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NNTP-Posting-Date: Thu, 29 Aug 2024 22:46:02 +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: Thu, 29 Aug 2024 15:46:19 -0700
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On 08/19/2024 12:27 PM, Ross Finlayson wrote:
> On 08/18/2024 09:56 PM, Jim Burns wrote:
>> On 8/18/2024 5:22 PM, Ross Finlayson wrote:
>>> 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,⩹⟩.
>>>>>>
>>>>>> 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.
>>
>>>> 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.
>>
>>> I'm really beginning to warm up to this idea of
>>> "finite" and "all orderings are well-orderings"
>>> being a thing.
>>
>> If you're referring to the idea of
>> ⎛ for finite,
>> ⎝ all orderings are well.ordered both ways
>> then I'm pleased to hear
>> that you're warming to the idea.
>> I wish you much future warming.
>>
>>> [...] that they're not "immediate" successors,
>>> thus it's delineated that they're "deferred" successors.
>>
>> Standardly, "successor" is "immediate successor".
>>
>> We have other ways to say "deferred successor".
>> For example, "after".
>>
>> Other than an opportunity to enmurken,
>> what does the use of "deferred successor" offer?
>>
>>> So, ordinals less than a limit ordinal are predecessors,
>>
>> To review:
>>
>>>>> 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.
>>
>> This model in which infinity isn't a successor of 0
>> by which you mean infinity doesn't come after 0
>> how would infinity not coming after 0 work, exactly?
>>
>>
>
> I mean it's a great definition that finite has that
> there exists a normal ordering that's a well-ordering
> and that all the orderings of the set are well-orderings.
>
> That's a great definition of finite and now it stands
> for itself in enduring mathematical definition in defense.
>
> Why is it you think that Stackel's definition of finite
> and "not Dedekind's definition of countably infinite"
> don't agree?
>
> The entire idea here that there's a particular _regularity_
> due dispersion and modularity only courtesy division down
> from a fixed-point, that "Peano's axioms" don't give integers,
> they only give increments, i.e. not necessarily constant increments,
> that there's more than one _regularity_, REQUIRED, is another
> little fact of mathematics missing from your neat little hedgerow.
>
>

...., REQUIRED, ....