| 单词 | axiom of choice |
| 释义 | Axiom of ChoiceAxiom of ChoiceAxiom of Choiceaxiom of choice[¦ak·sē·əm əv ′chȯis]Axiom of Choice(mathematics)If X is a set of sets, and S is the union of all the elementsof X, then there exists a function f:X -> S such that for allnon-empty x in X, f(x) is an element of x. In other words, we can always choose an element from each setin a set of sets, simultaneously. Function f is a "choice function" for X - for each x in X, itchooses 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 setin turn and pick an arbitrary element in some such way. Wecan 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 theconstruction or specification of such a choice function wouldnever end because we would have to consider an infinite numberof separate cases. For example, if we express the real number line R as theunion 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 whichdiffer by a rational, and Q+a == q+a : q in Q we cannot pick an element of each of these "copies" withoutAC. An example of the use of AC is the theorem which states thatthe countable union of countable sets is countable. I.e. ifX is countable and every element of X is countable (includingthe possibility that they're finite), then the sumset of X iscountable. AC is required for this to be true in general. Even if one accepts the axiom, it doesn't tell you how toconstruct a choice function, only that one exists. Mostmathematicians are quite happy to use AC if they need it, butthose who are careful will, at least, draw attention to thefact that they have used it. There is something a little oddabout Choice, and it has some alarming consequences, soresults which actually "need" it are somehow a bit suspicious,e.g. the Banach-Tarski paradox. On the other side, considerRussell'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) andZF(~C) (ZF with the negation of AC). This means that wecannot use ZF to prove or disprove AC. |
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