When you invoke the axiom of choice, it is because whatever you need to choose consists of small parts you technically have to choose one by one. So if I need to use the axiom of choice, but subsequently prove that no matter what I choose the end result is the same, is there really a need for the axiom of choice?

This is where I bumped into the question: You have a set of abelian groups $G_i$ each with a given subgroup $H_i$, with $i\in I$ for some index set $I$. Show that the following isomorphism holds: $$ \prod_{i\in I}\left(G_i/H_i\right) \approx\left(\prod_{i\in I}G_i\right)/\left(\prod_{i\in I}H_i\right). $$ And here's my proof:

An element in $g\in \prod_{i\in I}\left(G_i/H_i\right)$ is given by a (possibly infinite) tuple of elements of each quotient group. Each coordinate in the tuple is represented by an element $g_i$ in the corresponding $G_i$, which means that we can use the representatives to make an element in $\prod_{i\in I}G_i$ (this requires the axiom of choice if $I$ is big enough).

Now, the difference between any two such elements is a tuple in $\prod_{i\in I}H_i$, so in the end what choice we make doesn't matter, any choice will give the same element in the final quotient group. But I still used the axiom to show that a representative existed, so my intuition says that this needs the axiom of choice, although I'm not certain.

I use the word "tuple" here, because that's how I think of direct products. Representing the product as a function from the index set into the disjoint union (which is the proper way) will give a completely analogous reasoning.


Yes, there is still a need for the axiom of choice, even if any choice works. Actually, the axiom of choice is equivalent to the statement: "the Cartesian product of any family of nonempty sets is nonempty." So without the axiom of choice, it's not just that you can't pick the objects you want -- they might not exist.

Consider the situation with non-measurable sets. Any transversal of $\mathbb{R}/\mathbb{Q}$ gives rise to a non-measurable set. However, without choice, it is possible to have a model of set theory without non-measurable sets.

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  • $\begingroup$ Is "the Cartesian product of any family of nonempty sets is nonempty." also equivalent to the axiom of choice in the respect that for finite or countable index sets it's unnecessary? $\endgroup$ – Arthur Nov 11 '12 at 21:10
  • $\begingroup$ The axiom of choice is not unnecessary in the countable case -- it's just that there is a strictly weaker axiom, the axiom of countable choice, that gives us choice in those situations with a little less moral heartache. In the finite case, you don't need choice. Similarly, the Cartesian product of finitely many nonempty sets is always nonempty, without axiomatic considerations, and countable choice is equivalent to the Cartesian product of a countable family of nonempty sets being nonempty. $\endgroup$ – Elchanan Solomon Nov 11 '12 at 21:13
  • $\begingroup$ Let me see if I get this straight: If I assume the direct product is well defined, then I will implicitly have assumed the axiom of suitable choice, so there is no problem with choosing representatives? $\endgroup$ – Arthur Nov 11 '12 at 21:18
  • $\begingroup$ Well, the direct product is always well-defined. It just may be empty. But if you assume that it is not empty, then you can just take that element, and it will correspond to a choice. $\endgroup$ – Elchanan Solomon Nov 11 '12 at 21:55

Yes. You may still have to use the axiom of choice.

The best example, in my opinion, is the Vitali set. The matter what choice function from $\mathbb{R/Q}$ is used the result is non-measurable.

However you still need to use the axiom of choice to have such choice function (as it is consistent without the axiom of choice that there is no such choice).

Another, slightly less striking, argument is with vector spaces.

Suppose that $V\to W$ is a linear surjection, then using the axiom of choice it splits. No matter what is the choice of basis of the kernel, it will be the same space.

Without the axiom of choice it is consistent to have $V$ with a basis, $W$ with a basis, but the kernel of the surjection without a basis.

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