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Let $I$ be a non-empty set. $\kappa_i$ is non-zero cardinal number for all $i \in I$.

If without AC, then $\prod_{i \in I}\kappa_i=0$ seems can be true(despite I still cannot believe it).

But what property should $I$ and $\kappa_i$ have?

Can $\prod_{i \in I}\kappa_i\ne 0$ be proved without AC when $I$ and each $\kappa_i$ all is well-orderable?

Conversely if $I$ is not well-orderable, or if some $\kappa_i$ is not well-orderable, is $\prod_{i \in I}\kappa_i=0$ definitely holds?

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If each $\kappa_i$ is well-ordered, then the product is non-zero no matter what $I$ is; and if $I$ is finite, the product is non-zero as long as each $\kappa_i$ is non-zero. Otherwise the axiom of choice is required. –  Zhen Lin Nov 10 '12 at 14:09
@ZhenLin: $\kappa_i=2$ and $I=\omega$ may fail. –  Asaf Karagila Nov 10 '12 at 15:54
Note the difference between "well-orderable" (any finite set, for example) and "well-ordered". –  Andres Caicedo Nov 10 '12 at 16:43
@Andres: I'm not clear whether you made the comment towards me or not, I will add on your comment that when talking about cardinals we don't have concrete sets and well-ordered cardinals are simply cardinals of well-orderable sets. As this question is about cardinals, well-orderability need not play a particular role in assuring a choice function. –  Asaf Karagila Nov 10 '12 at 16:47
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4 Answers

up vote 2 down vote accepted

In its most general form, the axiom of choice can fail on a countable $I$ where $\kappa_i=2$ for all $i$. We can add assumptions, for example choice from countable families (countable choice), but then it could fail for a family of $\aleph_1$ disjoint pairs.

We could add the assumption that every family of finite sets has a choice function, but then it is still possible that a countable family of countable sets would have an empty product.

In some of the models, e.g. Cohen's first model, every well-ordered $I$ where $\kappa_i$ are well-ordered for $i\in I$ has a non-empty product. However, the axiom of countable choice fails. In fact there exists an infinite Dedekind-finite set of real numbers.

It is also known that we may assume that for every well-ordered family $I$ there exists a choice function, but this is consistent with the assumption that there is a set $A$ such that $\aleph_1\nleq|A|$ and $|A|\nleq\aleph_1$. We could then use $A$ to generate a family whose index set is not an ordinal, and whose product is empty.

Generally, the only properties you can deduce are these:

  1. $I$ is infinite.
  2. $\{i\in I\mid\kappa_i>1\}$ is infinite.

If you wish to point at specific sets, then you may deduce a bit more (e.g. there is no uniform to choose a structure on these sets), but generally just to assume something about the cardinalities is not sufficient.

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Should a product be empty where its index set is not an ordinal? –  Popopo Nov 12 '12 at 8:42
No. Of course not. It is consistent to have a choice function from which are not well-ordered. In such case we may arrange products of cardinals which are not empty over non-ordinal index sets. –  Asaf Karagila Nov 12 '12 at 8:47
I see, so $I$ is well-orderable is neither sufficient nor necessary condition of that such the product is non-zero. –  Popopo Nov 12 '12 at 9:05
Indeed. Unless it is finite we cannot deduce much without any form of AC. :-) –  Asaf Karagila Nov 12 '12 at 9:07
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Even if you don't assume the existence of AC, that doesn't mean the opposite is true. The existence of choice function when one is not constructible, or the existence of a set that is defined as having no choice function, are first-order statements that are independent from the rest of ZF.

If you can construct a theorem about sets without using AC, it can be used both in models that incorporate AC, and those that incorporate its negation. By discussing a zero product of non-zero sets, you already assume that such a construct exists. The property these sets would have, is the property you just mentioned. It's not necessarily intuitive. It doesn't stem from other properties you are familiar with.

Let me give you an example to demonstrate what I'm saying. ℤ can be seen as an extension of ℕ. In the set ℕ, we can say that there exists no number $n$ holding $n < 0$. In the set ℤ there exists such a number.

We could then prove theorems such as, $a + b ≥ a$. In ℤ, where the opposite holds, one can prove that there exist $a,b$ such as that $a+b<a$. These theorems are mutually exclusive given the other laws of ℕ and ℤ. However, you could also imagine a set $ℕ_?$ which would not posit anything at all about the existence of the number $n$ holding $n<0$. While in ℕ and ℤ either the statement or its inverse will hold, $N_?$ doesn't guarantee either.

Technically, you could define $ℕ_?$ as an extension to $ℕ$ in which the number 0 has zero or more predecessors, e.g. $\{-1,-2,-3\}∪ℕ$. It may or may not contain a number $n$ that satisfies $n < 0$, and there is no way of showing that it does, but it is unbounded from above, well-ordered, and has a least element, though there are serious limitations about how it can be used.

A proof in $N_?$ could be that for every number $n$ there exists a number $m$ such as that $m>n$, because that is independent of the existence of negative numbers, only on the fact that $N_?$ is well-ordered and is unbounded from above.

You could thus similarly ask that, given you know only ℕ, what sorts of qualities the numbers $a,b$ would have, if they hold $a+b<a$? The question is problematic. In ℤ we can say that negative numbers have certain definite qualities because that's how we define them. In $ℕ_?$ it's not possible to answer the question, except tautologically, because we don't know if any elements that satisfy the relation exist.

You could give a weaker version of AC in which, say, all countable sets have a choice function. In that case, you would definitely be able to say that if the product of two non-empty sets is empty, then one of them must be uncountable. Otherwise, you really cannot, by definition, derive anything at all about sets that satisfy the property.

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Er...I'm not quite understand, what is $N_?$? –  Popopo Nov 10 '12 at 15:28
I editted the answer to clarify what I was talking about. –  Greg Ros Nov 10 '12 at 16:05
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The question is based on presuppositions that might not be true in the absence of AC. Let's consider the simplest non-trivial case, the product of countably many copies of 2, that is, $\prod_{n\in\mathbb N}\kappa_n$ where $\kappa_n=2$ for all $n$. A reasonable way to define this product would be: Take a sequence of sets $A_n$ of the prescribed cardinalities $\kappa_n$, let $P$ be the set of all functions $f$ that assign to each $n\in\mathbb N$ an element $f(n)\in A_n$, and then define the product to be the cardinality of $P$. Unfortunately, the cardinality of $P$ can depend on the specific choice of the sets $A_n$.

On the one hand, it is consistent with ZF that there is a sequence of 2-element sets $A_n$ for which there is no choice function; that is, the $P$ defined above is empty. So these $A_n$'s lead to a value of 0 for the product.

On the other hand, we could take $A_n=\{0,1\}$ for all $n$, and then there are lots of elements in $P$, for example the constant function with value 0. Indeed, for any subset $X$ of $\mathbb N$, its characteristic function is a member of $P$. The resulting value for the product of countably many 2's would then be the cardinality of the continuum.

The moral of this story is that, in order for infinite products to be well-defined, one needs AC (or at least some special cases of it), even when the index set and all the factors in the product are well-orderable.

Digging into the problem a bit more deeply, one finds that the natural attempt to prove that "the cardinality of $P$ is independent of the choice of $A_n$'s" involves the following step. If we have a second choice, say the sets $B_n$, and we know that each $A_n$ has the same cardinality as the corresponding $B_n$, so we know that there are bijections $A_n\to B_n$ for all $n$, then we need to fix such bijections --- to choose a specific such bijection for each $n$. Then those chosen bijections can be used to define a bijection between the resulting two versions of $P$. But choosing those bijections is an application of the axiom of choice.

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Er... If $\kappa_n=2$, does this followed by $\kappa_n=\{0,1\}$ since $2=\{0,1\}$ in set theory? –  Popopo Nov 11 '12 at 7:17
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If all sets of reals are Lebesgue measurable, then the product $ \Pi_{x\in \mathbb{R/Q}}x $ , where $\mathbb{R/Q}=\{\{x\in\mathbb{R}|\exists y\in\mathbb{Q}.x+y=z\}|z\in\mathbb{R}\}$, is empty by Vitali's argument. If it is consistent for an inaccessible cardinal to exist, then ZF+"all sets of reals are Lebesgue measurable" + Dependent Choice(A weak form of the axiom of choice) is consistent, according to a theorem of Solovay.

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