I am reading "A Course in Analysis vol.3" by Kazuo Matsuzaka.
There is the following proposition in this book.
Does the author use the axiom of choice in the proof?
Proposition 4:
A subset $A$ of a metric space $X$ is open in $X$ if and only if $A$ is a union of open balls in $X$.
Proof:
An open ball is open, so a union of open balls is open. Conversely, if $A$ is open and $a \in A$, then there exists a positive real number $r(a)$ such that $B(a ; r(a)) \subset A$. Obviously $$A = \bigcup_{a \in A} B(a ; r(a))$$ holds.
I think the author's proof uses the axiom of choice as follows:
Let $S_a := \{x \in \mathbb{R} | x > 0, B(a ; x) \subset A\}$.
Since $A$ is open, for any $a \in A$, $S_a \ne \emptyset$.
So, by the axiom of choice, there exists a mapping $r$ from $A$ to $\bigcup_{a \in A} S_a$ such that $r(a) \in S_a$.
Best Answer
Yes. This proof does seem to use the axiom of choice, but we can fix that using the maxim "If you can't choose one, just take all of them".
$$A=\bigcup\{B(a,r)\mid a\in A, B(a,r)\subseteq A\}.$$
Another way to fix this is be more explicit about $r(a)$. We can assume it is a positive rational, since the rational numbers are dense and it has to be positive, so we can take the one with the smallest representation as $n/m$ where $(n,m)\in\Bbb{N\times N}$ ordered lexicographically.