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Could someone explain to me why it is important for a set to be non-empty when working with a choice function?

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    $\begingroup$ Because you can't choose a member of the empty set. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 24 '15 at 19:45
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Because a choice function, working on $\emptyset$, would be a function for which $f(\emptyset)\in\emptyset$, which is obviously untrue.

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Well, this depends on your definition of a choice function for a set $A$. I use the following definition, which I shall use in this answer:

A choice function for a set $A$ is a function $f:\mathcal P \left({A}\right)-\{\emptyset\} \rightarrow A$ such that $f(B) \in B$ for every non-empty subset $B$ of $A$.

Using this definition, there is indeed a choice function for the empty set, namely, the empty set.

If I totally misunderstood your question and you are asking why we can't have the empty set in the domain of a choice function, then the first post answers that.

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