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 infinitesets 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 Frankel set theory (ZF).

Godel 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.

(2003-07-11)