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

On 4/19/2024 11:15 AM, WM wrote:

Le 19/04/2024 à 00:09, Jim Burns a écrit :

On 4/18/2024 10:54 AM, WM wrote:

It is enough if you explain
where ω is in the lower line:

>
0, 1, 2, 3, ...,   ω
|  |  |  |  |||    |
0, 2, 4, 6, ..., ω*2
>

∀κ < ω: k⋅2 < ω

>
That means
the space between ω and ω*2 remains empty of
poducts 2k, and
not all natural numbers have been doubled
because new products have been inserted below ω.

Between ω and ω⋅2 is empty of
products k⋅2 from k < ω
Nothing is inserted anywhere.
Each finite even is finite and below ω
and is double an ordinal finite and below ω

∀κ ≥ ω: k⋅2 > ω

>
But is it also less than ω*2?
Are all products {ω, ω+1, ω+2, ω+3, ...}*2
less than ω*2?

https://en.wikipedia.org/wiki/Ordinal_arithmetic
|
| As an example,
| here is the order relation for ω·2
| 0₀ < 1₀ < 2₀ < 3₀ < … < 0₁ < 1₁ < 2₁ < 3₁ < …,
|
| which has the same order type as ω+ω.
| In contrast, 2·ω looks like this:
| 0₀ < 1₀ < 0₁ < 1₁ < 0₂ < 1₂ < 0₃ < 1₃ < …
|
| and after relabeling, this looks just like ω.
| Thus, ω·2 = ω+ω ≠ ω = 2·ω,
| showing that multiplication of ordinals
| is not in general commutative, c.f. pictures.
|
| Again, ordinal multiplication on the natural numbers
| is the same as standard multiplication.
It looks to me as though
the order relation for (ω+1)·2 is
0₀ < 1₀ < 2₀ < … < ω₀ < 0₁ < 1₁ < 2₁ < … < ω₁
and, after relabeling, this looks just like (ω·2)+1
0₀ < 1₀ < 2₀ < 3₀ < … < 0₁ < 1₁ < 2₁ < 3₁ < … < 0₂
(ω·2)+1 > ω·2
What's your point?