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

On 4/22/24 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.
 Regards, WM
 

Of course it does, as every number < ω is finite, and any finte number doubled is also finite and thus < ω.
This is the strange nature of unbounded sets that breaks finite logic.