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On 11/20/2024 2:56 PM, Ross Finlayson wrote:Countable choice goes without saying because it'sOn 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.
>
>
>
>
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