Talk:Choice Function for Set does not imply Choice Function for Union of Set

This implies the Axiom of Choice, since for any set $S$, $\{S\}$ is a set with a trivial choice function. But $\cup\{S\}$ is $S$. So if the hypothesis is true, every set has a choice function.

Therefore, this cannot be proved from the axioms of ZF given in the cited book; due to Axiom of Choice is Independent of ZF. --Cem (talk) 15:31, 7 April 2023 (UTC)


 * Okay let me go away and think about this, I need to review it. Are you saying that the statement as presented by Smullyan and Fitting is actually incorrect, or that it has been transcribed into this webpage wrong?


 * I will look at that properly in due course, unless one of my colleagues gets there first. --prime mover (talk) 15:47, 7 April 2023 (UTC)


 * It is indeed transcribed correctly. Smullyan and Fitting ask for a proof of this statement in an exercise following an introduction to the Zermelo-Fraenkel Set Theory, which is not possible. --Cem (talk) 16:46, 7 April 2023 (UTC)


 * Feel free to write this up in our formal style on Choice Function for Set does not imply Choice Function for Union of Set/Mistake. Many thanks. --prime mover (talk) 18:26, 7 April 2023 (UTC)


 * Done, thank you for the guidance! --Cem (talk) 15:29, 13 April 2023 (UTC)


 * Yes, good job. I think now we ought to go back and rename the parent page "Choice Function for Set does not imply Choice Function for Union of Set", offer up your page as a Proof by Counterexample, and reference or transclude that as appropriate from the "mistake" page.


 * But before we do this, we need to check the formal "errata" pamphlet which I have been sent on another thread and it's too late at night to think about right now. So all this is a job for another day (not necessarily tomorrow) as I have stuff to do. --prime mover (talk) 22:11, 13 April 2023 (UTC)


 * Well okay, done the rename :-) --prime mover (talk) 22:14, 13 April 2023 (UTC)