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How is it possible to reconcile the following...

In 1970, Solovay constructed Solovay's model, which shows that it is consistent with standard set theory, excluding uncountable choice, that all subsets of the reals are measurable.

A not too well known application of the Boolean prime ideal theorem is the existence of a non-measurable set[2] (the example usually given is the Vitali set, which requires the Axiom of Choice). From this and the fact that the BPI is strictly weaker than the Axiom of Choice, it follows that the existence of non-measurable sets is strictly weaker than the axiom of choice.

Also, a similar question regarding...

The ultrafilter lemma is equivalent to the Boolean prime ideal theorem, with the equivalence provable in ZF set theory without the axiom of choice. Many other theorems of general topology that are often said to rely on the axiom of choice are in fact equivalent to BPI. For example, the theorem that a product of compact Hausdorff spaces is compact is equivalent to it. If we leave out "Hausdorff" we get a theorem equivalent to the full axiom of choice.

Now, Janich's Topology gives a proof of the full blown Tychonoff theorem from the Ultrafilter Lemma (after showing Zorn's Lemma implies the Ultrafilter Lemma). But this should not be possible since Tychonoff is equivalent to AC and UL is weaker than AC. So do I need to read his proof more carefully...is he still using the full power of AC elsewhere?

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What are the sources for the cited parts? –  Arthur Fischer Dec 15 '12 at 12:40
I’ve not seen Jänich’s proof, so this is just a guess: does he assume that the product is non-empty? That would require full AC. –  Brian M. Scott Dec 15 '12 at 12:48
Ah, I think I found it. There is a lemma: the cartesian product X of compact spaces has a subbasis G with the property that every open cover of X by sets of G possesses a finite subcover. There is definitely AC needed here...I was just tricked because a couple pages later there is the mention of Zorn's Lemma (in connection with UL). I assumed this was the critical use of AC. –  Forever Mozart Dec 15 '12 at 12:59
Still confused about the first problem. I know Solovay's result is slightly more technical than stated, in that it requires an inacessible cardinal. Is anyone familiar with the non-meas. set via BPIT? –  Forever Mozart Dec 15 '12 at 13:14
What does that have to do with Solovay's model and non measurable sets? –  Asaf Karagila Dec 15 '12 at 13:15

2 Answers 2

up vote 5 down vote accepted

Free ultrafilters on $\mathbb N$ translate to non-measurable sets of real numbers.

Recall that the real numbers and the power set of the natural numbers, and the Cantor space. The Cantor space can be endowed with the product measure such that the full measure is $1$, and the measure is translation invariant (in the sense of finite modifications). Now it turns out that this is actually the completion of the Borel measure of the Cantor set and it is isomorphic, as a measure space, to the real numbers and the Lebesgue measure.

So what is an ultrafilter? It is a partition of $\mathcal P(\mathbb N)$ into two parts, and by translating a set of integers to its characteristics function we have a partition of the Cantor space into two parts. The full measure is one, and the complement map $A\mapsto\mathbb N\setminus A$, which is a bijection between the ultrafilter and its complement, is measure preserving. Therefore the Cantor set is the union of two disjoint sets of equal measure, which has to be $\frac12$. However by Kolmogorov's zero-one law we have that a free ultrafilter has to have either measure zero or measure one if it is measurable. Since $\frac12$ is neither zero nor one, we have to have that it is a non-measurable set.

So free ultrafilters make non-measurable sets. In a model where all sets are measurable, there are no free ultrafilters and so BPIT fails.

In Herrlich's The Axiom of Choice, Diagram 5.10 p. 122 has some nice principles which imply the existence of non-measurable sets. This is not a question of consistency, but rather actual implications. I believe that Gregory Moore's Zermelo's Axiom of Choice should have a similar diagram at the end of the book, but I cannot recall for sure (I will check and update tomorrow).

I will also add that the same argument shows that free ultrafilters cannot have Baire property (and therefore in models where every set of real numbers have the Baire property there are no free ultrafilters as well).

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Ok, very good I think I get it now. By the way, do you know of any texts where I can find a detailed construction of this set? –  Forever Mozart Dec 15 '12 at 13:42
@DavidL.: I wrote my answer from a cellphone, let me revise it a little. But generally there is no "special set" there are canonical bijections and injections and everything is generated from those (and the ultrafilter, of course). –  Asaf Karagila Dec 15 '12 at 13:50
So Solovay proved Con(ZF+Every set of reals is meas.) provided IC. Then we can go with ZF+BPIT to prove existence of a nonmeas. set, so long as we neglect IC? –  Forever Mozart Dec 15 '12 at 14:01
I'm not sure what IC means, but Solovay proved that Con(ZFC+There is an inaccessible cardinal) implies Con(ZF+DC+All sets are measurable). Sierpinski proved that ZF+BPIT proves there is a non-measurable set. The fact that AC proves there is a non-measurable set was already known due to Vitali's theorem. But the two sets are very different as sets. And there are more ways to create non-measurable sets from other choice principles. –  Asaf Karagila Dec 15 '12 at 14:03
I should add, @David, that it is consistent to have all sets are measurable without an inaccessible cardinal, but in those models it is often the case that all sets are Borel due to one pathology or another. It then makes it harder to develop a good sense of measure theory. –  Asaf Karagila Dec 15 '12 at 14:08

There are two steps in the ultrafiler proof of the Tychonoff theorem that require AC. You take an ultrafilter on the product and this ultrafilter induces along the projections an ultrafilter on every component space. You use the fact that these ultrafilters converge by compactness. Being able to do this step is equivalent to the BPI.

Now, convergence in the product topology is equivalent to coordinate-wise convergence, so you want to find a point in the product that the original ultrafilter converges to. You do that by picking for each coordinate a point the induced ultrafilter converges to. It is here that you need the full strenght of AC. For Hausdorff spaces, you can skip this step since in a Hausdorff space, a filter can not converge to more than one point. So you do not have to make arbitrary choices at that step.

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Nice. I am also glad you mentioned how easily BPI alone works if the spaces are Hausdorff. –  Forever Mozart Dec 15 '12 at 13:18

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