Re: Cantor and the binary tree
- From: David Kastrup <dak@xxxxxxx>
- Date: Mon, 04 Jul 2005 12:02:59 +0200
mueckenh@xxxxxxxxxxxxxxxxx writes:
>> Le q_0 be the first element of the well-ordering. Let q_1 be the first
>> rational, which is less than q_0.
>
> Virgil wrote:
> There can be no such q_1 by either the well ordering or by the standard
> ordering. In a well ordering. nothing precedes the first element, in
> the
> denseity of the standard rational ordering there is never any "next"
> ratinal to any rational.
>
> WM:
> 1/1, 1/2, 1/3, 1/4, 2/3, 1/5, ... is a well ordering
>
> q_0 = 1
> q_1 = 1/2 < 1
> q_2 = 1/3 < 1/2
> ...
> What is your problem, please?
>
> Virgil wrote: Since the rest of WM's daydream is based on the
> occuraence of an
> impossibility, it is irrelevant.
>
> It is based on the existence of infinitely many rationals and on the
> possibility to perform infinitely many transpositions. Cantor
> *explicitly* allowed for infinitely many transpositions. He added that
> a well order set remains well-ordered even after infinitely many
> transpositions.
You are confusing "arbitrarily" and "infinitely" again.
> Ich hebe noch folgendes hervor: wenn in einer wohlgeordneten Menge M
> irgend zwei Elemente m und m' ihre Plätze in der Rangordnung wechseln,
> so wird dadurch der Typus nicht verändert. Daraus folgt, daß solche
> Umformungen einer wohlgeordneten Menge die Anzahl derselben ungeändert
> lassen, welche sich auf eine endliche oder unendliche Folge von
> Transpositionen je zweier Elemente zurückführen lassen. (Letter to
> Laßwitz, 15.Feb.1884)
Looks like you are in good company here. Cantor appears obviously
wrong in this particular letter, judging from the limited context. I
suppose he realized this later. Nobody claimed all his utterings were
to be taken as gospel.
> It seems you are not fully aware of the theory you defend.
Which theory?
> Virgil wrote:
>
> But N cannot have a maximal member according to the Peano properties,
> so
> that WM's argument disproves itself.
>
> As it has not an infinite member, it has only finite members. Finite
> members count themselves. Therefore there cannot be infinitely many
> members.
That presumes that every set can be counted by one of its elements.
You never proved that pretty absurd assumption of yours.
> Virgil wrote:
> I repeat my unanswered question, where does it say that every set must
> be counted by some natural?
>
> WM:
> Every finite natural counts its initial set. If there are only finite
> naturals, then there are only elements, which count their initial sets.
> This implies that there is a finite set for each natural. And there is
> nothing surpassing an initial set.
Confusion of words again. There is no single finite subset that would
not be surpassed by some other initial set. And that is the _exact_
reason why the set of naturals can't be a finite subset.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
.
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