Is this statement equivalent to the axiom of choice?

Question

The statement is:

For every family $\{S_i\}_{i \in I}$ of mutually disjoint, non-empty sets, there exists a set $S$ which contains exactly one element from each $S_i$.

I know AOC implies this, but does it work the other way around?

Answer 1

Let $E$ be a set of non-empty sets on which we wish to define a choice function. Then the sets $\{S\} \times S$, for $S \in E$, are disjoint and nonempty. If $f$ consists of one element of each of these sets, then the required choice function can be selected to be $f$.

(If you don't have the replacement schema, you can show that the collection of sets $\{S\} \times S$ form a set. This collection is a subset of $P[P(E \times \bigcup E)].$)