Cantor Confusion
in [Math]
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From: mueckenh on 22 Feb 2007 03:52 On 21 Feb., 16:32, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171980199.401444.240...(a)p10g2000cwp.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > 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 > ... > > > > > > 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? > > As a subset, yes. > > > p(n) c T(n) > > p(oo) = Up(n) c UT(n) = T(oo), (n in N). > > Yes. > > > 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? Regards, WM
From: mueckenh on 22 Feb 2007 03:56 On 21 Feb., 16:34, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171980305.458947.244...(a)p10g2000cwp.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 19 Feb., 15:06, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > > Not in my opinion, and I think not in mathematics. If the meaning of > > > > > 3 is "three of something" how than do we calculate "three of > > > > > something" times "three of something"? Or "nine of something" > > > > > divided by "three of something"? Or more concrete, if I divide > > > > > nine apples by three apples what is the result? > > > > > > > > three of something, where the "something" here means the unit. > > > > > > What unit? > > > > I > > > > One of something. > > 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. Regards, WM
From: Virgil on 22 Feb 2007 04:05 In article <1172129749.611484.324300(a)p10g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. But this axiom allows us to speak of the set of all of them, which we cannot do without it, at least in ZF.
From: Virgil on 22 Feb 2007 04:16 In article <1172130169.610851.157050(a)v45g2000cwv.googlegroups.com>, mueckenh(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? > > 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. Since that extra "initial segment" is not one of the /finite/ initial segments, that does not pose any problems to anyone, except possibly WM himself. And, as usual, WM's attempt at a "proof" only proves him wrong again! >How could the infinite segment be > distinguished from the finite ones? Remember, in set theory sets are > to be distinguished by elements. For each finite segment, there is an element of the endless segment that is not a member of it, which is enough to distinguish the endless one from each of the others. As anyone at all familiar with sets would have been able to figure out for himself.
From: mueckenh on 22 Feb 2007 04:35
On 21 Feb., 16:37, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1171981192.821547.278...(a)p10g2000cwp.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 19 Feb., 15:19, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1171890516.474583.210...(a)p10g2000cwp.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > ... > > > > Before completing his first year, the being has the number of years > > > > which comes before 1. > > > > The first number drawn in lottery may be the 7. > > > > > > > > The first is that ordinal number which we start with. > > > > > > So your statement above: "In all set theory 0 is called the first ordinal > > > number, but in fact it is the zeroth one" was nonsense? > > > > No. The first is the first. The first is not the zeroest. It is > > nonsense to start counting by zero as is done in set theory. > > Indeed. So we agree. > But the first is not necessarily "1". The first element of the > ordered set {2, 3, 4} is "2". That is *not* the second element. Of course not. Compare my statement above: The first number drawn in lottery may be the 7. To be clear: In your example 2 is the first element, namley that element which is enumerated by 1. It is not the zeroest element or the element enumerated by 0, as Bourbaki would say and do. > > > > > > Indeed. There is only a definition of set. > > > > > > > > Not even that. > > > > > > <http://en.wikipedia.org/wiki/Set> > > > > LOL. Now, good old Cantor is good enough? > > Yes. Good old Cantor has written quite a bit of good stuff. I do not > agree, however, with everything he has stated. His definition(s) of set(s) lead(s) to contradictions, therefore it is no longer applied in modern set theory. Regards, WM |