9

I just encountered this:

"(Theorem of A. H. Stone) Every metric space is paracompact... Existing proofs of this require the axiom of choice... It has been shown that neither ZF theory nor ZF theory with the axiom of dependent choice is sufficient"

I understand that it's standard to mention whenever AC is used to prove any results. I also read here on this site that it is due to the fact that undesired stuff like the Banach-Tarski paradox happens when you assume AC.

My question is, as a non-set theorist, how should one think about results that depend on the axiom of choice?

I'm sorry if this question is too broad but please realize that part of the reason I'm asking it is because I'm not sure what is the real question here...

ADDED: There's a great thread on MO on that very question!

Community
  • 1
  • 2
    If you want links, there are dozens of link lists on answers of mine on this site. You can probably start from here and here and traverse the "Linked" list on the right hand side of the screen, and the actual links in the answer. – Asaf Karagila Dec 23 '13 at 20:21

2 Answers2

4

It depends on your philosophical bent, really. And what can't tell you how to think about things philosophically. You need to learn enough and figure it out on your own.

The axiom of choice is a tool. But it is a magical tool, like the summoning of Satanic, or Divine, or non-Judeo-Christian good/evil dichotomy related spirits.

Okay, I digress. The point of the axiom of choice is that it allows non-constructive results. It allows us to prove the existence of objects without specifying their exact "form". To some people it bothers, to others it bothers less.

What does it mean that a result depends on the axiom of choice? It really just means that with the usual foundation of $\sf ZF$ set theory we cannot point out at the objects that will prove that result (e.g. a choice function, or a well-order, or a basis for a vector space).

Of course sometimes we can prove limited and partial results without the axiom of choice. In fact, often we can prove the results that we want for well-behaved enough spaces (where "enough" depends on the result, of course). But assuming the axiom of choice allows us to prove much more general statements which are much easier to state. Let me give you an example.

Theorem. ($\sf ZF$) Suppose that $K$ is a field, and $K$ can be well-ordered, and suppose that $V$ is a vector space over $K$ such that $V$ can also be well-ordered, then there is a basis $B$ for $V$.

On the other hand...

Theorem. ($\sf ZFC$) Suppose that $V$ is a vector space over a field $K$. Then $V$ has a basis.

So it really just helps us to clean out a lot of assumptions which may or may not be true for spaces that we are interested in, and make them true in a broader and nicer generality. Since mathematics progresses towards infinitude, this is a good thing.

One last remark on the Banach-Tarski issue, which I often bring up, is that if we assume that all sets are Lebesgue measurable and the Banach-Tarski theorem fails, then we get something equally even more disturbing. There is a partition $P$ of $\Bbb R$ such that $|\Bbb R|<|P|$. That is, there are strictly more parts than elements!

How 'bout that for paradoxical?

Asaf Karagila
  • 393,674
  • 1
    (The sharp eyed reader will observe that the example was slightly an over exaggeration. If $V$ can be well-ordered, so can $K$... But my point is still valid!) – Asaf Karagila Dec 23 '13 at 20:01
  • 1
    I'd be interested in seeing a reference for your last remark regarding a paradoxical partition of $\mathbb R$. – Dustan Levenstein Dec 23 '13 at 20:04
  • Yes, what Dustan said. Does it come from that (what I am guessing) the typical surjection from $\mathbb{R} \to \mathbb{R}/\sim$ uses choice? –  Dec 23 '13 at 20:09
  • Can Banach-Tarski be averted without making all subsets of $\Bbb R^3$ measurable (and in particular without allowing a paradoxical partition)? – dfeuer Dec 23 '13 at 20:10
  • @AsafKaragila Thanks for the comprehensive answer! is that you? http://mathoverflow.net/users/5702/dr-strangechoice –  Dec 23 '13 at 20:11
  • @Dustan http://mathoverflow.net/questions/22927/why-worry-about-the-axiom-of-choice/22935#22935 it says here: "The statement is that either there is a nonmeasurable set or R has a surjection to a larger set...". –  Dec 23 '13 at 20:15
  • 2
    @Dustan: Shelah proved that $\sf ZF+DC+\aleph_1\leq2^{\aleph_0}$ implies that there is a non-measurable set; so in $\sf ZF+DC$ with all sets measurable we have $\aleph_1\nleq2^{\aleph_0}$. However there is a definable surjection from $\Bbb R$ onto $\omega_1$. Now use that surjection from $[0,1]$ onto $\omega_1$ and map the rest of the singletons onto themselves. The function defines a partition of size $2^{\aleph_0}+\aleph_1$ which is strictly larger than $2^{\aleph_0}$, as wanted. – Asaf Karagila Dec 23 '13 at 20:17
  • @user116457: That user is not me, unfortunately. The post was made when I was but a wee child in the ways of choice. – Asaf Karagila Dec 23 '13 at 20:18
  • 1
    @dfeuer: I don't quite know about that. It's a delicate situation, the BT paradox follows from quite weak assumptions like Hahn-Banach theorem, and in fact a weaker form of it. It might be possible to have $\sf ZF+DC+\aleph_1\leq2^{\aleph_0}$ with the choice principles related to the Boolean Prime Ideal theorem failing, and in particular the Banach-Tarski paradox. But I'm not quite sure about that. – Asaf Karagila Dec 23 '13 at 20:20
  • Thanks, that makes sense! I found the surjection $\mathbb R \twoheadrightarrow \omega_1$ here. – Dustan Levenstein Dec 23 '13 at 20:27
2

Math is a collection of creative games which together involve millions of players across the world and spanning all of human history. A subcollection of these games is called set theory. The most popular set theory game is called ZFC. ZFC has several popular relatives, one of which is ZF and one of which is ZF+DC. What you've stated is this: when playing the ZFC game, "metric spaces are paracompact" is a legal move. When playing the ZF game or the ZF+DC game, it is not a legal move.

In applications, these kinds of things tend not to come up. In most cases, something you need for an application can me proved without choice, but you may have to prove it in a more restricted setting and it may be more difficult to do so.

dfeuer
  • 9,069
  • In your parallel to games and (il)legal moves: there are more cases – the situation that some move is illegal is different from the situation when the rules of the game don't say anything about the move. – user87690 Dec 23 '13 at 20:12
  • @user87690, I think you may be misinterpreting the metaphor, which is of course imperfect anyway: the way I would phrase it is that the fact that a certain move is illegal does not imply that the "opposite move" must be legal. – dfeuer Dec 23 '13 at 20:47
  • Yes, I was just pointing out that “illegal move” can mean different things. – user87690 Dec 23 '13 at 21:22