[FOM] "Mathematician in the street" on AC
Vaughan Pratt
pratt at cs.stanford.edu
Mon Aug 31 16:41:45 EDT 2009
Bill Taylor wrote:
> The main idea is that one has a countable collection of sets which
> are each coverable by (necessarily countable) collections of intervals
> of arbitrarily small total finite length. Then when one comes to try
> to cover the union set with a collection of intervals, one must pick
> a countable (and small-lengthed) bunch of intervals from the countable
> union of countable sets. This requires making a countable number of
> choices to do so, hence AC.
Thanks, Bill. The reasoning that lured me to my incorrect conclusion (that a countable union of null sets must be null even without Choice) was that the property of having zero measure for a countable set was independent of the particular enumeration selected, whence one should be allowed to infer that the sum of a constantly zero series 0+0+0+... is its limit, namely zero.
The Feferman-Levy result shows that this argument cannot be made to work, by exhibiting a sequence of sets each of zero measure whose total measure is nonzero (say 1 in the case of the unit interval (0,1)). We then obtain 0+0+0+... = 1.
I seem to have committed a variant of the error made by Cauchy in 1821 when he proved that the pointwise limit of a sequence of continuous functions is continuous, by neglecting the possibility that the convergence might not be uniform in the sense appropriate to this variant.
In the variant at hand the source of the difficulty would seem to be the sensitivity of Lebesgue measure to the structure of the set being measured. One would naturally expect the Lebesgue measure of an increasing sequence of sets whose measures converge to a limit to be that limit; Feferman-Levy shows that this does not follow in the absence of Choice.
The Wikipedia article on Lebesgue measure mentions (more than once) that the existence of non-measurable sets depends on Choice. However in its list of 14 properties of Lebesgue measure on R^n it does not indicate which of those do not hold in the absence of Choice. It would appear from the above that property 2 (concerning the disjoint union of countably many measurable sets) does not. Which others fail?
(As an aside, is the Solovay result that every set could be Lebesgue-measurable consistent with the Feferman-Levy result that the reals could be a countable union of countable sets? I don't right away see how either possibility implies the negation of the other.)
For those who find these counterexamples unreasonably weird yet do not want to commit to the possibility of well-ordering the continuum, what does locale theory have to offer? Recently I argued for the naturalness of the axiom that the complete ultrafilters (aka complete homomorphisms to 2) of a complete atomic Boolean algebra (CABA) are its atoms, but I can see some being bothered by the choices implicit in the complete distributivity of CABAs, namely of infinite unions over infinite meets and vice versa. The advantage of frames (dual to locales in the same way Boolean algebras are dual to Stone spaces) over CABAs is that they allow infinite joins (of any cardinality) while avoiding the sorts of choices entailed by complete distributivity. Are there any locale theorists on this list who'd like to expand on the relationship between locales and choice? In particular what is the connection between Brouwer's proposal to drop excluded middle and Isbell's proposal half a century later to substitute frame homomorphisms for CABA homomorphisms (which is what the inverse-image operation used in defining continuity for topological spaces amounts to).
Vaughan Pratt
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