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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.

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Considering all your questions in the comments I suggest you read through and other questions tagged under [axiom-of-choice] and if you cannot find satisfactory answers you can and should ask another question. – Asaf Karagila Jul 14 '11 at 23:17
up vote 6 down vote accepted

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.

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Yes, thank you, i find this definition very enlightening. – Manos Jul 14 '11 at 22:00
@Manos: you're welcome. I added some formality to the answer. – Bruno Stonek Jul 14 '11 at 22:14
@Bruno: For sake of completeness I think that you should add that this product is possibly empty unless assuming the axiom of choice. – Asaf Karagila Jul 14 '11 at 22:33
When you say that "the set exists", you mean that the set is non-empty, right? But then we might have a subset of the power set that is empty!? Where does the axiom of choice mentioned in the other answer comes into play in your analysis? – Manos Jul 14 '11 at 22:39
@Manos: No, I really meant exists. Not everything you define naïvely is a set ("the set of all sets" is not a set! Russell's paradox), you actually have to prove from the very axioms of set theory that you have defined something which really is a set... For a classical example, if $A$ and $B$ are sets, then as I said above, $A\times B:= \{(a,b):a\in A,b\in B\}$ is a set, but you have to prove it. If you describe a set by $\{x:\varphi(x)\}$ where $\varphi$ is a formula, you have to prove it exists: it may not exist (again, Russell's paradox: $\{x: x\notin x\}$ is not a set). – Bruno Stonek Jul 14 '11 at 22:52

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.

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Could you please explain this a little bit more? – Manos Jul 14 '11 at 22:02
@Manos: you may consult – Andrea Mori Jul 14 '11 at 22:05
@Asaf: I feel that saying "not trivial" is somewhat misleading (one may be induced to think that a clever proof does indeed exist). In fact, the non-emptyness of the product cannot be proved at all and it needs to be introduced axiomatically. – Andrea Mori Jul 14 '11 at 22:08
@Andrea: I agree completely. As the time limit for editing comments have gone, I reposted my comment in a better phrasing. – Asaf Karagila Jul 14 '11 at 22:10
@Manos: When there exists a way to uniformly define a special element (least element, zero element, and so on) then there is in fact a choice function. As Russell said - it is always possible to choose from infinitely many pairs of shoes; it is not always possible to choose from infinitely many pairs of socks. If your sets are such that you cannot distinguish the elements in a "nice" way, then you cannot choose from infinitely many of them. A unique minimal element is a very nice way to distinguish $x_i\in S_i$ from the rest of the herd. – Asaf Karagila Jul 14 '11 at 22:21

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