Why is the axiom of choice controversial?

Question

In other words, what are the arguments for ZF over ZFC, and what philosophical issues have people raised against including it as a standard axiom of set theory?

^thanks Noah, you're right; my question is a duplicate. Thanks for pointing me the right way — Owain West, Jun 25 '16 at 22:29
It has to do with the theorems that are enabled by AC; they can seem a little bizarre. — zhw., Jun 25 '16 at 22:30
You might also want to read http://math.stackexchange.com/questions/1253650/is-the-axiom-of-choice-really-all-that-important and the many other long answers about why the axiom of choice leads to weird consequences (spoiler: it's not choice; it's infinity), or why the axiom is useful and what might happen without it. — Asaf Karagila, Jun 25 '16 at 22:35

change_picture · Answer 1 · 2016-06-25T22:37:16.683

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This axiom is controversial because although it seems like a relatively intuitive idea, there are still some issues. One of the best ways of understanding it is in this way: Take the set of all pairs of shoes and find a way to pick one shoe from each pair in order to form a new set. Easy! Just let our choice function be take every right shoe from each pair and you get the set of all right shoes. Consider this now: Take the set of all pairs of socks. The axiom of choice tells us there is a way to pick one sock from each pair to form a new set, but how you make that choice is not easy to describe. So although the axiom says a choice function always exists, what that choice function can be or how it is defined is not always apparent or even impossible to define in many cases.

Some more things: The axiom of choice is indeed an extremely useful axiom in many areas of math. However, it gives rise to the Banach-Tarski Paradox and the existence of nonmeasurable subsets of the real numbers. Also, the axiom of choice is equivalent to the statement that any set can be well-ordered, i.e., every nonempty set can be endowed with a total order such that every nonempty subset has a least element. Therefore, this means that $\mathbb{R}$ can be well-ordered. However, no one has ever been able to explicitly state how one does this, and if I'm not mistaken, it may be impossible to do so with the current axioms we have available. I also want to add that one person in the comments section mentioned, we need AC in order to talk about cardinalities of infinite sets.

edited Jun 25 '16 at 22:37

answered Jun 25 '16 at 22:30

change_picture

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...from the well-ordering theorem one may deduce the axiom of choice, but from the axiom of choice one cannot deduce the well-ordering theorem. – Count Iblis Jun 25 '16 at 23:20
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Are you sure about that @CountIblis ? I have never attempted the proof myself but many sources seem to agree the two are equivalent. – change_picture Jun 25 '16 at 23:36
I think they are equivalent when you assume some of the other axioms. – Count Iblis Jun 26 '16 at 00:04
Axiom of choice looks suspiciously similar to https://en.wikipedia.org/wiki/Gibbs_paradox
Does it feel like these are the sides of the same coin?
– Brian Cannard Jun 19 '20 at 05:59
Oh, if it gives rise to the Banach-Tarski Paradox, this set of slides may highlight a set of topics to learn: https://www.presentica.com/ppt/21289/a-top-down-take-a-gander-at-the-banach-tarski-catch-22 – Brian Cannard Jun 19 '20 at 06:02
It's well known that Gibbs paradox is resolved in quantum mechanics. Oh, no! Does it break the well-ordering theorem, and so the axiom of choice? – Brian Cannard Jun 19 '20 at 06:05

Why is the axiom of choice controversial?

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