Unprovable statements in ZF

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Advantage of accepting the axiom of choice
Advantage of accepting non-measurable sets

As you all know, Banach-Tarski paradox is solely a consequence of Axiom of Choice, and I think it is just absurd.

I'm trying to take ZF as my axiomatic model rather than ZFC. I wonder if there are some theorems really important understanding our 'Number system' that can be proved in ZFC, but not in ZF. (i.e., every vector space has a basis is equivalent to axiom of choice, but i think we actually don't need this strong theorem since we are always working on finite dimensions (maybe not! Please tell me if there are some branches of mathematics studying infinite dimension))

I wonder how many risks should I take when I'm removing AC.

You might be interested in the answers at http://math.stackexchange.com/questions/16020/equivalent-statements-of-the-axiom-of-choice?rq=1 — Gerry Myerson, Jul 21 '12 at 08:41
Several equivalents of AC are listed in the Wikipedia article on Axiom of Choice. — Martin Sleziak, Jul 21 '12 at 08:43
@Katlus: When you use a "standard" mathematical result, then, in many fields, there is a significant risk that you will use a result that in fact uses AC. And, as happened to some analysts who were early opponents of AC, you may end up using AC without being aware of it. It really is a "natural" principle of reasoning. But apart from this, there are no risks. The theorems one proves are just less general. — André Nicolas, Jul 21 '12 at 20:32

score 8 · Accepted Answer · edited Apr 13 '17 at 12:21

8

The axiom of choice, while seemingly having counterintuitive results is needed to ensure that infinite sets are well-behaved.

Of course if you would only like to work with finite sets then the axiom of choice is not needed, however some things which you think hold immediately would fail without the axiom of choice, and they may fail badly.

The union of countably many pairs may not be countable.
The real numbers may be a countable union of countable sets.
We may be able to partition a set into more parts than elements, in particular this set might be the real numbers.
In the real numbers continuity by sequences and by $\varepsilon$-$\delta$ are no longer equivalent.
Topology breaks down in acute manners, to horrid to begin to describe.

These three are far far more disturbing to me than Banach-Tarski, and these would hold in several models of ZF without the axiom of choice.

What more? Let's see, what else can fail badly

There might be no free ultrafilters, on any set.
In turn some fields might not have an algebraic closure; others could have two non-isomorphic closures, for example the rationals.
There could be a tree that every point has a successor, but there is no $\omega$-branch.
There could be a vector space which has two bases of different cardinality.
Functional analysis may stop working due to lack of Hahn-Banach, Krein-Milman, Banach-Alaoglu theorems.

If you wish to do some set theory, perhaps, it also becomes hard:

Cardinal arithmetics can fail for infinite sums and products.
In forcing the mixing lemma fails.
The partial ordering of cardinalities is not necessarily well-founded.
There may be no canonical representative for $|A|$, namely a function which returns a particular set of the cardinality of $A$ (like the $\aleph$ numbers).

This list can be made really quite infinite. Why? Because modern mathematics is very much about infinitary objects and for those to be well-behaved we really need the axiom of choice, or else a lot of bad things may occur.

It is also the case that most people are educated by choice-using people, so the basic intuitions about mathematics actually use the axiom of choice a lot more than you would think.

However there are still merits to working without the axiom of choice. For those, see my recent post: Is trying to prove a theorem without Axiom of Choice useless?

To read even more:

edited Apr 13 '17 at 12:21

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answered Jul 21 '12 at 09:16

Asaf Karagila

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I couldn't hold myself, despite voting to close - I still felt I had more to say. – Asaf Karagila Jul 21 '12 at 09:16
I suppose you could have added it to the earlier question. – Gerry Myerson Jul 21 '12 at 09:18
@Gerry: Which one? :-) With these sort of answers I usually stop at some point, otherwise I'll write a book in the answer. – Asaf Karagila Jul 21 '12 at 09:27
I was surprised that you left out that a product of nonempty sets might be empty. – MJD Jul 21 '12 at 12:36
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@Mark, well that is actually less useful in a direct way to mathematics, since most sets you encounter are such that they ensure someone the product is non-empty (e.g. vector spaces; open sets; etc.) furthermore the question suggested "fine, we can do without basis for vector spaces" so writing "products of sets may be empty" is kinda moot in my eyes. – Asaf Karagila Jul 21 '12 at 12:38
Your arguments remind me a lot of old arguments against non-Euclidean geometry. And we know how well those fared. Just as non-Euclidean geometry proves useful, so too does non-choice. And now that theories exist that conveniently enable variable foundations (e.g. Topos Theory) there are no barriers stopping these alternative universes from being explored. No barriers, except for bias, which is always naturally corrected in time. (btw "These three" probably should be "The above"). – Bill Dubuque Jul 21 '12 at 14:37
@Bill: Please point the place I said "it is useless to prove things without choice", here or elsewhere. I beg you to do find me saying it is useless to work without choice, and invalidate my own work as I do. And yes, there are new parts of mathematics which seem to be incompatible with the axiom of choice, but these are far from conventional mathematics yet; and far from being an established and accepted foundation to day-to-day mathematics. This seemed to be the question here, and this is what I answered. Feel free to contribute your answer if you wish. – Asaf Karagila Jul 21 '12 at 16:00
To demonstrate my point about day-to-day mathematics, it is provably difficult to develop a good sense of measure theory when the real numbers are a countable union of countable sets. You either have to give up on $\sigma$-additivity, or you have to start doing everything with Borel codes, which is a nice mechanism but it is a lot of work to carry it around everywhere. Similarly it if the Boolean Prime Ideal theorem is assumed, but choice is negated then it is much much harder to take ultrapowers of models; or it is harder to use Krein-Milman or Banach-Alaoglu arbitrarily. – Asaf Karagila Jul 21 '12 at 16:09
@Asaf My comment was merely a general remark on discussions of this sort, not about the views of any specific person. Perhaps I should have said "Arguments like these..." instead of "Your arguments...". Of course such arguments are very old (and oft-repeated). – Bill Dubuque Jul 21 '12 at 16:11
@Bill: I'm not sure how it fits as a comment on my answer at all, perhaps on the main question... – Asaf Karagila Jul 21 '12 at 16:19
@Asaf "Arguments like those" are not in the main question. – Bill Dubuque Jul 21 '12 at 16:51
@Bill: You either suggest that the comment was completely misplaced, or that I am not sure what the point of the comment was to begin with. What arguments? Against working with AC? Against working with mild versions of AC? Against not-working with AC? Against writing long comments that hardly anyone is going to read? – Asaf Karagila Jul 21 '12 at 17:23
@AsafKaragila: In several items on the list, the wording is perhaps ambiguous. For example, let's look at "The union of countably many pairs is not countable." This is indeed true in some models of ZF. But the absence of AC does not commit us to such a model. It merely restricts the generality with which we can prove certain useful things. – André Nicolas Jul 21 '12 at 20:15
@Andre, you are right. I did try to point out that this might happen without AC. To cut on repeated edits, please see if anything else pops out as badly phrased and I will edit those in a bit altogether. – Asaf Karagila Jul 21 '12 at 20:18

Unprovable statements in ZF

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