One of the standard statements of the axiom of choice is the following:

Let S be a given nonvoid `universal' set. To each i in a nonvoid set I associate a nonvoid subset X_i of S. (Other wordings are `` let (X_i:i\in I) be a family of subsets S)". Then there is a function f that maps each i to some element x_i in X_i (in other words the cartesian product \prod X_i is non void).

In my opinion this statement does not make sense, since in the hypothesis one already uses a special case of the conclusion. The statement "one associates X_i" is nothing but the assumption that there is a map iota from I into P(S), the power set of S. In other words, iota is an element of P(S)^I, the cartesian product of I copies of P(S). Thus a special case of the axiom of choice serves as the hypotheses for what is claimed....

How to go around in this version of the axiom of choice?


That is not a real problem. You are given a function which assigns indexes to a set. Not to mention, that if $A$ is any set at all, then it can be indexed by itself. For example $\Bbb N$ can be indexed using itself, where each element is its own index. This works for other sets as well (only you don't have a notion of first, second, and so on; and you don't need one either).

You could also say that it is a problem that when you talk about a limit of a uniformly convergent sequence of continuous functions being continuous, you're using the conclusion since each function in the sequence is a limit of such sequence on its own.

No. This is certainly not a problem. Suppose you are given such a family of non-empty sets. Then there is a function which chooses one element from each member of that family.

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    $\begingroup$ Indeed. Also, we can construct many families of non-empty sets, for which it is not obvious without AC that there is a choice function (e.g., we can easily construct the set of equivalence classes of $\mathbb{R}/\mathbb{Q}$, but finding a set of representatives is more difficult). $\endgroup$ – Carl Mummert Sep 18 '15 at 9:51

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