Can the indexing set of a cartesian product have the cardinality of the continuum?

Question

Let $I$ be any set and for any $i \in I$ let $S_i$ be a set. Then is the set-theoretic product $\Pi_{i \in I} S_i$ well defined even if $I$ has the cardinality of the continuum (uncountable)? In that case an element of the product would be a family $\left\{x_i\right\}_{i \in I}$ where $x_i \in S_i$. I don't see any problem with that, but i would appreciate an expert's opinion.

Answer 1

There is no problem with that. In fact, the rigorous definition of $\prod_{i\in I} S_i$ is:

$$\prod_{i\in I} S_i= \{f:I \to \bigcup_{i\in I} S_i | f \text{ is a function }, f(i)\in S_i \, \forall i\in I\}$$

The cardinality of $I$ changes nothing. Is this convincing enough?

ADDED: for more formality, in case you're not convinced that this set exists just yet. Observe that it is a set of functions $I\to \bigcup \{S_i: i\in I\}$. Therefore, it is a subset of $\mathcal{P} (I \times \bigcup \{S_i: i\in I\})$. Since the power set of a set exists (axiom), it remains to convince you that $I \times \bigcup \{S_i: i\in I\}$ is a set.

Well, the cartesian product of two sets exists: this is a good exercise on the axioms of set theory (replacement, power set, union, extension, comprehension.. I think that's it), and $I$ is a set by hypothesis, and $\bigcup \{S_i: i\in I\}$ is a set by the axiom of union.

Now, we have proven that $\prod_{i\in I} S_i$ is really well-defined, i.e. it really is a set, i.e. it exists.

But of course, it may very well be that $\prod_{i\in I} S_i = \emptyset$. Of course, if any of the $S_i=\emptyset$, then the product will be empty.

But what if $S_i\not=\emptyset$ for all $i\in I$? Well, if $I$ is a finite set, it is easily proven in ZF that $\prod_{i\in I} S_i \not= \emptyset$. What if $I$ is infinite? Here is where the axiom of choice comes into play.

One of the formulations of (AC) is: given any family of non-empty sets, their product is non-empty.

We know that (AC) is independent from ZF: that implies that, without (AC), we can't prove that for every $I$ and every $\{S_i:i\in I\}$ of non-empty sets, the product is non-empty, nor that it is empty. In general, we need (AC).

Of course, there are some infinite sets $I$ and families $\{S_i:i\in I\}$ of non-empty sets for which we don't need (AC) to build a choice function and prove that the product is non-empty. See Asaf's comment on Russell's comment on shoes and socks.

As a last comment, Cohen showed that the axiom of countable choice (which is like (AC) but for countable index sets), a weaker version of (AC), also is independent from ZF.

Answer 2

The indexing set $I$ can be just any set.

The only "problem" is that when $I$ is large enough you cannot show that $\prod_{i\in I}S_i$ is non-empty and you need to introduce the Axiom of Choice just to assure that.