I am reading through some course notes to prepare for an upcoming Real Analysis course I have to take in the fall, and found two definitions of the Axiom of Choice.

  1. [AC] If $\Lambda \neq \emptyset$ and for each $\lambda \in \Lambda$, $X_{\lambda}$ is a non-empty set, then $\prod_{\lambda \in \Lambda} X_{\lambda} \neq \emptyset$.
  2. [ACD] Suppose that $\Lambda \neq \emptyset$ and that

    • $\forall \ \lambda \in \Lambda$, $X_{\lambda}$ is a non-empty set, and

    • $X_{\lambda} \cap X_{\beta} = \emptyset$ if $\lambda \neq \beta \in \Lambda$

      Then $\prod_{\lambda \in \Lambda}X_{\lambda} \neq \emptyset$.

It wasn't immediately apparent to me that they were equivalent, so I started trying to prove it. I think that it is trivial to show that $1 \implies 2$, more or less just by saying that if every non-empty set has a choice function then every pairwise non-empty set will also have a choice function.

But I am stuck on trying to prove that $ 2 \implies 1 $ ... I can't seem to construct any meaningful way to use the disjoint version to prove the more general version.

Any tips?


As you've already noted: The implication $1 \Rightarrow 2$ is trivial, since [ACD] is a special case of [AC]. So let's prove $2 \Rightarrow 1$:

Let $\Lambda \neq \emptyset$ and for each $\lambda \in \Lambda$ let $X_{\lambda}$ be a non-empty set. Define $Y_{\lambda} := \{\lambda \} \times X_{\lambda}$. Note that, for any distinct $\lambda_{0}, \lambda_{1} \in \Lambda$, $Y_{\lambda_{0}} \neq \emptyset$ and $Y_{\lambda_{0}} \cap Y_{\lambda_{1}} = \emptyset$. So, by [ACD], we know that $\prod_{\lambda \in \Lambda} Y_{\lambda} \neq \emptyset$. Let $y \in \prod_{\lambda \in \Lambda} Y_{\lambda}$. Then, by construction of the sequence $(Y_{\lambda} \mid \lambda \in \Lambda)$, we have $y = ((\lambda, x_{\lambda}) \mid \lambda \in \Lambda)$ for some sequence $x = (x_{\lambda} \mid \lambda \in \Lambda)$. But now $x \in \prod_{\lambda \in \Lambda} X_{\lambda}$ witnesses that $\prod_{\lambda \in \Lambda} X_{\lambda} \not = \emptyset$. Q.E.D.

  • $\begingroup$ Ohh, clever, I didn't think to construct pairwise disjoint sets like that. Thanks! $\endgroup$ – Eric Hansen Jul 31 '16 at 14:07
  • $\begingroup$ @EricHansen You're most welcome! $\endgroup$ – Stefan Mesken Jul 31 '16 at 19:43

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