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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.

Asaf Karagila
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Manos
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    Considering all your questions in the comments I suggest you read through http://math.stackexchange.com/questions/15668/axiom-of-choice-examples 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

2 Answers2

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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.

Bruno Stonek
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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
  • @Manos: I'll add on the non-emptyness of the product to the answer, as per Asaf's suggestion. – Bruno Stonek Jul 14 '11 at 22:55
  • @Bruno: One final clarification to your post is that independence means that not only we cannot prove it is not empty, but we cannot prove it is empty either (directly from ZF that is). – Asaf Karagila Jul 14 '11 at 23:04
  • @Asaf: indeed, I added it for clarification, thank you. – Bruno Stonek Jul 14 '11 at 23:07
  • @Bruno: Ok, but as far as i know, there does not exist a definition of what a set is. So, i understand that i might define "something" that does not exist, but how will i be able to prove that something is a set, when i don't really have a formal definition of what a set is? – Manos Jul 14 '11 at 23:09
  • @Manos: A set in this context is an element of the universe of set theory, that is an object which is an element of its power set. – Asaf Karagila Jul 14 '11 at 23:11
  • @Manos: great question. How do you know it? Axioms! For example, $\mathbb{N}$ is a set (axiom of infinity). Also, the power set of any set is a set (axiom of power set). A subset of a set is a set (axiom of comprehension). The range of a function is a set (axiom of replacement). (I'm being deliberately non-formal). You can prove from the usual axioms of ZF that $\emptyset$ is a set. So, we know there are some things which are sets by axioms, and then we know how to build more sets from them, by axioms which allow us some constructions. – Bruno Stonek Jul 14 '11 at 23:14
  • I suggest you take a look at http://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory – Bruno Stonek Jul 14 '11 at 23:16
  • I really like this answer. – t.b. Jul 14 '11 at 23:20
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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.

Andrea Mori
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  • Could you please explain this a little bit more? – Manos Jul 14 '11 at 22:02
  • @Manos: you may consult http://en.wikipedia.org/wiki/Axiom_of_choice – Andrea Mori Jul 14 '11 at 22:05
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    @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: It is not necessary that the product of infinitely many non-empty sets is non-empty. The assertion that every product of non-empty sets is a non-empty set is called "The Axiom of Choice" and it is used thoroughly in modern mathematics. – Asaf Karagila Jul 14 '11 at 22:11
  • @Asaf: I am familiar with the Axiom of Choice, i have even attended a whole lecture on it. But the truth is i can not grasp its significance. For example: if we have a family of sets $\left{S_i\right}$ that is nonempty, then each of these sets contains at least an element $x_i$. So the product of the sets must contain an element $(x_i)$ and so it is non-empty. What am i missing? – Manos Jul 14 '11 at 22:15
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    @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
  • @Manos: exactly, you're choosing an element from each set. But how do you operate such choice? If there's some uniform way of choosing an element from each set you happily conclude. But if you have infinitely many "unrelated" sets to choose from, how do you complete your choice? You start choosing an element from a set, then another from a second set, and so on, but the procedure never stops and can never be completed. – Andrea Mori Jul 14 '11 at 22:24
  • @Asaf, Andrea: thank you people, i think i understand this better now. – Manos Jul 14 '11 at 22:28
  • @Andrea: but if we have a uniformal way of choosing our elements (e.g. left shoe), then the procedure of choosing the element from infinite sets is done automatically, by characterising the element, right? – Manos Jul 14 '11 at 22:34
  • @Asaf: So when you say "large enough" you mean infinite (both countable and uncountable), right? – Manos Jul 14 '11 at 22:44
  • Exactly. Actually, you can cook up many situations in which choices can be made. An important example is when all the $S_i$ are the same set $S$ in which case the elements of the product are the functions $I\rightarrow S$ which we know how to construct! – Andrea Mori Jul 14 '11 at 22:45
  • @Manos: I never said large enough, but yes. If the index set is infinite there is no guarantee for a choice function to exist (for arbitrary $S_i$, that is. As mentioned here before, if $S_i$ have some property in common it might be possible). Just to make things worse, there is can be a countable family of pairs with no choice function. – Asaf Karagila Jul 14 '11 at 22:52
  • @Andrea: So, to see if i get your point, for non-empty $I,S$ we can certainly define a function $f:I \rightarrow S$ and this is certainly an element of the cartesian product, which as a result it is non-empty. Right? – Manos Jul 14 '11 at 22:58
  • @Manos: For every $s\in S$ you have a function assigning $s_i = s$, so $S^I$ is never empty. The trouble begins, again, when $I$ is an infinite set and $S_i$ are sets which do not "play nice" with one another (that is - there is no uniform way to define an element from each $S_i$) – Asaf Karagila Jul 14 '11 at 23:10