Re: Which axiom prohibits this kind of construction?

On Oct 10, 10:13 am, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
On Oct 9, 7:17 pm, lwal...@xxxxxxxxx wrote:
infinite Zermelo ordinals
What do you mean by 'infinite Zermelo ordinals'?

Of course, we all know what _finite_ Zermelo ordinals are.

Let x be a set such that for every yex, either y is
the empty set, or there exists zex such that y = {z}.

We can all think of such sets with this property. For
example, the empty set (vacuously) has this property,
as does its singleton. Indeed, we find several sets
whose elements are (finite) Zermelo ordinals that
have this property:

0
{0}
{0,{0}}
{0,{0},{{0}}}
{0,{0},{{0}},{{{0}}}}

And the set of all (finite) Zermelo naturals itself
has this property.

The question is, are there any others?

What if we define a formula phi with a free variable
n as follows:

phi(n) <-> Ex (Ay (y = 0 or Ez (y = {z}))

Is it possible that there exists n such that phi(n)
yet n is not a (finite) Zermelo ordinal?

In ZFC, my guess is no. My guess is that:

ZFC |- An (n Zermelo natural <-> phi(n))

(stated without proof -- since this is just a guess).

But what if we have ZFC-Foundation/Regularity? My guess
is that En (phi(n) & ~(n Zermelo natural)) is actually
independent of ZFC-Foundation/Regularity, and therefore
ZFC-Foundation/Regularity+"En (phi(n) & ~(n Zermelo nat))"
is consistent (if ZFC is itself consistent). But once
again, this is only a guess, a conjecture.

If such a theory can be consistent, then I propose that
a set n such that phi(n), yet n is not a finite Zermelo
ordinal, be an infinite Zermelo ordinal.

Of course such an n is not infinite -- indeed, such an
n must be a singleton, just as all nonempty Zermelo
ordinals are singletons -- but it is intended to be the
analogue of extending finite Zermelo ordinals to the
infinite, just as we already know about extending finite
von Neumann ordinals to infinite von Neumann ordinals.

If this is all impossible -- and it very well may be --
then this should hopefully explain to Enkai why his
comparison of Zermelo to von Neumann ordinals is invalid.
.

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