Re: Replacement of Cardinality (infinite middle)

Liste des GroupesRevenir à s logic 
Sujet : Re: Replacement of Cardinality (infinite middle)
De : ross.a.finlayson (at) *nospam* gmail.com (Ross Finlayson)
Groupes : sci.logic sci.math
Date : 19. Aug 2024, 20:27:42
Autres entêtes
Message-ID : <jWSdneBt4MAqAV77nZ2dnZfqn_udnZ2d@giganews.com>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
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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.

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