Restatement of Axiom of choice

Question

Is the following a restatement of the axiom of choice?

Given an arbitrary collection $A$ of non empty sets there exists a set $C$ which contains precisely one element from each element of $A$.

According to Munkres, given a collection $\mathbb{B}$ of non empty sets there exists a choice function so that $c(B) \in B$ for each $B \in \mathbb{B}$. So why can’t I define my set to be the set containing $c(B)$ for $B \in \mathbb{B}$?

Answer 1

This can't work, for instance $A=P(X)\setminus\{\emptyset\}$ is a collection of non-empty sets but there can't be a set $C$ which contains exactly one element of each: it would have to contain all elements of $X$ because of singletons, so it would be equal to $X$... but it wouldn't contain exactly one element of $X$ which also is in $P(X)$.

On the other hand if you add the hypothesis that two elements of $C$ are always disjoint, you can probably make it work. By the axiom of choice, there would be a function such that $f(a) \in a$ and so you would build $C = \{f(a), a\in A\}$.

Conversely, if your property is true, then for any $A$ you can consider the set $A'=\{\{a\}\times a, a \in A\}$. Elements of $A'$ are clearly disjoint, so you can find $C$ containing exactly one element of each, of the form $(a,x)$. Finally you take $f:a \mapsto x$.

Answer 2

Here's a corrected version of your attempt to give something equivalent to the axiom of choice: the Cartesian product of any set of non-empty sets is non-empty. (It's the fifth bullet point under "Set theory" here.) Note that (i) we use products rather than a set containing elements of elements because these may repeat and (ii) it must be a set, and not arbitrary collection, of non-empty sets.