Is the following equivalent to the axiom of choice?
Let $A = \{a_i: i \in I\}$ be a collection of pairwise-disjoint non-empty sets indexed by $I$. Similarly, let $B = \{b_i : i \in I \}$. Further assume that for every $i \in I$, $|a_i| = |b_i|$. Then $|\bigcup A| = |\bigcup B|$
I'm interested in this question because the proposition seems like one of the more intuitively obvious ways to state the axiom of choice, but I'm getting stuck on proving that it actually is one!
It's easy to see that the axiom of choice implies the proposition. The argument is essentially to choose a bijection between $a_i$ and $b_i$ for every element $i \in I$, and combine them to form a bijection. In trying to show the reverse implication, I'm stuck on the fact that we might have a bijection between $\bigcup A$ and $\bigcup B$ that mixes up the partitioning sets.
I've also heard of Russell Cardinals, and that it is is consistent with $ZF$ to assume that there exists a countable union of countable sets that is itself uncountable.
The answer is that we don't know how to prove that this is equivalent to the axiom of choice.
In the above paper the author shows that this principle (which he names $\sf FB$) implies the partition principle (if there is a surjection from $A$ onto $B$, then there is an injection from $B$ into $A$). The question whether or not the partition principle implies the axiom of choice is open.
You can also find this principle in Gregory Moore's book about the axiom of choice. This is principle 1.4.12, and it is indicated to be open whether or not it is equivalent to the axiom of choice. However, it is also indicated that in conjunction with the following principle, we can prove the axiom of choice:
"If every member of an infinite set $A$ has the same cardinality of $A$ then $\bigcup A$ has the same cardinality as $A$."
So all in all, this question is another open question from the list of "very easy to formulate, very difficult to prove equivalent to the axiom of choice".