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

Le 23/04/2024 à 21:55, Jim Burns a écrit :

On 4/23/2024 2:57 PM, WM wrote:

Le 23/04/2024 à 05:51, Jim Burns a écrit :

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⦆

>
Use closed intervals: [0, ω]*2 = [0, ω*2]

 No.
1 ∉ ⟦0,ω⟧ᣔ×2
1 ∈ ⟦0,ω⋅2⟧
⟦0,ω⟧ᣔ×2 ≠ ⟦0,ω⋅2⟧

You are right. 1 belongs not in [0, ω]*2 and of ω we don't know it. But the first and the last elements are the same.

 

If [0, ω) --> [0, ω) and ω*2 --> ω*2,
then ω*2 is the only image point in
the interval (ω, ω*2].
Infinitely many points remain empty.

 Ordinals in ⟦0,ω⦆ double to ordinals in ⟦0,ω⦆
⟦0,ω⦆ᣔ×2 ⊆ ⟦0,ω⦆

That is crippled maths. And it is dyslogical if already all natnumbers are occupying all positions in ⟦0,ω⦆. Doubling a set of natnumbers yields elements which are not in the set.

Arithmetic of the familiar.

Nevertheless it is wrong because for every set {1, 2, 3, ..., n} coubling extends the set. All infinitely many natnumbers are finite n.

 ω in ⟦ω,ω⋅2⟧ doubles to ω⋅2 in ⟦ω,ω⋅2⟧
Arithmetic of the less.familiar.
 

Ugly mathematics.

 de gustibus non disputandum est
 

ω+i ⟼ 2⋅i

>
ω+i is not mapped.

 It is mapped in front of your eyes, sic: '⟼'
not your doubling.map, but
the map which show that ⦅0,ω⋅2⦆ fits ⦅0,ω⦆

Yo0u claim that ω+3 = 2*3?

 ⟦ω,ω×2⦆ maps to {even<ω}
⟦0,ω⦆ maps to {odd<ω}
⟦0,ω×2⦆ maps to ⟦0,ω⦆

No.
Regards, WM