An uncountable countable set
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From: mueckenh on 29 Jun 2006 06:33 Dik T. Winter schrieb: > I still suggest the following: I still emphasize the following: > when in a well-ordered set M somewhere two > elements m and m' change place in the ordering, the type is not changed. > From there we find that such changes do leave the ordinal number unchanged, > and that can be transmitted to any finite or infinite sequence of > transpositions. Because with finite sets every transformation can be > broken down to a sequence of transpositions is the reason that with > finite sets ordinal number and cardinal number are, in some sence, the > same, while here sets of the same "Valenz" (cardinality, size? Cantor, collected works, p. 387: Valenz = Cardinality. > > Note that there are still a few things missing from your statement that > Cantor thought that any infinite number of transpositions leave a > well-ordered set well-ordered. In this quote he talks about sequences > ("Folge"), implying countably infinite (and it is of ordering type w). > So your a priori explanation appears to be incorrect. No it does not. In my original quote Cantor talks about *sets* and not about *sequences* of transpositions. But I emphasized already that he might have meant "countable", and I utilize only countable sets of transpositions in my proof. Incorrect is only your remark: "Mueckenheim's statement was a complete misreading of the source he quoted in another article. ... (without a doubt)." Therefore, I would suggest, it is time for an excuse. Regards, WM
From: mueckenh on 29 Jun 2006 06:36 Virgil schrieb: > In article <1151442361.464301.177450(a)u72g2000cwu.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Daryl McCullough schrieb: > > > > > >f: 1 --> {1} and a --> { } with K_f = { }, > > > >g: 1 --> { } and a --> {1} with K_g = {1}. > > > >Here we have certainly no problem with lacking elements. > > > > > > The claim made by Cantor is that for any set A, there is > > > no surjection from A to P(A). Of course, if you construct > > > a new set B by adding extra elements to A, then there might > > > be a surjection from B to P(A). But why is that relevant? > > > > To show you that the condition K is impossible, independent of any > > surjectivity. > > Map |R (including |N) on P(|N) with the only condition that a natural > > number has to be mapped on that set K e P(|N) which contains all > > natural numbers which are not mapped on sets containing them. You see: > > It is impossible a condition. > > > There are all sorts of impossible conditions, most of which are > irrelevant to anything. > > What is the alleged relevance of this one? The fact that whole set theory rests upon an impossible condition like "no rule without an exception". Regards, WM
From: mueckenh on 29 Jun 2006 06:38 imaginatorium(a)despammed.com schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > <snip-snop> > > > If actual infinity exists, then my mapping comes to an actual end. > > I may have missed some context in which this makes some sort of sense > (though I doubt it, somehow)... > > Can I ask you then: is your concept of "actual infinity" the end of the > unending? "actual" means completed, finished, so much that starting from that base we can continue to count. Regards, WM
From: mueckenh on 29 Jun 2006 09:49 Daryl McCullough schrieb: > > The difference is that the claim that there is a surjection > from N to P(N) is provably false, "provably" by a false proof is not provably. while the claim that there > is a surjection from R to P(N) is provably true. Not that surjection which requires the set {k, K, f}. Regards, WM
From: mueckenh on 29 Jun 2006 09:52
Daryl McCullough schrieb: > mueckenh(a)rz.fh-augsburg.de says... > > >Daryl McCullough schrieb: > > > >> mueckenh(a)rz.fh-augsburg.de says... > > >> >There is a mathematical form containing all real numbers of [0, 1]. In > >> >binary representation it is given here. Why not call it a function in > >> >an extended sense? > >> > >> The question is this: does it map the naturals onto the real numbers > >> in [0,1] such that every real number is in its image? > >> > >> In particular, with your representation, which natural number maps > >> to the real 1/3? > > > >It is not necessary to know this natural number (which remains unknown > >or does not exist) in oder to prove that the infinite branches of the > >tree are countable. > > If you are claiming that your tree includes all real numbers in [0,1] > then yes, it is necessary to know that it contains the real number 1/3. Of course it contains the number 0.010101... > > Cantor's claim is that no countable set contains all real numbers. > To prove him wrong, you need to demonstrate both (1) that your tree > has countably many branches, and (2) that it contains every real number. It contains every real number o [0,1] by its construction. It has not more branches than nodes. The set of nodes is countable. Regards, WM |