Cantor Confusion
in [Math]
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From: Dik T. Winter on 21 Feb 2007 10:42 In article <1171981466.237613.54010(a)v33g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 19 Feb., 15:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > And if every set which contains all sets of the form {0,1,2,...,n} > > > contains N, > > > > As a subset. > > Fine. Every path of the tree is a special subset. (Not every subset is > a path.) But there are only countably many finite subsets and > countably many sets of subsets which belong to one and the same path. Yes, and so there are only countably many finite paths. > > > 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. > > So the union of finite trees U(T(n)) contains (as subsets) the path > p(oo) and all its co-paths q(oo), ..., i.e., Yes. I never said otherwise. Why do you think I said otherwise? All the p(oo) are subsets of U(T(n)). > it contains P(oo)? And that is wrong. Pray look close at what the elements of the different sets are: U(T(n)) has as elements nodes P(oo) has as elements paths, i.e. sets of nodes so P(oo) is neither an element of U(T(n)), nor is it s subset of it. If it where an element it should be a node. If it is a subset is should be a set of nodes, but P(oo) itself is *not* a set of nodes, it has sets of nodes as elements. -- 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 22 Feb 2007 02:35 On 20 Feb., 22:52, "MoeBlee" <jazzm...(a)hotmail.com> wrote: > On Feb 20, 1:40 pm, mueck...(a)rz.fh-augsburg.de wrote: > > > natural nunmbers alone do not bring together an > > infinite number of numbers. > > Right. In Z set theoreis, without the axiom of infinity, we can define > 'is a natural number' and prove the existence of any natural number, > but we can't prove the existence of a set that has as members an > infinite number of natural numbers. So we adopt the axiom of > infinity. This does increase the number of natural numbers in no way. Natural numbers existed before and exist independently of and will exist long after your axiom. Either their set has a cardinal number or it has not (it as not).That is independent of what you and this or that set theorist may say or believe. Regards, WM
From: mueckenh on 22 Feb 2007 02:40 On 20 Feb., 23:04, Franziska Neugebauer <Franziska- Neugeba...(a)neugeb.dnsalias.net> wrote: > mueck...(a)rz.fh-augsburg.de wrote: > > Note that the number of natural numbers can be mapped one-to-one on > > the natural numbers: > > > 1,2,3,...,n <---> n > > Applied brace elimination? As braces only lead to such confused opinions like "a set is more than its members" and "a singleton is more than its element" and "the empty set is more than nothing", it would be better to drop the braces once and for all --- or to replace it by another symbol, at least for a while, until this new symbol becomes a subject of adoration. Regards, WM
From: mueckenh on 22 Feb 2007 02:42 On 20 Feb., 23:26, "William Hughes" <wpihug...(a)hotmail.com> wrote: > On Feb 20, 4:40 pm, mueck...(a)rz.fh-augsburg.de wrote: > > 1,2,3,...,n <---> n > > i: every initial segment of the natural numbers can be mapped > to a natural number. > > ii: the set of all natural numbers does not contain an element > that is not in an initial segment of the natural numbers. > > iii: The set of all natural numbers can be mapped to a natural > number > > No i: and ii: do not imply iii: > Why not? But look here for a direct proof: Every natural number n can be mapped on its finite initial segment (i.e. that one where n is the largest number). Therefore, there are as many finite initial segments of natural numbers as are natural numbers (one-to-one). Conclusion: Should there be an infinite initial segment, we had one more initial segments than numbers. How could the infinite segment be distinguished from the finite ones? Remember, in set theory sets are to be distinguished by elements. Regards, WM
From: mueckenh on 22 Feb 2007 03:46
On 21 Feb., 16:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > 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*. > > What are you *talking* about? Simply about 2n - n < 2(n+1) - (n+1). > > > Provable by induction is: The cardinality is less than some numbers in > > every set of arbitrily many even natural numbers. > > That is not provable by induction. It is provable by induction for every > finitely many natural numbers. For every finite natural number it is provable. The mapping 1,2,3,...,n <--> n yields as many initial segments as natural numbers. In what respect does the infinite initial segment differ from the finie initial segments? Note,:segments are sets. Two sets differ by at least one elmement from one another, not by a an esoteric property, like being believed infinite. > > > (By the way, also > > the bijection n <--> 2n does not exist by definition, but by induction > > from n to n+1.) > > Wrong, there is no induction involved. Either induction or belief. Regards, WM |