist according to set theory. Why > > > should the limit {1}, {1,2}, {1,2,3}, ... = N not exist? > > > > There you go again. Set theory does *not* use limits. I know of *no* > > definition that gives the limit of a sequence of sets. So pray write > > down here how *you* define the limit of a sequence of sets. > > We need not get involved into a discussion about limits of sets or the > etymology of the limit ordinal number. That is irrelevant, you are using above undefined concepts. > Consider whether the following > definitions of N in your opinion are correct or not. > > N = {n | n e N} = {0,1, 2, 3, ...} > N = U {{0,1,2,...,n} | n e N} = U {{0}, {0,1}, {0,1,2}, ...} Both are correct. > If they are acceptable, then consider whether a set which contains all > sets of the form {0,1,2,...,n} also contains N. As a subset (because every set is a subset of itself), not as an element. > And if every set which contains all sets of the form {0,1,2,...,n} > contains N, As a subset. > why does the union of finite trees T(n) not contain an > infinite path? I have never said that. I have stated that it *does* contain infinite paths. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Carsten Schultz on 19 Feb 2007 11:46 mueckenh(a)rz.fh-augsburg.de schrieb: > On 19 Feb., 01:35, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: >> In article <1171816407.391676.216...(a)v33g2000cwv.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: >> >> ... >> > The limit {1,2,3,...} = N does exist according to set theory. Why >> > should the limit {1}, {1,2}, {1,2,3}, ... = N not exist? >> >> There you go again. Set theory does *not* use limits. I know of *no* >> definition that gives the limit of a sequence of sets. So pray write >> down here how *you* define the limit of a sequence of sets. > > We need not get involved into a discussion about limits of sets or the > etymology of the limit ordinal number. Consider whether the following > definitions of N in your opinion are correct or not. > > N = {n | n e N} = {0,1, 2, 3, ...} > N = U {{0,1,2,...,n} | n e N} = U {{0}, {0,1}, {0,1,2}, ...} > All of them are true, of course. > If they are acceptable, then consider whether a set which contains all > sets of the form {0,1,2,...,n} also contains N. > And if every set which contains all sets of the form {0,1,2,...,n} > contains N, why does the union of finite trees T(n) not contain an > infinite path? Please clarify at which instances you mean `contains as subset' and at which `contains as an element'. Carsten -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: mueckenh on 20 Feb 2007 08:47 On 19 Feb., 14:36, "William Hughes" <wpihug...(a)hotmail.com> wrote: > Take a property X. Take a potentially infinite set > A (say the union of all initial segments {1,2,3,...,n}). > Then, as you note ("The claim in its generality is clearly wrong") > the statements: > > i: Every initial segment {1,2,3,...,n} has > property X > > ii: Every element of A that can be shown to exist > is a natural number > > Do not imply > > iii: A has property X. > > Sometimes i and ii are true and iii is true. > Sometimes i and ii are true and iii is false. > Statements i and ii cannot be used to prove iii. They can be used in certain cases. But we have no common basis for discussion. You believe that if the chain a < b < c < d <... < z is only infinitely long, then a = z is possible. I deny that, because "infinitely long" means nothing but "never ending", in my opinion. Regards, WM
From: mueckenh on 20 Feb 2007 08:56 On 19 Feb., 14:53, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171889781.807587.262...(a)s48g2000cws.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 18 Feb., 15:53, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > > > You are right. The claim in its generality is clearly wrong, > > > > > > So stop using it. Stop claiming > > > > > > This holds for every initial finite segment therefore > > > it holds for the set. > > > > No. Then we must also stop claiming that the set which is the union of > > all initial segments {1,2,3,...,n} contains only natural numbers. > > That is not proven using induction. It follows from the definition of the > union. 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*. Provable by induction is: The cardinality is less than some numbers in every set of arbitrily many even natural numbers. (By the way, also the bijection n <--> 2n does not exist by definition, but by induction from n to n+1.) Regards, WM
From: mueckenh on 20 Feb 2007 09:03
On 19 Feb., 15:05, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171890114.237911.143...(a)k78g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 19 Feb., 00:57, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1171789678.561730.262...(a)t69g2000cwt.googlegroups.com> mueck...(a)rz.fh-augsburg.de > > > > The union of sets of alle finite paths is the union of all finite > > > > trees, and this union is the complete tree T(oo) if the union of all > > > > initial segmens of N is N. > > > > > > Wrong. The union of sets of paths is *not* the tree. > > > > The union of sets of paths is one set of paths. The union of this set > > of paths is a set of nodes. > > Yes. So your statement: > "The union of sets of all finite paths is the union of all finite > trees" > is false. Correct would be: > "The union of the union of sets of all finite paths is the union of all > finite trees". OK. I was a bit sloppy here because the tree accomplishes the union and the union of the union as well. > > > > There is not infinite path as an element *in* that union, that union *is* > > > the infinite path. Pray look at the difference. > > > > And this path belongs to the tree. > > Strawman. Is p(oo) it contained in the tree as N is in R? p(n) c T(n) p(oo) = Up(n) c UT(n) = T(oo), (n in N). And all nodes (as elements) and all finite paths (as special kind of subsets) of UT(n) are countable. Is p(oo) in the union of finite trees or not? Regards, WM |