Re: Replacement of Cardinality (infinite middle)

On 8/9/2024 3:17 PM, Ross Finlayson wrote:

On 08/09/2024 03:25 AM, FromTheRafters wrote:

Ross Finlayson explained :

On 08/08/2024 03:30 AM, FromTheRafters wrote:

on 8/8/2024, WM supposed :

>

There are not finitely many unit fractions.

>
Then stop assuming that
there is a first and last element.

>
Of course, you can start with a first and last element,
then make infinitely-many in the middle.
0 ... ( ... infinitely-many ... ) ... infinity

>
Sometimes you are as bad as he is. :)

>

Or, where do you think you're counting, to?

>
Not to infinity.
Whatever one counts to is not infinity.
That one can count to it means it is not infinity.
>
Consider
0, 1, 2, 3, ..., ℬ-3, ℬ-2, ℬ-1, ℬ
>
There are two cases to consider.
>
1.
There is a split
{0,1,2,3,...} ᵉᵃᶜʰ<ᵉᵃᶜʰ {...,ℬ-3,ℬ-2,ℬ-1,ℬ}
without any α in {0,1,2,3,...}
with α+1 in {...,ℬ-3,ℬ-2,ℬ-1,ℬ}
and
you can't count
from {0,1,2,3,...} to {...,ℬ-3,ℬ-2,ℬ-1,ℬ}
>
2.
Not 1.
There is no split
{0,1,2,3,...} ᵉᵃᶜʰ<ᵉᵃᶜʰ {...,ℬ-3,ℬ-2,ℬ-1,ℬ}
without any α in {0,1,2,3,...}
with α+1 in {...,ℬ-3,ℬ-2,ℬ-1,ℬ}
and
ℬ is finite.
>
Finite.
Not "it will take until the stars die to express".
>
"Finite", in its purposeful indefiniteness,
encompasses "until the stars die"
and more, some of which make that look small.
>

Then, drawing the ends apart with infinite in the middle,
has a little extra work and book-keeping to begin
instead of a usual "next",
yet it well expresses any matters of the "bounded",
for example, in any matters of the "unbounded".

>
Non.{} set C is bounded by
  non.0 ordinal k
  which has last.before k-1 and
  which each non.0 j < k has j-1
>
Set C must have least.upper.C.bound m
m-1 not.upper.C.bound
least.upper.C.bound m in C
max.C m
>
C is two.ended and
  each non.{} subset of C, also bounded,
  is two.ended.
>
Non.{} set C, bounded by
  non.0 ordinal k
  which has last.before k-1 and
  which each non.0 j < k has j-1
is finite.
>
0, 1, 2, 3, ..., ℬ-3, ℬ-2, ℬ-1, ℬ
with the ends drawn apart and infinity in the middle
fails at having last.before somewhere,
otherwise, there isn't infinity in the middle.
>
>