Re: Replacement of Cardinality (infinite middle)

On 8/18/2024 10:17 AM, Ross Finlayson wrote:

On 08/17/2024 02:12 PM, Jim Burns wrote:

Lemma 1.
⎛ No set B has both
⎝ finiteᵖᵍˢˢ order ⟨B,<⟩ and infiniteᵖᵍˢˢ order ⟨B,⩹⟩.
>
Definition.
⎛ An order ⟨B,<⟩ of B is finiteᵖᵍˢˢ  iff
⎜ each non.empty subset S ⊆ B holds
⎝ both min[<].S and max[<].S
>
A finiteᵖᵍˢˢ set has a finiteᵖᵍˢˢ order.
An infiniteᵖᵍˢˢ set doesn't have a finiteᵖᵍˢˢ order.
>
ℕ ℤ ℚ and ℝ each have infiniteᵖᵍˢˢ orders.
In the standard order,
ℕ ℤ ℚ and ℝ are subsets of ℕ ℤ ℚ and ℝ with
0 or 1 ends.
Thus, the standard order is infiniteᵖᵍˢˢ.
Thus, by lemma 1, no non.standard order is finiteᵖᵍˢˢ.
>
They do not have any finiteᵖᵍˢˢ order.
Whatever non.standard order you propose,
you are proposing an infiniteᵖᵍˢˢ order;
you are proposing an order with
some _subset_ with 0 or 1 ends.
>
One more time:
In a finiteᵖᵍˢˢ order,
_each non.empty subset_ is 2.ended.
Two ends for the set as a whole isn't enough
to make the order finiteᵖᵍˢˢ.

So, with "infinite in the middle", it's just
that the natural order
>
0, infinity - 0,
1, infinity - 1,
...
>
has pretty simply two constants "0", "infinity",
then successors,
and it has all the models where infinity equates to
one of 0's successors, and they're finite,
and a model where it doesn't, that it's infinite.

In the interest of of promoting understanding,
I think it would be better to call the second constant,
in models in which it's finite,
something other than "infinity".
Infinite is different from finite,
whether or not finite is called infinite.
Robinson arithmetic has non.standard models
with infinite naturals.
For example, {0}×ℕ ∪ ℚ⁺×ℤ
⎛ ⟨p,j⟩ <ꟴ ⟨q,k⟩  ⇔
⎝ p < q ∨ (p = q ∧ j < k)
⎛ Numbers ⟨p,j⟩ and ⟨q,k⟩ with p<q are
⎝ infinitely.far apart.
⎛ There are splits between ⟨p,j⟩ and ⟨q,k⟩
⎝ with no step from foresplit to hindsplit.
( ⟨p,j⟩ is not countable.to ⟨q,k⟩
( Not all subsets are 2.ended.

Then, also it happens that
there's the usual order of sucessors and predecessors
that happens to hold,
naturally enough those are both infinite also.

In the usual order of successors and predecessors,
which Robinson arithmetic isn't,
all the (usual) naturals are finitely.far apart.
I mention this mundane point because
I can't tell from what you've written
whether we agree or disagree here.
Do you prefer that I can't tell,
or would you like to clarify that?

At any rate, just identifying
even if just defining
the "predecessors of a limit ordinal"
as with no other facility than
"the successors of a limit ordinal",

There are two kinds of ordinals,
ordinals which are successors and
ordinals which aren't successors.
0 ω and the other limit ordinals are
the second kind of ordinal.
Not.being a successor, they not.have predecessors.
"Predecessor of a limit ordinal"
means pretty much the same as
"positive multiple of 0".

So, ..., "well-order the reals".

"An inaccessible ordinal exists"  ⇒
"The reals can be well.ordered"
https://en.wikipedia.org/wiki/Inaccessible_cardinal