axiom of choice [ ¦ak·sē·əm əv ′chȯis] (mathematics) The axiom that for any familyAof sets there is a function that assigns to each setSof the familyAa member ofS.
| (mathematics) | Axiom of Choice - (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 Fränkel set theory (ZF).
Gödel 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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