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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.

Sebastian Conybeare
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The answer is that we don't know how to prove that this is equivalent to the axiom of choice.

Higasikawa, Masasi "Partition principles and infinite sums of cardinal numbers." Notre Dame J. Formal Logic 36 (1995), no. 3, 425–434.

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".

Asaf Karagila
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  • Doesn't the statement in quotation marks imply the axiom of choice all by itself? Doesn't it imply that every infinite cardinal is equal to its square? Am I missing something? – bof Mar 30 '14 at 06:23
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    On paper, I agree it seems reasonable. But then again Sierpinski was the one to formulate this principle. I assume that himself and Tarski, and probably Moore too, would have noticed if that was the case. No, there must be something hiding in there to make it more difficult. – Asaf Karagila Mar 30 '14 at 06:28
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    Let $X$ be an infinite set. For $x\in X$ let $a_x={x}\times X$, and let $A={a_x:x\in X}$. Then $|a_x|=|X|=|A|$, and $\bigcup A=X\times X$, so $|X\times X|=|\bigcup A|=|A|=|X|$. Where's the mistake? – bof Mar 30 '14 at 06:40
  • @bof: I don't know where the mistake is. I do know that I have the habit of making these arguments, then returning to them hours later and finding some implicit assumption somewhere. It might not be the case here, and I did agree that it looks convincing (your argument is what immediately came to me when you pointed it out). But I still find it hard to believe that this would have been overlooked if it is so easy. So I will have to think about it quite hard to find out the problem. – Asaf Karagila Mar 30 '14 at 07:28
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    There is no possibility that you subtly misstated the proposition? – bof Mar 30 '14 at 07:43
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    @bof: The statement that Asaf gave is the same (modulo using "cardinality" instead of "power") as that given in Moore's book. (For a nice coincidence, there it is numbered (4.1.12).) In Moore the following sketch of (1.4.12)+(4.1.12) implies AC is given: Let $A$ be any infinite disjoint family of sets, each equipollent to $A$. By (4.1.12) $|A|=|\bigcup A|$. As $|a|=|A|$ for each $a \in A$, by (1.4.12) $|\bigcup A| = | \bigcup { A \times { a } : a \in A }|$, the latter set being $A \times A$. (Like Asaf, I cannot see any gap in your proof.) – user642796 Mar 30 '14 at 08:38
  • @Arthur: Thank you. (bof: No, there's no chance, I bought the book and I copied it from the book, as Arthur said.) – Asaf Karagila Mar 30 '14 at 08:42