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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.

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You might be interested in the answers at math.stackexchange.com/questions/16020/… –  Gerry Myerson Jul 21 '12 at 8:41
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Several equivalents of AC are listed in the Wikipedia article on Axiom of Choice. –  Martin Sleziak Jul 21 '12 at 8:43
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Function spaces are (generally) infinite-dimensional. –  Gerry Myerson Jul 21 '12 at 8:44
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@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
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marked as duplicate by Asaf Karagila, Zhen Lin, Gerry Myerson, Henning Makholm, t.b. Jul 21 '12 at 23:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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:

  1. Advantage of accepting the axiom of choice
  2. Advantage of accepting non-measurable sets
  3. Foundation for analysis without axiom of choice?
  4. Axiom of choice and calculus
  5. Number Theory in a Choice-less World
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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 9:16
    
I suppose you could have added it to the earlier question. –  Gerry Myerson Jul 21 '12 at 9:18
    
@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 9:27
    
I was surprised that you left out that a product of nonempty sets might be empty. –  MJD Jul 21 '12 at 12:36
    
@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
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