Re: Well Ordering the Reals

Daryl McCullough said:
> Tony Orlow says...
> >
> >Daryl McCullough said:
> >> Tony Orlow says...
> >>
> >> >Induction is stated axiomatically by Peano and used without regard to the
> >> >underlying nature of the method.
> >>
> >> That's true, but the axiom of induction is provable from the
> >> principles of set theory. So if you disagree with the way people
> >> do induction, then there must be some principle of set theory
> >> that you reject. Which one, and why?
> >
> >The axiom of induction is part of Peano's axioms regarding natural numbers, but
> >doesn't ultimately have that much to do with set theory itself.
>
> Except that it is provable in set theory that the finite ordinals
> satisfy all the Peano axioms. If you disagree with the way that
> people do induction, then which axiom of set theory do you disagree
> with?
>
> >> In set theory, one *defines* the naturals to be the finite
> >> ordinals, and it is *provable* that induction holds for them.
> >
> >According to the inductive axiom, but the naturals are not *defined* to be
> >finite.
>
> That's only because "finite" is not definable in the language
> of arithmetic. But it is definable in set theory, where it is
> provable that the set of finite ordinals satisfies all the
> axioms of Peano arithmetic.
>
> >> Let Phi(x) be a formula such that
> >>
> >> 1. Phi(0)
> >> 2. Forall x, Phi(x) -> Phi(x+1)
> >>
> >> Then we can prove that if n is some ordinal such that
> >> not Phi(n), then n must be infinite. Therefore, Phi(n)
> >> holds for every finite ordinal n.
> >
> >You mind proving that?
>
> Sure. Let me first rewrite equations 1 and 2 in
> terms of not Phi:
>
> 1. forall x, not Phi(x) implies x is not equal to 0.
> 2. forall x, not Phi(x+1) implies not Phi(x).
>
> Let n be any element for which not Phi(n) holds.
> Let S be the collection of all ordinals alpha less than n
> such that not Phi(alpha). Since the ordinals are well-ordered, then
> S has a smallest element. Call that element q. q cannot be
> 0 since that contradicts property 1. On the other hand, q cannot
> be a successor element: if for some x, q=x+1, then we
> have not Phi(x+1). Then by property 2, we have not Phi(x),
> which contradicts the assumption that q is the smallest
> ordinal for which not Phi(x) holds.
Well, you did it twice in two different ways. Thanks for the effort. I am
afraid I must make a similar comment to the last, though.
>
> >If this involves your limit ordinals, I am afraid it
> >won't be very convincing to me, but I wouldn't mind seeing
> >what it is that modern mathematicians think is acceptable,
> >if only to educate myself regarding the dominant viewpoint.
>
> If you reject the idea of limit ordinals, then which axiom
> of ZFC do you reject (since the existence of limit ordinals
> is provable)?

I will have to see what that proof depends on, but consider this. If there are
no limit ordinals, and all ordinals form a succession which extends to infinite
indexes into the set, then what this proves is that, if a fact is true for all
n in N, then the set of n in N for which the fact is false is empty, and the
only elements that could satisfy not Phi(n) must be outside the set of
successors. If you assume that S is nonempty, then all members are not
successors within that ordered set, and the concept of non-successor elements
arises. I agree with that, but that doesn't mean you have to have non-successor
elements. You could have no such elements at all, couldn't you? What proves
that any such elements exist within the ordered set?
>
> --
> Daryl McCullough
> Ithaca, NY
>
>
>
>

--
Smiles,

Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
.

Relevant Pages

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