Axiom of Choice and Cartesian Products

Question

According to Wikipedia one formulation of AC is

The Cartesian product of any family of nonempty sets is nonempty.

If I consider an cartesian product $\prod_{i} X_i$ of nonempty sets $X_i$, then there exists some $x_i \in X_i$ for each $i$ (simply by non-emptiness), and so $x := (x_i)$ is an element of the product $\prod_i X_i$ by definition. This seems quite trivial to me... imposed by the rules of logic, so why state it as an axiom? Indeed to me it appears as there isn't needed any axiom at all, by setting $x := (x_i)$ I have actually constructed the element?

Answer 1

Your proof is not a proof, but rather an intuition why the AC should be true.

Recall the precise(!) definition of the product of a family of sets: $\prod_{i \in I} X_i$ consists of functions $f : I \to \bigcup_{i \in I} X_i$ such that $f(i) \in X_i$ for all $i \in I$. Also recall the definition of a function $A \to B$ as a special subset of $A \times B$.

Now, given non-empty sets $X_i$, how do you define such a function, using the other ZF axioms? You say, for every $i \in I$ we choose some element $x_i \in X_i$. This works for every single $i$ at a time, but this doesn't define a function $i \mapsto x_i$.

Example: Let $I$ be the set of all non-empty subsets of $\mathbb{R}$, and $X_i = i$. Then an element $f$ in $\prod_{i \in I} X_i$ is a function which picks an element $f(T) \in T$ for every non-empty $T \subseteq \mathbb{R}$. How do you define such an $f$? If we would have $\mathbb{N}$ instead of $\mathbb{R}$, we could take $f(T)=\min(T)$, but this doesn't work for $\mathbb{R}$. Apparently, there is no canonical choice of an element in a non-empty set of real numbers. But the AC tells us that we don't have to worry about this, it gives us such a function, even if we cannot "write it down" (which means: construct it from the other ZF axioms).

By the way, if we let $I$ be to be the set of all non-empty open subsets of $\mathbb{R}$, then there is a choice function (provably in ZF): Choose any bijection $\tau : \mathbb{N} \to \mathbb{Q}$, and then assign to each open subset $\emptyset \neq U \subseteq \mathbb{R}$ the element $\tau(\min \{n \in \mathbb{N} : \tau(n) \in U\})$. This works since $U \cap \mathbb{Q} \neq \emptyset$.

Answer 2

The axiom of choice is, in fact, independent of the Zermelo-Fraenkel axioms.

The difficulty is to find a rule to select a real number out of ANY set.

But the axiom of choice is true for CONSTRUCTIBLE sets.

Answer 3

Sure, if you have five or six sets, then writing such a choice function, or an element of the product, is easily done. But the rules of logic only apply finitely many times.

What happens when you have an infinite number of sets? First of all, it's hard to talk about infinite objects explicitly with logic which is very finitary. So we need another framework, something like set theory works out just fine, because set theory is really just the study of infinite set so to speak.

And indeed the axioms of set theory (without the axiom of choice, that is) allow us to prove by induction that every finite product of non-empty sets is non-empty (and I am not even getting into the bog of standard and non-standard integers).

The role of axioms in set theory is to allow us to construct new sets. They assure us that from such and such conditions, we can ensure the existence of a particular set.

Nothing ensures that given an infinite number of non-empty sets, their product is non-empty. In fact, now we can even show that if set theory is at all consistent, then it is consistent that some products like that are empty.