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

Le 06/05/2024 à 22:16, Jim Burns a écrit :

On 5/6/2024 2:59 PM, WM wrote:

Le 05/05/2024 à 19:59, Jim Burns a écrit :

On 5/3/2024 4:31 PM, WM wrote:

 

If all could be counted to,
they would not remain after every counted number.

 Of those which CAN be counted.to,
each CAN be counted to.

Of course. But ∀n which can be counted to: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

Nevertheless almost all, namely ℵo, remain.

 Of those which CAN be counted.to,
each CAN be counted to,
and
none remain after all which CAN be counted.to

Count to one that has less than ℵo numbers which cannot be counted to. Fail.

 Of those which CAN be counted.to,
each CAN be counted to,
and
none remain after all which CAN be counted.to,

Stop your unfounded claims. Show that you can count to a number which has less than ℵo successors. Fail.
Regards, WM