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NNTP-Posting-Date: Sun, 18 Aug 2024 14:17:18 +0000
Subject: Re: Replacement of Cardinality (infinite middle)
Newsgroups: sci.logic,sci.math
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From: Ross Finlayson <ross.a.finlayson@gmail.com>
Date: Sun, 18 Aug 2024 07:17:06 -0700
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On 08/17/2024 02:12 PM, Jim Burns wrote:
> On 8/16/2024 10:28 PM, Ross Finlayson wrote:
>> On 08/13/2024 08:37 PM, Jim Burns wrote:
>>> On 8/13/2024 9:03 PM, Ross Finlayson wrote:
>>>> On 08/12/2024 09:25 PM, Jim Burns wrote:
>>>>> On 8/12/2024 8:28 PM, Ross Finlayson wrote:
>
>>>>>> It's like yesterday,
>>>>>> in this thread with the subject of it
>>>>>> talking about
>>>>>> "infinite in the middle and
>>>>>> always with both ends",
>
>>>>> "ALWAYS with both ends" is finite.
>>>
>>>> If it's infinite in the middle
>>>
>>> If
>>> it's infinite in the middle and
>>> its non.{} subsets always have both ends,
>>> then
>>> it's not infinite in the middle.
>
>> So, you seem to imply that
>> the integers by your definition,
>
> Paul Gustav Samuel Stäckelᵖᵍˢˢ
> (20 August 1862, Berlin – 12 December 1919, Heidelberg)
>
>> the integers by your definition,
>> by simply assigning 1 and -1 to the beginning,
>> then interleaving them,
>> and filling in as like a Pascal's Triangle of sorts,
>> or pyramidal numbers, that
>> that's, not, infinite?
>>
>> Or, the rationals in the usual assignment of
>> assigning them integer values
>> and all the criss-crossing and
>> from either end, building in the middle,
>> not, infinite?
>
> ℕ ℤ ℚ and ℝ are each infiniteᵖᵍˢˢ,
> each not.finiteᵖᵍˢˢ,
> in the sense of Stäckel's finiteᵖᵍˢˢ,
> by lemma 1.
>
> 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.
>
> ----
> Lemma 1.
> No set B has both
> finiteᵖᵍˢˢ order ⟨B,<⟩ and infiniteᵖᵍˢˢ order ⟨B,⩹⟩.
>
> ⎛ Assume otherwise.
> ⎜ Assume finiteᵖᵍˢˢ ⟨B,<⟩ and infiniteᵖᵍˢˢ ⟨B,⩹⟩.
> ⎜
> ⎜ When ordered by '<',
> ⎜ there is a first initial.segment[<] ⟨x₀,xⱼ⟩ᑉ
> ⎜ such that, when ordered by '⩹',
> ⎜ ⟨⟨x₀,xⱼ⟩ᑉ,⩹⟩ is infiniteᵖᵍˢˢ
> ⎜ and
> ⎜ ⟨⟨x₀,xⱼ₋₁⟩ᑉ,⩹⟩ is finiteᵖᵍˢˢ
> ⎜
> ⎜ However,
> ⎜ ⟨x₀,xⱼ⟩ᑉ  =  ⟨x₀,xⱼ₋₁⟩ᑉ∪{xⱼ}
> ⎜ By lemma 2,
> ⎜ there is no set B, no element x, no order '⩹'
> ⎜ such that
> ⎜ B is finiteᵖᵍˢˢ and B∪{x} is infiniteᵖᵍˢˢ.
> ⎝ Contradiction.
>
> Therefore,
> no set B has both
> finiteᵖᵍˢˢ order ⟨B,<⟩ and infiniteᵖᵍˢˢ order ⟨B,⩹⟩.
>
> ----
> Lemma 2.
> There is no set B, no element x, no order '⩹'
> such that
> ⟨B,⩹⟩ is finiteᵖᵍˢˢ and ⟨B∪{x},⩹⟩ is infiniteᵖᵍˢˢ.
>
> ⎛ Assume otherwise.
> ⎜ Assume ⟨B,⩹⟩ is finiteᵖᵍˢˢ
> ⎜ and ⟨B∪{x},⩹⟩ is infiniteᵖᵍˢˢ
> ⎜
> ⎜ Consider non.empty S ⊆ B∪{x}
> ⎜ Either a) b) c) or d) is satisfied, and,
> ⎜ in each case, S is 2.ended[⩹]
> ⎜
> ⎜ a)
> ⎜ x ∉ S
> ⎜ S ⊆ B
> ⎜ S ⊆ B∪{x} is 2.ended[⩹]
> ⎜
> ⎜ b)
> ⎜ x ∈ S
> ⎜ x = min[⩹].S
> ⎜ x ≠ max[⩹].S = max[⩹].(S\{x})
> ⎜ S ⊆ B∪{x} is 2.ended[⩹]
> ⎜
> ⎜ c)
> ⎜ x ∈ S
> ⎜ x ≠ min[⩹].S = min[⩹].(S\{x})
> ⎜ x = max[⩹].S
> ⎜ S ⊆ B∪{x} is 2.ended[⩹]
> ⎜
> ⎜ d)
> ⎜ x ∈ S
> ⎜ x ≠ min[⩹].S = min[⩹].(S\{x})
> ⎜ x ≠ max[⩹].S = max[⩹].(S\{x})
> ⎜ S ⊆ B∪{x} is 2.ended[⩹]
> ⎜
> ⎜ Each non.empty S ⊆ Bu{x} is 2.ended[⩹]
> ⎜ ⟨B∪{x},⩹⟩ is finiteᵖᵍˢˢ
> ⎝ Contradiction.
>
> Therefore,
> there is no set B, no element x, no order '⩹'
> such that
> ⟨B,⩹⟩ is finiteᵖᵍˢˢ and ⟨B∪{x},⩹⟩ is infiniteᵖᵍˢˢ.
>
>

So, with "infinite in the middle", it's just
that the natural order

0, infinity - 0,
1, infinity - 1,
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