Connection between the many different definitions of the Axiom of Choice

Question

I'm currently trying to write a paper on the Axiom of Choice. With my research I have found one very simple definition of the Axiom of Choice : "Let X be a non-empty set of non-empty sets. There exists a choice function for X." This seems intuitive to me and I feel like I can understand this from the classic shoes and socks example. But I am also seeing the Axiom of Choice defined as: "The Cartesian Product of a nonempty family of nonempty sets is nonempty." This seems less easy to understand at first glance, but reading further I understand how this could make sense. The issue is, I'm struggling to connect these two definitions. In my head I see them as two separate statements, each making sense individually. Is there an easy example to understand the Cartesian Product Definition, like can I relate it back to the sock and shoe example?

Answer 1

($X$ and $I$ will denote sets throughout.)
You are probably used to interpreting $X^{n}$ as the set of $n$-tuples $(x_{1}, \ldots, x_{n})$ such that each $x_{i}$ is in $X$.

However, an equivalent way of interpreting this is as the set of all functions $\{1, \ldots, n\} \to X$. Indeed, $(x_{1}, \ldots, x_{n}) \in X^{n}$ corresponds to the function $i \mapsto x_{i}$, whereas a function $f : \{1, \ldots, n\} \to X$ corresponds to the tuple $(f(1), \ldots, f(n))$.

In general, given sets $X_{1}, \ldots, X_{n}$, you can interpret $X_{1} \times \cdots \times X_{n}$ either as the set of $n$-tuples or as the set of functions $\{1, \ldots, n\} \to \bigcup_{i = 1}^{n} X_{i}$ such that $f(i) \in X_{i}$ for all $i \in I := \{1, \ldots, n\}$.

But this is precisely what a choice function is, for the collection $\{X_{1}, \ldots, X_{n}\} = \{X_{i} : i \in I\}$.

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Once you have the above in place, it is easy to see how one would define an arbitrary product of sets. Indeed, let $\{X_{i} : i \in I\}$ be a collection of sets, where $I$ is arbitrary (let us assume it to be nonempty). Then, one defines $$\prod_{i \in I} X_{i} := \left.\left\{f : I \to \bigcup_{i \in I} X_{i} \;\right\vert f(i) \in X_{i} \text{ for all } i \in I\right\}.$$ In other words, $\prod_{i \in I} X_{i}$ is precisely the set of all choice functions. It should now be clear how the two versions of choice are equivalent (this is actually an equivalence by definition, not by any mathematical work).

Answer 2

The motivation behind the formulation of AC involving choice functions is the following: If we have a well defined collection $C$ of non-empty sets there is a way of picking out exactly one member of $C$ such that the resulting collection is a set. Russell's example about socks is just a nice way to spell this out.

The same idea underlies the version of AC that refers to products: Given a non-empty family of non-empty sets there exists a function that sends every index $i$ to exactly one member of the set indexed by $i$. This function picks out a member of each non-empty set from the family such that the resulting collection is a set.

That both versions of AC are equivalent is immediate in the context of ZF: Assume the choice function formulation and consider the Product $\Pi _{i \in I} X_i$ for some familiy $(X_i)_{i \in I}$ with $I \not = \emptyset$ and $X_i \not = \emptyset$. Let $g$ be a choice function for the set of the $X_i$ and define $f$ by $f(i) = g(X_i)$, for every $i \in I$. Then $f \in \Pi _{i \in I} X_i$. Conversely, assume the product version of AC and consider some non-empty set $X$ of non-empty sets. If we let $I = X$, the identity map on $I$ is a non-empty familiy of non empty sets $(X_i)_{i \in I}$. So, there is some $f \in \Pi _{i \in I} X_i$, which is a choice function for $X$.

Answer 3

I'm not sure about the "sock and shoe" metaphor, but here's a simple way to think about things.

Let $X_1$ and $X_2$ be nonempty sets. Elements of $X_1 \times X_2$ are of the form $(x_1, x_2)$, with $x_1 \in X_1$ and $x_2 \in X_2$. In other words, an element of $X_1 \times X_2$ amounts to choosing an element for each factor set $X_1$ and $X_2$. That is, each element of $X_1 \times X_2$ is a choice function. Well, if you have a large collection of nonempty sets $X_i$ indexed over a set $I$, an element of $\Pi_{i \in I}X_i$ is a choice function, so the existence of a choice function is equivalent to this product being non-empty.

Answer 4

The direct sum in your example is a formal one. It's equivalent to a set of sets here. To pick an element of a direct sum is the choice function of your wording as a set of sets.

Hewitt/Stromberg, Real and Abstract Analysis, GTM 25 gave an equivalence proof for: $$ \text{AC }\Leftrightarrow\text{ Tuckey's Lemma }\Leftrightarrow \text{ Hausdorff maximality principle }\Leftrightarrow \text{ Zorn's Lemma }\Leftrightarrow\text{ well-ordering theorem}$$ in case you are interested in more perspectives.