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While looking at this problem, I was thinking about the more general statement:

For all sets $I,X$ and $U:I\to \mathcal P(X)$, the power set of $X$, there exists a $V:I\to \mathcal P(X)$ with the property that for all $i\in I$, $V_i\subseteq U_i$, for all $i,j\in I, i\neq j: V_i \cap V_j=\emptyset$ and $$\bigcup_{i\in I} U_i = \bigcup_{i\in I} V_i$$.

Obviously, if you have the axiom of choice, you can well-order $I$, and you get this result by the reasoning in the answer to that other question. But does this result imply choice, or is it weaker?

If it does imply choice, it seems like perhaps the most direct approach would be to show it implies Zorn's lemma. But I'm struggling to prove one way or the other.

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Thomas Andrews
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1 Answers1

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They are equivalent. Suppose that $\{ X_i \}_{i \in I}$ is a family of nonempty pairwise disjoint sets and let $X = \bigcup_{i \in I} X_i$. For each $x \in X_i$ define $A_x = \{ i \}$. Now take a pairwise disjoint family $\{ B_x \}_{x \in X}$ with $B_x \subseteq A_x$ and $\bigcup_{x \in X} B_x = \bigcup_{x \in X} A_x = I$. For each $i \in I$ pick the unique $x \in X_i$ such that $B_x \neq \emptyset$. This is your choice function for $\{ X_i \}_{i \in I}$.

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  • Nice and clean argument. – Asaf Karagila Mar 11 '13 at 20:40
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    @Asaf: Thanks. Coming from the math.SE's Emperor of Choice, I'll consider it high praise. – user642796 Mar 11 '13 at 20:43
  • Easier to read than this earlier version, but close enough that I was reminded that we’d seen the problem before. – Brian M. Scott Mar 11 '13 at 20:44
  • Arthur: first you call Brian Caesar, and now you call me Nero? – Asaf Karagila Mar 11 '13 at 20:48
  • @Asaf: Didn't know you played the fiddle. – user642796 Mar 11 '13 at 20:53
  • Yeah, that original problem uses a mix of upper and lower case in a way that makes the question and answer harder to read. It's one thing to realize that everything in set theory is a set, it's another thing to not try to distinguish between "sets" and "elements" in a context where it makes sense for human readability. :) @BrianM.Scott – Thomas Andrews Mar 11 '13 at 20:58
  • Arthur, I'm was burning ROM at one point. (Also, until I rewatched MacGyver a few years back and saw that episode with the army ants in the amazon, I didn't get the reference of that software Nero Burning ROM. Only when MacGyver was using a flamethrower to kill some ants and he said "Yeah, like Nero burning Rome" I thought "hey, they didn't have that software in 1988! Ohhhhh!!! ROM... ROME! I get it!") – Asaf Karagila Mar 11 '13 at 21:02
  • Caligula was also unpleasant. – Will Jagy Mar 11 '13 at 23:02
  • @Will: Not once I heard someone confuse my last name with his... but I never appointed a horse to the senate. – Asaf Karagila Mar 12 '13 at 18:42
  • @Asaf, give it time. You may not have reached your full potential. Hmmmm. This is embarrassing: I really thought Caligula was the son of Nero. But no, Claudius was emperor in between, and it is not a direct lineal descent anyway, Caligula was first, then Claudius, then Nero, but Nero was not Caligula's grandson. Oh, well. – Will Jagy Mar 12 '13 at 19:49