Re: because g⤨(g⁻¹(x)) = g(y) [1/2] Re: how

On 4/22/2024 1:56 PM, WM wrote:

Le 22/04/2024 à 19:32, Jim Burns a écrit :

On 4/22/2024 11:35 AM, WM wrote:

0, 1, 2, 3, ...,   w  = X
|  |  |  |  |||    |
0, 2, 4, 6, ..., w*2  = Y

>
The original set fits in (0, ω).
The doubled set doesn't.

>
⦅0,ω+ω⦆ fits ⦅0,ω⦆

>
No. ω+ω = ω*2.

Yes. ω+ω = ω⋅2
⦅0,ω+ω⦆ fits ⦅0,ω⦆
⦅0,ω⋅2⦆ fits ⦅0,ω⦆
ω+i ⟼ 2⋅i
i ⟼ 2⋅i+1
(k%2=0 ? ω+(k÷2) : (k-1)÷2)  ⟻  k
----
ω the.first.transfinite.ordinal stands between
different.sized.neighbor.haver ordinals
and same.sized.neighbor.haver ordinals.
Neighbors of different.sized.neighbor.havers
are different.sized.neighbor.havers.
Neighbors of same.sized.neighbor.havers
are same.sized.neighbor.havers.
... because g⤨(g⁻¹(x)) = g(y) exists.
The arithmetic of different.sized.neighbor.havers
is the familiar arithmetic.
Same.sized.neighbor.havers, ω among them,
have another, less.familiar arithmetic.