<
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 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(
Axiom of ChoiceC) (ZF with the negation
of AC). This means that we
cannot use ZF to prove or disprove AC.
(2003-07-11)