I have an exercise in a book that asserts that if a set $S$ has a choice function on it, then so does the union of all its elements $\bigcup S$ (without assuming the axiom of choice). I, however, have serious doubts that this statement holds and I even have an argument against it. However, since I am up against the word of a textbook, I cannot be sure. I may just be missing something obvious. Hence, I want to verify my hunch.
First a little clarification of my terminology:
- By a choice function $C$ on a set $S$, I mean a function $C : \wp(S)\setminus\{\emptyset\} \to S$ such that $C(\sigma) \in \sigma$.
- For a set $S$, $\bigcup S = \{x \in \mathcal{U} : \exists y \in S\text{ s.t. } x \in y\}$ ($\mathcal{U}$ is the universal class) i.e. it is union of all the elements of $S$.
Now, my argument against the book's assertion is this: if the axiom of choice (AOC) were false, then there would be a set $\Sigma$ which has no choice functions on it. Consider, now, the singleton set $\{\Sigma\}$; it clearly has a choice function on it but its union doesn't as $\bigcup \{\Sigma\} = \Sigma$. Hence, if AOC were false, then I can prove this statement false using the other axioms of set theory. Thus, contrapositively, if I could prove this exercise true using just the other axioms of set theory, I could prove AOC too which seems ridiculous! Is my reasoning correct? Or am I mistaken somewhere?
No. Of course not. Your proof is perfectly valid.
This is also mentioned in the errata for the book.