Re: Axiomatic Set Theory and the Uniqueness of the Natural Numbers

Adam Burley wrote:

> First of all, my idea is not to take the intersection of the set of all successor-inductive sets, but rather to take the intersection of the set of all successor-inductive sets which are subsets of a particular existential instantiation of the axiom of infinity (c, in your notation).
>
> Secondly, clearly I made a mistake in my previous post, I think what I meant to say was that we need to prove:
>
> (Aa)(Ez:[(z successor-inductive) & NOT(a e z)] -> Ez':[(z' successor-inductive) & NOT(a e z') & (z e P(c))]).
>
> I think this statement now is intuitively obvious and true.
>
> Is it possible to prove this one?

I didn't mean to say that you were claiming that we should take the
intersection of all successor-inductive sets. I was just saying that
with the method I described, one might want to do that, but since there
is no such set, we use a different trick instead.

In your new formula, did you leave an apostrophe off the last z? I.e.,
should it be z' e Pc?

Let me tighten some of your notation and see if I have it right:

(1) Aa(Ez(z successor-inductive & ~aez) -> Ez'(z' successor-inductive &
~aez' & z'ePc))

We already have:

(2) c successor-inductive.

To prove (1):

Suppose Ez(z successor-inductive & ~aez).

Let c' = {n | nec & Az(z successor-inductive -> nez)}.

c' satisfies all three of the conjuncts for z'.

So, to prove (1), I just used the same trick used in the usual method
of deriving a unique least successor-inductive set. But why do you want
to prove (1)? We get what we want - the unique least
successor-inductive set, which is the set of natural numbers - straight
from going from c to c'.

Regards,

MoeBlee

.