What is the advantage of accepting the axiom of choice over other axioms (for example axiom of determinacy)?
It seems that there is no clear reason to prefer over other axioms.
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3$\begingroup$ If you're fine with these consequences of not negating the axiom of choice, then no reason. $\endgroup$– lhfMar 16, 2012 at 13:56
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2$\begingroup$ @lhf: That looks like one consequence: you can't have the GCH. The rest are just choices that are open to you. $\endgroup$– CharlesMar 16, 2012 at 14:00
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$\begingroup$ @Charles the Wiki page also says "There is a set that can be partitioned into strictly more equivalence classes than the original set has elements, and a function whose domain is strictly smaller than its range. In fact, this is the case in all known models.", so as far as we know, a bad notion of cardinality could be a consequence as well. $\endgroup$– D.R.Apr 13 at 5:10
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6 Answers
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Well, there is quite the long list:
- Every set can be well ordered, so every two cardinalities are comparable.
- Continuity by $\epsilon$-$\delta$ is equivalent to continuity via sequences.
- There are free ultrafilters.
- Countable unions of countable sets are countable; similarly cardinal arithmetic become definable for infinite summations and products.
- Hahn-Banach theorem.
- Product of compact spaces is compact.
- Second countable implies Lindelof.
- Every vector space has a basis.
- Every unital ring admits a maximal ideal.
- Every set can be given a group structure.
The list goes on and on and on. In short, however, the main reason is that mathematics has grown to accept the axiom of choice and its consequences.
It really depends on what you want to do in mathematics, if you don't do set theory you're likely to benefit more from the axiom of choice than other axioms. If you are a set theorist (or planning on becoming one) then there will be places where the axiom of choice works naturally for your advantage and others where you will know to drop it and adopt other axioms instead.
Edit: Some of the implications above do not require the entire power of the axiom of choice. Indeed some of them follow by "mere" countable choice (choice for countable families) and other follow from weaker assertions like the Ultrafilter Lemma.
Furthermore, in some cases it is also possible to do things "by hand" and prove existence of ultrafilters, ideals, basis, etc. Most of classical analysis can get away with relatively weak principles like Dependent Choice, however it can require a bit more if you want to talk about $\ell^\infty$ kind of spaces (for example, without the axiom of choice it is possible for $\ell^1$ to be reflexive).
It is also important to notice that the universe can behave as though you have the full axiom of choice for sets which are so big that you will never go beyond their size - but after that thing breaks down badly, so while the axiom of choice is negated in a very strong way the universe that most mathematician like to think about has the axiom of choice in full.
Edit II: Having a few minutes to spare, I thought I should give my two cents why people prefer it (except for the fact they got used to it by now). One of the greatest set theorists of our time once told me during a break from a class he gave that a good axiomatic system is one that you use without noting.
For example, the axiom of extensionality makes a lot of sense because we want two sets to be equal exactly when they have the same elements. So does the axiom of power set.
The axiom of choice is very natural in the sense that it allows us to extend "finite definitions" to infinite ones (a good example is that every vector space has a basis, we can write one down for finite dimensional spaces, but we cannot "write it" for every infinite dimensional space). In this sense we are very accustomed to its gentle grace and willing to accept the few peculiarities like the Banach-Tarski paradox, or a well ordering of the real numbers.
This is why some of the mathematicians who were opposed to the axiom of choice (Lebesgue, Borel) actually used weak versions of it. You just don't notice that it's there.
Further reading on the site:
- Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice?
- Is there a non-trivial example of a $\mathbb Q-$endomorphism of $\mathbb R$?
- Does every set have a group structure?
- A question about cardinal arithmetics without the Axiom of Choice
- axiom of choice: cardinality of general disjoint union
- Cardinality of an infinite union of finite sets
- Axiom of choice and automorphisms of vector spaces
- Axiom of choice - to use or not to use
- Motivating implications of the axiom of choice?
answered Mar 16, 2012 at 14:21
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2$\begingroup$ I guess one should distinguish between uses of countable choice, uses of dependent choice, and uses of full choice, as the first (and second) are quite intuitive and most of classical analysis lives there, while full choice tends to be applied a bit farther away. For example, one can use Hahn-Banach to prove facts about convex subsets of ${\mathbb R}^n$, but the version of Hahn-Banach one needs here can be proved essentially by hand, without needing a strong appeal to choice. Also in many applications of Zorn's lemma, only countable choice is used when the sets involved are countable. $\endgroup$ Mar 16, 2012 at 14:32
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$\begingroup$ Andres, you are correct. I'll add some words on that. $\endgroup$– Asaf Karagila ♦Mar 16, 2012 at 14:33
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2$\begingroup$ I have added this question to my "favorites", for this answer. $\endgroup$ Mar 16, 2012 at 14:53
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$\begingroup$ I wonder whether the absence of Banach-Tarski from your list is a coincidence. Maybe we have a preference for cozy statements? $\endgroup$ Mar 16, 2012 at 15:37
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1$\begingroup$ @Marc: We definitely prefer cozy statements. I could add that "You cannot partition a set into strictly more parts than elements!" too! :-) $\endgroup$– Asaf Karagila ♦Mar 16, 2012 at 15:41
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One high level (non-mathematical) reason for accepting the Axiom of Choice over, say, the Axiom of Determinacy is that the Axiom of Choice is somehow more "basic." At some level it is more natural to think that all Cartesian products of nonempty sets are nonempty. It doesn't quite seem as natural to think that the game $G_A$ is determined for all $A \subseteq \omega^\omega$.
At least, I think that most mathematicians have come to grips with the idea that there are badly behaved subsets of $\mathbb{R}$: there are sets which are not Lebesgue measurable; there are sets which do not have the Baire property; etc. Why shouldn't there be sets of reals whose associated game is undetermined?
Of course, 100 years ago (or thereabouts) some mathematicians were up in arms over the possibility of well-ordering $\mathbb{R}$, and Zermelo's theorem was considered by some as a great argument against Choice. That is to say, attitudes may change.
answered Mar 16, 2012 at 14:21
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2$\begingroup$ It does seem natural, though, that De Morgan's law can be extended to infinitely many nested quantifiers (which is equivalent to AD). $\endgroup$ Dec 18, 2018 at 18:17
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A practical reason is that certain mathematical theorems are known to be equivalent to the axiom of choice. In particular, the fact that every vector space has a basis and the theorem of Tikhonov that arbitrary products of compact spaces are compact require the axiom of choice.
More importantly, the axiom of choice is equivalent to that the arbitrary product of nonempty sets is nonempty. This seems to be a consequence of our intuitive notion of sets. So even though we seldom use the full axiom of choice, it seems to be part of the way of thinking about sets.
answered Mar 16, 2012 at 14:19
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$\begingroup$ It's worth mentioning that the statement that the product of compact Hausdorff spaces is compact is strictly weaker: in fact it's equivalent to the ultrafilter lemma. And I think in practice the applications of Tychonoff's theorem that I've seen have been to compact Hausdorff spaces. $\endgroup$ Mar 16, 2012 at 16:32
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$\begingroup$ That is true. But I cannot imagine a person that dislikes AC because of its consequences, but thinks the ultrafilter lemma is acceptable. $\endgroup$ Mar 16, 2012 at 16:47
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1$\begingroup$ It is also worth mentioning that Ultrafilter Lemma implies Bananach-Tarski, so if the consequence troubling someone is that one the Ultrafilter Lemma is too much choice indeed! $\endgroup$– Asaf Karagila ♦Mar 16, 2012 at 21:44
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It allows the use of nonprincipal ultrafilters, which are needed for nonstandard analysis. (This doesn't actually need the full strength of AC, but you can't do it in ZF.)
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Another reason which might not be quite so accessible to you yet is that (a statement equivalent to) the axiom of choice is used to prove that every $R$-module admits an embedding into an injective $R$-module, which allows us to have a theory of sheaf cohomology in algebraic geometry. Sheaf cohomology is something which lots of people care about- not having choice around would present major problems for almost all of them.
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What is the advantage of accepting the axiom of choice over other axioms (for e.g. axiom of determinacy)?
The advantage of accepting the axiom of choice over other axioms is that if you accept the axiom of choice, you can use modern probability theory which is based on measure theory. The axiom of choice is necessary for it. For example, if you don’t accept the axiom of choice, it is possible that $\mathbb{R}$ is a countable union of countable sets (cf. Feferman-Levy model).
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$\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ Mar 26 at 6:20
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$\begingroup$ @Leucippus: Whether an answer or not, this certainly is nothing that should be posted as a comment. $\endgroup$ Mar 26 at 11:36
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$\begingroup$ The axiom of determinacy is consistent with DC, which implies a countable union of countable sets is countable. (The result that $\mathbb R$ is a countable union of countable sets may even be implied by AD without DC, since it e.g. proves countable choice for sets of reals... I'm not sure). $\endgroup$ Mar 26 at 15:46
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$\begingroup$ @spaceisdarkgreen: Since $\sf AD$ proves $\sf AC_\omega(\Bbb R)$, you don't need anything more than $\sf AD$ in order to get that $\Bbb R$ is not a countable union of countable sets. $\endgroup$– Asaf Karagila ♦May 21 at 12:53
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$\begingroup$ @AsafKaragila yes, certainly I meant $\mathbb R$ is not a CUCS. I didn't see it at the time but I guess the argument that $\mathsf{AC}_\omega(\mathbb R)$ implies a CUCS of reals is countable is an easy adaptation of the normal proof that CC -> a CUCS is countable, since the set of functions $\mathbb N\to \mathbb R$ is identifiable with $\mathbb R$ in ZF. $\endgroup$ May 21 at 20:52