Page "Axiom of choice" Paragraph 26
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In that case it is equivalent to saying that if we have several ( a finite number of ) boxes, each containing at least one item, then we can choose exactly one item from each box.
Clearly we can do this: We start at the first box, choose an item ; go to the second box, choose an item ; and so on.
The result is an explicit choice function: a function that takes the first box to the first element we chose, the second box to the second element we chose, and so on.
( A formal proof for all finite sets would use the principle of mathematical induction to prove " for every natural number k, every family of k nonempty sets has a choice function.
") This method cannot, however, be used to show that every countable family of nonempty sets has a choice function, as is asserted by the axiom of countable choice.
If the method is applied to an infinite sequence ( X < sub > i </ sub >: i ∈ ω ) of nonempty sets, a function is obtained at each finite stage, but there is no stage at which a choice function for the entire family is constructed, and no " limiting " choice function can be constructed, in general, in ZF without the axiom of choice.
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