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

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

Le 21/04/2024 à 20:08, Jim Burns a écrit :

On 4/20/2024 12:23 PM, WM wrote:

Le 19/04/2024 à 21:16, Jim Burns a écrit :

On 4/19/2024 11:05 AM, WM wrote:

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

>

I answer and you refuse to see it because
ω isn't what you (WM) want ω to be.

>
I assume that it is the first transfinite number,
following directly upon all natural numbers.

>
I answer and you refuse to see
what the first transfinite ordinal is.

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

⦅0,ω+ω⦆ fits ⦅0,ω⦆
ω+i ⟼ 2⋅i
i ⟼ 2⋅i+1
(k%2=0 ? ω+(k÷2) : (k-1)÷2)  ⟻  k
⦅0,ω⋅ω⦆ fits ⦅0,ω⦆
ω⋅i+j ⟼ (i+j-1)(i+j-2)÷2+i
ω⋅iₖ+jₖ  ⟻  k
sₖ = max{h: (h-1)(h-2)÷2<k}
iₖ = k-(sₖ-1)(sₖ-2)÷2
jₖ = sₖ-iₖ
Each ordinal before the first uncountable ordinal
fits ⦅0,ω⦆
----
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.
| Assume that ω has both
| different.sized.neighbor.haver neighbor ω-1
| and same.sized.neighbor.haver neighbor ω+1
|
| ω is a different.sized.neighbor.haver.
| ω is a same.sized.neighbor.haver.
| Contradiction.
Therefore,
ω does not have both ω-1 and ω+1
ω has ω+1 = ω∪{ω}
ω does not have ω-1
⎛ ω has a same.sized.neighbor.haver neighbor ω+1
⎝ ω is a same.sized.neighbor.haver.
----
The successor of a different.sized.neighbor.haver
is a different.sized.neighbor.haver.
The sum of different.sized.neighbor.havers
is a different.sized.neighbor.haver
The product of different.sized.neighbor.havers
is a different.sized.neighbor.haver
The arithmetic of different.sized.neighbor.havers
is the familiar arithmetic.
Same.sized.neighbor.havers, ω among them,
do not have the familiar arithmetic.