I have this form of the axiom of choice:
Suppose that C is a collection of nonempty sets. Then thre exists a function $f:C\rightarrow \cup_{A\in C}A$ such that $f(A)\in A$ for each $A \in C$.
I have this exercise:
Show that the axiom of choice is equivalent to the following statement : If $\{A_i\}_{i\in I}$ is any indexed collection of nonempty sets, then $\times _{i \in I}A_i \ne \emptyset$.
Now, having the axiom of choice, I think it is ok to show that the second statement follows from the first one. But assuming the second statement, and also assuming that we have the collection C, I have a little trouble showing the axiom of choice. Because I can not just assume that I can index the sets?, and create an index set?, or can I?
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$\begingroup$If you have a set $C$ which you want to index as $\{A_i\}_{i\in I}$, the natural choice would be to set $I=C$ and $A_c=c$. There's no rule that says you can't, and it certainly functions. (And one would only tend to care about the fact that an indexed collection is different from a set if they wished to include some members multiple times)
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