1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

Yes, the proof is fine. However, a few small remarks:

  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.

$\endgroup$
  • $\begingroup$ thank you for your answer. I've rewritten the Cartesian product only to conform to the notation used by Munkres. Just to make it look like his presentation. $\endgroup$ – Saaqib Mahmood Feb 3 '16 at 8:58
  • $\begingroup$ I can not understand why this problem is given in this chapter. This is supposed to be in the set theory chapter.@Asaf Karagila $\endgroup$ – cmi Aug 9 '18 at 10:57
  • $\begingroup$ @cmi: I didn't write the book. $\endgroup$ – Asaf Karagila Aug 9 '18 at 11:01
  • $\begingroup$ Do u agree with me?@AsafKaragila $\endgroup$ – cmi Aug 9 '18 at 11:02
  • 1
    $\begingroup$ @cmi: I never really went through Munkers' book. I can't quite tell you why it's there. I'm also not sure why it is important for me to agree with you on something that neither of us has any control over. $\endgroup$ – Asaf Karagila Aug 9 '18 at 11:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.