# Axiom of Choice

(redirected from*Choice axiom*)

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## axiom of choice

[¦ak·sē·əm əv ′chȯis] (mathematics)

The axiom that for any family

*A*of sets there is a function that assigns to each set*S*of the family*A*a member of*S*.## Axiom of Choice

(mathematics)(AC, or "Choice") An axiom of set theory:

If X is a set of sets, and S is the union of all the elements of X, then there exists a function f:X -> S such that for all non-empty x in X, f(x) is an element of x.

In other words, we can always choose an element from each set in a set of sets, simultaneously.

Function f is a "choice function" for X - for each x in X, it chooses an element of x.

Most people's reaction to AC is: "But of course that's true! From each set, just take the element that's biggest, stupidest, closest to the North Pole, or whatever". Indeed, for any finite set of sets, we can simply consider each set in turn and pick an arbitrary element in some such way. We can also construct a choice function for most simple infinite sets of sets if they are generated in some regular way. However, there are some infinite sets for which the construction or specification of such a choice function would never end because we would have to consider an infinite number of separate cases.

For example, if we express the real number line R as the union of many "copies" of the rational numbers, Q, namely Q, Q+a, Q+b, and infinitely (in fact uncountably) many more, where a, b, etc. are irrational numbers no two of which differ by a rational, and

Q+a == q+a : q in Q

we cannot pick an element of each of these "copies" without AC.

An example of the use of AC is the theorem which states that the countable union of countable sets is countable. I.e. if X is countable and every element of X is countable (including the possibility that they're finite), then the sumset of X is countable. AC is required for this to be true in general.

Even if one accepts the axiom, it doesn't tell you how to construct a choice function, only that one exists. Most mathematicians are quite happy to use AC if they need it, but those who are careful will, at least, draw attention to the fact that they have used it. There is something a little odd about Choice, and it has some alarming consequences, so results which actually "need" it are somehow a bit suspicious, e.g. the Banach-Tarski paradox. On the other side, consider Russell's Attic.

AC is not a theorem of Zermelo Fr?nkel set theory (ZF). G?del and Paul Cohen proved that AC is independent of ZF, i.e. if ZF is consistent, then so are ZFC (ZF with AC) and ZF(~C) (ZF with the negation of AC). This means that we cannot use ZF to prove or disprove AC.

If X is a set of sets, and S is the union of all the elements of X, then there exists a function f:X -> S such that for all non-empty x in X, f(x) is an element of x.

In other words, we can always choose an element from each set in a set of sets, simultaneously.

Function f is a "choice function" for X - for each x in X, it chooses an element of x.

Most people's reaction to AC is: "But of course that's true! From each set, just take the element that's biggest, stupidest, closest to the North Pole, or whatever". Indeed, for any finite set of sets, we can simply consider each set in turn and pick an arbitrary element in some such way. We can also construct a choice function for most simple infinite sets of sets if they are generated in some regular way. However, there are some infinite sets for which the construction or specification of such a choice function would never end because we would have to consider an infinite number of separate cases.

For example, if we express the real number line R as the union of many "copies" of the rational numbers, Q, namely Q, Q+a, Q+b, and infinitely (in fact uncountably) many more, where a, b, etc. are irrational numbers no two of which differ by a rational, and

Q+a == q+a : q in Q

we cannot pick an element of each of these "copies" without AC.

An example of the use of AC is the theorem which states that the countable union of countable sets is countable. I.e. if X is countable and every element of X is countable (including the possibility that they're finite), then the sumset of X is countable. AC is required for this to be true in general.

Even if one accepts the axiom, it doesn't tell you how to construct a choice function, only that one exists. Most mathematicians are quite happy to use AC if they need it, but those who are careful will, at least, draw attention to the fact that they have used it. There is something a little odd about Choice, and it has some alarming consequences, so results which actually "need" it are somehow a bit suspicious, e.g. the Banach-Tarski paradox. On the other side, consider Russell's Attic.

AC is not a theorem of Zermelo Fr?nkel set theory (ZF). G?del and Paul Cohen proved that AC is independent of ZF, i.e. if ZF is consistent, then so are ZFC (ZF with AC) and ZF(~C) (ZF with the negation of AC). This means that we cannot use ZF to prove or disprove AC.

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