Re: Which axiom prohibits this kind of construction?

On Oct 11, 7:49 pm, lwal...@xxxxxxxxx wrote:
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.

How do you define 'finite Zermelo ordinal'?

And in the rest of your post are you going to define 'infinite Zermelo
ordinal' or am I reading another of your posts just again to indulge
your inconclusive thinking aloud?

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

Again, what is the definition of 'finite Zermelo ordinal'?

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.

What is the definition of 'finite Zermelo natural'? After you've
answered that, then we can go on to try to see whether there exists a
set of all finite Zermelo naturals.

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}))

What? 'n' isn't free in the definens.

I'll leave off there.

MoeBlee



.

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