Re: Replacement of Cardinality (infinite middle)

On 9/1/2024 2:44 PM, Ross Finlayson wrote:

On 08/30/2024 02:41 PM, Jim Burns wrote:

On 8/30/2024 4:00 PM, Ross Finlayson wrote:

The reals actually give a well-ordering, though,
it's their normal ordering as via a model of line-reals.

>
No.
The normal ordering of the reals
is not a well.ordering.
In a well.ordering,
each nonempty subset holds a minimum.
In the normal ordering of ℝ,
(0,1] does not hold a minimum.
The normal ordering of ℝ is not a well.ordering.

Then, here is the great example of examples
from well-ordering the reals,
because
they're given an axiom to provide least-upper-bound,

Greatest.lower.bound property of standard ⟨ℝ,<⟩
For each bounded non.{} S ⊆ ℝ
exists greatest.lower.bound.S ∈ ℝ   🖘🖘🖘
Well.order property of standard ⟨ℕ,<⟩
For each (bounded) non.{} S ⊆ ℕ
exists greatest.lower.bound.S ∈ S   🖘🖘🖘
glb.S ∈ S = min.S
I threw in '(bounded)' for symmetry.
Each S ∈ ℕ is bounded by glb.ℕ = 0

"out of induction's sake",
then on giving for the axiom a well-ordering,
what sort of makes for a total ordering in any
what's called a space,
there are these continuity criteria where
thusly,
given a well-ordering of the reals,

If we are granted the Axiom of Choice,
then we can prove that
a well.ordering ⟨ℝ,◁⟩ of the reals exists.
That well.ordering ⟨ℝ,◁⟩ is NOT standard ⟨ℝ,<⟩

one provides various counterexamples
in least-upper-bound, and thus topology,
for example
the first counterexample from topology
"there is no smallest positive real number".

Ordered by standard order ⟨ℝ,<⟩
ℝ⁺ holds no smallest positive real number.
Ordered by well.order ⟨ℝ,◁⟩
ℝ⁺ holds a first positive real number.
They aren't counter.examples.
They are different orders.

Then the point that induction lets out is
at the Sorites or heap,
for that Burns' "not.first.false", means
"never failing induction first thus
being disqualified arbitrarily forever",

Not.first.false is about formulas which
are not necessarily about induction.
A first.false formula is false _and_
all (of these totally ordered formulas)
preceding formulas are true.
A not.first.false formula is not.that.
not.first.false Fₖ  ⇔
¬(¬Fₖ ∧ ∀j<k:Fⱼ)  ⇔
Fₖ ∨ ∃j<k:¬Fⱼ  ⇔
∀j<k:Fⱼ ⇒ Fₖ
A finite formula.sequence S = {Fᵢ:i∈⟨1…n⟩} has
a possibly.empty sub.sequence {Fᵢ:i∈⟨1…n⟩∧¬Fᵢ}
of false formulas.
If {Fᵢ:i∈⟨1…n⟩∧¬Fᵢ} is not empty,
it holds a first false formula,
because {Fᵢ:i∈⟨1…n⟩} is finite.
If each Fₖ ∈ {Fᵢ:i∈⟨1…n⟩} is not.first.false,
{Fᵢ:i∈⟨1…n⟩∧¬Fᵢ} does not hold a first.false, and
{Fᵢ:i∈⟨1…n⟩∧¬Fᵢ} is empty, and
each formula in {Fᵢ:i∈⟨1…n⟩} is true.
And that is why I go on about not.first.false.

least-upper-bound, has that
that's been given as an axiom above or "in" ZFC,

No, least.upper.bound isn't an axiom above or in ZFC.

that the least-upper-bound property even exists
after the ordered field that is
"same as the rationals, models the rationals,
thus where it's the only model of the rationals
it's given the existence",

No, the complete ordered field isn't
a model of the rationals.

Here then this "infinite middle"
is just like "unbounded in the middle"
which is just like this
"the well-ordering of the reals up to
their least-upper-boundedness",

If the well.ordering of the reals exists,
it is not the standard order of the reals,
which has the least.upper.bound property,
but is not a well.order.