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.