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Let $I$ be an infinite set. Suppose that for every $i\in I$, there exists a set $S_i$ satisfies a statement $\psi(S_i)$.

Here, is constructing a family of such $S_i$ (i.e. $\{S_i\}$ for $i\in I$) using Axiom of Choice??

For example, $\{V_i\}$ is given, suppose for every $i\in I$, there exists $G_i$ such that $V_i$ = $Y\cup G_i$. Then constructing $\{G_i\}$ is a result of AC?

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It varies.

If you can prove that there exists a unique $S_i$ then there is no need to use the axiom of choice.

On the other hand, think of $\psi(x) = \exists y(y\in x)$, then every non-empty set has this property, so one can easily construct an example where the axiom of choice is essential.

In the example, since $Y$ is a fixed parameter it is simple to define $G_i=V_i\setminus Y$. You can also take $G_i=V_i$ for all $i$.


For example, if $\cal A=\{A_i\mid i\in I\}$ is a family of non-empty and pairwise disjoint sets then $\varnothing$ can be in at most one element of the family. Without loss of generality, it is in none of the members. If $\cal A$ does not have a choice function then the sentence $\psi(x)=\exists y(y\in x)$ has at least one element in every $A_i$ which satisfy it, in fact $\{a\in A_i\mid \psi(a)\}=A_i$, so we cannot use this to choose from this family.

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Thanks, but I'm not really clear that if there exists a unique $S_i$, it's not a result of AC?? and how do i write down this process in mathematical language? –  Katlus Jul 24 '12 at 10:20
    
@Katlus: This really varies between the cases. It depends what are $V_i$ and what is $\psi$. It is impossible to just pull a general answer like that. –  Asaf Karagila Jul 24 '12 at 10:22
    
A standard way to avoid AC is to know which element you constructed. For example, if the sets in your collection are well-orderable without AC (e.g. any finite, N, Z, Q), then well-ordering them and taking the least element avoids AC. This is why we need more information on your property. –  trb456 Jul 24 '12 at 10:44
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@trb456: Another is simply a way where a canonical choice exists. For example in the product of $R$ modules you can always choose the zero element. –  Asaf Karagila Jul 24 '12 at 10:57
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@trb456: I should also remark that your comment is flat out wrong if you consider it closely. You agree that the number of ways to well-order a pair is finite. In fact it is exactly 2. However there can be a countable family of pairs that you cannot choose from. You need to be able and choose how to well-order the sets in a uniform way. –  Asaf Karagila Jul 24 '12 at 14:32

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