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On 8/28/2024 3:04 PM, Jeff Barnett wrote:Well "dense order" and "dense in itself" are well know terms inOn 8/28/2024 12:13 AM, Jim Burns wrote:On 8/27/2024 3:11 PM, WM wrote:Le 27/08/2024 à 02:06, Jim Burns a écrit :On 8/25/2024 3:45 PM, WM wrote:This function exists because
nothing contradicts its existence.
Except for the contradicting
consequences of its existence.
The function exists if
actual infinity exists.
The function does not exist if
only potential infinity exists.
¬∃ᴿx>0: NUF(x) = 1
Then NUF(x) does not exist
and infinity is not actual
and sets are not complete.
In a finiteⁿᵒᵗᐧᵂᴹ order ⟨B,<⟩
each non.empty S ⊆ B is 2.ended.
In a finiteᵂᴹ order ⟨B,<⟩
no one can say
what a setᵂᴹ is,
what an orderᵂᴹ is,
what finiteᵂᴹ is.
Perhaps, in 30 more years,
these question will have answers.
An infiniteⁿᵒᵗᐧᵂᴹ order ⟨B,◁⟩ is
trichotomous and not finiteⁿᵒᵗᐧᵂᴹ.
Trichotomous?
Did you actually mean to say "dense order",
"dense in itself", or something else akin?
I meant "trichotomous".
I could equally.well have said "total ⟨B,◁⟩"
but "trichotomous" seems less misunderstood by WM.
I didn't bother to say "trichotomous and finite ⟨B,◁⟩"
because trichotomy and transitivity follow fromeach non.empty S ⊆ B is 2.ended.
I've proved these elsethread:
⎛ if
⎜ ⟨B,◁⟩: each S ⊆ B: {} or 2.ended and
⎜ ⟨B,<⟩: NOT each S ⊆ B: {} or 2.ended
⎜ then
⎜ sets S, {x} exist such that
⎜ ⟨S,<⟩: each T ⊆ S: {} or 2.ended
⎝ ⟨S∪{x},<⟩: NOT each T ⊆ S∪{x}: {} or 2.ended
⎛ no sets S, {x}, trichotomous '<' exist such that
⎜ ⟨S,<⟩: each T ⊆ S: {} or 2.ended
⎝ ⟨S∪{x},<⟩: NOT each T ⊆ S∪{x}: {} or 2.ended
"Trichotomous" avoids x not being comparable to
the contents of S, and {x} being not.2.ended.
That is how I prove:
because
ℕ has ONE order with ONE non.2.ended non.{} subset
ℕ has NO order WITHOUT one non.2.ended non.{} subset.
That's why any all.2.ended.non.{}.subsets ORDER works
as a way to distinguish a finite SET from infinite.
I think that some of those in this thread
are skeptical of that result.
Thank you for the excuse to repeat it.
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