Cantor Confusion
in [Math]
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From: Dik T. Winter on 23 Feb 2007 08:35 > > > 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). Eh? Above you state (properly) that p(oo) is a subset of U(T(N)). Are you contradicting that now? And if we consider: N subset N, N contains one infinite initial subset. Or do you mean something different with U(T(N)) from U(T(n))? If so, what do you mean with it? > There are countably many unions of finite subsets of U(T(n)). Prove it. There are countably many *finite* unions of finite subsets, not countably many arbitrary unions of finite subsets. Consider N. Each subset of N is a union of finite subsets of N, namely the union of all singleton sets that contain an element of that set. > There are countably many unions like U(p(n)). Prove it. > Therefore there are countably many elements p(oo) = U(p(n)) of P(oo). > Therefore, P(oo) is a countable set. Wrong. -- 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 23 Feb 2007 12:26 On 22 Feb., 13:26, "William Hughes" <wpihug...(a)hotmail.com> wrote: > 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. There are times when it is false and there are times when it is true. The conclusion from the finity of finite segments on the finity of infinite segments is as false as the conclusion from the evenness on evenness, because all these are individual properties of segments. Another kind of property concerns the fact that every segment of even positive numbers contains numbers which are greater than the cardinal number.For sets which have a cadinal number which is n trichotomy with even positive numbers this holds. Otherwise you should explain how the set with n numbers {2,4,6,...,2n} can be filled up by other (namely greater) even numbers , without increasing the sizes of numbers. It is similar to create uncountably many paths in the binary tree withoutthe presence of uncountably many nodes or edges. Both tasks can only be managed by pure belief in these results and refraining from any attempt to understand it. > > > 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. I told you already. Potential infinity is not concerned, because there is no cardinal number, at least no cardinal number which could be in trichotomy with finite numbers. What distinguishes finite sets and pot. infinite sets? As every set can be extended by elements, every set is potentially infinite. Every finite set has this capability. If we say n, then we point to a finite set. If we say N, then we point to all those finite sets in general. Regards, WM
From: mueckenh on 23 Feb 2007 12:32 On 23 Feb., 05:31, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1172134012.414367.205...(a)j27g2000cwj.googlegroups.com> mueck...(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? The difference 2n - |{2,4,6,...,2n}| steadily inceases with incresing n. Only after infinitely many increasings it drops to -aleph_0. > > > The (only) infinite initial segment differs from each finite initial > segment by having an element that is not in the finite initial segment. How does it differ from all finite segments? Or is here an occasion where the simultaneous consideration of all (segments of) natural numbers is inappropriate (quite contrary to Cantor's diagonal proof)? > > > > (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? You need to find the position of such numbers as [pi*10^10^100] in the sequence of natural numbers, but without always having such an easily computable short hand as [pi*10^10^100]. There are at least [pi*10^10^90] natural numbers which cannot be determined other than by counting along the sequence of natural numbers. Therefore inductive counting is required. For reals the same is valid, further you have to count the digit positions behind the point. Regards, WM
From: mueckenh on 23 Feb 2007 12:33 On 23 Feb., 05:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1172134355.420444.86...(a)k78g2000cwa.googlegroups.com> mueck...(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, How can the set of finite subsets of a countable set be uncountable? Regards, WM
From: mueckenh on 23 Feb 2007 12:35
On 23 Feb., 05:42, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1172134605.880848.197...(a)q2g2000cwa.googlegroups.com> mueck...(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. Therefore in mathematics the unit "of something" is sufficient. > It is not > even a subset of it. Really not??? > What is the physical relevance of the factorisation > of large numbers? There is none. Nevertheless that is a quite important > field in mathematical research. And could not be done without physical means. No idea (of numbers, factors, relations or else) would have appeared without the possibility of physically visualizing its basics. Regards, WM |