Re: Replacement of Cardinality

On 8/28/2024 3:04 PM, Jeff Barnett wrote:

On 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 from

each 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.