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On 08/18/2024 10:50 AM, Jim Burns wrote: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,⩹⟩.
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Definition.
⎛ An order ⟨B,<⟩ of B is finiteᵖᵍˢˢ iff
⎜ each non.empty subset S ⊆ B holds
⎝ both min[<].S and max[<].S
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A finiteᵖᵍˢˢ set has a finiteᵖᵍˢˢ order.
An infiniteᵖᵍˢˢ set doesn't have a finiteᵖᵍˢˢ order.
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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.
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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.
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)
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⎛ 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.
I'm really beginning to warm up to this idea ofIf you're referring to the idea of
"finite" and "all orderings are well-orderings"
being a thing.
[...] that they're not "immediate" successors,Standardly, "successor" is "immediate successor".
thus it's delineated that they're "deferred" successors.
So, ordinals less than a limit ordinal are predecessors,To review:
This model in which infinity isn't a successor of 0So, 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.
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