[FOM] Eliminability of AC

Thomas Forster T.Forster at dpmms.cam.ac.uk
Mon Mar 24 18:07:34 EDT 2008



Does Shelah's work on the Whitehead problem count as a solution in 
anyone's book...?

On Mon, 24 Mar 2008, joeshipman at aol.com wrote:

> One month ago, I noted the standard result that any use of the Axiom of
> Choice could be eliminated from proofs of arithmetical statements, and
> indeed from proofs of Sigma^1_2 statements, and asked the question:
>
> What is the simplest example of a well-known open problem  in "ordinary
> mathematics" (that is, one of interest to mathematicians in general and
> not primarily of interest to logicians and set theorists) where there
> is a possibility some form of Choice is needed for any proof?
>
> No one was able to provide one that met all three criteria (well-known
> AND open AND outside of logic and set theory), so I conclude that
> mathematicians outside of logic and set theory do not care about the
> Axiom of Choice anymore -- they are only interested in questions that
> are sufficiently absolute that their truth value does not depend on AC.
>
> It is possible that this increased emphasis on concrete problems
> compared to several decades ago is a reaction to forcing and the
> independence proofs, combined with the failure to isolate sufficiently
> plausible or useful new axioms.
>
> Since AC is an axiom one may use without explicit mention and still
> have a publishable paper, I don't see any remarks in current
> mathematical literature outside of logic and set theory that proofs do
> or do not depend on AC. This is more evidence that mathematicians do
> not care about AC.
>
> But Shoenfield Absoluteness goes further: not only may AC be eliminated
> from the proofs of arithmetical or Sigma^1_2 statements; so may V=L, a
> much stronger axiom. Can anyone provide examples, particularly
> arithmetical ones, of theorems outside of logic and set theory which
> were first proven (or are most easily proven) by showing they follow
> from V=L and then applying Absoluteness?
>
> -- Joe Shipman
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