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Suppose there is a well-known theorem whose usual proof uses Axiom of Choice. Is trying to prove it without Axiom of Choice useless? What merits can such a proof have?

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mathoverflow.net/questions/22927/… –  user29743 Jul 15 '12 at 12:08
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This is an odd question to ask, after you have posted 8 questions about proving one theorem or another without the Axiom of Choice! Shouldn't you have decided whether it was useful first, before asking all those questions? –  Gerry Myerson Jul 15 '12 at 12:18
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@GerryMyerson Even if I know an answer, there can be other(or even many) answers that I don't know. –  Makoto Kato Jul 15 '12 at 14:19
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@Makoto: I think it is disingenuous to post a question like this and immediately post a response defending your other questions on the site. This kind of soapboxing is what blogs are for, not math.SE. –  Qiaochu Yuan Jul 16 '12 at 0:00
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If you would like to open a thread on meta you can register and thus gain quite a lot of privileges. Amongst them is voting and posting on meta. –  Asaf Karagila Jul 16 '12 at 7:27

1 Answer 1

It is often not useless at all. First let me give out a few reasons why it is useful:

  1. It gives us a better understanding of how well-behaved some objects are. For example, vector spaces are not well-behaved in general, but in particular cases they are well-behaved (e.g. finitely generated ones).

    On the other hand, compact metric spaces are quite well-behaved and many of their properties remain valid even without the axiom of choice.

  2. It can be the case that a proof without the axiom of choice is a much more constructive one, and allows us to examine the features of an object which was proved only to exist if the axiom of choice were used.

    One example coming to mind is the compactness of closed and bounded subsets of $\mathbb R$, another is the uniform continuity of a continuous function from a compact metric space into a metric space.

  3. If a proof fails it allows us a better understanding of how much choice is needed for a certain assertion. Dependent Choice for Baire's Category theorem; Countable Choice for the equivalence between different topological properties; etc.

On the other hand, it is somewhat useless to try and prove a theorem without the axiom of choice if other parts of the theory already assume it. If you already assume that every vector space has a basis, showing that a certain proposition relying on this property holds without the axiom of choice is moot.

One example for this is the ultrafilter theorem which, together with the Krein-Milman theorem, imply the axiom of choice [4], so if you end up using both these propositions there is no use in avoiding choice anymore. It's there.

Regardless of the above, one should remember that not all things in modern mathematics are true in the absence of choice (that is, aside from the axiom of choice itself) and often reformulation and distinction between equivalent definitions is required. In those cases one can, and should, ask themselves how much choice is needed for a particular result.

For example, the assertion "$\mathbb R$ is not a countable union of countable sets" requires the axiom of choice, but it is provable from a vastly weaker statement, "countable unions of countable sets are countable".


Further reading:

  1. Why worry about the axiom of choice? (MathOverflow)
  2. Axiom of choice - to use or not to use
  3. Advantage of accepting the axiom of choice
  4. Is Banach-Alaoglu equivalent to AC?
  5. Motivating implications of the axiom of choice?
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I wish to remind the downvoters that I am not a mindreader, and constructive comments will be welcomed. –  Asaf Karagila Jul 15 '12 at 12:45
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Some people downvoted this?! In my opinion, this is a great answer. Very interesting,+1! –  Malik Younsi Jul 15 '12 at 13:08
    
The grammar and phrasing was a bit strange. I've suggested a reasonably conservative edit, but please let me know if I've mutated your meaning anywhere. –  Ben Millwood Jul 15 '12 at 23:21
    
@Ben: Thanks a lot! –  Asaf Karagila Jul 15 '12 at 23:21

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