Re: how

On 4/16/2024 10:33 AM, WM wrote:

Le 15/04/2024 à 23:06, Jim Burns a écrit :

On 4/15/2024 7:59 AM, WM wrote:

That is wrong if
all natnumbers are present already such that
no further natnumbers fits below ω.

If
  [0,n) is different.in.size
  from [0,n+1) and [0,n-1)
then
  [0,n+n) is different.in.size
  from [0,n+n+1) and [0,n+n-1)
>
Would you (WM) like me to explain that to you?

>
It is enough if you explain
where ω is in the lower line:
>
0, 1, 2, 3, ...,   w
|  |  |  |  |||    | 0, 2, 4, 6, ..., w*2

∀k < ω: |⟦0,k⦆| < |⟦0,k+1⦆|
∀k ≥ ω: |⟦0,k⦆| = |⟦0,k+1⦆|
∀k < ω: k+k < ω
∀k ≥ ω: k+k > ω
∀k <≥ ω: k+k ≠ ω
For no step, visibleᵂᴹ or darkᵂᴹ, is
|⟦0,k⦆| < |⟦0,k+1⦆| = |⟦0,k+2⦆|
...because,
if
|⟦0,k+1⦆| = |⟦0,k+2⦆|
then
exists 1.to.1 f: ⟦0,k+2⦆ ⟶ ⟦0,k+1⦆  and
f() can be edited to f⤨() such that
1.to.1 f⤨: ⟦0,k+1⦆ ⟶ ⟦0,k⦆ so that
|⟦0,k⦆| = |⟦0,k+1⦆|
Proof: Let f⤨(f⁻¹(k+1)) = f(k+2)
If,
for any sum k+n
|⟦0,k⦆| < |⟦0,k+1⦆|
|⟦0,n⦆| < |⟦0,n+1⦆|
|⟦0,k+n⦆| = |⟦0,k+n+1⦆|
then
for some _first_ sum k+m+1
|⟦0,k+m⦆| < |⟦0,k+m+1⦆| = |⟦0,k+m+2⦆|
But there isn't such a first sum.
Therefore,
there isn't any sum k+n
|⟦0,k⦆| < |⟦0,k+1⦆|
|⟦0,n⦆| < |⟦0,n+1⦆|
|⟦0,k+n⦆| = |⟦0,k+n+1⦆|