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 implies Choice Function for Union of Set/Mistake. Many thanks. --prime mover (talk) 18:26, 7 April 2023 (UTC)