A choice function on simply picks one element from each member of . So in order for to have a choice function, every member of must be a nonempty set. The Axiom of Choice (http://planetmath.org/AxiomOfChoice) (AC) states that every set of nonempty sets does have a choice function.
Without AC the situation is more complicated, but we can still show that some sets have a choice function. Here are some examples:
If is a finite set of nonempty sets, then we can construct a choice function on by picking one element from each member of . This requires only finitely many choices, so we don’t need to use AC.
If every member of is a well-ordered nonempty set, then we can pick the least element of each member of . In this case we may be making infinitely many choices, but we have a rule for making the choices, so AC is not needed. The distinction between “well-ordered” and “well-orderable” is important here: if the members of were merely well-orderable, we would first have to choose a well-ordering of each member, and this might require infinitely many arbitrary choices, and therefore AC.
If every member of is a nonempty set, and the union is well-orderable, then we can choose a well-ordering for this union, and this induces a well-ordering on every member of , so we can now proceed as in the previous example. In this case we were able to well-order every member of by making just one choice, so AC wasn’t needed. (This example shows that the Well-Ordering Principle, which states that every set is well-orderable, implies AC. The converse is also true, but less trivial — see the proof (http://planetmath.org/ProofOfZermelosWellOrderingTheorem).)
|Date of creation||2013-03-22 14:46:26|
|Last modified on||2013-03-22 14:46:26|
|Last modified by||yark (2760)|