# Prob. 9, Sec. 19 in Munkres' TOPOLOGY, 2nd edition: Equivalence of the choice axiom and non-emptyness of Cartesian product

The Axiom of Choice is as follows:

Given a collection $\mathcal{A}$ of disjoint non-empty sets, there exists a set $C$ consisting of exactly one element from each element of $\mathcal{A}$; that is, a set $C$ such that $C$ is contained in the union of the elements of $\mathcal{A}$, and for each $A \in \mathcal{A}$, the set $C \cap A$ contains a single element.

Now here is Prob. 9, Sec. 19 in the book Topology by James R. Munkres, 2nd edition:

Show that the choice axiom is equivalent to the statement that for any indexed family $\{ A_\alpha \}_{\alpha \in J}$ of non-empty sets, with $J \neq \emptyset$, the Cartesian product $$\prod_{\alpha \in J} A_\alpha$$ is not empty.

My effort:

Suppose the choice axiom holds. Let $\{A_\alpha\}_{\alpha \in J}$ be an indexed family of non-empty sets, with the index set $J \neq \emptyset$. For each $\alpha \in J$, let's define the set $A^\prime_\alpha$ as follows: $$A^\prime_\alpha \colon = \left\{ \ (\alpha, a ) \ \colon \ a \in A_\alpha \ \right\}.$$ Now let $\mathcal{A}$ be the indexed family $\{A^\prime_\alpha\}_{\alpha \in J}$. Since $J \neq \emptyset$ and since each set $A_\alpha$ is non-empty, therefore each set $A^\prime_\alpha \neq \emptyset$.

Moreover, for any $\alpha, \beta \in J$, where $\alpha \neq \beta$, we also have $$A^\prime_\alpha \cap A^\prime_\beta = \emptyset.$$ Thus, $\mathcal{A}$ is a collection of disjoint non-empty sets. So, by the choice axiom, there exists a set $C$ such that $C$ consists of exactly one element from each set in $\mathcal{A}$. That is, for each $\alpha \in J$, the set $C$ consists of exactly one ordered pair $(\alpha, a)$ such that $a \in A_\alpha$. Then the set of all these ordered pairs in $C$ constitutes an element of the Cartesian product $\prod_{\alpha \in J} A_\alpha$.

Am I right?

Conversely, suppose that, for any indexed family $\{A_\alpha\}_{\alpha \in J}$ of non-empty sets, with the index set $J \neq \emptyset$, the Cartesian product $$\prod_{\alpha \in J} A_\alpha$$ is not empty. Let $\mathcal{A}$ be a collection of disjoint non-empty sets. Then, assuming that the collection $\mathcal{A}$ is non-empty, the Cartesian product $$\prod_{A \in \mathcal{A}} A$$ is not empty. Let $$\mathcal{A} \colon= \left\{ \ X_i \ \colon \ i \in I \ \right\},$$ where the index set $I$ is non-empty and, for each $i \in I$, the set $X_i \neq \emptyset$. Then $$\prod_{A \in \mathcal{A}} A = \prod_{i \in I} X_i.$$ Let $x$ be an element of this Cartesian product. Then, by definition,$x \colon I \to \cup_{i \in I} X_i$ (i.e., $x$ is a function with domain $I$ and images in the union $\cap_{i \in I} X_i$) such that $x(i) \in X_i$ for each $i \in I$. Let $C$ be the set $$\left\{ \ x(i) \ \colon \ i \in I \ \right\}.$$ Since for each $i, j \in I$ with $i \neq j$, we have $X_i \cap X_j = \emptyset$, therefore the set $C$ consists of exactly one element from each set in the collection $\mathcal{A}$.

Is the above proof correct?

1. If you really want to be thorough, you need to argue why $A_\alpha'\cap A_\beta'=\varnothing$ (namely, each set has ordered pairs with different left coordinate). It's not a big deal, I'm just throwing it out there.
2. It's not entirely clear why you had to "rewrite" $\cal A$ as a collection of $X_i$'s. If $\prod_{A\in\cal A}A$ is non-empty, and $f$ belongs to that product, then $\operatorname{rng}(f)=C$ is your wanted transversal set. There's no need to go through a renaming process.