Trying to work out for myself the solution to Exercise 5.11 in Goldrei's book "Classic Set Theory for Independent Study" (which is excellent!). There is no answer provided in the book itself so wanted to see if I got it right.
Exercise 5.11
R is said to be a relation between sets A and B if $R \subseteq A \times B$. The domain of R is the set $\{a \in A : (a,b) \in R \text{ for some } b \in B\}$.
Recall that a function F is a set of ordered pairs such that for each a there is at most one b for which $(a,b) \in F$.
Show that AC is equivalent to the statement: Suppose that R is a relation between non-empty sets A and B. Then there is a function F with the same domain as R such that $F \subseteq R$.
My solution:
(i) In the first direction, assuming AC, the result is pretty straight forward. Construct a family of sets S consisting of all sets-of-tuples of the form $S_i = \{\{a_i, b_{i1}\}, \{a_i, b_{i2}\}, ...)$ e.g. each such set enumerates all the relations holding for a particular $a_i$. then by AC we can define a choice function on this family S that picks exactly one element (e.g. $\{a_i, b_{ij}\}$ tuple) from member of the family. This function defines our function F.
(ii) in the second direction, given a family of sets S=$\{_i\}_{ ∈ }$, take A = S and B = $\cup_i(S_i \in S)$. Define the relation R to be subset of the cartesian product AxB that maps each element of A, $S_i$, to all elements of B which are also elements of $S_i$. then by the statement assumption, there is a function F with the same domain as R such that $F \subseteq R$. This function is therefore a choice function on the family of sets S, QED.
Is this valid?