Re: Contradiction of bijections as a measure for infinite sets

On 4/4/24 5:33 AM, WM wrote:

Le 03/04/2024 à 15:59, FromTheRafters a écrit :

WM presented the following explanation :

Le 02/04/2024 à 17:51, Jim Burns a écrit :

On 4/2/2024 3:36 AM, WM wrote:

>

If your assumption leads to "no bijection",
but there is a bijection,
then your assumption is wrong.

>
My trick proves that there is no bijection.
Or could you explain why first bijecting n and n/1 should destroy an existing bijection?

>
Your 'trick' only fails to demonstrate a bijection. Failing to demonstrate a bijection does not mean that there is no bijection, only that your 'trick' doesn't work to that end.

 Explain why first bijecting n and n/1 should destroy an existing bijection!
 Regards, WM

It doesn't, Bijections are always between two DISTINCT sets, not a set and a piece of itself thought of as a set.
Not following directions breaks a lot of things.