# If a set $S$ has a choice function, does $\bigcup S$ have one too?

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?

• What book is this from? Jul 6, 2015 at 13:12
• Jul 6, 2015 at 13:15
• Thanks for the confirmation. I will cross out that exercise in the book. For everyone's reference, it is exercise 4.3 of Chapter 4. Jul 6, 2015 at 13:19
• See the edit to my answer. Jul 6, 2015 at 13:23

No. Of course not. Your proof is perfectly valid.

This is also mentioned in the errata for the book.