Can someone please clear my doubts regarding a result concerning the axiom of choice:
Prove that (i) implies (ii) where:
(i) For every nonempty set whose elements are non empty sets there exists a choice function.
(ii) If $\{a_i\}$ is a family of nonempty sets indexed by a nonempty set $I$, then there exists a family $\{x_i\}$ with $i \in I$ such that $x_i \in a_i$ for each $i \in I$.
Here is the proof which is in my book: Let $A$ be a collection of disjoint sets. We have $A \subset \mathcal{P}(\cup_{i \in I}a_i)$. ($\mathcal{P}$ is the power set). By (i) there exists a choice function $f$ on $\mathcal{P}(\cup_{i \in I}a_i)$. Let $b$ be the image of $A$. Pick an element $a \in A, f(a) \in a \cap b$ since $f(a) \in a$. Let $y \in b$ where $y \neq f(a)$ thus we have $y = f(a')$ where $a'\neq a$, and thus $y \in a'$. Since $a$ and $a'$ are disjoint $y \notin a$. Thus the only element of $b \cap a$ is $f(a)$.
My doubts are:
Why do we need to prove this? Won't the choice function $f$ give us the family ${f(i)}$ anyway.
How is the existence of $A$ guaranteed?
The choice function $f$ has what domain and co-domain? If its co-domain is $\mathcal{P}(\cup_{i\in I} a_i)$ then how is that set indexed? Or is the indexing set $\mathcal{P}(\cup_{i\in I} a_i)$?
How does $b \cap a$ being $f(a)$ guarantee that (ii) holds.
Thanks a lot.