Re: Contradiction of bijections as a measure for infinite sets

WM explained on 4/4/2024 :

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!

You still seem to think that sets change. If you mean 'n' is an element of the naturals then of course N bijects with the naturals as embedded in Q. Also, the complement of the naturals over one in Q is the same size as the proper subset you created. No sets (read also functions) were destroyed.