Motivating implications of the axiom of choice?

Solution 1:

Each of the following is equivalent to the Axiom of Choice:

  • Every vector space (over any field) has a basis.

  • Every surjection has a right inverse.

  • Zorn's Lemma.

The first is extremely important and useful. The third is used all the time, especially in algebra, also very important and useful.

You could write an entire book on important consequences (and equivalents) of the Axiom of Choice.

Unfortunately, any publisher worth his salt would reject it, since both have already been written:

  1. Rubin, Herman; Rubin, Jean E. Equivalents of the axiom of choice. North-Holland Publishing Co., Amsterdam 1963 xxiii+134 pp.

  2. Rubin, Herman; Rubin, Jean E. Equivalents of the axiom of choice. II. Studies in Logic and the Foundations of Mathematics, 116. North-Holland Publishing Co., Amsterdam, 1985. xxviii+322 pp. ISBN: 0-444-87708-8

  3. Howard, Paul; Rubin, Jean E. Consequences of the axiom of choice. Mathematical Surveys and Monographs, 59. American Mathematical Society, Providence, RI, 1998. viii+432 pp. ISBN: 0-8218-0977-6

These books are probably not the best place to start, though; the first book is okay, listing some of the most important equivalents as they were known before Cohen's work, but at least the last is pretty difficult to slough through.

If you want a good introduction to the Axiom of Choice and some idea of its uses, Horst Herrlich's Axiom of Choice, Lecture Notes in Mathematics v. 1876, Springer-Verlag (2006) ISBN: 3-540-30989-6 is pretty good, discussing some of the bad things that happen if you don't have AC, some of the bad things that happen if you do have AC, and some alternative axioms that contradict AC but lead to very nice theorems.

Solution 2:

Axiom of Choice $\iff$ A non-empty product of non-empty sets is non-empty.

Solution 3:

The Axiom of choice is required to prove for example that every ring has a maximal ideal (or a minimal prime), or the existence of an algebraic closure.

Solution 4:

There Lebesgue measure is non-trivial - there are non-measurable sets in $\mathbb{R}$.