Re: Orlow cardinality question

In article <MPG.1d29efd2227077c4989eea@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:

> Jan de Vos said:
> > In sci.math, Tony Orlow wrote:
> > >> *Sigh* all right, I'll dig up the reference...
> > >>
> > >> Your words:
> > >> "By induction, the set size is ALWAYS the same as the maximal
> > >> number."
> > >> See http://groups.google.nl/group/sci.math/msg/15b7551204790e00?hl=en
> > >>
> > >> So if you call the size of N (the set of naturals) N, then N should
> > >> also be the maximal element of N. However, by definition, N+1 is also
> > >> in N, and since N+1>N, N is no longer the maximum element of N.
> > > Uh huh. That's a problem isn't it? Well if you declare the size to be N
> > > then
> > > you can always add another element, and get a set of size N+1 can't you?
> > > The
> > > same problem exists for both maximal element AND size for the finite
> > > naturals,
> > > which Cantor has "resolved" by falsely caliming the finite naturals
> > > constitute
> > > an infinite set.
> > >>
> > >> Let us call the size of N 's', for a while, to avoid clutter. Again,
> > >> in short:
> > >>
> > >> |N| = s => s is the maximum element of N => s \in N => s+1 \in N
> > >> => s is not the maximum element of N (since s+1 > s).
> > >>
> > >> This is a contradiction, plain and simple. Surely even you won't deny
> > >> this?
> > > I never claimed there was a maximal element to the set of finite naturals
> > > as a
> > > whole, just as I never claimed it has any specific size. However, if the
> > > maximal element or upper bound is finite, then the same is true of the
> > > set
> > > size, as I have shown in three different ways, and if the set size is
> > > infinite,
> > > then the maximal element or upper bound is also infinite.
> > >>
> > >> Note that I only used your theorem and the definition of the natural
> > >> numbers. That means that your theorem /has to be false/. Which,
> > >> whooptydoo, means that your 'induction' really can't prove anything
> > >> for 'infinite numbers' or infinite sets.
> > >
> > > Uh, no, as usual you applied YOUR theorem that every finite set MUST have
> > > a
> > > maximal element, no matter how indeterminately defined it is. I don't
> > > accept
> > > this theorem in cases like this, but rely on actual formulas and axioms
> > > of
> > > finiteness, such as, for finite S and finite L, S^L is finite. You don't
> > > disagree with THAT do you?
> >
> > Please re-read my entire post. I am not speaking about any finite
> > set, I am speaking about N, the set of natural numbers. It is defined
> > by the peano axioms, and your theorem, in the post I referenced, is
> > about that set.
> >
> >
> > Jan
> >
> >
> Yes, I understand exactly what you are talking about, the set of finite
> natural
> numbers, which you claim is infinite, but which I say is finite. We are
> talking
> about the same set and disagreeing, because you claim to prove it is infinite
> since it has no maximum element, but I say that is irrelevant

But if a well-ordered set, and the naturals are well-ordered, does not
have a maximum member, it is easy to construct an injection from that
set to a proper subset.

If TO rates such sets, being injectionable into proper subsets, as
finite, just what TO require of a set to rate it as infinite?
.

Relevant Pages

  • Re: infinity
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  • Re: Orlow cardinality question
    ... > In sci.math, Tony Orlow wrote: ... >> same problem exists for both maximal element AND size for the finite naturals, ... >> size, as I have shown in three different ways, and if the set size is infinite, ...
    (sci.math)
  • Re: Orlow cardinality question
    ... >>> proof you gave for any finite set having a maximal element. ... If the set is infinite, and has a maximal element E, then ... If an ordered set has a maximal member then appending any ... >> naturals is itself not finite, and thus need not have any maximal ...
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  • Re: infinity
    ... naturals has basically an infinite element. ... that the set of ordinals does contain its maximal element. ... calling that collection a set breaks their set theory. ...
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  • Re: Orlow cardinality question
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    (sci.math)