Cantor Confusion
in [Math]
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From: mueckenh on 22 Feb 2007 06:08 On 21 Feb., 16:42, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171981466.237613.54...(a)v33g2000cwv.googlegroups.com> mueck...(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 and it has paths as subsets. > 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. Every element of P(oo) is a path and as such a subset of U(T(n)), as you say. U(T(N)) has only finite paths as subsets (as N has only finite initial segments --- both U(T(n)) and N have infinitely many such subsets, but these subsets are finite). There are countably many unions of finite subsets of U(T(n)). There are countably many unions like U(p(n)). Therefore there are countably many elements p(oo) = U(p(n)) of P(oo). Therefore, P(oo) is a countable set. Regards, WM
From: William Hughes on 22 Feb 2007 07:26 On Feb 22, 2:42 am, mueck...(a)rz.fh-augsburg.de wrote: > 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? > Because, as you have noted, there are times when i and ii are true and iii is false. > 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). So set of initial segments of natural numbers is potentially infinite. > > Conclusion: Should there be an infinite initial segment, we had one > more initial segments than numbers. There is no infinite initial segment (statement i). This does not tell you whether the set of initial seqments is potentially infinite or not. > How could the infinite segment be > distinguished from the finite ones? What infinite initial segment? The fact that there are only finite initial segments does not mean that the set of initial segments is not potentially infinite. - William Hughes
From: Dik T. Winter on 22 Feb 2007 23:31 In article <1172134012.414367.205920(a)j27g2000cwj.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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). I do still not understand. What are you *talking* about? > > > 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. Yes, by induction. > > 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? Except that you have lost some braces. The major difference it that an infinite initial segment differs from a finite initial segment by being infinite. > 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. The (only) infinite initial segment differs from each finite initial segment by having an element that is not in the finite initial segment. > > > (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. I think you are on "induction and belief". What induction is there in f(n) = 2 * n? Moreso, what induction is there in (when considering the reals) f(x) = x^3? -- 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 22 Feb 2007 23:34 In article <1172134355.420444.86660(a)k78g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 21 Feb., 16:32, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > 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? > > > > Yes. Did I ever argue otherwise? That is why I said" "strawman". > > > > But that does *not* show that P(oo) is countable. > > No? P(oo) = {p(oo), q(oo), r(oo), ...} > > p(oo) = Up(n), q(oo) = Uq(m), r(oo) = Ur(i) > > How many unions Up(n), Uq(m), Ur(i), ... can be in a union of finite > trees? Uncountably many, -- 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 22 Feb 2007 23:42
In article <1172134605.880848.197100(a)q2g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 21 Feb., 16:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > So "nine apples" divided by "three apples" is "three I"? A strange > > formulation. What is "nine oranges" divided by "three apples"? > > In mathematics we usually do not use apples and oranges. > In physics, the result is 3 orange/apple. You are starting to see the light? Mathematics is not physic. It is not even a subset of it. What is the physical relevance of the factorisation of large numbers? There is none. Nevertheless that is a quite important field in mathematical research. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |