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Do we need Axiom of choice(or weaker version axiom of countable choice) to say countable Cartesian product of countable sets is nonempty? I think yes. I read somewhere answer no giving argument: each countable set can be well ordered and after well ordering each countable set we choose least element in each to prove their Cartesian product is non empty. But I see gap in this argument because there are many ways a countable set can be well ordered. So which way we will well order sets?

Asaf Karagila
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Sushil
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  • It is false in general that the countable product of countable sets is countable. Take for example $\prod_{\alpha\in \aleph_0}{2}=2^{\aleph_0}=|\mathbb{R}|$. – Hayden Nov 26 '14 at 15:08
  • @Hayden Sorry there was mistake in typing which completely changed the meaning of question. Now see it makes sense or not – Sushil Nov 26 '14 at 15:12
  • Ah, that makes much more sense. The argument you give does require the axiom of choice (you have to choose for each set a well-ordering, i.e. a bijection with $\mathbb{N}$). – Hayden Nov 26 '14 at 15:16
  • But bijection with N also we can define in many ways and I doubt in uncountable no. of ways – Sushil Nov 26 '14 at 15:18
  • Choosing a single bijection isn't an issue, even though there are many ways to do so. Finite choice only refers to the indexing set, not the sets being indexed, and this is provable without AC. It's only when you move to the infinite indexing set that AC (or one of its weaker versions) is needed. – Hayden Nov 26 '14 at 15:20
  • @Hayden Yes this I know finite choice don't need any axiom. But let me give argument now as: we first well ordered each set(without using AC or any other weaker version of AC) then we define choice function. So giving argument like this in 2 steps is okay? And so countable Cartesian product of countable set is nonempty(without AC or any weaker version) – Sushil Nov 26 '14 at 15:25
  • The problem is that you can't well-order a countable number of sets without at least some form of choice (in this case countable choice). – Hayden Nov 26 '14 at 15:27
  • possible duplicate of The cardinality of a countable union of countable sets, without the axiom of choice – Nate Eldredge Nov 26 '14 at 15:27
  • You are right. You need some amount of choice even if all sets have size two. See here. – Andrés E. Caicedo Nov 26 '14 at 15:28
  • @Hayden No but by definition of countable set there will exist a bijection of set with N. And N well ordered so set will get well ordered – Sushil Nov 26 '14 at 15:31
  • @AndresCaicedo So to say countable cartesian product of countable sets is nonempty(we need AC or weaker version of AC) – Sushil Nov 26 '14 at 15:41
  • @Sushil Yes. A version of the axiom of countable choice suffices, but some such assumption is needed. – Andrés E. Caicedo Nov 26 '14 at 15:48

2 Answers2

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Even when free the sets have a natural well order to them, the countable union of countable sets is not necessarily countable.

For example, in some models of $\sf ZF$ the first uncountable ordinal, $\omega_1$ is the countable union of countable ordinals.

And no, the countable product of finite sets doesn't have to be non-empty without choice, let alone that of countable sets. Not only that, it is true that the statement "every countable product of countable sets is non-empty" is strictly weaker than the axiom of countable choice.

In fact! It can Bethe case that the countable product of countable sets are non-empty, but there is a countable family of countable sets whose union is not countable. Because in the proof of the latter we choose from sets of size continuum, not just countable sets.

Asaf Karagila
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  • And 'Every countable product of finite set is nonempty' must be strictly weaker. Am I right? – Sushil Nov 26 '14 at 15:57
  • and just for reference axiom: 'every countable product of countable sets is non-empty' would it imply dedekind finite and finite sets are same? – Sushil Nov 26 '14 at 16:05
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    Yes to the first, no to the second (In Cohen's first model every countable family of countable sets has a non-empty product; but there is an infinite Dedekind-finite set of real numbers). – Asaf Karagila Nov 26 '14 at 16:31
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If you have a product $\prod_{i\in I}x_i$ of sets $x_i$ indexed by some set $I$, then an element in that product is exactly a choice of an element $y_i \in x_i$ for each $i \in I$. The axiom of choice (and its weaker counterparts) will therefore tell you exactly that such an element always exists.

Note that there are products that can be proven to be non-empty even without the axiom of choice. For instance, if all the $x_i$ are ordinals or subsets of ordinals.

And yes, choosing a well-ordering for each $x_i$ is (at least) as difficult as choosing a single element, logically speaking. So unless there's a distinguished well-ordering you can choose (for instance, if the $x_i$ are ordinals, then $\in$ defines a well ordering for all of them, or if the union $\bigcup_{i \in I}x_i$ admits a well-ordering, or $I$ is finite), then you're back where you started, and in need of choice.

Arthur
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    I think the OP is asking if, when $x_i$ is countable for each $i$ and $I$ is also countable, the result continues to hold even if we don't assume choice, or if it requires some weaker variant like countable choice. – Hayden Nov 26 '14 at 15:18
  • @Hayden Yes exactly – Sushil Nov 26 '14 at 15:26