Axiom of choice equivalence problem.

Question

I ran into this problem :

Prove that the axiom of choice is equivalent to the proposition that for every pair of sets A,B, and for every relation R such that $R \subseteq A \times B$ and $\text{dom}(R) = A$, there exists a function $f : A \to B$ such that $f \subseteq R$.

So, first off I'm not sure how to prove equivalence, but I assumed it would be similar to an if and only if kind of proof where you show both directions lead to each other, here's what I did for the first direction:

Let us assume the axiom of choice, and let $A,B$ be two sets.

Let $R$ be a relation such that $R \subseteq A\times B$ and $\text{dom}(R) = A$ .

Let $B' \subseteq B$ such that $B' = \{b \mid \exists a (\langle a,b\rangle \in R)\}$, let $I$ be a set of indices such that for every $i \in I$, there exists a set $B_{i}$ and an element $a_{i}$ such that $B_{i} = \{ b \mid \langle a_{i}, b\rangle \in R\}$.

Since $B' = \bigcup_{i\in I}B_{i}$, from the axiom of choice there exists a function $f : \bigcup_{i\in I}B_{i}\to B'$ such that $f( B_{i}) = b_{i} \in B_{i}$ for every $i$ in $I$.

Therefore we may define a set $A_{i}$ such that $A_{i} = \{ a \mid \langle a, b_{i}\rangle \in R\}$ for every $i$ in $I$.

Since $B' = \bigcup_{i\in I}B_{i}$ and $A = \bigcup_{i\in I}A_{i}$, we may define a function $g: A\to B$ such that for every $a \in A_{i}$, $g(a) = b_{i} \blacksquare$.

I'm not sure that I used the axiom of choice the way I was supposed to, but more importantly I have no idea how to go about the other direction, how do deduce the existance of an axiom?

Any help would be much appreciated, thanks!

Answer 1

The direction you have looks like the right idea, but you could clean up the indexing a little bit. For example, use $A$ itself as an index set: Given $a\in A$ define $B_a=\{b\in B:(a,b)\in R\}$. Then each $B_a$ is nonempty since $dom(R)=A$. Now continue in the same way and apply the axiom of choice to the collection $\{B_a\}_{a\in A}$.

For the other direction: To prove the axiom of choice, you want to take a collection $\{B_i\}_{i\in I}$ of nonempty sets and use the assumption to construct a choice function. So we need some relation $R$ to apply this assumption to. You can reverse engineer the idea from the proof you have. Before, we used the domain $A$ as an index set. So here we use the index set $I$ as a domain. Let $B=\bigcup_{i\in I}B_i$. Define a relation $R$ on $I\times B$ so that $(i,b)\in R$ iff $b\in B_i$. Now apply the assumption to $R$ to get a choice function for the original collection.