Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-standard)

On 11/20/2024 2:56 PM, Ross Finlayson wrote:

On 11/20/2024 11:19 AM, Jim Burns wrote:

I am satisfied that using the other definition
which you mentioned isn't bait.and.switch.ing.

Dedekind.finite with countable.choice is
equivalent to Stäckel.finite.
>
⎛ Countable.choice:
⎜ ∃S: ℕ→Collection: ∀k∈ℕ:S(k)≠{}  ⇒
⎝ ∃ch: ℕ→⋃Collection: ∀k∈ℕ:ch(k)∈S(k)

Therefore,
if
P is a finite sequence of claims, each claim of which
  is true.or.not.first.false,
then
P is a finite sequence of claims, each claim of which
  is true.
>
That conclusion is the telescope which
finite beings use to observe the infinite,
because
each claim in finite.length.P is true whether.or.not
it is a claim referring to one of infinitely.many.

I am satisfied that using the other definition
which you mentioned isn't bait.and.switch.ing.

Are you, though?

I moved my "I am ..." several dozen lines
in order to make sense of your question.
I can only hope it is the sense you intend.
⎛ Sometimes, Ross,
⎜ I get a very unsettling feeling that
⎜ you expect me and others to
⎜ literally read your mind.
⎜
⎜ That might be the reason that you (RF)
⎜ seem to be allergic to answering questions.
⎝ If you have thought it, it's been answered.
Anyway, no kidding, I'm satisfied with
both Dedekind.finite and Stäckel.finite.

With regards to choice and countable choice,
the weaker form that goes without saying anyways,

"Goes without saying" probably means "accept".
Thank you.
It goes without saying that
claims which go without saying
can be in a finite sequence of claims,
each claim of which is true.or.not.first.false.
There, in that sequence, some claims might be
the opposite of going.without.saying,
might even deny our intuition.
Nonetheless,
because of that claim.sequence, in part
because of those going.without.saying claims,
we accept no.less.securely the truth of
the intuition.denying claims.
I'm just saying.

the existence of a choice function being [implies?]
a bijection [between?] any given set, and
an [at least one?] ordinal's elements lesser ordinals,
making a well-ordering of the set,

which I read as
"the Choice axiom implies the Well.ordering axiom"
-- which I accept.

has that,
>
well-foundedness
and
well-ordering
>
sort of result dis-agreement.

Well.ordering requires well.founded.ness.
There is no infinite descent from any ordinal.
Each ordinal is an ordinal.without.infinite.descent.
⎛ Assume otherwise.
⎜ Assume there is an ordinal.with.infinite.descent.
⎜
⎜ The ordinals are well.ordered.
⎜ There is a first ordinal.with.infinite.descent ψ₀
⎜  ⟨ ψ₀, ξ, ζ, ... ⟩ is an infinite descent from ψ₀
⎜
⎜ ξ before ψ₀ is an ordinal.without.infinite.descent
⎜  ⟨ ξ, ζ, ... ⟩ = ⟨ ψ₀, ξ, ζ, ... ⟩⟨ ψ₀ ⟩
⎜ is a descent from ξ, necessarily finite.
⎜  ⟨ ψ₀, ξ, ζ, ... ⟩ is necessarily finite.
⎝ Contradiction.
Therefore,
there is no infinite descent from any ordinal.