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

On 4/24/2024 12:57 PM, WM wrote:

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

Arithmetic of the familiar.

>
Nevertheless it is wrong because
for every set {1, 2, 3, ..., n} doubling extends the set.

Doubling doesn't extend it to or beyond ω
The arithmetic of ω is not the familiar arithmetic.

All infinitely many natnumbers are finite n.

Each ⟦0,n⦆ is finite.
{n: ⟦0,n⦆ is finite} is infinite.
"Infinite" does not mean "humongous".
"Finite" describes certain ordinals,  and
the familiar arithmetic operates within
what "finite" describes.
ω stands between the finites and the others.
Doubled finite numbers are finite.
Doubled familiar numbers are familiar.
×2: ⟦0,ω⦆ ⟶ ⟦0,ω⦆
"Infinite" does not mean "humongous".

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⦆

>
You claim that ω+3 = 2*3?

I defined a map for which ω+3 ⟼ 2⋅3
It would explain a lot of your (WM's) posts,
if you don't know what a function/map is,
what familiar addition is,
what familiar multiplication is.
It would also raise the question of
who is responsible for putting you (WM)
in front of a classroom of students.

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

>
No.

...according to one who thinks
ω+3 ⟼ 2⋅3  means  ω+3 = 2⋅3