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

On 4/20/24 12:31 PM, WM wrote:

Le 20/04/2024 à 00:24, Jim Burns a écrit :

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

 How do the ordinal numbers ω+1, ω+2, ... come into being?

They are not "Ordinal Numbers" but "Transfinite Ordinal Numbers", only reached by Transfinite (past finite) counting.
Something your logic doesn't seem to understand.

>
Nothing is inserted anywhere.
>
Each finite even is finite and below ω
and is double an ordinal finite and below ω

 Then that one directly before ω is not multiplied. Or it is not existing. But what exists directly before ω?

But there isn't one "directly before" ω, as you don't seem to understand what the ... means. The ... contains a set with no upper bound.
The ... shows a sequence of a countably infinite number of digits, with no end. You pass it when you do a Transfinite step in your counting.

 Regards, WM