Re: Replacement of Cardinality (infinite middle)

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,
...
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.
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.
About induction, there's still always for each case
both "greatest yet" and "not greatest overall", given
one constant "least", in the ordering, this simply enough
has two constants, and that there's a well-ordering
of between them, and no end, and a well-ordering from
either end, and no end.
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", and
just that all the predecessors of a limit ordinal are
great than all the successors of a previous limit ordinal,
does indeed arrive at those are both well-orderings, and,
that via a definition of infinite as having the same cardinal
as the integers or an inductive set, does arrive at that's infinite.
So, a rather stronger "all orderings are well-orderings" might
make for the usual of "finite". What you got there is just
some assistance to ignore some things about ordering theory
in set theory.
So, ..., "well-order the reals".