Cantor Confusion
in [Math]
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From: David Marcus on 20 Feb 2007 21:32 mueckenh(a)rz.fh-augsburg.de wrote: > On 20 Feb., 16:11, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > You wrote (then snipped) > > > > M: [t]he property that every set of even natural numbers must contain > > numbers > > M: larger than its cardinal number, is correct, unless the set > > contains > > M: unnatural numbers. > > > > As I noted this is false even in Wolkenmueckenheim. > > You may note what you want. The above statement is true if the > statement is true that even in an infinite chain we can conclude from > a<b<c<...<z that a is not larger than z. > > (Should you continue with personal insults, I will stop this > discussion with you.) What "personal insults"? If you tell us how to make you stop talking, we would all be grateful! -- David Marcus
From: Dik T. Winter on 21 Feb 2007 10:10 In article <1171979760.586857.221470(a)p10g2000cwp.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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*. 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. > (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. -- 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 21 Feb 2007 10:32 In article <1171980199.401444.240520(a)p10g2000cwp.googlegroups.com> mueckenh(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. -- 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 21 Feb 2007 10:34 In article <1171980305.458947.244550(a)p10g2000cwp.googlegroups.com> mueckenh(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"? -- 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 21 Feb 2007 10:37
In article <1171981192.821547.278230(a)p10g2000cwp.googlegroups.com> mueckenh(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. 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. > > > > 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |